Content uploaded by Bhoj R Singh
Author content
All content in this area was uploaded by Bhoj R Singh on Mar 19, 2016
Content may be subject to copyright.
1
Microbial Genetics Questioned to Understand
Table of Contents
Chapter
No.
Title Pages
1 True False in Microbial Genetics
Questions 1-370
Answers 1-370
3-16
17
2 Multiple choices in Microbial Genetics
Questions 1-400
Answers 1-400
18-41
42
3 Fill in the blanks of Microbial Genetics
Questions A1-185
Answers A1-185
Questions B1-120
Answers B1-120
Questions C1-190
Answers C1-190
43-51
52-53
54-59
60
61-70
71-72
4 Short Questions in Microbial Genetics
(220 questions with answers)
73-142
5 Bacteriophages (81 questions with answers) 143-167
6 Plasmids and Conjugation (69 questions with answers) 168-187
7 Mutation and Mutagenesis (88 questions with answers) 188-215
8 Transformation Genetics (30 questions with answers) 216-224
9 Bioinformatics in Microbial Genetics
(29 questions with answers)
225-237
For our mothers, teachers and students
2
Microbial Genetics
Genetics is the study of inheritance and variability in life including genes, genetic materials,
genetic code, genetic information in the code, expression of the genes or genetic information and
understanding the generation and inheritance of variations (mutations). Besides, it also studies
physical, chemical and biological nature of the genomes as arrangement of genes, genotype and
their interaction the environment to be expressed into a certain phenotype. A particular organism
may possess alternate forms of one or more genes known as alleles and leads to genetic
polymorphism. Genome (the whole DNA or sometimes RNA) is stored in form of chromosomes,
which are chains of double stranded DNA in all living being except the virions which may also
posses other forms of genome made of ssRNA, dsRNA, ssDNA. Within DNA, genes exist as
sequences of nucleotides which code for functional proteins. The genetic material of bacteria and
plasmids is dsDNA but of bacteriophages it may be dsRNA, dsDNA, ssRNA, or ssDNA. The two
major and essential functions of genetic material are replication and expression in all organisms.
Microbial genetics means genetics of microbes (bacteria, Archaea, viruses, including bacterial
viruses i.e., bacteriophages and unicellular or mycelial eukaryotes including yeasts, other fungi,
algae and protozoa). Microbial genetics has applications in medicine, veterinary, agriculture, food
and pharmaceutical industries.
Because of simplicity in physiology microbes have been exploited for understanding interaction
of biochemical, physiological and genetic components of the life. It has successfully furnished
details of genetic code and the regulation of gene function. Besides, microbes being both useful and
pathogenic, genetic studies have helped in understanding pathogenesis of infectious as well as
genetic and non-infectious diseases, and variability in pathogens. The first chapter on microbial
genetics started by George W. Beadle (1903–1989) and Edward L. Tatum (1909–1975) while
investigating genetics of tryptophan metabolism and nicotinic acid synthesis in Neurospora, a
fungus postulating the "one gene one enzyme" hypothesis. However, studies on bacterial genetics
started in 1947(Joshua Lederberg) with demonstration of exchange of genetic factors in
Escherichia coli through conjugation, mediated through plasmids, “fertility factors”. Later on
process of transformation, transduction and chromosomal gene mobilization lead to genome
(chromosome) mapping in bacteria. These techniques combined with restriction enzyme analysis
lead to sequencing, cloning and expression of several genes (prokaryotic and eukaryotic) in
microbes.
3
Chapter.1. True False in Microbial Genetics
1. Phenotype of cell may vary without change in genotype.
2. All mutations cause genotypic changes.
3. All genotypic changes induce phenotypic variation.
4. Examination on cells for many generations at time scale under varying environment can
elucidate the nature of phenotypic variation.
5. In bacteria only one replicon is present.
6. Chromosome is the largest replicon of a cell.
7. Antibiotic resistant variants can emerge even in absence of any antibiotic selection pressure.
8. Mutations can be induced to occur at much faster rate.
9. For selection of auxotrophic mutants by penicillin enrichment just one cycle of enrichment is
sufficient for having about 10% cells with auxotrophy.
10. Replica plating can be employed for selecting auxotrophic cells.
11. Point mutations are in general revertible.
12. Transition occurs when one purine base is replaced by the purine or one pyrimidine is replaced
by another pyrimidine.
13. The genes transcribed to form one mRNA are said to be in one operon.
14. Most chemically induced base substitutions are transitions.
15. Transversion occurs when one purine base is replaced by the pyrimidine or one pyrimidine is
replaced by purine.
16. Transition mutations are less common than transversions.
17. There are many chemicals to induce transversion but only few to induce transition.
18. Most common alkylating reagent used for inducing mutation s in bacteria is EMS (Ethyl
methane sulfonate).
19. Bifunctional alkylating reagents are more powerful mutagen than nitrous acid.
20. There are mutagens which can induce exclusively transition mutations.
21. There are mutagens which can induce exclusively transversion mutations.
22. Acridine dyes induces frame shift mutations.
23. It is easier to detect reversion of an auxotrophic mutation than its appearance.
24. Frame-shift mutations can be induced to revert by base analogues, nitrous oxide and
hydroxylamine.
25. Frame-shift mutations can be induced to revert by acridine dyes.
26. Proflavine is the poor mutagen for normal cells of E. coli.
27. Proflavine is strong mutagen for T4 phage infected cells of E. coli.
28. All frame shift mutagens are intercalating agents.
4
29. All intercalating are frame shift mutagens.
30. Reversion frequencies are comparatively lower, than mutations.
31. The basis for the Ames test is analysis of reversion because it is easier to detect reversion.
32. Polar mutation affect the gene(s) down stream to the gene in which mutation has occurred,
while mutated gene is affect a little.
33. Pseudorevertants are more common than true revertants.
34. Unless repaired pyrimidine dimers are always lethal.
35. The product of hin of Salmonella enterica, gin of Mu phage and cin of P1 phage are
functionally interchangeable for induction of site specific recombination.
36. P1 phage can suddenly change its host specificity to infect Bacteria.
37. Mu (µ) and D128 are transposing bacteriophages.
38. Mu (µ) is a bacteriophage which uses a form of transposon for replication.
39. Transposons have ability to insert into any replicon.
40. Signature Tagged mutagenesis (STM) is a modified method of transposon mutagenesis.
41. Signature Tagged mutagenesis (STM) is method of positive selection of mutants for
identification of virulence genes.
42. IVET is method of positive selection of mutants for identification of genes expressed in vivo.
43. All of the integrons carry sulI gene encoding resistance to sulphonamides.
44. All the four methods of gene transfer i.e., Transformation, Transduction, Conjugation and cell
fusion are common to both G+ve and G-ve bacteria.
45. Conjugation is common in both G+ve and G-ve bacteria.
46. The gram positive bacteria which usually shows the property of conjugation is Enterococcus
faecalis.
47. Conjugative function of most plasmids is mediated through proteins encoded by an operon
containing about 24 genes.
48. Conjugative pili can also act as receptor for a few bacteriophages.
49. Some bacteriophages can be used to determine presence of some conjugative plasmids in
bacteria.
50. Some plasmids also carry transposons.
51. Phage mediated conjugation occurs in S. aureus
52. In high copy number plasmids, having relaxed replication control, replication control is not as
rigorous as in low copy number plasmids.
53. For high copy number plasmids ratio of plasmid: chromosome DNA is usually low.
54. Plasmid partition encoded by par functions of plasmid are limited to low copy number
plasmids.
5
55. Large, low copy number plasmids are inherited in much better way than high copy number and
smaller plasmid.
56. Post segregation killing is the mechanism of maintaining the plasmids in the host.
57. Penicillins and cephalosporins can kill those bacteria which have receptors on there cell wall.
58. β-lactamases responsible for penicillins and cephalosporins resistance are chromosomally
encoded in many bacteria.
59. Plasmids are often responsible for drug resistance in Mycobacterium tuberculosis.
60. For production of most of the antibiotics, plasmids are responsible.
61. For Hg++ resistance in Enterobacteriaceae members’ genes are mediated through large
transposons.
62. For emergence of Hg++ resistance in Enterobacteriaceae members is associated with the use of
dental amalgams.
63. Plasmids inducing production of related/ similar pili are always in the same incompatible group.
64. F pili are thick and flexible.
65. Plasmid producing rigid pili conjugate better on solid surfaces.
66. Production of rigid pili is constitutive while F pili and other flexible thin or thick pili are not
expressed under normal growth conditions.
67. Plasmids of same incompatibility group share a similar host range.
68. Antibiotic resistance transferred by plasmids of same incompatibility group is also similar.
69. Twort-d’ Herelle phenomenon is nothing but action of lytic phages.
70. For optimum lysis MOI should be around 1:10.
71. Lysis from without phenomenon occurs when MOI is <10
72. Small phages tend to produce larger plaques.
73. Plaque size is larger when bacteria are grown in nutritionally limited conditions.
74. Cytosine is not present in DNA of T-even phages.
75. DNA of some Bacillus subtilis phages has uracil in place of thymine.
76. DNA of some phages often contains introns.
77. DNA of some bacteria often contains introns.
78. Structure of chromosome of bacteriophages is often more complex than plasmids.
79. DNA of bacteriophages often contains more than 100 genes.
80. Interaction of bacteriophages and bacteria in adsorption phase is reversible.
81. Phages specific for R-core region are often narrow host range phages.
82. E. coli lacking OmpC protein are resistant to T4 phages.
83. E. coli lacking L-gala-D-manno-heptose are resistant to T3, T4, and T7 phages.
84. Lysozyme and EDTA treated E. coli are less susceptible to some Bacteriophages.
6
85. Rate of adsorption of phages onto bacteria depends on temperature.
86. ΦX174 phage has a linear chromosome.
87. Only ΦR73, an ssRNA phage of E. coli, replicates similar to retroviruses through integration
into chromosome.
88. Coliphages (RNA phages) viz., R17, f2, MS2 and QB replicate similar to retroviruses through
integration into chromosome.
89. Centrally turbid plaques are characteristic of lysogenic phages.
90. Centrally turbid plaques are characteristic of temperate phages.
91. You can isolate mature phage particles from broth culture of an established lamda (λ) lysogen.
92. Lysogenized phage can revert to lytic cycle by change in nutrient conditions.
93. Recipient of R plasmid can transfer plasmid to other bacteria just for a very short time is a
closed system but indefinitely in an open system.
94. In most of the F plasmids finO gene is inactivated to make them more competent for
conjugation at any stage.
95. puc and pBR plasmids can coexist if are having resistance genes for different antibiotics.
96. Solution of DNA if have 280/260 absorbance ratio equal to 1.0 then sample of DNA is said to
be pure.
97. Genome size of mitochondria in different organisms does not vary to great extent.
98. Genome size of chloroplast in different plants and green algae is almost similar.
99. Mitochondrial genome of human cells has no introns.
100. Mitochondrial genome of yeast cells has about 8 introns.
101. Most of the prokaryotic genomes are about 5 Mb in size.
102. Genome of plasmid may be small piece of linear or circular DNA.
103. Archaea are also bacteria found in more adverse environment and they also have DNA
packaging proteins similar to HU proteins of eubacteria.
104. DNA is a linear un-branched polymer as the proteins.
105. Primary structure of proteins depends only on peptide bonds.
106. Zinc finger proteins which determine the secondary structure of DNA are common in
pathogenic bacteria.
107. Promoter structure determines the basal level of transcription initiation.
108. The strong promoter may direct productive transcription initiation about one thousand times
more than a weak promoter.
109. Binding specificity of σ subunit of RNA polymerase is dependent on DNA sequence.
110. Gene silencers and enhancers are the chief modulators of gene expression in bacteria.
111. A single silencer or enhancer can control expression of many genes.
7
112. Bacterial mRNA does not undergo any considerable form of processing before translation.
113. In bacteria translation starts even before completion of transcription.
114. In bacteria there is more than one kind of RNA polymerases.
115. Bacteria have RNA polymerase I, II and III for synthesis of different kinds of coding and
non-coding RNA molecules.
116. All tRNA must end with a tri-nucleotide 5’-CCA-3’.
117. Codone-anti-codone recognition is complicated by wobble.
118. Non-standard base pairs can form between the third nucleotide of the codone and the first
nucleotide of the anti-codone.
119. G-U is the permitted base pair in codone anti-codone pairing.
120. Inosine can base pair with any of the four nucleotides.
121. Inosine is a modified version of guanosine.
122. A-U is a non-permitted base pair in codone anti-codone pairing.
123. Super wobble can base pair with any of the four nucleotides in codone anti-codone pairing.
124. Some amino acids can bind to appropriate codone of RNA even in absence of tRNA.
125. E. coli consensus sequence for ribosome binding is 5’-AGGAGGU-3’.
126. E. coli consensus sequence for ribosome binding is 10 bases ahead of start codone.
127. Kozak consensus sequence 5’-ACCAUGG-3’ is found in all bacteria.
128. For formation of peptide bond the required enzyme called peptidyl transferase is a ribozyme.
129. For formation of peptide bond the required enzyme called peptidyl transferase is part of 23s
rRNA in E. coli.
130. Kozak consensus sequence always contains start (initiation) codone.
131. About 97% of human genome can be mutated without any significant change/ effect.
132. All those mutation which occur in extragenic DNA are virtually silent.
133. Synonymous change in codone has significant change on coding function of genome.
134. Non-synonymous change has a little significant change in function of protein coded.
135. Molecular clock is used to measure time in respect to gene divergence which led to
speciation.
136. At pH 7.0 DNA is an acid because it contains net negative charge.
137. Nucleotide polymer containing only ≤50 nucleotides are named oligonucleotides.
138. DNA synthesis proceeds in a 5’ to 3’ direction and is semi-discontinuous.
139. Lagging strand is synthesized in opposite direction to movement of forks.
140. Some aminoacyl tRNA synthetases have proofreading ability to select the correct amino acid
to be attached with the specific tRNA.
141. During translation, polypeptide synthesis begins at the amino terminal end.
8
142. DNA in the bacteria always remains attached to a point on plasma membrane.
143. Points of attachment of replicating bacterial DNA do lie on either side of the plane of future
cell cleavage.
144. In yeast nuclear membrane vanishes during mitosis similar to plants and animals.
145. A structure similar to centromere is also present in bacteria to help the proper segregation of
dividing DNA.
146. Yeast cells may be diploid or haploid.
147. Haploid cells of yeast can reproduce as efficiently as diploid cells.
148. All bacteria are haploid in nature.
149. Z DNA is better substrate than B DNA for DNA polymerase.
150. Tertiary structure is more commonly seen in Z form of DNA.
151. There is about one Z DNA segment in E. coli for each 18 kb DNA.
152. Runs of G-C may results into formation of Z DNA.
153. Mutation rate is often high in slow growing cultures.
154. Frame shift hotspots are often some specific repeats.
155. Most bases in DNA are in keto form.
156. Replication of cross linked DNA (induced by mitomycin or dimethyl psoralen) yields
daughter DNA molecules having gaps however these molecules normally replicate further to
produce mutants.
157. Long patch repair method of repair of UV damaged DNA is more effective against the
pyrimidine-(6-4)-pyrimidone structure.
158. When SOS repair pathway is ineffective to repair DNA damage short patch repair pathway is
induced to restore the life of the affected cell.
159. T series of phages have members numbered from T1 to T7.
160. All members of T series phages are quite similar in structure and action.
161. DNA of some phages is also present in their tails too along with head.
162. Hydroxy-methyl-cytosine in T4 phage offers a reactive site for attachment of glucose
molecule with DNA.
163. rT4 phage infection of E. coli is rapidly susceptible to lysis inhibition and thus produces
turbid plaque.
164. Lytic phages are also called intemperate phages.
165. Yeast and fungal viruses are transmitted through only cell fusion and rarely have an
independent state of existence.
166. Similar to T4 phage T2 and T6 phages of E. coli has their all hydroxy-methyl-cytosine
residues conjugated with glucose.
9
167. Nearly all glucosylated hydroxy-methyl-cytosine residues of T2 phage have only one glucose
molecule is attached with each residue in α configuration.
168. Ff group of phages infect only male bacteria.
169. Ff phages insert their DNA into bacteria through F pili.
170. Ff phages have ssDNA and are filamentous in structure.
171. Ff phages replicate through rolling circle mechanism.
172. RNA phages are male specific phages.
173. Bacteria infected with DNA phages are resistant (immune) to ssRNA phages.
174. Reversion rate for mutation in RNA viruses in quite high for RNA phages than for DNA
phages.
175. Φ29 phage of Bacillus subtilis is one of the smallest DNA phage in terms of both head size
and DNA size.
176. Cryptic prophage produces turbid plaques.
177. Temperate phages produce turbid plaques.
178. ssRNA phages produce turbid plaques due to incomplete lysis of the infected bacteria.
179. Nicked DNA circles resulting from the presence of the cohesive ends in phage DNA is called
Hershey circles.
180. Regardless of whether infection is destined to be lytic or temperate, Mu (µ) DNA always
integrates into the host chromosome.
181. After induction, Mu replicates as an autonomous molecule.
182. Generalized transducing particles may be the progeny of both virulent and temperate phages.
183. Specialized transducing particles may be the progeny of both virulent and temperate phages.
184. Specialized transducing particles may be the progeny of only integrative temperate phages as
λ.
185. Specialized transducing particles always contain bacterial DNA bound to one end of phage
DNA by covalent linkages.
186. About 90% of all generalized transducing DNA introduced into a host cell fail to recombine
and remain as a persistent, circular, non replicating DNA.
187. Abortive transductants may also express the transduced genes.
188. Abortive transductants always breed pure.
189. Both specialized and generalized transducing particle originate due to mistake by enzyme
system responsible for excision and packaging phage DNA.
190. Plasmid or chromosomal DNA added into non-sterile soil is degraded slowly and some times
may transform Pseudomonas stutzeri present there.
191. Most naturally competent bacteria can regulate their competence.
10
192. Competence of Bacillus subtilis is not necessarily associated with sporulation.
193. Both competent and non-competent cells can bind dsDNA on their cell wall.
194. Divalent cations are necessary for processing the surface bound transforming DNA.
195. During transformation with chromosomal DNA, it is incorporated into host DNA by
transposition.
196. Recombination process bocks the replication.
197. Transformed cultures are more stable to detrimental effects of UV radiation.
198. Both electrotransformation and chemical transformation of E. coli make majority of the cell
nonviable.
199. All genetically transformable cells can undergo transfection.
200. It is more difficult to transform naturally competent bacteria with linear DNA then with
plasmid DNA.
201. Excision of F plasmid DNA from chromosome of Hfr strains results into loss of sex pili.
202. R65 plasmid of IncP1 can be transferred to any of the gram negative bacteria.
203. Conjugative FP plasmids of Pseudomonas spp. also code for a specific sex pili.
204. During pheromone mediated conjugation usually occurring in Enterococcus faecalis
pheromone is produced by plasmid carrying bacteria.
205. Presence of plasmid in Streptomyces is often associated with pock formation.
206. Presence of plasmid in Streptomyces is often associated with inhibition or delayed formation
of mycelia and spores.
207. Ti plasmid of Agrobacterium is usually a non-conjugative plasmid.
208. Archaea also have conjugative plasmids.
209. Some colicin plasmids are conjugative.
210. Cloning of DNA in a plasmid may retard or even stop plasmid’s replication process.
211. Intercalating dyes stops replication of F plasmid but leave chromosomal DNA unaffected.
212. Large plasmids are independent in their host for initiation of their DNA replication.
213. Smaller plasmids uses more host functions than larger ones for their replication.
214. Col E plasmids do not code for any function of their DNA replication.
215. When F plasmid is inserted near oriC in host chromosome only then they can maintain
correct DNA/cell mass ratio to activate its replication origin.
216. R100 and ColE1 undergo θ replication having only unidirectional replication.
217. The single strand origin of replication is essential for replication of plasmid.
218. Partitioning is the act of segregating DNA molecules of only selected type so that there is at
least one copy of DNA molecule for each daughter cell.
11
219. Method of partitioning is same for R1 and F as for bacterial chromosome because they are
low copy number plasmids.
220. sop (stability of plasmid) is the gene whose function used by F plasmid for partitioning.
221. par (partition) is the gene whose function used by R1 plasmid for partitioning.
222. Common fimbriae are more strongly attached to bacterial cells than sex pili.
223. SDS can easily remove common fimbriae than F pili.
224. Broad host range plasmids are conjugative ones.
225. DNA from Bacillus cereus cloned in E. coli is easily expressed but E. coli genes are not easy
to be expressed in B. cereus.
226. Bacterial DNA cloned in Ti plasmid is not expressed well in plants unless an appropriate
plant promoter is fused with it.
227. Phages like T7 and T3 which code for their own RNA polymerase recognizes its own
particular promoter only.
228. Phages like T7 and T3 RNA polymerase does not recognizes bacterial or other phage
promoters.
229. Sensitivity of Salmonella system is more than that of E. coli systems in Ames mutagenicity
tests.
230. Both E. coli and Salmonella systems of Ames tests use reversion of histidine mutation as
measure of mutagenic potency.
231. Ampicillin and cycloserine kills only growing bacteria.
232. Intercalating agents mostly induces frame shift mutations.
233. SOS induction can induce base substitution, frame shift, deletion and insertion mutations.
234. Base analogs are used to induce frame shift mutations.
235. Mutator genes are the mutated genes encoding DNA repair functions.
236. Frequency of spontaneous mutations depends on growth rate.
237. Competent cells made by PEG-DMSO protocol maintains their efficiency of transformation
longer than those made with CaCl2 method at -70oC.
238. In electrotransformation partially purified plasmids can also be transported to different
bacteria.
239. In electroporation number of transformants increases with amount of DNA used for
transformation but is little affected by cell density.
240. In electroporation number of transformants is not affected by temperature of transformation
within normal range of growth temperature.
241. Number of transformants in electroporation increases by 100 times if you use precooled (0-
4oC) cuvettes.
12
242. When you observe >50 bp bands during dideoxy DNA sequencing in all four lanes or in
more than one lane at the same level it indicate that your DNA may be either contaminated or
having secondary structures.
243. Non-specific products in PCR are more when too little amount of temple DNA is used.
244. Frequency of true reversion for deletion mutations is around 10-12.
245. Nine genes are needed in his operon, the strain having merodiploid genotype hisG+D-C+H+A-
F+I+/ hisA+F-I+ wil be histidine auxotroph.
246. Phenotype of a merodiploid leu(Ts)/ leu + will be a temperature sensitive phenotype.
247. In gradient separation of different species macromolecules sucrose plays an important direct
role.
248. In gradient separation of different species macromolecules CsCl2 plays no important direct
role.
249. The most common type of electrophoresis in molecular genetics is zonal electrophoresis.
250. Supercoiled DNA migrates faster than linear DNA for the same molecular weight.
251. Lysogenic infection of E. coli with P1 phage always induces non-integrative lysogeny.
252. During Lysogenic infection of E. coli with P1 phage, the phage DNA exists in circular form.
253. For a normal dsDNA phage burst size is between 50 to a few hundred.
254. Only dsDNA phages can undergo lysogenic cycle of growth.
255. Phage receptor are usually proteins or carbohydrate molecules on the surface of bacteria
which has some normal function to bacteria rather than just being the phage anchors.
256. After infection with virulent phages bacteria is often unable to synthesize its own DNA or
RNA.
257. RNA phages always codes for their replication enzymes.
258. Lysogen is resistant to infection with the phage which made it lysogen.
259. More than 90% of known phages are temperate.
260. More than 90% of known phages are dsDNA type.
261. Phages grow more rapidly when bacteria are suddenly exposed to starvation.
262. Dormant bacteria have very little or no mRNA.
263. Virulent phages have more stable head and tail structure than temperate phages.
264. In bacteria most of the genes are transcribed from only one strand of DNA.
265. In bacteria most mRNA molecules are destroyed within few minutes of their synthesis.
266. In bacteria mostly primary transcript of DNA act as mRNA.
267. In Archaea primary transcript of DNA act as mRNA.
268. Translation of RNA starts from 5’ terminus itself and ends at 3’ terminus.
269. Translation of RNA starts from 3’ terminus itself and ends at 5’ terminus.
13
270. Polypeptides synthesis starts from amino terminus to carboxy terminus.
271. There are 61 codones corresponding to amino acids in bacteria.
272. There are only three tRNA bearing anti-codone for stop codones in bacteria..
273. There is only one rRNA molecule in ribosome of bacteria.
274. Some E. coli RNA transcripts undergo post processing to produce different RNA molecules.
275. Synthesis of enzymes which are required continuously throughout the life of an organism
they are under strict coordinated control.
276. It is often essential for a repressor gene to be present near operator for regulation of gene.
277. In anti-termination RNA polymerase is altered to ignore the termination sites.
278. Attenuator is specific protein binding site on DNA.
279. All proteins translated from a single polycistronic mRNA necessarily made in a fixed
proportion
280. The DNA damage to bacteria can be detected using survival curve studies.
281. SOS repair is responsible for tolerance of damage done by mutagens in DNA.
282. SOS repair system is a bypass system that does not allow DNA chain growth across damaged
segment at the cost of fidelity.
283. Unprepared DNA induces SOS mechanism.
284. SOS repair is the major cause of mutagenesis by UV radiation and chemicals.
285. ‘Rats deserting a sinking ship’ also apply to the lysogenic induction of lysogens.
286. UV irradiated λ phage produce more infective centres in lawn of healthy E. coli than in lawn
of UV irradiated E. coli.
287. On UV irradiation thymine dimers are formed between thymines present on opposite strand
leading to cross linking.
288. Incision step is part of the excision repair system.
289. In E. coli a repair endonucleases identifies thymine dimers and make incision to cleave away
the bases only.
290. C to U conversion often leads to mutation while C to T conversion rarely leads to mutation.
291. During growth of ts-mutants at high temperature the affected gene is not transcribed and
translated into the desired protein.
292. Cold sensitive mutations are more common than the ts-mutants.
293. Cold sensitive mutations are less common than the ts-mutants.
294. Cold sensitive mutants are often defective in molecular interaction required for the reaction
than function per se.
295. Ts-mutants and cs-mutants are often not found in Bacteriophages.
296. Presence of BU-NTP inhibits synthesis of dCTP.
14
297. EMS affects only G and T.
298. EMS affects more to A & C than G & T.
299. Alkylating agents leads to depurination which is the most important cause of their mutagenic
effect.
300. Treatment of phages at pH 4 may cause depurination which often leads to transversion.
301. Hydroxylamine specifically reacts to cytosine.
302. Hydroxyl amine induced mutations can be reverted with hydroxyl amine.
303. Base pair change always leads to mutation.
304. Base pair change always leads to phenotypic change.
305. ts-mutants are often produced as result of missense mutations.
306. 5-Bromouracil causes mutation due to its tautomerization effect leading to mis-incorporation
of bases.
307. Frame shift mutations always cause a phenotypic change.
308. Deletion mutations are irreversible.
309. Chain termination mutations are easily reverted through intragenic substitution.
310. Most tRNA genes are multicopy genes.
311. Nitrous acid causes both AT to GC and GC to AC transition.
312. Revertants are often temperature sensitive mutants.
313. Plasmids are often dispensable to their host.
314. Unlike phages, plasmids often don’t depend on their host for most metabolic and
reproduction.
315. R plasmids are often high copy number plasmids.
316. Cloning vector plasmids are usually low copy number plasmids.
317. DNA of plasmid transfers through hollow of the conjugative pilus (F, R pilus).
318. Motive force for conjugal transfer of plasmid DNA is DNA replication.
319. Most of the R plasmid containing bacterial cells are competent donors.
320. F plasmid can integrate into chromosomal DNA through homologous recombination.
321. There may be a plasmid even without a single coding sequence (gene).
322. G loop affects the host range of Mu but is not relevant to transposition.
323. Mu DNA is never found free in the cell.
324. The DNA in the Mu phage particles rarely contains terminal bacterial DNA.
325. Bacterial cells usually contain a stable number of copies of a transposon.
326. Mutations that lead to reduced activity of lysozyme lead to formation of turbid plaques.
327. Mutations that permit bacteria to grow on medium containing inhibitory compounds like
acridine leads to formation of turbid plaques on specified medium.
15
328. In bacteria gene flow is both horizontal and vertical.
329. Bacterial chromosome is about 1000 times longer than the bacterium and occupies about 1%
of space in bacterial cell.
330. Bacterial DNA is uncoiled by an enzyme Gyrase.
331. In genetic mapping of three gens in bacteria we need not to know about recessiveness or
dominance of the alleles.
332. During conjugation oh Hfr and F- strains part of bacterial chromosome is transferred.
333. λ phage has operon and regulatory proteins to regulate for lytic or lysogenic cycle in bacteria.
334. In Moraxella bovis change is pilin protein is mediated through inversion of a 2 kb
chromosomal DNA between two ORF of pilins.
335. E. coli vir pilin is encoded by 10 kb region in chromosomal DNA.
336. Enterohaemolysin gene in E. coli is a phage encoded fuction.
337. All the important virulence factors of Bacillus anthracis are encoded in genes on plasmids.
338. Most of the virulence of Salmonella enterica serovars is due to vir plasmids.
339. Smooth and rough phenotypes in Shigella sonnei depends on the presence of 120 MDa
plasmids.
340. tox gene of Clostridium tetani is a plasmid encoded gene.
341. lcr plasmid is found in Bacillus anthracis which get expressed as low calcium response.
342. Capsular antigen of Yersinia pestis is of plasmid origin.
343. Pili of Dichelobacter nodosus is encoded through plasmid gene.
344. Length of O-side cahins of a. S. Dublin, is a plasmid encoded factor.
345. Serum resistance, complement killing resistance, invasiveness and multiple drug resistance
genes of Salmonella are on plasmids.
346. Genes for Congo red dye binding factor of Shigella are on chromosomes while for
Shigatoxins are on plasmids.
347. Pesticin of Yersinia pestis is bacteriocin encoded by a 10 kb plasmid which also has genes for
Pla (plasminogen activator) antigen.
348. Plasmid encoding 15-17 kDa antigen of Rhodococcus equi is is similar a vir plasmid of
Yesrsinia pestis.
349. Virulence plasmids of Salmonella, E. coli and Yersinia pestis get lost on growing bacteria at
42oC.
350. Clostridium perfringens type D causing enterotoxemia due to epsilon toxin gene for the toxin
is on a megaplasmid.
351. Plasmids of sporeforming bacteria are unusually large in size.
352. All but botulinum toxin A, B, E and F are plasmid mediated characters.
16
353. BoNT/E gene of Clostridium botulinum is plasmid encoded while the same gene in C.
butyricum is on chromosome.
354. Insect toxin gene of Bacillus thurgiensis is on a megaplasmid.
355. Food poisoning strains of Clostridium perfringens type A human type and of animal origin
produces enterotoxin encoded by cpe gene which present on chromosome in first and on
plasmid in the second is a part of composite transposon having IS 1470 element.
356. Among all the linear plasmids ends are covalently closed for stability.
357. Killer phenotype of yeasts (Kluyveromyces lactis) is mediated through linear plasmids.
358. Similar to hairpin plasmids of prokaryotes some animal viruses also have covalently closed
termini.
359. Incorporation of centromere or telomere DNA in plasmids of eukaryotes can make the
plasmids more stable in absence of selection pressure.
360. Genes for synthesis of most of the β-lactam antibiotics by Actinomyces are located on linear
plasmids.
361. All linear plasmids present in mitochondria shows maternal inheritance similar to
mitochondrial genome.
362. Some linear plasmids may be associated with mitochondrial genome rearrangements also and
may induce phenotypic effects.
363. Linear mitochondrial plasmids may have one to six ORFs encoding unknown proteins.
However ORFs encoding DNA and RNA polymerases are often constant finding.
364. All linear plasmid have a common structural feature called an invertron.
365. Linear mitochondrial plasmids are present in many fungi and in some plants, and in most
animal cells.
366. Plasmids in Mycobacterium xenopi, M. branderi, and M. avium have been reported to have
plasmids with linear topology.
367. Invertrons of linear plasmids show hypervariability in structure and sequence among
members of Actinomycetales (Mycobacterium, Rhodococcus, and Streptomyces).
368. Invertron of linear plasmids of Actinomycetales always have hairpin morphology.
369. Long TIRs (terminally inverted repeats) are absolutely necessary for replication and
maintenance of linear plasmids.
370. Smaller linear plasmids are easier to identify than the larger linear plasmid.
17
Answers. 1. T, 2.T, 3. F, 4. T, 5. F, 6. T, 7. T, 8. T, 9. T, 10. T, 11. T, 12. T, 13. T, 14. T, 15. T, 16.
F, 17. F, 18. T, 19. T, 20. T, 21. F, 22. T, 23. T, 24. F, 25. T, 26. F, 27. T, 28. T, 29. F, 30. T, 31. T,
32. F, 33. T, 34. T, 35. T, 36. T, 37. T, 38. T, 39. T, 40. T, 41. F, 42. T, 43. F, 44. T, 45. F, 46. T,
47. T, 48. T, 49. T, 50. T, 51. T, 52. F, 53. F, 54. T, 55. T, 56. T, 57. T, 58. T, 59. F, 60. T, 61. T,
62. T, 63. F, 64. T, 65. T, 66. T, 67. T, 68. F, 69. T, 70. T, 71. F, 72. T, 73. F, 74. T, 75. T, 76. T,
77. F, 78. T, 79. T, 80. F, 81. F, 82. T, 83. T, 84. T, 85. T, 86. F, 87. T, 88. F, 89. T, 90. T, 91. T,
92. F, 93. T, 94. T, 95. F, 96. F, 97. F, 98. F, 99. T, 100. T, 101. T, 102. T, 103. F, 104. T, 105. T,
106. F, 107. T, 108. T, 109. T, 110. F, 111. T, 112. T, 113. T, 114. F, 115. F, 116. T, 117. T, 118. T,
119. T, 120. F, 121. T, 122. F, 123. T, 124. T, 125. T, 126. T, 127. F, 128. T, 129. T, 130. T, 131.
T, 132. T, 133. F, 134. T, 135. T, 136. T, 137. T, 138. T, 139. T, 140. T, 141. T, 142. T, 143. T,
144. F, 145. F, 146. T, 147. T, 148. F, 149. F, 150. T, 151. T, 152. T, 153. F, 154. T, 155. T, 156. F,
157. F, 158. F, 159. T, 160. F, 161. T, 162. T, 163. F, 164. T, 165. T, 166. F, 167. T, 168. T, 169. F,
170. T, 171. T, 172. T, 173. T, 174. T, 175. T, 176. F, 177. T, 178. F, 179. T, 180. T, 181. F, 182.
T, 183. F, 184. T, 185. T, 186. T, 187. T, 188. F, 189. T, 190. T, 191. T, 192. T, 193. T, 194. T,
195. F, 196. T, 197. T, 198. T, 199. T, 200. F, 201. F, 202. T, 203. F, 204. F, 205. T, 206. T, 207. F,
208. T, 209. T, 210. T, 211. T, 212. F, 213. T, 214. T, 215. F, 216. T, 217. T, 218. F, 219. F, 220.
T, 221. T, 222. T, 223. F, 224. T, 225. T, 226. T, 227. T, 228. T, 229. F, 230. F, 231. T, 232. T,
233. T, 234. F, 235. T, 236. T, 237. T, 238. T, 239. F, 240. F, 241. T, 242. T, 243. T, 244. F, 245. T,
246. F, 247. F, 248. F, 249. T, 250. T, 251. T, 252. T, 253. T, 254. T, 255. T, 256. T, 257. T, 258.
T, 259. T, 260. T, 261. F, 262. T, 263. T. 264. F, 265. T, 266. T, 267. F, 268. F, 269. F, 270. T, 271.
T, 272. F, 273. F, 274. T, 275. F, 276. F, 277. T, 278. F, 279. F, 280. T, 281. T, 282. F, 283. T,
284. T, 285. T, 286. F, 287. F, 288. T, 289. F, 290. F, 291. F, 292. F, 293. T, 294. T, 295. F, 296. T,
297. F, 298. F, 299. F, 300. T, 301. T, 302. F, 303. T, 304. F, 305. T, 306. T, 307. T, 308. T, 309. F,
310. T, 311. T, 312. T, 313. T, 314. F, 315. F, 316. F, 317. F, 318. F, 319. F, 320. T, 321. T, 322. T,
323. T, 324. F, 325. T, 326. T, 327. T, 328. T, 329. F (~100 times longer and occupies ~10%
space), 330. F, 331. T, 332. T, 333, T, 334. T, 335. F, 336. T, 337. T, 338. F, 339. T, 340. T, 341. F
(Y. pestis), 342. T, 343. F, 344. T, 345. T, 346. F, 347. T, 348. T, 349. F, 350. T, 351. T, 352. F (C
& D are phage mediated), 353. F (order is reverse), 354. T, 355. T, 356. F, 357. T, 358. T, 359. T,
360. F, 361. F (e.g. in Brassica 11.6-kb plasmid of mitochondria does not show maternal
inheritance), 362. T, 363. T, 364. T, 365. F, 366. T, 367. F (conserved), 368. F, 369. F, 370. T
18
Chapter. 2. Multiple choices in Microbial Genetics
1. Biomolecules first arose by a. genetic evolution, b. chemical evolution, c. biological evolution,
d. randomly.
2. According to Oparin’s theory source of energy for different chemical reaction leading to
formation of ‘primordial soup’ was a. heat from volcanoes, b. electrical discharges, c. solar
energy, d. all three.
3. Biological evolution began about a. 3 million years ago, b. 3 billion years ago, c. 3 trillion years
ago, d. 30 million years ago.
4. Translation and transcription is generally coupled in a. Bacteria, b. viruses, c. plants, d. animals.
5. Many proteins involved in genome expression are in complex form called a. primary structure,
b. secondary structure, c. tertiary structure, d. quaternary structure.
6. Melting of DNA/RNA involves disruption of a. Hydrogen bonds between pairing bases, b.
disruption of hydrophobic interactions between stacking bases, c. none, d. both.
7. The replicon encoding genes essential for the cell survival is called, a. genome, b. chromosome,
c. codone, d. proteome.
8. Different replicons in bacterial cells are a. Plasmid DNA, b. Bacterial chromosome, c. Phage
DNA, d. all.
9. mRNA is a. double stranded, b. single stranded, c. triple stranded, d. none.
10. During transcription newly synthesized RNA forms base pairing with coding sequence a. for
short distance, b. for whole length, c. for no length, d. half length.
11. During synthesis of daughter DNA strands in vivo primers are a. short RNA strands, b. short
DNA strands, c. both a and b, d. None.
12. Mating pair in conjugation can be separated by a. on a shaking platform used to grow bacteria,
b. pipetting, c. vortaxing, d. none.
13. Each DNA has a. 1, b. 2, c. 3, d. 6 reading frames.
14. DNA strands are always synthesized in a. 5’to 3’ direction, b. 3’ to 5’ direction, c. both
directions, d. it varies strand to strand.
15. DNA of T-even phages of E. coli does not have a. adenine, b. cytosine, c. thymine, d. guanine.
16. To protect DNA of T-even phages and some Bacillus subtilis phages in host cells their DNA is
a. methylated, b. glucosylated, c. brominated, d. acylated.
17. The special base present in DNA of T-even phages is a. uracil, b. bromouracil, c.
bromothymine, d. 5-hydroxymethylcytosine, e. 5-hydroxymethyluracil.
18. DNA of some Bacillus subtilis phages do not have a. Adenine, b. cytosine, c. thymine, d.
guanine.
19
19. DNA of some Bacillus subtilis phages may have special base, it is, a. uracil, b. bromouracil, c.
bromothymine, d. 5-hydroxymethylcytosine, e. 5-hydroxymethyluracil.
20. Chromosome of most phages is a. linear, b. circular, c. linear paired.
21. Terminal redundancy is found in a. T-even phages, b. λ phages, c. T1-T3 family of phages, d.
P22 phages.
22. Which one has closed circular chromosome? A. T-even phages, b. lamda (λ) phages, c. T3
family of phages, d. ΦX174 phages.
23. To protect DNA of T-even phages and some Bacillus subtilis phages in host cells their DNA is
glucosylated, the glucose groups are attached with a. 5-hydroxymthyl residue of cytosine/
uracil, b. adenine, c. cytosine, d. guanine.
24. Introns may frequently be present in DNA of a. bacteria, b. plasmids, c. transposons, d. phages.
25. Which one is not a flagellotropic phage a. χ1, b. PBS1, c. T4, d. None.
26. M13 phage has extensively been used for cloning purpose because it a. has circular DNA, b. has
dsDNA, c. replicates via a dsDNA intermediate, d. replicate via rolling circle mechanism.
27. The phage which behaves like a transposon is a. χ1, b. ΦX174, c. Mu, d. λ phage.
28. Course of one step growth cycle of lytic phages can not be modified by a. change in nutritional
conditions, b. with >1MOI, c. addition of protein synthesis inhibitors, d. presence of 5-methyl
tryptophan, e. none.
29. Burst size of lytic phages can be increased by a. reducing nutritional supply, b. with >1 MOI, c.
reinfection of infected bacteria, d. none.
30. Burst size of lytic phages can not be increased by a. optimizing nutritional supply, b. by
increasing MOI, c. reinfection of infected bacteria, d. addition of 5-methyltryptophan, e. by
addition of chloramphenicol.
31. Primary encounter between bacteria and phages in most cases is through a. Chemotaxis, b.
random collision, c. ionic attraction, d. hydrophobic surface interactions.
32. For high copy number plasmids ratio of plasmid: chromosome DNA is usually a. 2-5, b. 5-10,
c. 10-15, d. 15-20.
33. Lysis from without means, a. lysis of bacteria without phage infection, b. lysis of bacteria
releasing large number of mature phages, c. releasing of only few mature phages, d. releasing
immature non infective phages.
34. T-even phages DNA is injected into E. coli at the rate of a. 3 kb/s, b. 7 kb/s, c. 9 kb/s, d. 1 mb/s.
35. During the study on λ lysogeny it may be observed that a few plaques lack central zone of
turbidity, it may be due to mutation in a. phage, b. bacteria, c. both, d. none.
20
36. During the study on λ lysogeny it may be observed that a few plaques lack central zone of
turbidity, due to mutation in phage genes named a. cI, cII, cIII , b. intI, intII, intIII, c. both a &b,
d. none.
37. A 30 kDa acidic protein called immunity repressor or immunity substance is coded by a. cI, b.
cII, c. cIII, d. cIV.
38. During vegetative replication of T-even phages in E. coli one found many replication forks for
rapid multiplication of phage DNA, their number may reach up to a. 30, b. 60, c. 90, d. 120.
39. For protection and replication of λ phage DNA in host it is a. circularized, b. glycosylated, c.
methylated, d. glucosylated.
40. For replication of DNA in host λ phage opt for a. θ replication, b. σ replication, c. rolling circle,
d. all.
41. Proportion of pfu to total count of morphologically normal particles is called, a. MOI, b. EOP,
c. PFU, d. CFU.
42. We can induce reversion of lysogenized phage to the lytic cycle by exposing the culture to a.
penicillin, b. >1% glucose, c. alkylating reagent, d. UV light, e. mitomycin C.
43. Exposure of a bacterial culture to sub-lethal doses of UV light increases levels of a protein more
than 17 times, this protein is a. RecA, b. RecB, c. Helicase, d. RecC.
44. Ability of cI+ product to act as immunity substance in a lysogen is abolished on exposure to
small amount of UV due to a. RecA, b. RecB, c. Helicase, d. RecC.
45. Okazaki fragments are synthesized by a. ribosome, b. primosome, c. DNA polymerase III, d.
DNA polymerase I.
46. Bifunctional alkylating reagents are more powerful mutagen because they cause a. deamination,
b. cross linking of bases in same strand, c. cross linking of bases in opposite strands, d.
oxidative deamination.
47. Okazaki fragments are synthesized in direction a. opposite, b. same, c. any d. no relation of the
replication fork.
48. Synthesis of Okazaki fragments is regulated by a. ribososme, b. secondosome, c. primosome, d.
mesosome.
49. Okazaki fragments are formed for synthesis of a. lagging strand, b. leading strand, c. coding
strand, d. non-coding strand.
50. In PCR which type of DNA polymerase is used a. type I, b. Type II, c. Type III, d. Type IV.
51. A new nucleotide can be added in to a DNA strand at a. 3’ hydroxyl group, b. 5’ hydroxyl
group, c. 5’ phosphate group, d. none.
52. Okazaki fragment length may be about a. 100 nucleotide, b. 1000 nucleotide, c. None, d. both.
21
53. Every new nucleotide added to growing strand is held in position by a. newly formed
phosphodiester bond, b. hydrogen bonds with its partner in template strand, c. base stacking
interactions with neighbours, d. all three
54. During replication the possibility of error nucleotide inclusion is one in a. 109-10, b. 1011-12, c.
105-6, d. 107-8.
55. Considering the fidelity of DNA replication there is only one possibility of mutation in E. coli
per a. 100-200 replications, b. 1000-10000 replications, c. 500-900 replications, d. 105 to 106
replications.
56. Replication of DNA starts at an origin of replication and replication takes place a. left to origin,
b. right to the origin, c. in both directions.
57. In one minute DNA polymerase I can add about a. 150 bases, b. 300 bases, c. 600 bases, d.
1200 bases.
58. Proofreading ability is present in DNA polymerase a. type I, b. Type II, c. Type III, d. all three.
59. Proofreading by most polymerases takes place in a. 3’ to 5’ direction, b. 5’ to 3’ direction, c.
both direction, d. Varies.
60. 5’ to 3’ exonuclease activity is present in DNA polymerase a. type I, b. Type II, c. Type III, d.
all three.
61. Most rapid polymerization is done by DNA polymerase a. type I, b. Type II, c. Type III, d. type
IV.
62. Maximum processivity (activity) resides with DNA polymerase a. type I, b. Type II, c. Type III,
d. type IV.
63. Most slow activity of polymerization is in DNA polymerase a. type I, b. Type II, c. Type III, d.
type IV.
64. When 5 to 3’ exonuclease domain of DNA polymerase I is removed by mild protease action the
remaining fragment of DNase possess a. polymerization activity, b. proof reading activity, c.
both, d. none (c) and the remaining fragment is called a. Klenow fragment, b. Raziu fragment,
c. replisome, d. all.
65. Besides DNA polymerase DNA replication requires a. Helicase, b. topoisomerases, c. primases,
d. ligases, e. all.
66. RNA primers in E. coli are removed from DNA strand by DNA polymerase a. type I, b. Type
II, c. Type III, d. type IV.
67. In E. coli origin also called as oriC is consisted of a. 145 bp, b. 245 bp, c. 345 bp, d. 45 bp.
68. Replication fork is formed by the action of a. Helicase, b. topoisomerase, c. ssb, d. all.
69. DNA replication is a. conservative, b. semi conservative, c. non-conservative, d. all.
22
70. Leading strand synthesis begins with the action of a. DNA polymerase type I, b. DNA
Polymerase Type II, c. Primase, d. RNA polymerase.
71. Length of RNA primers is usually a. 6-60, b. 10-60, c. 10-100, d. 3-13 nucleotides.
72. Synthesis of lagging DNA strand requires DNA polymerase a. type I, b. Type II, c. Type III, d.
a & b, e. a & c, f. b & c.
73. Lagging strand DNA synthesis is initiated by a. DNA polymerase type I, b. DNA Polymerase
Type III, c. Primase, d. RNA polymerase.
74. For separation of two circular DNA molecules after completion of replication takes place with
the help of a. DNA polymerase type I, b. DNA Polymerase Type II, c. Primase, d.
topoisomerase I, e. topoisomerase IV.
75. Origin of replication in eukaryotes is called a. oriA, b. ARS, c. oriC, d. RFC.
76. ARS (autonomously replicating sequences) are found in chromosomes of a. eukaryotes, b.
viruses, c. bacteria, d. plasmid.
77. To screen potential carcinogens Ames test is done in a. bacteria, b. viruses, c. protozoa, d.
yeasts.
78. Mismatch repair system discriminate between old and new strands through tagging the old
DNA with methyl groups at N6 position in GATC sequence of a. adenine, b. guanine, c.
cytosine, d. thiamine.
79. In process of DNA hybridization in mild denaturing buffer (formamide) maximum hydrogen
bonding is obtained at a. 30oC, b. 40 oC, c. 50 oC, d. 60 oC.
80. Methyl directed mismatch repair system can correctly repair mismatches up to a distance (from
hemi-methylated GATC) of a. 100 bases, b. 1000 bases, c. 10000 bases.
81. SOS response brings to halt a. DNA synthesis, b. RNA synthesis, c. Protein synthesis, d.
Carbohydrate synthesis.
82. SOS responses results in to increased levels of a. DNA, b. lipids, c. Protein, d. Carbohydrate.
83. Which of the following is DNA recombination event a. Homologous genetic recombination, b.
site-specific recombination, c. DNA transposition, d. all?
84. Hopping genes, first observed in maize by Barbara McClintock in year 1950 were originated as
the result of a. Homologous genetic recombination, b. site-specific recombination, c. DNA
transposition, d. all.
85. Genetic transfer in bacteria is a. unidirectional, b. bidirectional, c. multidirectional, d. none.
86. False diploids stage in bacteria formed during transfer of genetic material is called a. zygote, b.
merozygote, c. merizygote, d. haplozygote.
87. Holliday intermediates are formed during a. Homologous genetic recombination, b. site-specific
recombination, c. DNA transposition, d. all.
23
88. Immunoglobulin genes are assembled by a. Homologous genetic recombination, b. site-specific
recombination, c. DNA transposition, d. programmed recombination.
89. Site specific recombination requires an enzyme recombinase which identifies a unique DNA
sequence of a. 2-20 bases, b. 20-200 bases, c. 200-2000 bases, d. None.
90. Integration of lamda phage into E. coli chromosome takes place with the help of lamda
integrase this integration is the example of a. Homologous genetic recombination, b. site-
specific recombination, c. DNA transposition, d. all.
91. Which of the following has both informational and catalytic functions a. proteins, b. RNA, c.
DNA, d. nucleolipids, e. nucleoproteins?
92. RNA polymerase is most active in presence of a. double stranded DNA, b. single stranded
DNA, c. coding strand, d. all.
93. RNA synthesis direction is a. 3’ to 5’, b. 5’ to 3’, c. both, d. none.
94. RNA polymerase does not require primers but for initiation it needs specific sequences called a.
oriR, b. initiators, c. promoters, d. hologenes.
95. RNA synthesis usually starts with a. GTP or ATP, b. ATP and GTP, c. UTP & CTP, d. CTP or
UTP.
96. Template strand is called a. – strand, b. + strand, c. coding strand, d. replicating strand (a).
97. The non templet strand is also called a. – strand, b. duplicating strand, c. coding strand, d.
replicating strand.
98. Promoters are sequences responsible for start of RNA synthesis their distance from point of
initiation varies from a. 10-15 bp, b. 10-25 bp, c. 10-35 bp, d. 10-45 bp.
99. For most promoters in E. coli and related bacteria, the consensus sequence for the -10 region
also called Pribnow box is a. TATATA, b. TATAAT, c. ATATAT, d. ATATTA (b) and at -35
regions is a. TTGAGA, b. TTGACA, c. GACATT, d. TTCAGA.
100. Which of the following inhibit DNA dependent RNA synthesis by jamming the zipper
includes a. actinomycin D, b. acridine, c. rifampicin, d. alpha aminitin?
101. Which of the antibiotic inhibit RNA synthesis by binding to beta unit of RNA polymerase a.
a. actinomycin D, b. rifampicin, c. polymyxin B. d. alpha aminitin?
102. During conjugation transfer of DNA starts with a nick produced at a. oirA, b. oriC, c. oriN,
d. oriT.
103. During conjugal transfer of DNA a pore (channel) complex is formed between donor and
recipient with the help of a protein called a. pra, b. tra, c. hin, d. ret.
104. Rolling circle replication initiate at 3’ OH and proceed in direction a. 5’ to 3’, b 3’ to 5’, c.
any, d. none.
24
105. Donor and recipient bacteria are brought together with the a. formation of sex fimbriae, b.
extension of sex fimbriae, c. hydrophobic forces, d. retraction of sex pili.
106. In conjugation transfer of DNA is in form of a. ssDNA, b. dsDNA, c. both.
107. Which strand enters the recipient a. leading, b. lagging, c. templet.
108. For treatment of bacterial diseases phage therapy can be used in phage therapy we can use a.
lytic phage, b. nonlytic phage, c. both, d. none.
109. In amperometric assay for pathogen analysis in a system one can use a. lytic phage, b.
nonlytic phage, c. both, d. none.
110. Bacteriophages are a. obligate intracellular parasite, b. obligate extra cellular parasite, c.
facultative intracellular parasite, d. facultative extra cellular parasite.
111. Bacteriophages can be used for a. diagnosis of bacterial diseases, b. for treatment of
bacterial diseases, c. for prophylaxis, d. all.
112. The simplest Bacteriophages have nucleic acid to code about a. 70-100 genes, b. 3-5 genes,
c. 7-10 genes, d. 15-17 genes only.
113. Length of bacteriophages range between a. 24-200 nm, b. 24-200µm, c. 2.4 – 20 µm, 0.2-.24
mm.
114. T4 is among the a. smallest phages, b. largest phages, c. medium sized phages, d. spiral
phage.
115. Icosahedral head of phages have a. 8 sides, b, 12 sides, c 18 sides, d. 20 sides, e. 24 sides.
116. Bacteriophages attaches to bacteria through receptors these receptors may be a. proteins on
the outer surface of the bacterium, b. LPS, c. pili, d. lipoprotein, e. all, f. a & c.
117. Unit of bacteriophage count is a. cfu, b. pfu, c. bfu, d. vfu.
118. Number of Bacteriophages released after lysis of the bacterial cells may be as high as a. 10,
b. 1000, c. 1, d. 10000.
119. During eclipse phage components are synthesized which of the component is synthesized
first a. a. Structural proteins, b. phage DNA/RNA, c. both together.
120. During life cycle of a lysogenic phage the phage DNA in this repressed state is called a.
prophage, b. lysophage, c. lysogen, d. Qpahge.
121. The Bacteria containing a lysogenic phage in quiescent state is called a. prophage, b.
lysophage, c. lysogen, d. Q phage.
122. Phages those can either multiply via the lytic cycle or enter a quiescent state in the cell are
called a. lysogenic, b. temperate, c. lysogen, d. a &b, e. b &c, f. a & c.
123. Which one is not a lysogenic phage a. T4, b. lamda, c. p22, d. all?
25
124. A phage coded protein, called a repressor, which binds to a particular site on DNA, called
the operator, and shuts off transcription of most phage genes. The operator is part of a. bacterial
chromosome, b. phage chromosome, c. plasmid DNA, d. may be of all 3.
125. Immunity to super infection with the same phage is due to activation of a. repressor, b.
operator, c. sigma factor, d. lamda signal.
126. The stimulants which favour the termination of the lysogenic state include: a. desiccation, b.
exposure to UV or ionizing radiation, c. exposure to mutagenic chemicals, d. all.
127. To end the quiescent stage of lysogenic phage a protease is required called a. recA, b. recC,
c recT, d. rec E.
128. To end the quiescent stage of lysogenic phage a protease required is encoded by a. bacterial
chromosome, b. phage chromosome, c. plasmid DNA, d. may be of all 3.
129. The decision for lambda phage to enter the lytic or lysogenic cycle when it first enters a cell
is determined by the concentration of the a. repressor protein, b. cro protein c. both, d. none.
130. Toxin production by Corynebacterium diphtheriae is mediated by a gene carried by a.
phage, b. plasmid, c. transposon, d. bacterial chromosome.
131. The CTX phage has received special attention because it was the first a. filamentous phage,
b. RNA phage, c. lytic phage, d. lysogenic phage found to transfer toxin genes to its host.
132. µ phage have characters of a: a. transposon, b. plasmid, c. transposon, d. all.
133. During transformation the best size of DNA to give maximum transformation efficiency is
a. 5×105 Dalton, b. 5×106 Dalton, c. 5×107 Dalton, d. 5×108 Dalton.
134. A plasmid having Mu phage genes is used to construct operon and gene fusions and for in
vivo cloning is called a. Puc, b. Mud, c. Pvd, d. mudJ.
135. In principle, it is possible to find transposon insertions in only a. nonessential gene, b.
essential genes, c. virulence genes, d. toxin genes.
136. Insertion mutants can be recovered at high frequency after a. low-level, b. medium level, c.
high level, d. all types of mutagenesis.
137. Transposons can be delivered to bacteria through a. phages, b. plasmids, c. over expression
of transposase in trans, d. all three.
138. Insertion sequences are not present in a. IS elements, b. composite transposons, c.
noncomposite transposons d. none.
139. Genes for the degradation of hydrocarbons are carried by a. lytic transposons, b. catabolic
transposons, c. degradative transposons, d. non-composite transposons.
140. Wu formula is used for calculation of a. transposition frequency, b. transduction frequency,
c. mapping the genome, d. cotransduction frequency e. degree of supercoiling.
26
141. Twist and Writhe formula given by David Stump and Peter Watson is used to determine a.
transposition frequency, b. transduction frequency, c. mapping the genome, d. cotransduction
frequency e. degree of supercoiling.
142. Termination of transcription using GC rich hair pin loops occur in a. rho dependent, b. rho
independent, c. none, d. both types of termination of transcription.
143. tRNAs have a 7 bp stem that includes the 5'-terminal nucleotide and may contain non-
Watson-Crick base pairs, GU. Arm carrying this portion of the tRNA is called the acceptor arm
its role is to a. to carry specific amino acid, b. to identify the code in mRNA, c. to attach with
rRNA, d. none.
144. Two methods of termination of transcription are: rho dependent and rho independent. a.
true, b. false, c. neither true nor false.
145. Chances of mutations occurring over a short time period are the maximum in a.
bacteriophages, b. E. coli, c. Leptospira, d. Mycobacterium paratuberculosis.
146. Rate of mutation is expected to be low in a. RNA viruses, b. DNA viruses, c. enveloped
viruses, d. nonenveloped viruses.
147. Chances of mutations occurring over a short time period are the minimum in a.
bacteriophages, b. E. coli, c. Leptospira, d. Mycobacterium paratuberculosis.
148. Classical recombination involves breaking of covalent bonds within the nucleic acid,
exchange of genetic information, and reforming of covalent bonds. This kind of break/join
recombination is common in bacteriophages having a. DNA, b. RNA with a DNA phase
(retroviruses), c. negative stranded RNA viruses, d. a & b, e. a & c.
149. Which of the following has the smallest chromosome? a. Mycobacterium, b. E. coli, c.
Salmonella, d. Mycoplasma, e. Haemophilus.
150. Which of the following has the largest chromosome? a. Mycobacterium, b. E. coli, c.
Salmonella, d. Mycoplasma, e. Haemophilus.
151. As a general rule, prokaryotes tend to have very little junk DNA, typically it is about a.
25%, b. 15%, c. 35%, d. 45% of the genome.
152. Amount of junk DNA in chromosome of bacteria increases depending on their a. cell size,
b. adaptation to specific niche, c. pathogenicity potential, d. toxigenicity potential.
153. Majority of tRNA contain a. 26, b. 46, c. 56, d. 76 nucleotides.
154. There exist at least a. 20, b. 21, c. 22. d. 19 different aminoacyl-tRNA synthetases.
155. The endonuclease Ceu-I recognizes a 19 bp sequence. How often would you expect this
enzyme to cut the 4.8 x 106 bp chromosome of Salmonella Typhimurium? a. 3, b. 7, c. 11, d.
15.
27
156. Pseudomonas aeruginosa loses its restriction ability when grown at least for 5 generations at
a. 250C, b. 430C, c. 320C, d. 370C.
157. If a bacterium losses its endonuclease activity it is easier to induce a. transformation with
foreign DNA, b. Conjugation, c. transduction, d. none.
158. Nucleosomes are present in a. prokaryotes, b. eukaryotes, c. both, d. none.
159. While cloning it is more advantageous to digest (cut) the cloning vector and target DNA
producing single strand overhangs with a. one restriction enzyme, b. two enzymes, c. 3
enzymes, d. 4 enzymes.
160. During transformation with double strand DNA recombination of DNA results into
exchange of a. single strand, b. double strand, c. partly single and partly double.
161. Even those bacteria which are naturally resistant for transformation can take up foreign
DNA when they are in a. protoplast phase, b. spheroplast phase, c. exponential phase, d. lag
phase.
162. Efficiency of transformation in E. coli can be correlated with increasing concentration of a.
poly-B-hydroxybutyric acid, b. cAMP, c. cations, d. polyhydroxy-quinolone in cell membrane.
163. Electroporation is more efficient method than chemical transformation and its efficiency is
greater by a. 2-20 fold, b. 10-100 fold, c. 50-500 fold, d. 3-5 fold.
164. During electroporation pulse length is very important it should range between a. 1-2 msec,
b. 2-3 msec, c. 3-4 msec, d. 4-5 msec.
165. During electroporation voltage potential difference per cm, between two electrode is a.
1.25-2 Kv, b. 12.5-20 Kv, c. 125- 200 Kv, d. 400-500 Kv.
166. F and R plasmids are self transferable but efficiency of transfer is more in case of a. F, b. R,
c. equal, d. can not be compared.
167. Naturally transformable bacteria take up DNA in form of a. circular, b. linear, c. both, d.
single stranded.
168. The enzyme which causes failure to transform E. coli with linearized DNA is, a.
Exonuclease I, b. Exonuclease IV, c. Exonuclease III, d. Exonuclease V.
169. E. coli can be transformed with a. circular DNA, b. linear DNA, c. Circular nicked, d.
supercoiled DNA.
170. To transform E. coli with linear DNA it should have mutation in a. recA, b.recB, c. recC, d.
sbcB.
171. By which method you can transform bacteria with largest size of plasmids, a. chemical
transformation, b. electroporation, c. conjugation, d. transduction.
172. In conjugation DNA transfer occurs in a. one way, b. Two way, c. both way ie. a or b, d.
none.
28
173. DNA responsible for F pili when conjugated in DNA of host chromosome it is called a. F’,
b. Hfr, c. F, d. all.
174. DNA responsible for F pili when found as free plasmid in host cells it is called a. F’, b. Hfr,
c. F, d. all.
175. During conjugation DNA is transferred as a. linear dsDNA, b. linear ssDNA, c. circular
dsDNA, d. circular ssDNA.
176. During conjugation DNA synthesis is necessary for efficient transfer of plasmid in a. donor,
b. recipient, c. none, d. both.
177. In Agrobacterium tumefaciens efficiency of transfer is more when Ti plasmid is present in a.
donor and recipient both, b. only donor, c. only recipient, d. none.
178. In Agrobacterium tumefaciens conjugation efficiency more at a. high cell density, b. low
cell density, c. medium cell density.
179. F plasmids are a. broad host range, b. narrow host range, c. medium host range.
180. Heat shock response genes are not present in a. Thermus aquaticus, b. E. coli, c. Haloferex
volcanii, d. Drosophila spp., e. none.
181. Mitochondrial genome is about a. 16.6 kb, b. 16.6 Mb, c. 1.66 Mb, d. 1.66 kb.
182. Out of three types of restriction enzymes which group has strict control over the site of cut?
a. Type I, b. Type II, c. Type III, d. all.
183. Cohesive ends or the sticky ends of DNA produced by RE digestion get attached by a.
themselves, b. DNA ligase, c. Taq polymerase, d. none.
184. DNA ligase can join strands with a. cohesive ends, b. blunt ends, c. non-cohesive ends d.
all.
185. For assembling of sequencing data with shotgun method you don’t require a. any prior
knowledge of genome, b. any genetic map, c. any physical map, d. all three.
186. Shotgun approach for assembling a genome was first used for a. Salmonella Typhimurium,
b. E. coli, c. Mycobacterium tuberculosis, d. Haemophilus influenzae.
187. In genetic engineering most common type of plasmids used for transformation are a.
conjugative, b. mobilizable, c. nonmobilizable, d. nonconjugative non mobilizable.
188. For selection of auxotrophic mutants by penicillin enrichment for having about 10% cells
with auxotrophy you need about a. 1-2, b. 4-5, c. 9-10, d. 99-100 cycles of enrichment.
189. Individual suppressor can translate a nonsense mutation to the extent of a. 5-15%, b. 15-
30%, c. 35-45%, d. 60-75%.
190. tRNA frame shift suppressors have a. insertion, b. deletions, c. alterations in anticodone
sequence.
29
191. Anticodone in tRNA frame shift suppressors usually increase their size by, a. 1 base, b. 2
bases, c. 3 bases, d. 4 bases.
192. For repair of UV irradiated DNA best light repair takes place at wavelength, a. 250 to 350
nm, b. 300 to 400 nm, c. 350 to 450 nm, d. 400 to 500 nm.
193. Photolyase and hydrolyses a. cyclobutane ring, b. phosphodiester bond, c. N-glycosidic
bonds, d. hydrogen bonds.
194. In DNA, base is attached to 1’ C of the sugar by a. β-N-glycosidic bonds, b. phosphodiester
bond, c. α-N-glycosidic bonds, d. hydrogen bonds.
195. In DNA, deoxyribose sugar is a derivative of ribose whose one –OH group is replaced by –
H at carbon position a. 1’, b. 2’, c. 3’, d. 4’, e. 5’.
196. In DNA, phosphate group is attached to deoxyribose sugar at carbon position a. 1’, b. 2’, c.
3’, d. 4’, e. 5’.
197. In poly nucleotides, nucleotides are attached with phosphodiester bonds between their sugar
carbon number a. 1’-3’, b. 2’-4’, c.1’ -4’, d. 3’-5’, e. 2’-5’.
198. Stability to double helix is brought about by base stacking it is called, a. alpha-alpha
interaction, b. alpha-beta interaction, c. beta-beta interaction, d. phi-phi interaction, e. pie-pie
interaction.
199. Pie-pie interaction is between adjacent bases it involve a. hydrophobic interaction, b. Van
der Waals interaction, c. hydrophilic interaction, d. hydrogen bonding.
200. Drug resistance in Mycobacterium tuberculosis is due to a. R-plasmids, b. transposons, c.
integrons, d. chromosomal genes, e. phage mediated.
201. Production of antibiotics by microbes is mostly controlled by a. plasmids, b. transposons, c.
integrons, d. chromosomal genes, e. phages.
202. For Hg++ resistance in Enterobacteriaceae members’ genes are mediated through a.
plasmids, b. transposons, c. integrons, d. chromosomal genes, e. phages.
203. Resistance to cadmium, lead, antimony, arsenic, tellurium and silver compounds in
Staphylococcus spp. is mediated by a. plasmids, b. transposons, c. integrons, d. chromosomal
genes, e. phages.
204. On a F plasmid containing bacteria you can find pili there number/ cell averages between a.
0.4 to 1.7, b. 1.4 to 2.7, c. 0.1 to 1.9, d. 1 to 19.
205. Nonconjugative-Mobilizable plasmids lack a. oriT, b. tra cassette, c. oriV.
206. Non-transmissible plasmids lack a. oriT, b. tra cassette, c. oriV.
207. Mitochondria and chloroplast might have been once free living a. protozoa, b. bacteria, c.
yeasts, d. algae.
30
208. A bacteria similar to a. Chlamydia, b. Rickettsia, c. Salmonella, d. Brucella, might have
gave rise to mitochondria.
209. A bacteria similar to a. Cynobacteria, b. Archaea, c. Pseudomonas, d. Rhodobacteria, might
have gave rise to chloroplast.
210. Photosynthetic structures (cyanelles) of a protozoa Cyanophora paradoxa resembles in
structure to a. Cynobacteria, b. Archaea, c. Pseudomonas, d. Rhodobacteria, might have gave
rise to chloroplast.
211. σ subunit of RNA polymerase which bind to consensus promoter sequence is called a. σ54,
b. σ70, c. σ59 d. σ32.
212. σ subunit of RNA polymerase produced during heat shock by bacteria is called a. σ54, b. σ70,
c. σ59 d. σ32.
213. For transcription of nitrogen fixation genes bacteria need a. σ54, b. σ70, c. σ59 d. σ32.
214. Superscript number on σ subunit of RNA polymerase indicates a. temperature of activity, b.
sedimentation coefficient, c. molecular weight in kDa, d. it is an arbitrary number.
215. Lactose operon is induced by a. lactose, b. allolactose, c. beta lactose, d. D-lactose, e. L
lactose.
216. Allolactose activate lac operon by a. binding to the repressor, b. modulating the RNA
polymerase, c. by binding to the operon, d. by binding to the promoter.
217. Lactose operon is switched on in a. presence of lactose, b. in absence of lactose, c. not
dependent on lactose, d. none.
218. Tryptophan operon is switched on in a. presence of Tryptophan, b. in absence of
Tryptophan, c. not dependent on Tryptophan, d. none.
219. Catabolic activator protein activates a. RNA transcriptase; b. binds to site upstream to
operon to increase efficiency of RNA transcriptase, c. form direct contact to RNA polymerase,
d. Initiate RNA polymerase.
220. The most abundant type of RNA is a. tRNA, b. rRNA, c. mRNA, d. mtRNA.
221. The RNA which is not present in bacteria but found in Eukaryotes a. snoRNA, b.tmRNA, c.
tmRNA, d. scRNA, e. all, f. None.
222. mRNA editing is done by a. end modification, b. splicing, c. cutting events, d. chemical
modifications.
223. Population of RNA in a cell a. changes with time, b. remains constant, c. degrades with
increasing age, c. increases with age.
224. Transcriptome a. never changes, b. changes during development and differentiation, c.
changes during sporulation, c. changes during storage of cells at liquid nitrogen.
31
225. Transcriptome of every cell is a. synthesized de novo, b. partly received from its parent, c.
total received from parents, d. changing.
226. Transcriptome of a cell is maintained by replacing of degraded a. DNA, b. mRNA, c. tRNA,
d. rRNA.
227. For yeasts, average half life of mRNA is a. 1-5 min, b. 5-10 min, c. 10-20 min, d. 20-30
min.
228. Inosine base is found in a. tRNA, b. mRNA, c. rRNA, d. all, e. none.
229. Uridine can base pair with a. A & G, b. A & C, c. G & C, d. A & C.
230. Bacteria can decode their mRNA with as few as a. 32, b. 30, c. 61, d. 20, e. 22 tRNAs.
231. Mitochondria of human cells can decode their mRNA with as few as a. 32, b. 30, c. 61, d.
20, e. 22 tRNAs.
232. Exteins are sequences which are combined form active proteins, these are founds in a.
proteins, b. mRNA, c. tRNA, d. rRNA.
233. In Archaeae translation is matching more to a. Bacteria, b. Yeasts, c. Eukaryotes, d.
mitochondria.
234. Archaeal ribosome is a. 60s, b. 70s, c. 80s, d. 90s.
235. Archaeal mRNA is a. capped, b. polyadenylated, c. nonmodified, d. a & b.
236. The methionine carried by Archaeal initiator tRNA is a. N-formylated, b. non-N-
formylated, c. N-methylated, d. a & c.
237. Origin of replication in Archaea is a. similar to bacteria, b. similar to Eukaryotes, c. hybrid
between two, d. not yet known.
238. In Eukaryotes DNA replication is started with DNA polymerase α , it can a. synthesize
RNA primers, b. it can extend RNA primer with nucleotides of DNA, c. none, d. both a & b.
239. E. coli have known specific terminator sequence(s), till now a. 1, b. 3, c. 5, d. 7 terminator
sequence(s) are known which provide anchor to Tus, a sequence specific DNA binding protein.
240. Which one is not a mutagen a. UV radiation, b. ionizing radiation, c. photoradiation, d.
infrared radiation?
241. The UV radiation is most lethal at wavelength a. 250 nm, b. 260 nm, c. 270 nm, d. 280 nm.
242. Different pyrimidine dimers formed in DNA as UV lesion, in their decreasing order of
frequency, are a. TT, CT, TC, CC, b. AT, CT, TC, CC, c. TT, AT, TC, CC, d. CC, CT, TC, TT.
243. Cyclobutyl dimers formed in DNA due to UV light is adjacent pyrimidine form covalent
bonds between carbon number a. 3,6, b. 4, 6, c. 3, 5, d. 4, 9.
244. Operational taxonomic unit (OUT) is a unit in a. unit of genetic data, b. unit of phenotypic
data, c. a gene or nucleotide sequence, d. a protein or polypeptide.
32
245. The positions of purine rings which do not form Watson and Crick base pairs but have
potential to form hydrogen bonds are a. 1, 3, 5, b. 3, 7, 9, c. 3, 5, 7, d. 5, 7, 9.
246. Thermophilic bacterial DNA has more a. GC content, b. more AT content, c. no relation
with GC or AT.
247. The largest molecule found in a cell is a. DNA, b. RNA, c. Protein, d. Lipid, e.
polysaccharide.
248. For fin al separation of two completed circular DNA molecules essential enzyme required is
a. Topoisomerase I, b. Topoisomerase II, c. Topoisomerase III, d. Topoisomerase IV.
249. Aminoterminal end amino acid of most bacterial protein is a. N-formylmethionine, b. N-
lysine, c. a or b, d. none.
250. A synthetic DNA segment containing recognition sequences for several restriction
endonucleases in cloning region of a vector is called a. cloning site, b. polylinker, c. linker, d.
clonacone.
251. UV irradiation can not induce mutation in a. Haemophilus influenzae, b. Streptococcus
pneumoniae, c. both, d. none.
252. Hydroxy methyl cytosine in T4 phage DNA offers a reactive site for attachment of glucose
molecule through α and β linkage the ratio between both linkage is a. 7:3, b. 3:7, c. 1:1, d. 2:5,
e. 5:9.
253. Glucose moieties present in T4 DNA protect it from the action of a. bacterial exonucleases,
b. phage excinucleases, c. bacterial polymerases, d. bacterial restriction endonucleases.
254. At any time phage can not infect (rarely infect) bacteria if the culture has bacteria or phage
density below a. 106/ ml , b. 109/ ml, c. 104/ ml, d. 102/ ml.
255. If a T4 infected E. coli is super infected with T6 phage with an MOI ~100 you will get
particles (complete or incomplete) of a. T4, b. T6, c. both, d. none.
256. Lysis from within is caused by a. high MOI, b. completed phage infection, c. lysis of
infected bacteria with chloroform, d. lysis of bacteria on super infection with another phage.
257. You can determine the ghost particles of phages using, a. electron microscope, b. one-step
growth experiment, c. both, d. none.
258. If a T4 phage infected E. coli is infected with another T4 phage before lysis, the lysis
process will be a. quickened, b. delayed for about 1 hr, c. not affected, d. delayed for days
together.
259. If a T4 phage infected E. coli is infected with another T4 phage before lysis, burst size will
a. increase slightly, b. increase enormously, c. decrease slightly, d. decrease enormously.
33
260. For transcription and translation bacteriophage a. produces their own ribosomes and RNA
polymerases, b. modifies the specificity of host ribosomes and RNA polymerases, c. uses host
ribosomes and RNA polymerases as such, d. do not need ribosomes and RNA polymerases.
261. Ff phages produces plaques due to a. lysis, b. lysogeny, c. difference in growth rate of
infected and non-infected host, d. none.
262. You can observe Ff plaques a. only in early hours of incubation (16-18 hr), b. after extended
period of incubation, c. Only in late hours of infection, d. at all time.
263. Ff phages are released from bacteria by a. lysis, b. budding, c. extrusion through cell
membrane, d. all.
264. ΦX174 phages are released from bacteria by a. lysis, b. budding, c. extrusion through cell
membrane, d. all.
265. Reversion rate for mutation in RNA viruses in quite high, it is about a. 1 %, b. 0.1%, c.
0.01%, d. 0.001%.
266. It is difficult to prepare DNA map of RNA phages because a. Reversion rate for mutation in
RNA viruses in quite high, b. they have linear chromosome, c. they have changing genome, d. it
is difficult to sequence RNA.
267. After induction, Mu replicates as a. an autonomous molecule, b. as plasmid, c. as replicative
transposon, b. as prophage.
268. Mu phage undergoes heedful mechanism of DNA packaging starting at fixed left end and
may carry host DNA up to a. 0.5- 1 kb, b. 1-2 kb, c. 2-4 kb, d. 3-6 kb, e. 4-8 kb.
269. Some temperate phages do not insert into the host chromosome and remain as plasmids e.g.
are a. Mu and P2, b. P1 & P4, c. λ 7 P2, d. λ & Mu.
270. Concatameric DNA molecules for heedful packaging of phage heads are produced through
a. rolling circle replication, b. recombination event, c. both a & b, d. theta replication.
271. On B. subtilis cells dsDNA can bound to DNA binding sites, the number of site is about a.
10, b. 20, c. 30, d. 40, e. 50.
272. In B. subtilis cells transforming dsDNA strat to enter cells from a. 3’ end, 4. 5’ end, c. as
circular DNA, d. all three.
273. Pock formation on plate of Streptomyces is associated with a. viral infection, b. plasmid in
the cells, c. bacteriocins, d. porins.
274. Transfer of Ti plasmid occurs only when cell density of host is a. high, b. low, c. medium;
d. cell density has no effect.
275. Ti plasmids can transfer DNA from a. prokaryotes to plants, b. plants to prokaryotes, c. both
ways, c. can not transfer, e. cloned eukaryote DNA to plants.
34
276. Sulfobus islandicus, an Archaea posses some conjugative plasmids, the process of
conjugation requires about a. 5 min, b. 5 hrs, c. 5 days, d. 5 seconds.
277. Sulfobus islandicus, an Archaea posses some conjugative plasmids, the process of
conjugation requires a. sex pili, b. a pheromone, c. both, d. none.
278. It is better to store purified DNA at 4oC than at -20oC because a. storage at -20oC causes
nicking of DNA, b. at -20oC DNA get crystallized and less useful for cloning work, c. a & b
both, d. DNA stored at 4oC can be used without wait for thawing.
279. DNA and RNA get precipitated in 70% ethanol when a. temperature is low, b. there are
monovalent cations, c. when there are divalent cations, d. ions soluble in ethanol.
280. Ammonium acetate is used to augment precipitation of DNA with ethanol because a. it can
be removed easily from DNA being insoluble in Ethanol, b. it can be removed easily from DNA
being soluble in Ethanol, c. it is more efficient, d. it is cheaper.
281. Ammonium acetate is often not preferred to precipitate for use in a. ligation with T4 ligase,
b. RE analysis, c. to be modified for other uses, d. cloning, e. all.
282. DNA is precipitated better with ethanol chilled at a. -70oC, b. -20oC, c. 0oC, d. 4oC, e.
equally good at all 4 temperatures.
283. After precipitation of DNA you need to centrifuge at a. RT, b. 20oC, c. 0oC, d. 4oC, e. it
does not matter much.
284. For precipitation of DNA from samples with DNA concentration lower than 0.1µg/ml you
should add in solution a. glucose, b. glycogen, c. isopropanol, d. methanol.
285. The DNA is purer on precipitation with a. ethanol, b. isopropanol, c. chloroform, d. isoamyl
alcohol.
286. Agarose with low EEO mean a. agarose with low melting point, b. agarose with low organic
impurities, c. agarose with high melting point, d. agarose with low anionic impurities.
287. Common anionic impurities are a. Cl-; b. SO4--; c. CO3-; d. pyruvate; e. all four; f. a, b; g. b,
c; h. b, d.
288. Standard agarose does not dissolve well to give uniform gel at a. >1%, b. >1.5%, c. >2%, d.
>2.5%, e. >3%.
289. When one of the two primers is added in excess into PCR mix there will be a. no or little
reaction, b. more specific and less non-specific product will be there c. only non-specific
product will arise, d. results into ssDNA product.
290. In asymmetric PCR limiting primer is added in concentration of about a. 1 pmole, b. 0.1
pmole, c. 2 pmole, d. 0.2 pmole.
291. In asymmetric PCR excess primer is added in concentration of about a. 1 pmole, b. 10
pmole, c. 2 pmole, d. 20 pmole.
35
292. Frequency of reversion is the maximum for a. deletions, b. point mutations, c. insertions, d.
double mutations.
293. After infection with virulent phages bacteria is often unable to synthesize its own DNA or
RNA however this shutdown is less common with infection of a. ssDNA, b. dsDNA, c. dsRNA,
d. ssRNA.
294. Synthesis of phage mRNA from phage DNA is always initiated by a. bacterial RNA
polymerase, b. modified bacterial RNA polymerase, c. phage RNA polymerase, d. none.
295. The structural protein of Bacteriophages are encoded by a. early mRNA, b. late mRNA, c.
both, d. none.
296. Some Bacteriophages encode for their genome replication enzymes they are a. ssRNA
phages, b. dsRNA phages, c. ssDNA phages, d. all.
297. Adsorption of phage to bacterial cell follow a. Z-distribution, b. Poisson distribution, c. T-
distribution, d. F- distribution.
298. Infective centre of a plaque is formed by a. single phage, b. a bacterial cell infected with
phage but not lysed before plating, c. Many bacteria infected with a single phage, d. a bacterium
having multiple infection.
299. Lytic phages multiply a. more rapidly than host bacteria, b. at the same rate as the host
bacteria, c. slower than the host bacteria, d. can not be compared.
300. High concentration of phage can be prepared by growth a. solid defined medium, b. liquid
defined medium, c. solid rich medium, b. liquid rich medium.
301. P22 phage causes infection to a. S. Typhimurium, c. S. Typhi, c. E. coli, d. S. Gallinarum, e.
all.
302. Open promoter complex is formed through initial binding of RNA polymerase with DNA
followed by a. detachment of RNA polymerase, b. local unwinding of DNA close to binding
site of RNA polymerase, c. synthesis of RNA primer, d. none.
303. RNA polymerase moves along DNA in direction of a. 3’ to 5’, b. 5’ to 3’, c. both directions
depending on gene orientation, d. both directions according to need of the time.
304. Promoter down mutations have a. poor expression of down stream genes, b. excessive
expression of upstream genes, c. poor expression of upstream genes, d. excessive expression of
down stream genes.
305. Promoter up mutations have a. poor expression of down stream genes, b. excessive
expression of upstream genes, c. poor expression of upstream genes, d. excessive expression of
down stream genes.
36
306. Coordinated regulation of a function in bacteria is brought about by regulation of synthesis
of polycistronic mRNA regulated through a. a single signal, b. double signal, c. many signals
coordinated by Rho, d. many signal molecules regulated by sigma factor.
307. Some times you can see mRNA molecules growing in both directions of dsDNA of
bacterial chromosome; it is due to fact that a. RNA polymerase can work in both directions, b.
Opposite strands of a DNA sequence always code for two different genes, c. genes can be
transcribed from opposite strands, d. none.
308. All amino acids have 2 or more codones coding for them except a. methionine, b.
tryptophan, c. both, d. none.
309. Anticodone is found in a. antisense strand, b. negative strand of DNA, c. tRNA, d. mRNA,
e. a & c.
310. Certain anticodones can pair with several codones if the first base of anticodone is a. U, b. I,
c. G, d. any of the three.
311. Different rRNA molecules present in bacterial ribosomes are a. 23s rRNA, b. 5s rRNA, c.
16s rRNA, d. all.
312. Shine-Dalgarno sequences are present in a. mRNA, b. DNA, c. rRNA, d. all three.
313. One E. coli rRNA transcript is known to produce a. 5s rRNA, 16s rRNA, 23s rRNA, b. 5s
rRNA, 16s rRNA, 23s rRNA, tRNATrp, tRNAAsp, tRNAAla, tRNAIle, c. tRNATrp, tRNAAsp,
tRNAAla, tRNAIle, d. none.
314. Repressor regulated enzymes of Lac operon of E. coli are a. β-galactosidase, b. lactose
permease, c. transacetylase, d. all three.
315. In dark repair mechanism which of these does not take place, a. excision of damaged bases,
b. reconstruction of functional DNA molecule from undamaged fragments, c. cell permits some
degree of damage to DNA, d. none.
316. During growth of ts-mutants at high temperature a. the affected gene is not transcribed, b.
the affected gene is neither transcribed nor translated into the desired protein, c. the affected
gene is transcribed but not translated into the desired protein, d. The desired protein is produced
but that is inactive.
317. Alkylating agents cause mutagenesis due to a. depurination, b. mispairing of G with T, c.
causing transition, d. causing transversion, e. all.
318. Mutation caused by Hydroxylamine can not be reverted with a. an alkylating reagent, b.
hydroxylamine, c. Nitrous acid, d. none.
319. Nitrous acid primarily converts a. a keto group to amino group, b. an amino group to keto
group, c. carboxyl group to amino group, d. nitro group to amino group.
320. Which base pair change are transition a. AT to TA, b. AT to GC, c. GC to TA, d. none.
37
321. Which base pair change are transversion a. AT to TA, b. AT to GC, c. GC to TA, d. none.
322. Base analogs are used to produce a. deletion, b. transition, c. transversion, d. insertion.
323. A wrong amino acid is introduced as a result of a. nonsense mutation, b. missense mutation,
c. both, d. none.
324. Nitrogen mustard [bis (b-chloroethyl) sulphide, (ClCH2CH2)2S] causes ring cleavage
followed by hydrolysis of N-glycosidic bond of a. guanine, b. adenine, c. thymine, d. cytosine.
325. Suppressor sensitive mutations are often a. conditional, b. reversible, c. permanent, d.
always nonsense.
326. Suppressor sensitive mutations are often a. nonsense i.e., chain termination type, b.
missense or amino acid substitution type, c. none, d. both type.
327. Single base substitution in a. 6 codones, b. 18 codones, c. 12 codones, d. 24 codones can
give rise to origin of chain termination mutation codone UAG/ UAA/UGA.
328. Rarest class of revertants are a. frame shift revertants, b. base substitution, c. true revertants,
d. suppressor sensitive revertants.
329. The tRNA molecule which has anticodone CUA is called a. suppressor tRNA, b. mutator
tRNA, c. none, d. both.
330. Suppressor tRNA to AUG can substitute nonsense codone with codones for any of the a. 6,
b. 7, c. 8, d. 5 amino acids.
331. A particular suppressor mutant can not suppress all UAG chain terminating mutations
because a. some times suppressor tRNA do not pair with UAG, b. it often fails to incorporate an
amino acid, c. the inserted amino acid may fail to produce the active protein, d. All the three.
332. In presence of suppressor tRNA, termination is suppressed in about a. 10%, b. 20%, c.
30%, d. 40% cases.
333. Nitrogen mustard cleaves guanine and the empty space serves as templet for base a. C, b. T,
c. U, d. A.
334. Nitrogen mustard induces a. transition, b. transversion, c. frame shift, d. all.
335. F pilus is made up of pilin which is made of a. single hydrophobic protein, b. one
hydrophobic and one hydrophilic peptide, c. two hydrophobic peptides, d. two hydrophilic
peptides, e. single hydrophilic protein.
336. Conjugative Pili get detached on agitation of the culture and can a. regenerate within 1 hr, b.
regenerate in 0.5 hr, c. regenerate in 12 hr, d. not regenerate.
337. Male specific Phages can inhibit conjugation but only those which attaches to a. shaft of
pilus, b. Tip of pilus, c. base of pilus, d. all three types, e. none.
338. During conjugation transfer of DNA starts with a nic produced in a. single strand, b. both
strands, c. none.
38
339. Only few R plasmid containing bacterial cells in a culture are competent donor, their
proportion is usually a. 0.2%, b. 0.02%, c. 0.002%, d. 2%.
340. To cause gall disease Ti plasmid genes integrate into a. plant chromosome, b.
Agrobacterium tumefaciens, c. both, d. none.
341. The phenomenon which is not mediated through transposable elements is. A. replication, b.
fusion, c. deletion, d. inversion, e. transposition, f. mutation, g. none of these.
342. When Mu phage infects the bacterium its DNA a. totally integrates into bacterial
chromosome, b. whole but terminal parts of DNA integrates into host, c. whole but central part
of phage DNA integrates, d. maintains as plasmid.
343. Progeny from a particular burst during a single burst experiment with Mu phage a. have
same sequence, b. may have variable sequence.
344. During conjugation F+ or F cells transfer--------------DNA than Hfr strains, a. more, b. less,
c. equal, d. variable.
345. During conjugation complete plasmid DNA is donated by a. F+, b. Hfr, c. both, d. None.
346. To form a large plaque burst size of phage should a. smaller than 10, b. 50-100, c. 10-40, d.
4.
347. A phage having burst size equal to 4 will form a. a small plaque, b. large plaque, c. medium
sized plaque, d. no plaque.
348. For effective recombination of two phages in bacterium MOI should be a. high, b. low, c.
medium, d. MOI has no role.
349. Size of F plasmid is a. 94.5 kb, b. 54.5 kb, c. 14.5 kb, d. 74.5 kb.
350. Specialized transduction is usually brought up by a. lytic, b. lysogenic, c. circular, d. spiral
phages.
351. Out of a. K88, b. K99, c. 987P, d. F1 and e. vir pili of E. coli which is not encoded by genes
on plasmids.
352. Which of the virulence factor of Bacillus anthracis is not a plasmid encoded fuction: a.
capsule, b. O-side chains, c. lethal toxin, d. edema factor.
353. Which of the Clostridium perfringens type D toxin is plasmid encoded fuction: a. epsilon
toxin, b. alpha toxin, c. delta toxin, d. haemolysin.
354. Which of the E. coli virulence factor is not encoded by plasmids: a. alpha haemolysin, b.
contact haemolysin, c. enterohaemolysin, d. capsular antigen.
355. Size of Salmonella virulence plasmid ranges from a. 50-100 kb, b. 120-200 kb, c. 5-9.6 kb,
d. 15-30 kb.
39
356. The Hershey-Chase used ---------------- to prove that DNA was the genetic material causing
the transformation observed in Griffith's and Avery's experiments. A. bacteriophages, b.
cosmids, c. plasmid, d. transposons.
357. Fraenkel-Conrat experimented with tobacco mosaic virus and Holmes rib grass virus to
conclude that some viruses use --------------- as their genetic information. a. DNA, b., RNA, c.
DNA as well as RNA, d. proteins.
358. Fraenkel-Conrat dissociated the RNA from the protein coats of the viruses and intermixed
them to produce a. hybrid virues, b. hybrid proteins, c. infective particles, d. none of these.
359. Rosalind Franklin used ------------experiments for revealing that DNA is a helical molecule.
a. ultrasound, b. X-ray diffraction, c. electron microscopy, d. al of the three.
360. RNA polymerase binds to one strand of a DNA molecule at a site ---------------a. initiator, b.
promoter, c. start codone, d. Okazaki point.
361. Activators are regulatory proteins that assist in the -----------of the double helix in several
genes. a. straightening, b. uncoiling, c. unzipping, d. activating inactive RNA polymerase.
362. DNA loops near the promoter called----------is positioned to regulate the promoter. A.
activator, b. regulator, c. repressor, d. enhancer.
363. Mutations that make reading of the gene out of register are known as----------- mutations. a.
point, b. null, c. neutal, d. frameshift.
364. The active substance which transformed harmless bacterium into a virulent bacterium in
Avery's experiment a. DNA, b. RNA, c. protein, d. OMP, e. LPS.
365. Chargaff's rule of base pairing revealed that how the molecules maintain. a. constant
thickness, b. double helical structure, c. thermoresistance, d. histone in it.
366. The replication of DNA molecules during cell division take place while a. cells have started
dividing, b. after the formation of daughter cell boundary, c. before the initiation of any
apparent cell division, d. after cell division.
367. During transcription one of the DNA strands serves as a source of information for the
synthesis of mRNA which is known as a. coding strand, b. Sense strand, c. template, d. all the
three terms.
368. Each amino acid is specified by----------codon. a. one, b. two, c. three, d. one or more than
one.
369. In ribosomes the rRNA is located in the---a. large unit, b. small subunit, c. both a & b, d.
none.
370. During translation ------------of tRNA molecule binds to a specific codon on the mRNA
molecule. a. Y loop, b. codon loop, c. anticodon loop, d. complementary loop.
40
371. On reaching a codon on the mRNA molecule that does not specify any of the 64 tRNA
anticodon, the ribosome, a. releases the newly formed protein molecule, b. separate in to
subunits, c. translocation of the newly synthesized protein, d. a &b, e. a & c.
372. When we add allolactose in growth medium the lac operon is a. activated, b. repressed, c.
not affected, d. over transcribed.
373. In eukaryotes, after transcription, the mRNA molecule must have its-------------- removed
before it is translated into a protein molecule in the posttranscriptional processing. a. exons, b.
introns, c. SNPs, d. transposons.
374. If one end of a DNA molecule contains the AGTAAGCCG nucleotide sequence the other
end will contain a. TCATTCGGC, b. TCAUUCGGC, c. AAAAAAAAA tail, d. not fixed.
375. Which of the following is not a part of an operon? a. protein-encoding genes, b. an
operator, c. a promoter, d. none.
376. Genes that can move about from one location on a chromosome to another are called—a.
entron, b. exons, c. transposons, d. transferons.
377. The regulatory site at the beginning of each gene controls the expression of the gene,
specific regulatory proteins within the cell bind to these sites, turning transcription of the gene
off and on, these proteins are may act as a. repressor, b. activator, c. both a & b, c. none.
378. The process of horizontally acquiring a gene from another organism is called a. mutation, b.
transformation, c. transversion, d. all the three.
379. Chargaff's rule suggests that the purines and pyrimidines in the DNA molecule combine in a
a. nonrandom pattern, b. randomly, c. depends on environment.
380. The two sides of the double helix of DNA are kept together with a. carbon-carbon bonds, b.
hydrogen bonds, c. Vander Wall’s forces, d. electrorepusive forces, e. hydrophobic interactions.
381. Using genetic engineering genes from bacteria can be moved to a. bacteria, b. virus, c.
fungi, d. plants, e. animals, f. to all.
382. When any restriction enzyme produces sticky ends it cut the nucleotide sequence a. at the
centre, b. subterminally, c. off center at a specific cutting site.
383. Sanger earned his second Nobel Prize for development of a technique called a. chain
termination, b. PCR, c. sequencing E. coli genome, d. sequencing human genome.
384. Glyphosate prevents plants from growing by stopping them from –a. transcription, b.
translation, c. protein translocation, d all the three.
385. During genetic engineering cuts in DNA strands are sealed with a. Taq polymerase, b. PFU
polymerase, c. gyrase, d. ligase.
386. DNA can be made from RNA templet with the help of a. DNA polymerase, b. RNA
polymerase, c. transcriptase, d. reverse transcriptase, e. any of the four.
41
387. When you received a sample for disease diagnosis you can not culture it because the
bacteria there might have died or injured you can still have chance to diagnose with the help of
–a. PCR, b. immunological techniques, c. isolation of DNA and then doing RE analysis, d. any
of the three.
388. While cloning eukaryotic DNA in bacterial cells you require a. PCR product of the gene, b.
cDNA of the gene, c. mRNA, d. all three.
389. Ti plasmid of Agrobacterium has revolutionized plant genetic engineering but it can not be
used for transforming a. cereals, b. pulses, c. dicots.
390. In stead of probes you can use--------while screening transformants for successful gene
transfer process. a. PCR, b. Western blotting, c. both, d. none.
391. Parasites of unicellular organisms (UOPs) do not include a. the cyanophage of
cyanobacteria, b. the actinophage of Actinomycetes, c. bacteriophages of bacteria, d.
Bdellovibrio spp. of negative bacteria, .e. none.
392. Which is not corect about UOPs? a. obligate intracellular parasites, b. have two phase life
cycle, c. have both extracellular and intracellular replication, d. all.
393. Two phases of life cycle of UOPs are a. adsorption and intracellular replication, b.
intracellular and extracellular replication, c. adsorption and dispersion pahses, d. mitosis and
meiosis.
394. Most abundant calss of organisms on earth are a. insects, b, plants, c. bacteria, d. Parasites
of unicellular organisms (UOPs), e. Parasites of multi-cellular organisms (UOPs).
395. Which is not an example of bacteriophage? A. Actinophage, b. Archaeophage, c. Coliphage,
d. Cyanophage, e. none.
396. Which is not a non viral parasite of unicellular organisms? A. Bdellovibrio, b. Legionella, c.
Vampirococcus, d. none
397. Phage therapy has been used for treating a. plant diseases, animal diseases, c. topical
infections, d. gastrointestinal infections, e. all.
398. Phages can not be used for a. phage therapy of diseases, b. as an indicator of faecal
pollution, c. model for viral disinfection, d. decontamination of surfaces, e. none.
399. Commonly Mycoplasma phages have a. dsRNA, b. ssDNA, c. dsDNA, d. ssRNA, e. all.
400. Virions of family Phycodnaviridae infect a. plants, b. bacteria, c. unicellular green algae, d.
multicellular green algae.
42
Answers. 1. b, 2.d, 3. b, 4. a, 5. d, 6. d, 7. b, 8. d, 9. b, 10. a, 11. a, 12. c, 13. d, 14. a, 15. b, 16. b,
17. d, 18. c, 19. a,e, 20. a, 21. a, 22. d, 23. a, 24. d, 25. c, 26. c, 27. c, 28. e, 29. c, 30. b, 31. b,
32. d, 33. d, 34. a, 35. a, 36. a, 37. a, 38. b, 39. a, 40. d, 41. b, 42. d,e, 43. a, 44. a, 45. c, 46. c,
47. a, 48. c, 49. a, 50. a, 51. a, 52. d, 53. d, 54. a, 55. b, 56. c, 57. c, 58. d, 59. a, 60. a, 61. c, 62.
c, 63. b, 64. a, 65. e, 66. a, 67. b, 68. d, 69. b, 70. c, 71. b, 72. e, 73. c, 74. e, 75. b, 76. a, 77. a,
78. a, 79. d, 80. b, 81. a, 82. c, 83. d, 84. c, 85. d, 86. b, 87. d, 88. d, 89.b, 90. b, 91. b, 92. a, 93.
a, 94. c, 95. a, 96. a, 97. c, 98. c, 99. b, 100. a,b, 101. b, 102. d, 103. b, 104. a, 105. d, 106. a,
107. b, 108. c, 109. a, 110. a, 111. d, 112. b, 113. a, 114. b, 115. d, 116. e, 117. b, 118. b, 119. b,
120. a, 121. c, 122. d, 123. a, 124. b, 125. a, 126. d, 127. a, 128. a, 129. c, 130. a, 131. a, 132. d,
133. a, 134. a, 135. a, 136. a, 137. d, 138. c, 139. b, 140. d, 141. e, 142. b, 143. a, 144. a, 145. a,
146. b, 147. d, 148. d, 149. d, 150. a, 151. b, 152. b, 153. d, 154. b, 155. b, 156. b, 157. a, 158.
a, 159. b, 160. a, 161. a,b, 162. a, 163. b, 164. d, 165. b, 166. a, 167. b, 168. d, 169. a,c,d, 170.
b,c,d, 171. c, 172. a, 173. b, 174. a, 175. b, 176. d, 177. a, 178. a, 179. b, 180. e, 181. a, 182. b,
183. b, 184. a,b, 185. d, 186. d, 187. b, 188. b, 189. d, 190. a, 191. a, 192. b, 193. a, 194. a, 195.
b, 196. e, 197. d, 198. e, 199. a, 200. d, 201. a, 202. b, 203. a, 204. b, 205. b, 206. a,b, 207. b,
208. b, 209. a, 210. a, 211. b, 212. d, 213. a, 214. c, 215. b, 216. a, 217. a, 218. c, 219. b, 220.
b, 221. e, 222. d, 223. a, 224. b, 225. b, 226. b, 227. c, 228. a, 229. a, 230. b, 231. e, 232. a, 233.
c, 234. b, 235. d, 236. b, 237. d, 238. d, 239. d, 240. c, 241. b, 242. a, 243. b, 244. b, 245. b,
246. a, 247. a, 248. d, 249. a, 250. b, 251. c, 252. a, 253. d, 254. a, 255. a, 256. b, 257. a, 258. b,
259. b, 260. b, 261. c, 262. a, 263. c. 264. d, 265. b, 266. a, 267. c, 268. b, 269. b, 270. c, 271. e,
272. a, 273. a,b, 274. a, 275. a,e, 276. b 277. d, 278. d, 279. b, 280. b, 281. e, 282. e, 283. e,
284. b, 285. a, 286. d, 287. h, 288. b, 289. d, 290. d, 291. d, 292. b, 293. a,d, 294. a, 295. b, 296.
d, 297. b, 298. b, 299. a, 300. c, 301. a, 302. b, 303. b, 304. a, 305. d, 306. a, 307. c, 308. c, 309.
c, 310. d, 311. d, 312. a, 313. b, 314. d, 315. d, 316. d, 317. e, 318. b, 319. b, 320. b, 321. a,c,
322. b, 323. b, 324. a, 325. a, 326. d, 327. d, 328. c, 329. a, 330. b, 331. c, 332. a, 333. d, 334. b,
335. e, 336. b, 337. b, 338. a, 339. b, 340. a, 341. g, 342. b, 343. b, 344. b, 345. a, 346. b, 347.
d, 348. a, 349. a, 350. b, 351. d, 352. b, 353. a, 354. d, 355. a, 356. a, 357. b, 358. a, 359. b, 360
b, 361. c, 362. d, 363. d, 364. a, 365. a, 366. c, 367. d, 368. d, 369. b, 370. c, 371. d, 372. b, 373.
b, 374. d, 375. d, 376. d, 377. c, 378. b, 379. a, 380. b, 381. f, 382. c, 383. a, 384. b, 385. d, 386.
d, 387. a, 388. b, 389. a, 390. c, 391. e, 392. c, 393. a. 394. d, 395. e. 396. d, 397. e, 398. e, 399.
b, 400. c.
43
Chapter: 3. Fill in the blanks of Microbial Genetics
A. Fill in the Blanks
1. E. coli can be transformed using linear DNA of ---------------------.
2. In Haemophilus, selectivity of transforming DNA is dependent on presence of ----------------
--------------------.
3. Vesicular transformasomes are formed in -------------------.
4. Generally transformasomes have -----------------------------DNA.
5. In general, competence is commonly found is ------------phase of growth.
6. During conjugation DNA is transferred as ------------------ DNA and it ----------end of DNA
which enters the recipient first.
7. If the presence of plasmid in bacterial cell has no phenotypic effect on host cells is called----
-------------------------.
8. During multiplication, two identical double strands of DNA may become intertwined at a
point called-----------------------------. At both side of the junction pint -----------------complex
clamp the strands to hold them so that DNA can be copied correctly.
9. The 2u circle is plasmid found in nucleus of ------------------------.
10. Common methods of transfer of genetic material in bacteria are a.--------------------b.---------
------------c.------------, d.--------------e.---------- & f.----------------.
11. Competence of Streptococcus pneumoniae due to a ---------------------------antigen of
molecular weight about--------------present on cell surface.
12. An essential feature of replicative transposition is formation of-----------------.
13. The complete Mycobacterium genitalium genome is only 0.58 Mbp (580 Kbp) and is
predicted to encode ------------- proteins.
14. Agrobacterium tumefaciens carry a plasmid responsible for genes coding for gall formation
in plants, plasmid is named as----------plasmid.
15. Ti plasmid is ------------plasmid with molecular weight equals to----------.
16. During transformation gram +ve bacteria take up--------------------------------while gram –ve
bacteria take up -----------------------DNA.
17. Prerequisites of homologous recombination in bacteria are-----------------,----------------,------
---------- and------------------------------.
18. The smallest free-living bacteria have a genome size of about ---------------.
19. For example, M. genitalium lacks the genes required for ----------------------,------------------,-
-----------------------------,and ---------------------------.
20. Ribosome read the mRNA in -----------------------direction. (5’ to 3’).
44
21. Protein synthesis starts from ------------------terminus.
22. An example of eukaryotic plasmid is ---------------- of Saccharomyces cerevisiae.
23. Many transposons carry antibiotic resistance genes, but some transposons carry other genes
example of such transposons are -------.
24. Some times phage genes can modify the Salmonella O antigen it is called- -------------------.
25. Why bacteriophages need to inject their DNA/RNA while animal viruses do not have such
mechanism? --------------------------------------------.
26. Viruses get mutations due to a.-----------------------b. ----------------------------.
27. Size of the bacterial chromosome ranges from-----------------to------------------.
28. Size of the eukaryotic chromosomes range from ------------------to---------------.
29. In gram –ve bacteria major method of plasmid gene transfer is -----------------------
(conjugation) while in gram positive bacteria plasmids are transferred by ----------------------
------and-----------------------.
30. In gram –ve bacteria conjugation is mediated through------------------------while in G+ve
bacteria it is through-------------------------------.
31. ORF stands for------------------------------- while URF stands for------------------------.
32. Knockout mouse have---------------------------.
33. Knockin mouse have----------------------------.
34. Electroporation is the technique for--------------------, it is very successful with---------------
while comparatively less successful in--------------.
35. Nuclease cut DNA at every-------i.e. one cut per-------------.
36. Regular structure to bacterial DNA is lend by a histone like protein called-------, it is an
abundant protein each cell contain about---------------copies.
37. In presence of histone like proteins, nuclease cut bacterial DNA at every--------it indicates
that these proteins produces----------in DNA.
38. The histone like proteins of E. coli that confer higher order structure to DNA are--------------
----------------------------------------------------.
39. R100 and ColE1 undergo -------replication, they replicate in ---direction.
40. Sigma replication differs from rolling circle replication in one sense that it does not-----------
------------------.
41. There are-------------different modes of plasmid replication namely, -----------------------------
-------------------.
42. Some groups of wide host range plasmids are -----------------------------------.
43. ----------------------------------are some of the plasmids of group----------which can be
transferred to Mycobacteria and Bacteroides.
45
44. Transformation of IncP group plasmids is often better ---------------than----------media.
45. Conjugative plasmids mut synthesise------------------------to form mating aggregates.
46. Incompatibility is determined by the genes of same loci as the ---------------------.
47. When donor cell form aggregate with recipient cell a nick in plasmid DNA is produced at---
---------and DNA is displaced into recipient by the enzyme------------.
48. Membrane bound cytoplasmic ATPase is known as----------------.
49. Chaperonins are ---------------proteins which try to---------------------------------using ATP as
source of energy.
50. Expression of heat shock proteins is controlled by-------------------which act as new-----------
-------for RNA polymerase holoenzyme complex.
51. Molecular weight of σ70 is-------------------
52. For transcription of rpoH at 37oC requires -----------&-------------sigma factors while >50oC
it requires----------------------.
53. Fill ups
Bacteria Gene product Phenotype Gene source
a. E. coli CT/LT toxin Cholera like disease
b. Vibrio cholera CT Cholera
c. E. coli Shigalike toxin Haemolytic diarrhoea
d. Clostridium botulinum Botulinum toxin Food poisoning
(Botulism)
e. Corynebacterium
diphtheriae
Diphtheria toxin Diphtheria
f. Streptococcus pyogenes Erythrogenic toxin Scarlet fever
g. E. coli Heat stable toxin Persistent diarrhoea
54. SOS repair pathway is the member of --------------------------network.
55. Synthesis of each new sigma factor induces synthesis of ---------------------to control the specific
biochemical events to occur in a defined sequence.
56. If the recombination always occurs at the same site on both duplexes it is classified as------------
-----------------------.
57. Models for generalized recombination are based on----------------, these sites have directionality
and cause exonuclease -------to produce the nicks needed to initiate recombination.
58. When recombination occurs at a specific site on one DNA duplex but randomly on the other, it
is called ------------------------------. It is common in case of---------------integration.
59. All proved carcinogens are----------------------but converse is not true.
46
60. In Salmonella system of Ames test for carcinogens we select for the --------------------of----------
------mutation in Salmonella------------------.
61. Pseudo gene is a---------------------------of gene.
62. Micro satellites in genome are ----------------in tandem.
63. Two common micro satellites in human DNA are repeats of -------------------&----------.
64. Salmonella contains about---------------------genes --------% of which encode for virulence
factors.
65. In Salmonella large chunk of virulence genes are clustered near ----------------minutes on
linkage map.
66. Analysis of a sequence is complicated by the fact that each dsDNA may have ---------reading
frames, --------- in one direction and -----------reverse direction.
67. Spontaneous nature of mutations was first shown by------------------------in 1943 by using --------
----test.
68. Concentration of a mutagen and time of incubation for mutagenesis can be determined sensibly
as those conditions under which about-----------------of the input cells survive.
69. Auxotrophic mutants in a culture can be enriched by the method of------------------and------------
------------- enrichment.
70. Nitrous acid can cause--------------mutations by converting amine group of a base to hydroxyl
group called------------------deamination. During this process adenine, thymine and guanine are
converted respectively into-------------------------------------&--------------.
71. Alkylating reagents are powerful mutagens which mostly induce-------------mutations but can
also induce -------------------&--------------to some extent.
72. Acridine dyes induce ----------------mutations, due to -------------------------------.
73. When second mutation reverts the effect of first mutation, it is called------------------. effect the
DNA
74. One of the best acridine dyes is proflavine because it is a planer molecule which mimics the
structure of--------------------------------------- which can intercalate between the stacked bases.
75. Proflavine affect DNA which is----------------------------------.
76. Two commonly used frame shift mutagens are--------------------&---------------------.
77. The principle of Ames test is----------------------------------------------------, it is used to detect-----
--------------------------------------------.
78. Introduction of a --------------mutation into an early gene of an operon cause----------------.
79. Polar effect is more when mutation has occurred---------------------------the mutated gene.
80. When one mutation is nullified by the second mutation in the same gene at another point or in
the other gene the mutant is called-------------------------.
47
81. Intergenic suppression operates at the level of------------------, it is mostly due to appearance of
altered---------------which can read-----or------------ nonsense codones.
82. Four types of macro lesion mutations are----------,----------------,------------------&--------.
83. Duplication and deletions are mostly caused by error in ------------and is called------------.
84.--------------during replication cases duplication and------------during replication causes deletions.
85. Deletions and inversions may be produced as a result of of -----------------------&-----------------
recombination.
86. Recombination between homologous but inverted sequences results into----------------.
87. Most common lesion induced by UV in DNA is-------------------.
88. Pyrimidine dimers are formed by joining of two pyrimidine residues by a------------------.
89. Different types of pyrimidine dimers are------------------,-----------------------&-----------------,
however the most common is---------------------.
90. Mutation in genes encoding for the components of DNA repair mechanism leads to production
of -----------gene.
91. By site specific recombination mechanism specific sequence of DNA get inverted to turn off
one gene while turn on the other alternative gene it is called------------------mechanism.
92. When reversion frequency of a certain mutation could not be enhanced by any of the known
mutagen, it suggests that mutation is neither--------------------nor-------------.
93. Three methods of transposition are ---------------, --------------------&------------------.
94. Rolling circle transposition is a common phenomenon in ---------------------element.
95. The IS elements which lack the terminal repeat is---------------.
96. The Is elements found in gram positive bacteria are---------------.
97. In Tn1 transposition-------------% transposition is through direct pathway (cut and paste) and
rest is through replicative pathway.
98. Cointegrates are formed during------------------------pathway of transposition.
99. The size of linear dsDNA fragments which can be transformed into amplicon of Bacillus
subtilis, S. pneumoniae, H. influenzae is --------------,-------&------------------, respectively.
100. During transformation in Bacillus subtilis, and S. pneumoniae there is a period after DNA
uptake in which the acquired DNA is incorporated into the genome, it is called-----.
101. A bacterial gne of 900 base pair long will encode a protein containg----------------amino acids.
102. Both strands of λ phage DNA act as -----------strand and can be read independently for----------
------------------ one is -----------------towards--------------and other towards---------.
103. Three pahses that regulate lytic cycle of λ phage are--------, -------------and----------phase.
104. Out come as lytic or lysogenic cycle is pivoted on concentration of --- and --- proteins, the first
one initiate lysogenic and second the lytic cycle.
48
105. Jumping genes were first discovered by Barbara McCintock in------------------.
106. In conjugation male bacterium conjugate with female bacterium after conjugation there is -----
---female bacterium.
107. During conjugation some bacteria incorporate F factor in their chromosome to become----------
-strains and when such strains conjugate with other bacteria they transfer part of------------to
female bacteria and transfer is ---------------.
108. In transduction certain bacterial genes (markers) are more prone for transfer than other
because a phage gene product recognizes ----------- sequences in bacterial chromosome.
109. Sexduction is a specific form of transduction mediated through-----------factor.
110. During conjugation of F+ and F- strains the DNA which is most likely to be transferred is -----
-------.
111. Gene for bacteriocin production by Clostridium perfringens is present on-------------------------.
112. TxeR gene regulating toxin A and toxin B production in Clostridium difficile is on ------------.
113. cpe gene od Clostridium perfringens type A causing food poisoning in human being is present
on ---------------and is associated with a ---------------------.
114. Urease gene present in Clostridium perfringens type D and E is on the same plasmid which
carries genes for--------------------------.
115. A plasmid known as ------------in carries genes for neurotoxin (TeNT) of Clostridium tetani,
its size is about---------------.
116. Genes for botulinum toxin (BoNT) type C and D are on-----------------------.
117. In Yesinia pseudotuberculosis highest concentration of plasmid is detected in bacteria grown
in glucose containg medium when incubated on---------------platform.
118. --------------------and----------------------are a few antibiotics which can be used for eliminating
drug resistance mediating plasmids.
119. Plasmids in Archaea are generally------------in size.
120. Archaea are also infected by phages, most of the phages are having---------------as genetic
material.
121. Linear plasmids may be --------, --------- or ------------ for the host similar to-------------
plasmids.
122. Linear plasmids often encode for production of --------------, --------------------, ------------, ------
------- and ------------------------------.
123. CEN plasmids found in eukaryotes are plasmids which contain----------------------------- and
more----------------------to loss during ---------------------reproduction.
124. Plasmids of eukaryotes are maintained to greater extent during--------------- division than
during--------------------multiplication.
49
125. Linear plasmids without CENs are about ---------stable-------circular plasmids in mitosis.
126. Circular recombinant DNA plasmids that contain -------------------------------are maintained in
extrachromosomal form in transformed yeast cells.
127. Pseudomonas putida and Ochrobactrum strains isolated from hazardous waste sites often
contain-------- plasmids for -------------------to get carbon and energy from-------------------.
128. Linear plasmids are often-----------in size than circular plasmids.
129. Linear plasmids those have --------------------------- are named as hairpin plasmids, these are
common in --------------------------.
130. Borrelia burgdorferi, causative agent of Lyme disease harbours a 49 kb linear plasmid for
encoding ------------------------------through genes ospA and ospB.
131. There are two types of linear plasmids exist, hairpin plasmids with --------------and those with -
------------bound to their 5′ termini.
132. Cryptic hairpin plasmids are often detected in --------------of the plant pathogenic fungus
Rhizoctonia solani.
133. Most of the linear plasmids have ---------------------- while ----------------------are found only in
few pathogens.
134. Majority of linear plasmids from eukaryotes are--------------------, with only a few exceptions.
135. In Neurospora spp. and in Podospora anserina, -----------------and-----------respectively are
correlated with the prsence of linear plasmids.
136. Linear mitochondrial plasmids are present in many ------and in some -----------, but they seem
to be absent from most ------------cells.
137. Common structural feature of linear plasmid called an -------------is characterized by the
presence of terminal inverted repeats and proteins covalently attached to their 5′ termini.
138. Linear mitochondrial plasmids possess -------------ORFs that can encode ------------- proteins
but they often have ORF coding for the----------------------.
139. Although the functions of most linear plasmids in plant mitochondria are unknown, some
plasmids may be associated with ----------------------rearrangements and may have ----------------
-effects due to their integration into mitochondrial genome.
140. The Brassica 11.6-kb plasmid, one of the linear mitochondrial plasmids in plants, shows a -----
---------inheritance, in contrast to --------------genomes.
141. The genome was first sequenced in 1977 by-----------.
142. ----------------used microarrays to locate mutations in cancer victims.
143. The part of genomes that encode proteins or gene is known as
144. The order of events in sequencing a DNA is
145. The -------------- step of DNA sequencing is skipped while using shotgun sequencing method.
50
146. There are about ------------times as many kinds of mRNA as there are genes in eukaryotic
genome because genes are fragmented and use -------------- splicing.
147. Genes for typical single-character Mendelian traits are called ----------------------.
148. The genes encoding rRNAs are ---------------------------------of genes.
149. Around centromere in eukaryotes ------------------------------or---------------------- sequences
tend to localize.
150. In higher animals and plants about------------of genome is actual protein encoding ORFs while
rest is --------------------is structural DNA.
151. -------------- transposon jump is most likely associated with a harmful mutation.
152. Long terminal repeats are also called ----------------------.
153. --------------% of human genome constituted by repeated sequences.
154. Transposons that have lost their replication machinery are called-----------transposons.
155. In higher eukaryotes as Human, --------------of genome is made of transposable elements
156. At genetic level it has been deduced from comparison of different genomes that organisms are
quite----------------to each other as ------------% of genes of fruit fly Drosopholia have human
counterparts.
157. --------------is the most effective method to analyze variation at the whole genome level.
158. Certain genes of bacteria are present in the human genome but are not present in the genomes
of roundworms or fruit flies and those bacterial genes might have become part of human
genome through---------------------transfer.
159. -----------------------is the field of study involving the sequencing of the genomes of organisms.
160. ----------------- stands for cataloging and analyzing every protein of an organism in order
understand the genome.
161. Microarray gene chips are used to screen for mutations leading to cancer, to identify ------------
of genetic diseases and to identify probable behavioral traits.
162. The identification of the function of a gene in a genome can be accomplished using ------------.
163. Labeling a stretch of DNA according to its function is called ------------.
164. Bioinformatics combines --------------- with molecular genetics to determine what protein a
particular nucleotide sequence encodes.
165. CpG islands and codon bias are tools used in eukaryotic genomics to --------------------.
166. With the increase in complexity of an organism gene density-------------- and -------------
number increases.
167. -----------------are proteins that are translated off of one mRNA. They are then cleaved ---------
-------------------into individual proteins.
51
168. Proteomics is mainly concerned with ------------------- and little with DNA as it pertains to
translation of proteins.
169. ------------------------emerge out of gene duplication and mutation events while only gene
duplication alone would create only copies of a gene.
170. On genome sequencing of the first free-living organism, (Haemophilus influenzae), it was
found that about ----------------- of its genes have some known function.
171. The "shotgun sequencing" method was outlined by----------------------.
172. In the Sanger method of sequencing DNA, the four nucleotide bases are differentiated by ------
----------------and results are read on-------------------.
173. Groups of distinctly different genes that often occur together in a cluster are called --------------
--------families.
174. The whole block of genes that has been copied from one chromosome to another is known as--
-------------------------.
175. The conservation of blocks of genes moved to different chromosomes is referred as -------------
-------.
176. Polyploidy is common in -----------------------and results due to gene duplication.
177. A fast-growing new field of science predicting the structure of a protein from its nucleotide
sequence is called is-----------------------.
178. Production of two daughter cells by simple binary fission in most of the prokaryotes is known
as -------------------cell cycle.
179 In eukaryotes----------- protein monitors DNA and destroys cells with damaged DNA.
180. In eukaryotes duplicated DNA strands held together by a centromere are called---------.
181. DNA and histones bind together to -------------------------.
182. Adenine and guanine have nitrogen-containing bases with --------- ring(s) while thymine and
cytosine have nitrogen-containing bases with ------ ring(s)
183. Piggyback vaccines work by triggering the body's defense system to produce antibodies
against antigens carried on a harmless -----------------virus/ bacteria.
184. Interferons are used to disrupt the reproduction of --------------------.
185. In icrobial genetics LUCA stands for---------------------------------------.
52
Answers: A. 1. E. coli, 2. 11bp unique sequence, 3. Haemophilus, 4. linear, 5. stationary, 6. linear
ss, 5’, 7. cryptic plasmid, 8. Holiday Junction, RuvAB, 9. Saccharomyces serevisiae strains, 10.
a. transformation, b. transduction, c. transposition, d. conjugation, e. protoplast fusion, f.
electroporation, 11. protein, 10 Kda, 12. co-integrate, 13. 480, 14. Ti, 15. conjugal, 140-250 kb,
16. single stranded, double stranded, 17. recA, B and C and homology between the DNA's
involved, 18. 1Mbp, 19. synthesis of all 20 amino acids, to synthesize a cell wall, enzymes
required for the TCA cycle, and many other biosynthetic genes, 20. 5’ to 3’, 21. N-terminus, 22.
2µ, Ti, 23. Catabolic transposons, 24. lysogenic or phage conversion, 25. Due to the inability
of bacteria to engulf materials, 26. mutations, recombinations, 27. 0.58 Mbp to 10 Mbp, 28. 2.9
Mbp (Microsporidia) to over 4,000 Mbp, 29. transduction and conjugation, 30. F pilus,
adhesion molecules, 31. open reading frame, unidentified reading frame, 32. disruption in
specific gene, 33. human gene cloned in their genome, 34. Transformation, G-ve bacteria, G+ve
bacteria, 35. 10 bp, turn of helix, 36. HU, 2 × 104 to 105, 37. 8.5 bp, overwinding, 38. H-NS,
IHF-integration host factor, Lrp-lecine responsive regulatory protein, 39. θ, one, 40. produce
concatemers of DNA, 41. three, theta, sigma, rolling circle, 42. IncP, IncN, IncW groups, 43.
RP1, RP4, RK2, R68; IncP, 44. on solid, liquid, 45. surface pili, 46. copy number, 47. oriT,
helicase, 48. ABC transporter, 49. heat shock, refold the denatured proteins, 50. rpoH, sigma
factor, 51. 70 KDa, 52. σ70 and σE, σE, 53. a. plasmid, b. phage, c. phage, d. phage, e. phage, f.
T12 phage, g. plasmid, 54. global regulatory, 55. yet another sigma factor, 56. double site
specific, 57. Chi sites, V, 58. single site specific, Mu and Tn10, 59. mutagens, 60. reversion,
histidine, Typhimurium, 61. non-functional copy of gene, unreadable copy of the gene, 62.
short repeats 2-4 bases, 63. 16 repeats of GA, i.e.
GAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGA, 6 repeats of TATT, i.e., TATT
TATT TATT TATT TATT TATT, 64. 5000, 4%, 65. 58-60, 66. 6, 3, 3, 67. Luria and
Delbruck, Fluctuation test, 68. 5%, 69. penicillin and radioactive bases, 70. transition,
oxidative, hypoxanthine, uracil, xanthine, 71. transition, transversion, frame shift, 72. Frame
shift, insertion or deletion of one or a few bases, 73. suppressor, 74. purine-pyrimidine base
pair, 75. undergoing recombination, 76. Proflavine and acridine orange, 77. it is easier to detect
revertant of auxotrophic mutants than auxotrophic mutants, mutagenic potential of a chemical ,
78. Nonsense, polarity, 79. near to the start of , 80. pseudorevertant, 81. protein synthesis,
tRNA, 1 more, 82. additions, deletions, inversions and duplication, 83. replication, copy error,
84. Jump back, Jump ahead, 85. intra-replicon, inter-replicon, 86. inversion of intervening
sequences, 87. formation of pyrimidine dimmers, 88. cyclobutane ring, 89. thymine dimers,
cytosine dimers, cytosine-thymine dimers, thymine dimmers, 90. mutator, 91. flip flop, 92.
deletion, point mutation, 93. conservative or Cut and paste, replicative transposition, rolling
53
circle transposition, 94. IS91, 95. IS91, 96. IS256, 97. 5, 98. semi conservative/ replicative,
99. 15-30 kb, 8-9 kb, ~18 kb, 100. Eclipse phase, 101. 300, 102. sense, transcription,
transcribed, left, right, 103. immediate early, delayed early, late, 104. cI, Cro, 105. maize, 106.
no, 107. Hfr, chromosomal DNA, unidirectional, 108. specific base, 109. F, 110. F factor, 111.
megaplasmid, 112. megaplasmid, 113. chromosome, composite transposon (Tn5561), 114.
enterotoxin (etx) iap and ibp, 115. pE88, 75 kb, 116. bacteriophages, 117. stationary, 118.
Quinolones, novobiocin, 119. small, 120. DNA (single or double stranded), 121. beneficial,
neutral, detrimental, circular, 122. protein toxins, specific catabolic capabilities, antibiotic
resistance, pathogenicity factors, senescence induction, 123. centromere DNA of host, resistant,
meiotic, 124. mitotic, meiotic, 125. as, as, 126. autonomously replicating sequences (ARSs),
127. linear, bidegradation, nondegradable substrates, 128. bigger, 129. covalently closed ends,
spirochetes (Borrelia), 130. outer membrane proteins, 131. covalently closed ends, proteins,
132. mitochondria, 133. protein bound to their 5’ end, hairpin, 134. cryptic, 135. senescence,
longevity, 136. fungi, plants, animals, 137, invertron, 138. 1 to 6, unknown, DNA and RNA
polymerases, 139. mitochondrial genome, phenotypic, 140. non-maternal, mitochondrial, 141.
Frederick Sanger, 142. Todd Golub and Eric Lander, 143. open reading frame (ORF), 144.
amplify DNA fragment of unknown sequence; mix with primer; add four different chain-
terminating chemical tags; heat; gel electrophoresis; computer analysis, 145. mapping, 146.
four, alternative, 147. single-copy genes, 148. tandem clusters, 149. constitutive
heterochromatin, structural DNA, 150. 1%, 20%, 151. ALU, 152. retroposons, 153. 3, 154.
dead, 155. 45%, 156. similar, 157. single nucleotide polymorphisms (SNPs), 158. lateral
(horizontal), 159. genomics, 160. proteomics, 161. carriers, 162. functional genomics, 163.
annotation, 164. computer analysis, 165. identify ORFs, 166. decreases, chromosome, 167.
polyproteins, posttranslationally, 168. RNAs, 169. pseudogenes, 170. 50%, 171. Craig Venter,
172. unique radio label, X-ray films, 173. multigene, 174. segmental duplications, 175.
conserved synteny, 176. plants, 177. bioinformatics, 178. simple, 179. p53, 180. sister
chromatids, 181. to neutralize the charge, 182. two, one, 183. vector, 184. virus, 185. Last
Universal Common Ancestor (from which all modern life forms are supposed to be originated),
54
B. Fill in the Blanks
1. For transformation natural competence works well with------------------------------while artificial
competence works well with---------------------------.
2. Bacterial conjugation was first discovered by---------------------------------in -------------.
3. A plasmid can integrate into bacterial chromosome by----------------&------------at specific sites
as some Is elements common in bacterial chromosome and plasmid replicon.
4. Three types of conjugation are a----------------, b.--------------------------& c. . -------------.
5. Plasmids are ----------------endosymbionts of bacteria.
6. Drug resistance in Mycobacterium tuberculosis is acquired through -----------------.
7. Four phases in lytic cycle of bacteriophages are ----------------------------&-------------------.
8. The classic type of investigation for determining the outline of four different stages in growth of
phages is--------------------------.
9. For best lysis by lytic phages the MOI should be----------------------.
10. Number of phage particles released after lysis of phage infected bacteria is designated as---------
--------.
11. Mu phage can undergo a inversion process involving a short piece of DNA to change its host
specificity (similar to flagellar gene of two phases of Salmonella), this inversion in one
orientation allows it to grow on--------------------while in another orientation to infect-------------
&----------------.
12. The dilution at which phages reveals their best specificity is called the -----------------.
13. Phages used in phage typing of Salmonella Typhi are derivatives of--------------and they use ----
----------antigen as their receptors.
14. Promiscuous DNA is one --------------------------------------as-----------------------------.
15. The best method of chromosome sequencing for bacterial genomes is----------------.
16. Archaea are bacteria found in more adverse environment and they also have DNA packaging
proteins --------------------------.
17. There are about-------------------operons in E. coli.
18. Mini-satellite DNA cluster are up to ----------in size while micro satellites are usually of----------
-----.
19. Repeat units are usually ---------long often repeated in a genome.
20. Transposition which involves an RNA intermediate is called----------------.
21. Two main types of secondary structures in proteins are-----------------&-------- stabilized by------
------bonds.
22. Four types of non-covalent bonds found in most biomolecules are -----------------------------------
-----------------------------&--------------.
55
23. Proteins which help to fold the other proteins or molecules are called----------------.
24. The DNA binding proteins which work together or singly to increase or decrease gene
expression are called-----------or------------------------, respectively
25. Transcription termination factor-1 (TTF-1) regulates activity of RNA polymerase I and
activate-------------------------------.
26. During -------activation occurs when the extra cellular signalling compound is able to enter the
cells, while in---------activation the extra cellular signalling compound is not able to ll
membrane, instead they bound to surface receptors to transduce the signals.
27. Two types of transcription terminators in bacteria are------------------and---------------
terminators.
28. Rho is a ----------which breaks base pairs of templet and transcript to terminate the transcription.
29. Two types of transcription termination controls are ------------------------------&------------.
30. Attenuation type of termination control is used for control of transcription of genes coding for--
------------------------------.
31. Most common type of intrinsic terminators are---------------.
32. In E. coli transcription of tryptophan operon is controlled by ------------------&------------.
33. In bacteria all different kinds of noncoding RNA molecules are synthesized by------------.
34. All tRNA must end with a trinucleotide 5’-CCA-3’ which either have lost it or do not have it ,
have the motif added by-------------------------.
35. Two common types of introns in pre-mRNA of eukaryotes are-------------&----------types.
36. Self splicing introns can be classified into-------------------------------------&---------, they get
spliced by autocatalytic process.
37. 5’ end of Bacterial tRNA is created by------------------------------.
38. tRNAPh undergoes self catalytic cleavage in presence of-------------. ()
39. Archaeal bacteria contain introns in noncoding RNA as --------------------&----------------.
40. Introns are absent in ------------------pseudo genes.
41. The pseudo genes can be transcribed i.e. may be readable if they have origin from genes
transcribed---------------------------------because they have internal promoters.
42. There is rapid turn over of bacterial mRNA because they need rapid changes in their protein
profile to meet requirement of their short generation time, their mRNA’s half life is usually------
----.
43. Bacterial mRNA is degraded by a multienzyme complex called--------------------which include
at least an -----------and an-------------------.
44. Stability of mRNA is associated with absence of------------------and presence of------------ to
protect from endo and exo nucleases, respectively.
56
45. The wobble which can recognize all four different codone for a family due to redundant
nucleotide in the wobble position in tRNA is called--------------------.
46. Inteins are intervening sequences found in-------------------------.
47. Special subunits of σ factors present in Bacillus subtilis during sporulation are-------&------
while during vegetative growth are---------------------------&-----------------.
48. Plantibodies are-------------------------------.
49. Ionizing radiation may also cause mutation and it induces lesions like-------------------------&----
-----------, it may also cause direct lesions due to formation of reactive molecules as-----------.
50. Heat may also cause lesions in DNA as --------------, it is due to cleavage of --------------------
bond which attaches sugar component to nucleotide.
51. In each human cell about-----------------AP sites are created daily but efficiently repaired but
may result to mutation when ------------response is activated.
52. If phosphodiester bond has lost between adjascent nucleotides with any damage to 5’ phosphate
and 3’-OH group, this type of nick is often results due to the effect of ------------------radiation,
it is easily repaired by----------------------.
53. Ada enzymes can rapidly repair DNA damage caused by----------------.
54. Deoxyribopyrimidine photolyase is stimulated by the light with the wavelength of ---------------
to---------.
55. The Central player in induction of cell cycle arrest and apoptosis in eukayotes is a protein
called---------.
56. β-n-glycosidic bond between a damaged base and sugar component is caused by---------------,
thereafter the AP site is cut by AP endonucleases viz., ----------------&---------------.
57. Dam converts --------------to------------------in sequence 5’-GATC-3’ and Dcm converts-----------
-------to---------------in sequences 5’-CCAGG-3’ and 5’-CCTGG-3’.
58. Two or more DNA which have their origin from a single common ancestral DNA are called-----
--------------.
59. A clade is a group of --------------------DNA sequences that consists all sequences included in
the analysis of descendance or ancestral history.
60. Incubation of DNA at pH 3.0 selectively removes ---------------and the resulting derivative is
called---------------.
61. In response of hormones and other chemical signals--------is the common second messenger.
62. --------------------------- is an enzyme of bacterial origin which can catalyze the formation of
polynucleotide of RNA without ant templet and lead to formation of -----phosphodiester bonds.
57
63. The common mechanism of inhibiting progression of HIV is use of some dNTP analogues
which are more readily used by HIV reverse transcriptase than by the T cells, some examples
are --------------------------------&-----------------.
64. Transcription in E.coli proceeds at the rate of -------------------bases per second.
65. Partially diploid cells are called------------.
66. Runs of A-T pairs can also form Z forms of DNA is---------------------.
67. Replication of cross linked DNA (induced by mitomycin or dimethyl psoralen) yields daughter
DNA molecules having gaps of the size is equal to one or more------------------.
68. --------------------------method of dark repair of UV damaged DNA is more effective against the
pyrimidine-(6-4)-pyrimidone structure.
69. Hydroxy methyl cytosine in T4 phage DNA reacts with glucose molecule through --& --------
linkages.
70. In phage infected bacteria some in vitro protein synthesizing systems cannot translate the
replicase gene without first translating the coat protein it is called------------------------.
71. Lysogen usually produces -------------phage particles/ ml in mid-exponential phase culture
containing about 2 x 108/ ml.
72. A completely inactive prophage unable to produce any infectious particle is known as------------
------..
73. The region of DNA that includes the cI repressor binding site is referred as the --------------------
-region.
74. ---------is a non inducible temperate phage and also the most efficient transposon.
75. Packaging of phage DNA can be through----------------------------as in ---, ----&-----or via use of
------------------that demarcate the region to be encapsidated as in ---, ----&-----.
76. To find out the evolutionary relation between phages the most commonly used tool is ------------
--------.
77. Phages P1 and Mu can change their host range; the determinants of host range are in-------------
region of their genome.
78. Generalized transducing particles are the progeny from -------------infection of bacteria.
79. About 90% of all generalized transducing DNA introduced into a host cell fail to recombine and
remain as a persistent, circular, non replicating DNA these types of cells are called----------------
------------.
80. Most naturally competent bacteria except-------------------can regulate their competence.
81. In Bacillus subtilis competence is the maximum when the doubling time ranges between---------
----&--------.
82. During transformation with chromosomal DNA, it is incorporated into host by-------------.
58
83. Recombination process bocks the replication for giving---------------------------.
84. Haemophilus influenzae as---------------------copies of an 11 bp sequence that is required for
uptake of DNA for transformation.
85. F plasmid of E. coli is about ---------in size and during conjugation it takes about-------------for
transfer of plasmid DNA into recipient cells.)
86. If we construct a map of S. Typhimurium in similar way as of E. coli K12, then size of E. coli
chromosome is 100 min and that of Salmonella is ------------------.
87. The major difference in E. coli and Salmonella chromosome is a region of inversion, this
regions encompasses about--% of total genome and lies between ----------&--------at ------------
to----------minutes of map.
88. Pheromone mediated conjugation usually occurs in ------------------------.
89. Integration of SCP1 into Streptomyces genome requires ------------located near main plasmid
DNA and right side inverted repeat and another copy near agarase gene.
90. Colicins are bacteriocins produced by E. coli to kill another E. coli and are divided into --------
groups named-----&-------, E. coli resistant to a bacteriocin become resistant to colicins of that
group.
91. Some colicin plasmids are conjugative for e.g. -------------.
92. The molecular weight of Colicin Ia and Ib is about --------------and producing strains are---------
-----immune.
93. Some colicin plasmids are non-conjugative but mobilizable for e.g. -------------.
94. Plasmid R100 is in incompatibility group IncF1 and thus can coexist with---------plasmid.
95. Plasmid R 100 is a conjugative plasmid, conjugation is mediated through-----------.
96. Even plasmids which are not incompatible can interact for example transfer of F plasmid is
inhibited by presence of other related plasmids; this inhibition is mediated through----------------
--system.
97. Addition of chloramphenicol to a culture gradually stops replication of bacterial as well as
plasmid DNA, however some plasmids e.g.,------------------can continue their replication and at
much enhanced level because they-----------------------------.
98. Many standard cloning vector based on ColE3 plasmids multiply much better to produce
innumerable copies when ------------------------is added into culture.
99. When F plasmid is inserted in host chromosome near----------------has excess of copies than in
those strains where it is inserted near oriC.)
100. Some plasmids like----&----replicate in theta mode but will shift for making linear
concatamers in host that are -------------------and --------.
59
101. An enzyme that functions as “biochemical scissors” to cut bacterial DNA at a specific site is
known as------------------while those trims the DNA from ends are called-----------.
102. Conjugation, transduction and transformations are best studied method of gene transfer in------
--------------.
103. After replication of the bacterial chromosome, the new progeny bacterium contains two DNA
strands,------------newly synthesized.
104. A plasmid containing a bacterial DNA and a monkey DNA is called---------------.
105. The process in which bacteria take up naked genetic material from the surrounding
environment is called--------------------.
106. Transformation, transduction and conjugation are all examples of----------- gene transfer.
107. Anticodon is present on-----------.
108. A purine may base pair with--------------------.
109. The 3 nucleotide group that codes for one amino acid is called-----------------.
110. If a nucleic acid contains the bases A-U-G-G-C, it would have to be a molecule of------.
111. Circular plasmids in eukaryotes have been identified in ------------, --------------and the ----------
--------of numerous fungi and plants.
112. Plasmids are-------------in parasitic protozoans.
113. Polycistronic transcription is ----------------in equkaryotes.
114. The term plasmid was first introduced by-------------------in------------.
115. Vectors are the plasmid commonly used for transformation in--------------------------.
116. In plasmids we can clone maximum gene length of about -------------------, if we want to clone
longer sequences tha we should use--------------------, --------------------, ---------------------- or----
---------------------------------.
117. Sex pheromones are produced by some bacteria--------------------plasmid, chemically they are -
----------------------.
118. One of the largest chromosome in Archaea has been sequenced is of---------------------- its size
is around-----------.
119. Archaea are genetically distinct from bacteria and eukaryotes, with up to ------of the proteins
encoded by any one archaeal genome being unique to the Archaea. However, most of unique
genes have no known function.
120. Among Bacteria and Archaea large differences in ------ genes and their ------------------------.
60
Answers. B. 1. linear DNA, covalently closed circular DNA, 2. Lederberg and Tatum, 1946, 3.
transposition, recombination, 4. Phage mediated, pheromones mediated & sex pili mediated conjugation,
5. obligate, 6. mutation in genes encoding for drug receptors, 7. adsorption, penetration of phage
nucleic acid, intracellular development and release, 8. one step growth cycle, 9. 0.1, 10. burst size, 11. E.
coli K12, E. coli C, Citrobacter freundii, 12. Routine test Dilution, RTD, 13. Vi phage II, Vi,
14.Transferred from one organelle to other in the same cell, mitochondrial genes found in chloroplast, or
genes from chloroplast or mitochondria in plant chromosome, 15. shotgun approach, 16. histones of
eukaryotes, 17. 600, 18. 20 kb, <150 bp, 19. 4bp, 20. retro transposition, 21. alpha helix, beta sheet,
hydrogen, 22. hydrogen bonds, van der Waals forces, electrostatic interactions, hydrophobic
interactions, 23. chaperones, 24. enhancers, repressors, 25. RNA polymerase I and II, 26. direct,
indirect, 27. intrinsic, Rho dependent, 28. helicase, 29. antitermination, attenuation, 30. enzymes
involved in amino acid biosynthesis , 31. A-U base pairs in a stretch, 32. attenuation, repressor, 33. RNA
polymerase, 34. tRNA nucleotidyl transferase, 35. GU-AG, AU-AC, 36. groupI, II, III, 37. ribonuclease
P, 38. divalent lead ions, 39. pre-tRNA and pre-rRNA, 40. processed, 41. RNA polymerase III, 42. a
few minutes, 43. degradosome, endonuclease, exonuclease, 44. target sequence, hairpin loop on 3’ end,
45. super wobble, 46. some proteins, 47. σF, σE, σA, σB, 48. Antibodies grown in GM plants, 49. point
mutation, insertion, deletion, H2O2, 50. AP sites/ baseless sites or apurinic/apyrimidinic sites, β-n-
glycosidic, 51. 10 000, SOS, 52. ionizing, ligase, 53. alkylation, 54. 300 nm, 500 nm, 55. p53, 56.
DNA glycosylase, exonuclease III, endonuclease IV, 57. adenine, 6-methyladenine, cytosine, 5-
methylcyosine, 58. monophylectic, 59. monophylectic, 60. Purine bases, apurinic acid 61. cAMP, 62.
polynucleotide phosphorylase, 3’, 5’, 63. AZT, 3’-azido-2’,3’-dideoxythymidine; DDI, 2’,3’-
dideoxyinosine, 64. 50, 65. merodiploid, 66. flanked by 3-5 G-C pairs, 67. Okazaki fragments, 68. short
patch repair or nucleotide excision repair, 69. α & β, 70. translational coupling, 71. 106, 72. cryptic
prophage, 73. immunity, 74. Mu, 75. heedful mechanism, P1, P22, Mu, cohesive ends, λ, P2, P4, 76.
heteroduplexing, 77. invertible, 78. lytic, 79. abortive transductants, 80. Neisseria gonorrhoeae, 81. 150
minutes, 390 minutes, 82. homologous recombination, 83. time to repair, 84. about 600, 85. 94.5 kb, 1
min, 86. 135 min, 87. 12, purB, aroD, 25, 37, 88. Enterococcus faecalis, 89. IS46 6, 90. 2, A & B, 91.
ColIa, ColIb, 92. 80 kDa, not cross, 93. Col E & Col A, 94. F plasmid, 95. specific pili which is thinner
than F pili, 96. fin, 97. Col E3 family, can use existing replication enzymes, 98. antibiotics like
chloramphenicol, 99. trp/ terC, 100. ColE1, F, recBCD, sbc, 101. endonuclease, exonuclease, 102.
bacteria, 103. one is, 104. chimera, 105. transformation, 106. horizontal, 107. tRNA, 108. pyrimidine,
109. codon, 110. RNA, 111. cytoplasm, nucleus, organneles, 112. wide spread, 113. unusual
(uncommon), 114. Joshua Lederberg, 1952, 115. genetic engineering, 116. 10 kb, λ phage without
lysogeny genes, cosmids, bacterial artificial chromosome or yeast artificial chromosome, 117. not
carrying, small peptides, 118. Methanosarcina acetivorans, 5.75 Mb, 119. 15%, 120. tRNA, aminoacyl
tRNA synthetases.
61
C. Fill in the blanks
1. Plasmids often need host --------------------for primer synthesis.
2. There are five 19 to 22 bp direct repeats in repA-repE region of F and P1 plasmids, these are
called introns, it has been seen that number of introns is ---------------proportional to copy
number.
3. A plasmid defective in -------------------,----------------------------&---------------------will not
persist in its new host.
4. Broad host range plasmids are those which are----------------and can stabilize in new host.
5. Genes from Bacillus cereus cloned in E. coli are easily expressed but E. coli genes are not easy
to be expressed in B. cereus it is because E. coli promoter does not require specific base
sequence out side of the ----------&------------regions, while B. cereus requires a specific ------
base sequence upstream to------site.
6. Eukaryotic genes are not easily expressed in bacteria because of-----------------.
7. Phages like T7 and T3 which code for their own RNA polymerase recognizes its own particular
promoter because of difference in the promoters at-------------site.
8. Cryptic genes are those which are never expressed they are also found in some bacteria namely-
-----------------------------------------------------------&--------------.
9. Bacillus subtilis sacB gene encodes for----------------------------------, gram negative bacteria
expressing sacB gene can---------------------in presence of 5% sucrose but exhibit no phenotype
in absence of sucrose.
10. The wavelength to measure culture density is either-----------------or----------, at low wavelength
sensitivity is -------but linear range of measurement is ------------- and at higher wavelength
sensitivity is -------but linear range of measurement is -------------.
11. Klett units are used to measure -------------------, these are photoelectric colorimeters which take
measurements at----------.
12. Salmonella Typhimurium LT2 genome contains about-----bp long ------pseudo pack sites for
P22 phage, while P22HT mutant transduce at much higher rate because it recognizes pseudo
pack site as efficiently as the bacteriophage genome due to alteration in specificity of enzyme---
--------------. Altered system recognizes------------pack sites both in phage and chromosomal
DNA with equal efficiency producing large number of -------------particles, and thus HT
mutants are the tool of choice for generalized transduction.
13. mutD is an alternative name to --------gene which encode for-----------of DNA polymerase III
required for its-----------------activity.
62
14. Frequency of spontaneous mutations depends on composition of-----------, -----------------and ---
----------------. Faster growth conditions result into---------------fold------------mutation
frequencies.
15. Most of the transposons used contain strong-------------and insertion of transposons is thus
associated with----------.
16. Polar mutations affect down stream genes in ---------------operon.
17. Nonpolar insertions usually consist of ---------------cassettes that contain-----------------------------
--------------------.
18. Extent of supercoiling is maintained by--------------encoded----------&--------genes which adds --
--------------supercoiling to DNA, and by ---------------encoded by -------gene which ---------------
---supercoiling.
19. Studies suggest that expression of about-------------% genes are affected by degree of
supercoiling.
20. Bacterial histone like proteins as HNS regulates expression of many genes of bacteria by
binding to ----------DNA.
21. Integration host factor (IHF) regulates many genes by-------------and causing-------------.
22. Mutations that eliminates IHF and HNS in Salmonella Typhimurium result into----------
virulence.
23. Phage that adsorbed--------------make larger plaques than those adsorbed---------------.
24. When a host bacterium is partly resistant to a phage the plaque formed will be-----------.
25. When a temperate phage infects a population of exponentially growing cells, each phage
produce a plaque with ---------------------morphology i.e.,----------------.
26. Lytic growth of phage is favoured when rate of bacterial growth is---and MOI is---- whereas
lysogeny is favoured when rate of bacterial growth is---and MOI is----.
27. Purified DNA is stored in Tris-EDTA (TE) buffer because it chelates heavy metals which may
cause-------------------and also chelate divalent cations which are required by most---------------
which can deteriorate DNA.
28. The major draw back with isopropanol precipitated DNA are------------------------------------------
------------------------------&--------------------.
29. Isopropanol gives more pure DNA at -------------------and after washing the precipitate with------
---------------.
30. Two common methods to remove impurities from DNA are a.------------------------and b.---------
----------------.
31. An OD260 with dsDNA is equivalent to about ----------/ml solution and ratio of absorbance to
DNA concentration is linear to approximately----------OD.
63
32. For pure dsDNA, ratio of absorbance of a solution at 260 and 280 nm should be between-----&-
-------, higher ratio is due to ---------contamination and lower ratio is due to-------------
contamination.
33. Impure DNA preparation’s concentration is not easy to measure with spectrophotometric
method but it should be ---------------------------and compare the band intensity on agarose gels
after treatment with--------------.
34. Agarose to give good resolution of low molecular weight fragments of DNA we use--------------
agarose which contains agarose with ---------------------along with standard agarose.
35. Pure ethidium bromide is ----------------times -------------fluorescent than DNA-ethidium
bromide complex.
36. Sensitivity of ethidium bromide staining is more for----------------.
37. Primers in PCR are complementary to -----------ends of the opposite strands.
38. Taq polymerase has optimum temperature for activity between ----------&-----oC can cause
extension of DNA strands up to----------------, however efficiency is much better up to------------
.)
39. Frequency of misincorporation of nucleotides using Taq polymerase is----------------, it is
because it lacks----------.
40. If we increase the concentration of Taq polymerase in PCR mix it might laed to-----------.
41. Non specific products in PCR can--------- when too little amount of templet DNA is used.
42. Primers are used usually at ---------------concentration in PCR.
43. When too much templet DNA is used efficiency of PCR decreases because of--------------.
44. When 200 µM of each of the 4 dNTPs are used they can yield ------------µg of DNA at the level
of 50% utilization.
45. On decreasing the dNTPs below 10-20 µM frequency of -------------increases.
46. Excess of Mg++ increases---------------------and thus decreases the --------&--------of desired
product.
47. Synthesis of a specific PCR product goes on until concentration reaches--------.
48. In 30 PCR cycle you may have as many as----------------- DNA molecules of specific product.
49. By increasing the cycles too much (above 30) in PCR you may get more--------------.
50. In asymmetric PCR the proct increases -- -----------.
51. The ratio of molecular velocity to centrifugal force is called-------------------------, its unit is------
--- or -----------------------it is equivalent to-------------------.
52. Zonal centrifugation also a kind of -----------------, in which density of the solution of the
molecule to be separated is adjusted--------------than that of density of solution of the gradient
fluid which -------------from top to bottom, e.g. -----------gradient method.
64
53. Equilibrium centrifugation also a kind of -----------------, in which density of the solution of the
molecule to be separated is adjusted so that it is -------------- that of density of solution of the
gradient fluid but its density changes with centrifugation creating a density gradient e.g. ---------
----density gradient.
54. When you use sucrose gradient and centrifuge for quite long there will be-----------separation
while on using caesium chloride gradient increase in time will have---------effect on separation.
55. For linear DNA electrophoresed on agarose gels distance moved, depend roughly on
logarithmic value of its molecular weight and can be expressed as-------------------------.
56. Terminals of nucleic acids are made of--------------&------------------chemical groups.
57. There are ----------phosphate groups per base in DNA and RNA.
58. If the cell wall of a bacterium is completely stripped it is called it is called-----------and when it
is stripped partly it is called-------------.
59. The doubling time of a culture growing in 75 ml chemostat with flow rate of 0.3 min is-----------
----------.
60. A bacteriophage capable of causing only lytic growth is called-------------.
61. The -------------------------converts intracellular phage DNA into a form appropriate for
packaging in phage particle.
62. Filamentous phages often release their progeny continuously from their host by----------of cell
wall; this process is called---------------------.
63. EOP for most phages is nearly--------- except for those which form----------plaques.
64. Storage of phage at ----------rapidly reduces its EOP; it can be protected by adding----------------
---------------------------or-------------.
65. Lysate of phages grown in lawn of soft agar is called-----------------.
66. When a toothpick is stabbed into a plaque it can pick up about-----------------sufficient enough
to cause----------------------.
67. Two common conditions which stimulate lysogenic conversion of a temperate phage are---------
------------&-----------------.
68. Both excision and insertion of temperate phages into bacterial chromosome are examples of-----
-----------------------recombination.
69. Excision and insertion of temperate phages into bacterial chromosome are catalysed by-----------
----&--------------------, respectively.
70. One can ensure a lytic cycle of temperate phages by using ---------------------------&---------------
-------.
71. Temperate phages often form plaque with inner turbid and outer clear zone; the turbidity in
inner zone is due to---------------------.
65
72. RNA polymerase binds to a region in dsDNA consisting of specific base sequence of about------
-------------------bp long called-----------------, and the RNA polymerase bound to dsDNA is
called---------------------------.
73. Turning on and off a set of genes as a unit, simultaneously or by a single signal is called--------.
74. ----------------------------regulation involves removal of inhibitor.
75. It is not required for a repressor gene to be present near operator for regulation of gene because
repressor is often----------------------.
76. In presence of glucose level of cAMP is -------------.
77. The DNA damage to bacteria can be detected using survival curve studies, survival curve
analysis is done using ---------------------.
78. The most effective wave length for killing of bacteria is same as for---------------------and also
matches with the absorption spectrum of---------------.
79. The dark repair can be accomplished by three mechanism called a.--------------, b.-----------------
--& c.------------------.
80. UV irradiated λ phage produce more infective centres in lawn of UV irradiated E. coli than in
lawn of healthy E. coli this phenomenon is called--------------------or---------------.
81. In E. coli a repair endonucleases identifies thymine dimers and make incision to remove---------
--------, in the process two cuts are made one---------------nucleotide to 5’ and one at ----------
nucleotide to 3’ of thymine dimmer.
82. C to U conversion rarely leads to mutation because cells posses -------------.
83. C to T conversion often leads to mutation because -----------------------system can not correct the
misplaced thymine as the cells do not have--------------------.
84. Sites within genes where frequency of mutation is high are called-----------------.
85. In ts-mutants the affected gene is transcribed and translated into the desired protein, but the
protein is inactive because of-------------------------------------.
86. Due to presence of complex regulatory pathway for dNTP synthesis BU-NTP inhibits synthesis
of --------------------, which leads to disturbed ration of------------and promote misincorporation
of------------in opposition of---------during elongating DNA.
87. The maximum effect of alkylation of ethyl methane sulfonate (EMS) is on -----------&------------
----, which impairs normal-----------leading to mispairing.
88. Alkylation of DNA impairs normal------------------of thymine and guanine.
89. Hydroxylamine reacts specifically with--------------to convert it into------------which than pairs
with-------leading to-------------.
90. Nitrous acid converts ---------------------&----------to-------------------&----------------, which form
base pairs with---------&------------, respectively.
66
91. When hydroxylamine is used in vivo, it produces-----------------to damage DNA, this damage
induces-----------------which results into mutations.
92. Donor conjugal DNA synthesis is to -----------------------.
93. Recipient conjugal DNA synthesis is to -------------------------.
94. Uptake of purified DNA by bacteria is known as-----------------.
95. DMSO make the bacterial cell membrane permeable to the-------------------------.
96. During chemical transformation using CaCl2 as inducing agent naked DNA is adsorbed onto
the cell as -------------------.
97. The optimal electric field for electroporation of E. coli and Salmonella spp. is usually between--
-----------to-------------.
98. In renatured DNA of phage Mu there may be some unpaired region called--------.
99. Some times some region of phage gets inverted in the life cycle of Mu then normal and inverted
configurations are called ------------&-------------, respectively, annealing of strands from the G+
and G- phages yields----------.
100. The function of counter selective marker in an Hfr × F- mating is -----------------------.
101. Two activities of recA protein are---------------------&-------------------.
102. T4 phage DNA does not have---------------------base, instead it has--------------.
103. T4 nucleases digest the host DNA its target is-------------------containing DNA.
104. Although single molecule of templet DNA may yield product in PCR, best results are
obtained when ------------copies of templet DNA are present.
105. In practice PCR is a self limiting process and amplification factor usually ranges between --
--------to---------.
106. Taq, Pfu and Vent DNA polymerases are isolated from -------------,------------------and--------
----------, respectively.
107. Trangenge or Transgene are genes transferred into an organism to create---------------.
108. For quantification of DNA or RNA absorbance is measured at----------while for detecting
protein at-----------------.
109. DNA concentration (µg/µl) in a purified sample is determined by multiplying OD at 260nm
with dilution factor and----------.
110. RNA concentration (µg/µl) in a purified sample is determined by multiplying OD at 260nm
with dilution factor and----------.
111. OD ratio at 260/280nm between --------and ----------indicate purity of DNA and RNA,
respectively.
112. To calculate appropriate amount of PCR product in to a ligation reaction while cloning in a
vector the formula is-----------.
67
113. ---------of a PCR primer is crucial in deciding its false priming efficiency.
114. F plasmids are about -------------in size and bacteria can be freed of them through----------------
-with acridine orange.
115. Presnece of cryptic plasmid has----------effect on cell physiology.
116. F+ strains of bacteria usually posses F plasmid in -------------------form.
117. Hfr strains have F plasmid in ------------------form.
118. Chromosomal gene transfer along with F plasmid is common in --------bacterial strains.
119. In transposition genes moves from one-------------to other -----------in --------bacterium.
120. IS elements are about -------kb in length.
121. Transposable elements are of thre types including------------, --------------and--------------.
122. In gene mapping, conjugation is used for mapping of a ------------of chromosome while
transduction is commonly employed for mapping of ---------- of chromosome.
123. An engineered plasmid is ---------------plasmid having--------------region for-----------.
124. Translation begins at------------------and ends at-----------------------------------.
125. Three major classess of enzymes as per their genetic control are 1. metabolism enzymes which
are expressed at a fixed rate, 2. --------------and 3. -------------------enzymes.
126. Mutations which occur in absence of any mutagen are named as---------------mutations.
127. Point mutations are of three types on functional basisnamely, -----------------, -------------and----
------------------ mutation.
128. Frame shift mutations may be due to----------------or -------------------mutations.
129. UV radiation induces formation of ----------dimers while --------------- repair revert this
dimerization.
130. Mutagens increases mutation rate by about ----------------------times.
131. -----------------and-----------------mutations permit bacteria to suvive in natural environment.
132. Dissimilation plasmids encodes for enzymes for catabolism of ---------------------.
133. Pili are encoded by genes on--------------------plasmids.
134. The precise control of ‘when’, ‘where’ and ‘what’ genes are to be switched ‘ON’and ‘OFF’ is
called -------------------.
135. The paradigm of Gene Regulation begins with the seminal work of Francois Jacob, Jacques
Monod and their colleagues which led to the concept of -------------of gene regulation.
136. Gene regulation can take place at any of the five stages in the pathway of information flow
including----------, -------------, ---------------, ------------------- and ------------------.
137. Three types of commonly required genes by microbes are------------, ---------- and-------.
138. ------------------------genes are continuously been kept On (transcribed) because their products
are needed throughout life span.
68
139. ------------------are switched on (transcribed) only when some signals is present.
140. -------------------are generally On but get switched off in the presence of the signal.
141. Genes can be induced (switched ON) or repressed (switched OFF) by using either a ---------
control or a ----------- control mechanism. In the first, presence of signal molecule switches -----
---and in the second it switches------- the transcription.
142. Signals regulators of transcription may be of two types viz., ----------------and-------------.
143. A repressible system using a negative control envisages the presence of a negative regulator
(repressor) which is normally inactive, the signaling molecule which make -----repressor active
is called------------------------.
144. A polycistronic mRNA is one which codes information for more than one------------.
145. For all genes invoved in carrying out a function in a given pathway of anabolism or catabolism
are switched ON or OFF simultaneously under control of a ------------ or the same-----------------
----- are said to be in one-----------------.
146. Although the presence of lactose induces the lac operon, ---------------is the true inducer as its
binding to the repressor molecule de-represses the operon.
147. The lac operon, an inducible operon, is under ---------and ------------ regulation as for the
operon to be switched ON the ---------should be inactivated and ------------complex should bind
to the promoter for its activation.
148. One of the control points of the trp operon is the repressor protein which is encoded by the
trpR., however fot total control the trp operon also employs another control mechanism called -
-------------.
149. In arabinose operon (ara) regulation AraC protein has a dual function. In the presence of
arabinose it acts as a ------------ while in its absence acts as-----------------.
150. Arabinose operon is an example of an --------------operon which uses a positive regulator
(AraC protein) to turn ON the operon when arabinose is present while in absence of arabinose,
action of positive regulator is inhibited and the operon got turned OFF. The operon is further
kept repressed by the formation of a--------------.
151. Colicins are antimicrobial compounds produced by members of Enterobacteriaceae as defense
mechanism against other related bacteria; genes for colicins are present on---------.
152. Mutations are the --------------changes in DNA.
153. Spontaneous mutation rate for the average gene is-------------------.
154. Transposons in eukaryotes carry --------------useful genetic information and are often called----
-------------------.
155. Insertion elements carrying some additional gene than those required for transposition are
called----------------------.
69
156. A codon is a series of 3 nucleotide bases that code for an amino acid, it is known as-------------
-code.
157. Process of building protein molecules from the genetic information in mRNA is called ------.
158. Proteins are produced through process of polymerization of---------------------.
159. Nucleic acids are made through process of polymerization of---------------------------.
160. ------------------was discovered by Francis Griffith while experimenting with Streptococcus
pneumoniae.
161. The term “DNA duplexes” was first used by --------------to define double strand of DNA.
162. In the process called ------------replication each daughter duplex consists of one strand
conserved from the original duplex from parent and one newly-synthesized chain.
163. Bacteria don't have sex. Normal mechanism of cell division is ------------------.
164. When one cell produces many identical offspring it is called--------- cell division.
165. Agrobacterium tumefaciens can introduce a part of its Ti plasmid DNA called T-DNA into
plant cells with cell-to-cell contact, this phenomenon is called---------------------.
166. Transcription and translation in archaea have more similarity with these processes in -----------
than in------------.
167. Archaea reproduce asexually by---------,--------, --------- and--------; there is no-----------, so
even if a species of archaea exists in more than one form, all forms will have the same genetic
material.
168. In archaea chromosomes are replicated from -------------- using ----------------- that resemble
the equivalent eukaryotic enzymes.
169. Spores are made by many -------- and -----------, but are not formed in any of the known -------.
170. Linear plasmids of ----------------are contemporary manifestations of ancient viruses, which
have adjusted to two cellular compartments during evolution.
171. Plasmids of filamentous fungi are ---------------------routinely residing in the mitochondria, in
which they underwent coevolution with their hosts. In addition to the archetypical viral B-type
DNA polymerase, they exclusively encode a viral---------------------.
172. Most of the yeast linear plasmids are located in-----------------------------.
173. Some linear plasmids of yeasts encode protein toxins, which benefits the respective host while
competing with other yeasts i.e., conferring them --------------phenotype.
174. Two factors produced by killer yeast are ----------&-------- are encoded by genes on -----------.
175. Nicked Open-Circular DNA of a plasmid has one cut in-------------.
176. The CTX phage has received special attention because it is the first -----------------found to
transfer toxin genes to its host.
70
177. Trypanosoma brucii (agent of African trypanosomiasis) have a plasmid which have--------
transcription.
178. Corynebacterium diphteriae, the bacteria causing diphtheria, doubles every hour in vitro, but
in that same hour a lytic corynephage can produce -----------converting phages.
179. If phage-associated VF genes are to be disseminated, then ------------------and-------must occur
in situ.
180. The term allelopathy is used to describe the ability of some organisms to -------------------the
growth of plant competitors.
181. In process of--------------, phages released from induced bacterial lysogens block the growth of
competing bacteria by infecting and subsequently killing them. Bacteriocins can also lead a
similar competition but, unlike phages, are not equipped to ---------an effective toxin dose.
182. Hot start PCR is used to curtail the amplification at------------ and -----------------------formation
which may occure with low fidelity priming. It improves the -----------------, ---------------and
product yield in PCR.
183. Some oligonucleotides can also act as inhibitor for amplification of DNA in PCR there utility
is predicted in ----------------PCR because there action is ------------dependent.
184. Similarity in the structure and terminal nucleotide sequence -----------------of Mycobacterium
linear plasmids (pCLP) and linear plasmids of Streptomyces and Rhodococcus species,
indicating a -------------these elements within the Actinomycetales.
185. Large plasmids in Mycobacterium xenopi, M. branderi, M. avium and M. celatum have a -----------
-----topology. Studies have confirmed long TIRs are --------------------necessary for -------------
and maintenance of linear plasmids.
186. Several linear plasmids have been identified in Streptomyces, but these are all relatively -------
because ------ linear plasmids cannot be separated from ---------------DNA by conventional
techniques.
187. There are several -----------evidence suggests that a character is encoded by a plasmid but no
plasmid can be physically detected it is because the coding plasmids are ------------------- and
they can not be differentiated from---------------------DNA.
188. For isolation and identification of linear plasmids the best technique is one of the --------------.
189. In mitochondria of some fungi and plants youcan detect---------types of plasmids with-----------
-------------------------morphology.
190. Similar to vaccinia virus in several linear plasmids origin of replication reside near---------
terminii.
71
Ans. C. 1. RNA polymerase, 2. inversely, 3. replication, copy number control, partitioning, 4.
conjugative, 5. -10, -35, 8-9, -10, 6. introns, 7. -11, 8. Lactobacillus, Bacillus, Escherichia &
Salmonella, 9. levan-sucrase or sucrose:2,6-β-D-fructan 6-β-D-fructosyltransferase, not grow,
10. 420nm, 650nm, high, narrow, low, wide, 11. cell density of a culture, ~660nm, 12. 8-12, 7,
terminase, cryptic, transducing, 13. dnaQ, ε-subunit, proof reading, 14. growth medium, salt
concentration, growth temperature, 10-100, increase, 15. terminators, polar mutations, 16. the
same, 17. antibiotic resistance, strong outward directed promoter, 18. DNA gyrase (Type II
topoisomerase), gyrA, gyrB, negative, Topo I (Type I topoisomerase), topA, removes negative,
19. 50, 20. curved, 21. Binding to DNA, DNA bending, 22. decreased, 23. early, late, 24.
uniformly turbid, 25. bull’s eye, turbid centre surrounded by a ring of clearing, 26. high, low,
slow, high, 27. nicking, nucleases, 28. it may contain salts, it is difficult to evaporate
isopropanol from DNA which interfere with enzyme reactions with DNA, less suitable for
electroporation, 29. room temperature, 70% ethanol, 30. drop dialysis, and spin columns, 31. 50
µg, 2, 32. 1.7, 1.9, RNA, protein, 33. stained with ethidium bromide, RNase, 34. blended, low
melting point, 35. 50, less, 36. dsDNA, 37. 3’, 38. 75-80, 10 kb, 3 kb, 39. 10-4 to 10-6
nucleotide per cycle, 3’-5’ exonuclease activity, 40. increase in non-specific products, 41.
increase, 42. 0.1 -0.5 µM, 43. impurities in DNA, 44. 12.5, 45. misincorporation 46. non-
specific annealing of primers, yield and specificity, 47. 0.3 – 1 pmole, 48. 104-105, 49. non-
specific products, 50. arithmetically, 51. sedimentation coefficient, s, Svedberg, 10-13, 52.
gradient centrifugation, lower, decreases, sucrose, 53. gradient centrifugation, equal, caesium
chloride, 54. no, no, 55. D= a-b log M, where a and b are constant depending on gel
concentration, buffer and temperature, 56. -3’-OH and 5’-phosphate group, 57. one, 58.
protoplast, spheroplast, 59. 75/0.3 = 250 min, 60. virulent, 61. maturation, 62. out-folding,
extrusion, 63. 1, very small, 64. ≥4oC, glycerol, cations, proteins, 65. plate lysate, 66. 106,
confluent lysis of a bacterial lawn, 67. depletion of nutrients in growth medium, high MOI, 68.
site specific, 69. excisionase, integrase, 70. low MOI and rich growth medium, 71. growth of
lysogens, 72. 20-40, promoter, closed promoter complex, 73. coordinated regulation, 74.
negative, 75. diffusible, 76. low, 77. target theory, 78. mutagenesis, DNA, 79. excision repair,
recombinational repair, SOS repair, 80. UV –reactivation, W-reactivation, 81. sugar-phosphate
back bone, 8, 4-5, 82. ausat glycosylase, 83. mismatch repair, thymine glycosylase, 84. hot
spots, 85. failure of proper folding, 86. dCTP, dTTP/dCTP, dTTP, dGTP, 87. guanine, thymine,
hydrogen bonding, 88. hydrogen bonding, 89. cytosine, N-hydroxycytosine, adenine,
transition, 90. cytosine, adenine, uracil, hypoxanthine, adenine, cytosine, 91. free radicals, SOS
system, 92. replace the transferred DNA strand, 93. convert the transferred single stranded
DNA to double stranded DNA, 94. transformation, 95. PEG-DNA complex, 96. Ca-DNA
72
complex, 97. 12.5, 20 kV/cm, 98. G loops, 99. G+, G-, G loop, 100. to prevent the Hfr to
produce a colony, 101. DNA binding, proteolysis, 102. Cytosine, 5-hydroxymethylcytosine,
103. cytosine, 104.105, 105. 105, 109, 106. Thermus aquaticus, Pyrococcus furiosus,
Thermococcus lioralis, 107. transgenic organisms, 108. 260nm, 280 nm, 109. 0.05, 110. 0.04,
111. 1.7-1.9, 1.8-2.1, 112.
vectorofsizekbproductPCRofsizekbvectorofngproductPCRofng /)( ×= , 113. Stability or End
stability, 114. 100 kb, curing, 115. no, 116. non-integrated, 117. integrated, 118. Hfr, 119.
place, place, the same, 120. 1, 121. Insertion sequences, transposons, transposable phages (µ
phage), 122. particular region, finer structure, 123. an artificial, cloning, foreign genes, 124.
start codon (AUG), stop codon (UAA, UAG, UGA), 125. repressible, inducible, 126.
spontaneous, 127. silent, missense, nonsense, 128. insertion, deletion, 129. thymine, light
enzymes, 130. 100-10000, 131. beneficial, neutral, 132. unusual compounds, 133. conjugative,
134. gene regulation, 135. Operon model, 136. transcription, post-transcription RNA
processing, the stability of mRNA, translation ausation and the post-translational
modification of proteins, 137. constitutive, inducible, repressible, 138. constitutive, 139.
inducible, 140. repressible, 141. positive, negative, On, Off, 142. negative, positive, 143. apo-,
co-repressor, 144. protein, 144. single promoter, regulatory protein, operon, 145. allolactose,
146. negative, positive, repressor, CAP-cAMP, 148. attenuation, 149. positive regulator,
negative regulator, 150. inducible, DNA loop, 151. Col-plasmids, 152. permanent, 153. 10-6 to
10-7, 154. no, insertion sequences, 155. transposons, 156. triplet, 157. translation, 158. amino
acids, 159. nucleotides, 160. transformation, 161. William Hayes, 162. semi-conservative, 163.
clonal, 164. clonal, 165. transformation, 166. eukaryotes, bacteria, 167. binary, multiple fission,
fragmentation, budding, meiosis, 168. multiple origins of replication, DNA polymerase, 169.
prokaryotes, eukaryotes, archaea, 170. eukaryotic microbes, 171. selfish DNA elements, RNA
polymerase, 172. cytoplasm, 173. killer, 174. tRNase, DNA-damaging factor, linear plasmids,
175. one strand, 176. filamentous phage, 177. polycistronic, 178. 30–60, 179. adsorption and
infection with phage, prophage induction, 180. chemically block, 181. lysogen allelopathy,
disseminate 182. room temperature, primer dimer, sensitivity, specificity,183. hot, temperature,
184. invertrons (TIRs), conservation, 185. linear, not absolutely, replication, 186. small, large,
chromosomal, 187. Genetic, linear, chromosome, 188. pulsed field gel electrophoresis, 189.
two, circular and linear, 190. hair pin.
73
Chapter. 4. Short Questions in Microbial Genetics
Q.1. What are the major differences between Eukaryotic and Prokaryotic genomes?
Ans.
Character Prokaryote Eukaryote
Introns Absent Present
% of noncoding DNA Low (~11%0 More (>50%)
Packaging proteins HU Histones
Operons Common Less common
Tandemly repeated DNA or satellite DNA Rare Common
RNA transposons or retrotransposons Absent Present
DNA transposons Common Rare
Q. 2. What do you mean by genetics?
Ans. Science of heritance (heredity) and materials associated with inheritance i.e physicochemical
properties of the hereditary material.
Genetics is the study of genes, their structures, organization and expression.
Q.3. What is chromosome? Give a brief account of genetic material of common microbes.
Ans. A chromosome is a structure in which indispensable hereditary material (usually DNA
molecules containing the genes are packaged). Genetic material in different microbes in brief is
tabled as under:
Organism Mode of life Chromosome
s
Type of
genetic
material
Remarks Codon uses
Most bacteria Saprophytes/
parasites
1, haploid ds DNA Covalently
closed
circular
Rickettsiae Intracellular
parasites
1, haploid ds DNA Covalently
closed
circular
Chlamydia Intracellular
parasites
1, haploid ds DNA Covalently
closed
circular
Start codons –AUG
(methionine),
rarely GUG
(valine)
Stop-UAA (amber),
UAG (ochre),
UGA (opal,
sometimes for
selenocystine)
Mycoplasma Parasites/
saprophytes
1, haploid ds DNA Covalently
closed
circular
As above except that
UGA is not a stop
codon but codes for
tryptophan as in
mitochondria of
most eukaryotes
but plants
Leptospira Parasites/
saprophytes
2, haploid ds DNA Covalently
closed
circular
As in most bacteria
Treponema Parasites/
saprophytes
1, haploid ds DNA Covalently
closed
circular
As in most bacteria
74
Borrelia Parasites/
saprophytes
>1, haploid ds DNA 100 kb-linear
numerous
circular and
linear small
molecules
As in most bacteria
Agrobacterium
tumefaciens
Symbiotic 2 ds DNA 3000 Kb
circular
chromosome,
a 2100 Kb
linear
chromosome
As in most bacteria
All plant and a
few bacterial
viruses
Parasite 1, haploid dsRNA or
ss RNA
open
Most animal and
bacterial
viruses
Parasite 1, haploid dsDNA or
ss DNA
Open or
circular
covalently
closed
As in most bacteria
During replication
most viruses have
double stranded
circular molecule
Some ciliated
protists
Free living 1, haploid ds DNA Covalently
closed
circular
UAA and UAG both
(stop codons)
codes for
glutamine
UGA- not a stop codon, codes for tryptophan, AUA for methionine
rather then isoleucine, AGA & AGG are stop codons then coding for
arginine
Same as above but AGA & AGG codes for serine
Same as in most bacteria
Mitochondria-
vertebrates
invertebrates
Plants
Yeast mould
&Trypanosom
es
1 chromosome of
dsDNA
Covalently
closed circular
UGA- not a stop codon, codes for tryptophan, AUA for methionine in
yeast but not in trypanosomes and moulds, CUN codes for threonine
in yeasts but not in mould and tryps where it codes for Leucine as in
other organisms
Plasmids Found in
bacteria, yeasts
and moulds
1 dsDNA Covalently
closed
circular
Stp codons =nonsense (stop) codons, initiation codon= start codon
All amino acids are coded by more than one codon (called degeneracy of the genetic code)
except methionine (AUG) and tryptophan (UGG), however no codon specifies more than 1
amino acid.
Q. 4. Four types of non-covalent bonds found in most biomolecules are hydrogen bonds, van der
Waals forces, electrostatic interactions, hydrophobic interactions, what are their characteristics
and strengths?
Ans. Hydrogen bonds: Weak electrostatic attraction between an electronegative atom (as Oxygen,
Nitrogen) form a comparatively longer bond than covalent bond and is much weaker having
bond energy ~1-10 kcal per mol compared to about 90 kcal per mol of a covalent bond, they
play their role in formation of double RNA/ DNA helix and secondary structure of protein.
Van der Waals forces: Can be attractive and repulsive, having energy of about 0.1-0.2 kcal per
mol, it is due to fluctuations between electron charge densities of adjacent atoms. They play
role in secondary folding of proteins. Repulsive forces occur when two atoms come too close
and play in deciding limit of degree of packaging of a protein.
75
Electrostatic interactions: Ionic bonds between charged groups, the energy is low in aqueous
environment (~ 3 kcal per mol) because of the shielding effect of water. They help in
attachment of proteins to DNA; they bring together R groups of amino acids.
Hydrophobic interactions: Hydrogen bonded structure of water forces hydrophobic groups to
internal part of the protein to provide it a tertiary structure. These are not true bonds.
Q. 5. How can you locate the protein binding site on a DNA molecule?
Ans. There are a few methods as;
1. Slowing of mobility on gel-electrophoretic: Naked DNA runs faster than the one with
nuclear protein, it can be done after restriction digestion to get more accurate position of
protein bound site.
2. Foot printing or Protection assay: Protein bound portion is resistant to digestion by RE.
3. Modification interference method: If the bases interacting with nuclear proteins are
modified (methylated) or changed may not bind to protein and thus exact site of binding can
be found out.
Q. 6. Name some bacteria which have linear chromosome?
Ans. Most of the bacteria have circular chromosomes, however, Streptomyces coelicolor, S.
lividans, Rhodococcus fascians and Borrelia spp. Have linear chromosome.
Q. 7. Name the bacteria which have both linear and circular chromosomes.
Ans. Agrobacterium tumefaciens biovar 1.
Q.8. Name the bacteria which have two circular chromosomes.
Ans. Rhodobacter ausation (3 million bp and 0.9 million bp), Brucella melitensis, Leptospira
interog.
Q.9. Name the bacteria which have three chromosomes.
Ans. Rhizobium meliloti and Burkholderia cepacia.
Q.10. Name the bacterium which is not haploid, how many sets of chromosomes are there?
Ans. Dienococcus radiodurans, a highly radiation resistant bacterium, is never haploid instead it is
tetraploid, containing 4 sets of same DNA molecule.
Q.11. How to classify the endonucleases?
Ans. Endonucleases are classified in four broad groups:
76
a. Type I: Occurs only in enteric bacteria, are of three types A, B and C. A & B are coded by
chromosomal genes (bsd loci) while family C is plasmid borne. They require ATP,
magnesium ion and S-adenosyl-methionine (SAM) for their activity and cuts only non-
methylated DNA. They cut randomly, usually at a distance of 1000 nucleotides from
original recognition sequence and they are non-specific as far as cutting site is concerned.
b. Type II: Always cuts at a definite site, always have non-associated methylase and restriction
proteins. Recognition sequences are generally 4-7 bases in length. If the recognition
sequences contain odd number of bases the centre base is non-specific. They are mostly
symmetric i.e. recognition site is even number of bases and palindromes. More than 600
enzymes are known.
C. Type III: Modification proteins alone or complexed but endonuclease must be complexed to
be functional thus, modification proteins confer the recognition ability to these. These are
specific in their action. Only 4 enzymes are known in this category e.g. EcoP1 (of P1 phage,
recognize 5 bp site and cut 24-26 bp away from 3’ end of recognition site)
D. Type IV: they cuts methylated DNA e.g. MspI.
Q.12. Why genome projects are important?
Ans. 1. To attain the sequences of target genome. 2. For the development of molecular life sciences
(molecular genetics, cell biology, cell chemistry and physiology). 3. To understand various
gene functions and their regulation, 4. Isolation and utilization of genes responsible for
understanding and cure of hereditary diseases, 5. Possibly to understand role of noncoding
sequences, 6. To solve the puzzle of unknown i.e., the challenge of the unknown.
Q.13. What are the different functions of nucleotides?
Ans. They are: a. subunits of nucleic acid.
b. Carriers of chemical energy in cells,
c. They act as complement of enzyme cofactor as coenzyme A, NAD, FAD.
d. Important intermediates in cellular communication as second messenger as cAMP, cGMP,
ppG, ppC (produced in bacteria in response to slowing down of protein synthesis in
bacteria)
Q.14. What do you mean by mini and macro chromosomes?
Ans. Mini-chromosomes are smaller in size but rich in genes while macro chromosomes contains
large amount of DNA but very few genes e.g. out of 39 chromosomes in chicks, 6 macro
chromosomes contains 66% of DNA but only 25% of genes.
77
Q.15. What do you mean by B chromosome?
Ans. An additional chromosome possessed by a few individuals in a population, it may affect the
phenotype of an individual and may reduce viability.
Q.16. What do you mean by genetic and physical mapping of genome?
Ans. Genetic mapping is done by using genetic techniques to construct maps for showing position
of different genes in genome, e.g., cross breeding, pedigree studies.
Physical mapping is achieved by molecular biology techniques as sequencing.
Q. 17. What do you mean by DNA markers?
Ans. Mapped features of DNA which are not genes are called DNA markers, e.g., restriction sites
for Res (RFLPs), simple sequence length polymorphism (SSLPs), single nucleotide
polymorphisms (SNPs), and dynamic allele –specific hybridization (DASH) etc.
Q.18. Give name of important markers used in construction of genetic maps.
Ans. 1. Known genes and their alleles e.g., requirement of specific conditions for growth, various
resistance markers, fermentation abilities and some non-coding DNAs or patterns e.g., RFLPs,
SNPs, SSLPs, DASH, are often used as marker genes.
Q. 19. What are different methods of genetic mapping?
Ans. Linkage analysis by cross breeding and selected breeding, microarray analysis and pedigree
analysis are some of the common methods used in genetic mapping.
Q.20. What are different methods of physical mapping of genome?
Ans. Restriction mapping, FISH (fluorescent In Situ hybridization), STS (sequence tagged site
mapping) are some of the methods of physical mapping.
Q.21. Which of the two methods of genome mapping is more accurate and gives more resolution?
Ans. It is Physical mapping.
Q.22. MspI and HpaII are isoschizomers in the sense that they recognize same sequence; however
they differ in their activity, why?
Ans. MspI is a class four enzyme and cuts only methylated DNA while HpaII cuts non-methylated
DNA, thus they differ in activity.
78
Q. 23. How are different banding patterns produced in chromosomes and what they indicate?
Ans.
Technique Procedure Banding pattern
G-banding Chromosome is Proteolysed before
Giemsa staining
Dark bands –AT rich regions
Light bands- GC rich regions
R- Banding Chromosome is denaturated by
heating before Giemsa staining
Dark bands – GC rich regions
Light bands- AT rich regions
Q- banding Chromosome is stained with
quinacrine
Dark bands –AT rich regions
Light bands- GC rich regions
C-banding Chromosome is denaturated by
Ba(OH)2 before Giemsa staining
Dark bands are constitutive heterochromatin (DNA
having no genes as most of the Y chromosome
of human being
Q.24. What are the benefits and limitations of Restriction mapping? How one can overcome these
problems?
Ans. Restriction mapping is easy, rapid and provide detailed information but its accuracy is limited
by the size of restriction fragments; it is useful for smaller DNA molecules rather than large
molecules. To overcome this problem one can use enzymes with 7 or 8 nucleotide recognition
sequences (SapI, 5’ GCTCTTC 3’ and SgfI, 5’ GCGATCGC 3’) and by use RE whose motifs
are rare.
Q.25. Why 5’CG3’ motifs are rare in human genome? How it is exploited?
Ans. 5’CG 3’ is rare in human genome because an enzyme present in human cells add methyl
group to carbon 5 of cytosine giving rise to 5-methylcytosine which in turn is replaced by
thymine on deamination, over long period of evolution most of the C from CG motifs have been
replaced by TG. This character has been used to construct RFLP maps of human genome using
enzymes like SmaI (5’ CCCGGG 3’), cuts every 78 kb, BssHII (5’ GCGCGC 3’), cuts every
390 kb and NotI (5’GCGCGCGC 3’), cuts human DNA very rarely, once every 10 Mb)
Q.26. How one can monitor the products of DNA after cutting with a rare cutter?
Ans. Its is usually done by the pulsed field gel electrophoresis (PFGE) or its variants viz.,
Orthogonal field alteration gel electrophoresis (OFAGE)- electric field alternate between two
pairs of electrodes each positioned at 45o to the length of the gel; contour clamped homogenous
electric field (CHEF) and field inversion gel electrophoresis (FIGE). It can also be done with
optical mapping.
79
Q. 27. What do you mean by optical mapping?
Ans. Optical mapping is used to monitor the restriction fragments of DNA after digestion with rare
cutting RE. In this technique fragments are stretched into threads for optical mapping either
through gel stretching or by molecular combing. For gel stretching DNA suspended in agar is is
placed on microscope slide as the gel cools it solidifies and DNA molecule get stretched. The
slide used for the purpose is coated with RE but RE remains inactive in absence of Mg++. Once
the agarose is solidified slide is washed with solution containing solidified MgCl2. There after a
fluorescent dye is added (DAPI, 4,6-diamino-2-phenylindole di-hydro-chloride) to stain the
DNA fibres and make them visible under fluorescence microscope. At the cut sites gaps can be
seen as cut releases the stretch and molecule shrinks.
Q.28. What is fluorescent In Situ hybridization (FISH)? What are its advantages and
disadvantages?
Ans. FISH is a optical mapping technique for DNA. Here the method is same as used for optical
mapping of restriction fragments but instead of using RE here we hybridize the DNA with
fluorescent probes for defined markers. Instead one can also use radioactive probes.
It can be used to map large DNA with high resolution but it is complex to carry out and a slow
process as only 2-3 markers can be used at a time.
Q.29. What is sequence tagged site (STS) mapping?
Ans. It is one technique which provides most detailed maps of large genomes. STS is short (100-
500 bp) DNA sequence. For the application overlapping fragments from entire DNA are needed
and then fragments of DNA are analysed to determine that which fragment contains which STS
and b detecting number of STS on a fragment you can deduce the closeness of two sequences.
STS can be selected either from expressed sequence tags (ESTs) i.e., obtained by analysing
complementary DNA (cDNA) clones made from mRNA, or SSLPs through examining micro
satellites or random genomic sequences.
Q.30. What are different methods of DNA sequencing?
Ans. There are two methods for sequencing a. Chain termination sequencing of Sanger and b.
chemical degradation method of Maxim and Gilbert.
Q.31. What are the basic requirements for chain termination method of DNA sequencing?
Ans. A single stranded DNA templet, a labelled (radio labelled or fluorescent labelled) primer to
anneal, dNTPs, dideoxynucleotides (ddATP, ddCTP, ddGTP, ddTTP), polyacrylamide gel
80
electrophoresis (it can differentiate between strands differing by one nucleotide base, specila
DNA polymerase (lacking 3’ to 5’ and 5’ to 3’ exonuclease activity).
Q.32. How can you get single stranded DNA to sequence by chain termination method?
Ans. To have single stranded DNA for sequencing one can clone it in a. a plasmid (cloning vector)
and then get a single strand by denaturation, b. M13 vector (a specific phage designed to
produce single stranded phage DNA which in E. coli turns to dsDNA but excreted as ssDNA),
c. phagemid (contains origin of replication from a plasmid and M13 or other phage. It can
produce ssDNA of 10 kb and constructs are stable rather than unstable as on cloning in M13.
Q.33. What do you mean by thermal cycle sequencing?
Ans. In thermal cycle sequencing (the common method used now) dsDNA can be used as templet
because DNA polymerase can act at high temperature and are thermostable. Another advantage
is that only small amount of templet DNA is required and need not to be cloned before
sequencing. Here also one primer is used in the same fashion as chain termination sequencing
in a reaction mix containing all 4 dNTPs with small amount of a ddNTP. Here product
accumulates in linear fashion, not exponentially (as in real PCR).
Q. 34. What is pyrosequencing?
Ans. It is the newer method which does not require electrophoretic separation and DNA sequence
id deduced by a little different method not using the ddNTPs but the dNTPs are used. During
sequencing addition of each of the new nucleotide is detected by chemiluminescent signal
originated from pyrophosphate (released with addition of each nucleotide) by enzyme
sulfurylase. Addition of any of the four dNTP release same signal therefore to differentiate all
four dNTPs are added one by one and then exposed to nucleotidase to degrade the unused
dNTP. The technique has potential of automation as the primer synthesis.
Q.35. What is chemical degradation method of DNA sequencing?
Ans. The technique developed by Maxam and Gilbert (1977), involve cutting of DNA by using
specific chemicals cutting at specific nucleotide. Here the target DNA is first converted to
ssDNA, and is radio or luminescent labelled. Reactions are set up separately for 4 dNTPs. As
for G reaction, DNA is treated with limited amount (just sufficient to modify one G per strand)
of dimethyl sulphate which attaches to G. Then second chemical, piperidine is added to cleave
at the site of modified ring of G. The set of strands will be produced for G as the ddGTP
terminated chains in chain termination method. The other reactions used are A+G, C and C+T.
81
All different reactions are electrophoresed in four lanes and read as in chain termination
method.
Q.36. What are isoschizomers?
Ans. These are the restriction endonucleases of class II, they have same recognition site but of
different origin i.e. duplicate enzymes.
Q.37. When we want to introduce some restriction sites at ends of PCR products for cloning, we
often do not chose EcoRII, why?
Ans. For activity EcoRII must bind to a non-cleaved site about 1 kb away from the cleaved site,
thus the ends having EcoRII sites are refractory for digestion.
Q. 38. How bacteria having linear chromosomes protect the free ends of their DNA?
Ans. In eukaryotes telomeres protect the ends of DNA but in bacteria may use different methods,
e.g., Borrelia DNA loops backs to become the complementary strand (a hairpin) while
Streptomyces protects its DNA ends using covalently attached proteins (having probable role of
primer as well).
Q. 39. While cloning it is more advantageous to digest (cut) the cloning vector and target DNA
producing single strand overhangs with two enzymes, why?
Ans. When we cut the DNA for cloning with two enzymes producing two different sticky ends they
do not self-ligate and the cloning vector also do not produces blank vectors on relegation.
Moreover, we can control the direction of ligation in the vector to achieve the best results.
Q. 40. How many kinds of fragments can be produced by endonuclease activity?
Ans. Three types, viz.,
a. blunt end (cut in the middle of recognition site),
b. sticky ends; single strand tailed cut with 5’ end hangings (cut left to the centre of recognition
site), 3’ end overhangings (cut right to the centre of recognition site).
c. Unique end: are produced by asymmetric cutting enzymes e.g. Hga I produces fragments
with single base overhang
Q. 41. Where from the terms southern and northern blotting came?
82
Ans. DNA hybridization technique was first invented by Earl Southern using a DNA gel and
therefore it was named Southern blotting. If an RNA gel is used the technique is named
Northern blotting by the simple convention i.e. opposite to south is north.
Q. 42. What are the limitations of the custom synthesis of DNA?
Ans. In the process bases are added one by one to the anchored first base on solid support, the
completeness of the addition is the first limitation as chemical reaction never 100% complete,
some DNA molecules have incorrect base sequence and proportion of incorrect bases increases
with the length of the DNA. Another limitation is the length of DNA molecule, usually the limit
is about 100 bases.
Q.43. How can you identify the protein binding sites on a DNA molecule?
Ans. It is done by Foot printing technique (the protein covered regions of DNA are protected fron
Dnase I action while protein free DNA is easily cut) another technique is Gel mobility shift
analysis (if a DNA molecule is bound with protein its shape will differ and thus the mobility in
an agarose gel during electrophoresis).
Q. 44. What are there different components of bacterial nucleotides?
Ans. It is made up of 60% DNA, 30% RNA and 10% protein.
Q. 45. What do you mean by RNA world theory?
Ans. Probably RNA molecules which can act as catalyst in their own formation were the first genes
as well as the first organic catalysts (enzymes). The origin of life is hypothesized by
development of self replicating RNA, and then the variants of RNA i.e. DNA and then the
proteins to provide stability to the replicating molecule resulted in origin of life, this theory is
called “RNA world theory”. The first and final class of molecules are thus the nucleotides.
Q. 46. How and why did RNA world develop into DNA world?
Ans. Probably the development of protein enzymes led the way to DNA world as they replaced
ribozymes probably due to more efficient catalytic action and inherent flexibility of folded
polypeptides then the rigidly base paired RNA. It demanded radical change in protogenome
leaving ribozymes as the key molecule to synthesize the coding molecules or becoming coding
molecules themselves (RNA).
Q. 47. What is cosmos theory?
83
Ans. Theory which explains that origin of life on earth has come from outer space through
asteroids and meteorites. But the question of origin of life remains unsolved. How it originated
in outer space?
Q. 48. What do you mean by ploidy?
Ans. After division of cell into daughter cells, if daughter cells contain one copy of each
chromosome then the daughters are called haploid if the two copies of each chromosome are
present then cells are called diploid, i.e., number of copies of the chromosomes present in cells
is cells’ ploidy.
Q. 49. Does bacteria can form diploid cells? If yes, give example and explain the mechanism.
Ans. Usually not but under certain circumstances yes. The protoplast fusion technique can be used
to produce non-complementing stable diploid cells. Protoplasts are produced by the use of some
enzymatic or antibiotic treatments. Protoplasts are fused in presence of propylene glycol. E.g.,
Bacillus subtilis, Successful diploids have been produced in Actinoplanes, Brevibacterium,
Bacillus, Mycobacterium, Providencia, Staphylococcus and Streptomyces.
Q. 50. What do you mean by protoplasts and what are their uses?
Ans. Bacteria devoid of cell wall are called protoplasts. They are used to produce diploid bacterial
cells and for transformation work as these can easily uptakes naked or plasmid DNA.
Q. 51. What is the number of chromosomes in Chromosomes in different organisms: Bacteria, fruit
fly, red clover, garden pea, yeast, honey bee, corn, frog, hydra, fox, cat, mouse, rat, rabbit,
human, chicken, pig, sheep, cow, horse, dog.
Ans. Chromosomes in Bacteria-1, fruit fly-8, red clover-14, garden pea-14, yeast-16, honey bee-16,
corn-20, frog-26, hydra-30, fox-34, cat-38, mouse-40, rat-42, rabbit-44, human-46, chicken-78,
pig-38, sheep-54, cow-60, horse-64, dog-78
Q. 52. What is a gene? Define different two general types of genes. Give some characteristics of
bacterial genes.
Ans. Genes are the DNA segments coding for a polypeptide chain or RNAs, these are also called
the Structural genes, encode the sequence of some final gene product eg. A polypeptide, RNA,
or an enzyme. Other group of genes is Regulatory sequences- that denote signals for beginning
and end of structural genes, turning on and off the transcription of structural genes or act as
point of initiation for replication or recombination.
84
Most bacteria has only one copy of a gene except of some rRNA genes (repeated several
times) and genes are collinear with amino acid sequence or RNA sequence for which it codes
i.e. they lack introns (intervening sequences as is common in most eukaryotic genes, coding
sequences are called exons, introns are usually bigger than exons and can contribute up to 85 %
of total sequence length)
Q. 53. What is supercoiling, how it is determined?
Ans. DNA in all organisms except in very low life forms is in super coiled form irrespective of
closed or open strand of DNA. DNA is in coiled form when a coil undergoes further coiling, it
is called super coil. In purified form, DNA is usually supercoiled while in cellular form it is
underwound (B-form, I.e. 10.5 bases per helix resulting in to relaxing rather than in
supercoiling. When purified closed circular dNA get strained and result in to super coiling.
It is calculated by Twist and Writhe given by David Stump and Peter Watson. The
mathematical formula Lk=Tw+Wr can be used to describe this
process. Lk, the linking number, represents the number of times one strand winds around the
other, Tw is the twist or the amount of rotation about the centre line and Wr, the writhe,
describes how hard it is to straighten out the curve. When the curve is straightened out the
writhe, Wr, is zero and the twist, Tw, is high. You can feel the elastic band trying to untwist.
When the elastic band is relaxed it supercoils. The twist Tw is now very small and the writhe
Wr is high.
Q. 54. What is linkage number, how it is determined?
Ans. Number of helical turns in a closed circular DNA in absence of any supercoiling, the number
for a closed circular DNA molecule can be calculated by dividing number of bases by 10.5. It
can be changed by topoisomerases, by nicking only 1 of the two strands and then joining it after
rotating broken end by 360o around the intact strand. Change takes place by +-1 with every
360o rotation.
Topoisomerase type 1-acts by transiently breaking one of the two strands and then it
rotating around unbroken strand (in presence of ATP, as source of energy). Lk changes in
increments of 1.
Topoisomerase type 2- breaks both the strands and change Lk in increments of 2.
(Topoisomerases can be of some use in profiling different plasmids as different topoisomers get
separated on agarose even those having difference of only one in Lk value. Different plasmids
can be treated for a fix period with fixed amount of Topoisomerase type I or II and then
electrophoresed.
85
Q. 55. Give brief account of nomenclature of different nucleotides.
Ans. Nomenclature of nucleotides
Major Bases Nucleoside ( base without
phosphate but pentose
Nucleotide (Nucleoside + phosphate)
ribonucleotides or
deoxyribonucleotides
Nucleic acid
Purines
Adenine Adenosine/ deoxyadenosine Adenylate/ deoxyadenylate RNA/ DNA
Guanine Guanosine/ deoxyguanosine Guanylate/ deoxyguanylate RNA/ DNA
Pyrimidines
Cytosine Cytidine/ deoxycytidine Cytidylate/ deoxycytidylate RNA/ DNA
Thymine Thymidine/deoxythymidine Thymidylate/ deoxythymidylate RNA/ DNA
Uracil Uridine Uridylate RNA
Minor bases
Found in DNA: Methylated forms as N6-methyladenine (in bacterial DNA), 5-methylcytosine (in animal and
plant DNA), N2-methylguanine, Hydroxy methylated forms as 5-hydroxy-methyl-cytosine (in bacteriophage
infected bacterial DNA)
Found in RNA (mostly in tRNA): 7 methyl guanine, 4-thiouracil, hypoxanthine, pseudouracil (attaches with C-
5 to 1’-C of pentoses rather than using its N-1 position), inosine
Q. 56. Why purines and pyrimidines are called bases? What are their other characteristics?
Ans. Purines and pyrimidines are called bases because in free form they are weakly basic
compounds, and can exist in 2 or more tautomeric forms depending on pH of solution, the
normal structure shown is usually that predominate at pH 7.
Bases absorb UV light due to resonance of their molecules all bases strongly absorb UV
light of 260 nm. When hydrophobic forces bring them to be stacked together (due to wander
Waals and dipole-dipole interactions in bases) in form of nucleotide polymers and absorb less
UV light resulting in to “Hypochromic effect”.
DNA is a double helix, has hydrophilic backbone made or alternating deoxyribose and
phosphate (-) on outside of double helix and bases (hydrophobic) are stacked inside almost at
perpendicular plane to the axis of the double helix. The two chains or strands are in anti-parallel
position to each other i.e. their 5’, 3’ phosphodiester bonds run in opposite direction and are
complementary to each other as A-T, G-C pairing takes place. All bases have hydrophobic
nature; therefore at acidic or alkaline pH they are more soluble in water.
Phosphate-phosphoester bond (with 5’ C of pentose)-Pentose-N-glycosyl bond (1’ carbon of
pentose with N1 of pyrimidines or N9 of purines)
Two nucleotides are attached with 3’-5’ phosphodiester bond between pentoses attached
with bases, 5’ end lacks base at 5’ position while 3’ end at 3’ position.
86
Oligonucleotide: nucleotide polymer containing 50 or <50 nucleotides beyond this limit
they are named polynucleotides.
Q. 57. Name different forms of DNA and their characteristics.
Ans. B form or Watson and Crick structure-most stable, right handed, rise per base 0.34 nm,
number of bases per helical are 10.5.Common form found in life, i.e. in cells.
A form: In solutions relatively devoid of water, right handed helix, rise per base is 0.23 nm,
number of bases per helical turn are 11, have greater diameter than B form.Its existence in
cells is not known.
Z-form: Left handed helical rotation, 12 bases per turn, rise per base is 0.38, backbone takes a
zigzag appearance, and short stretches of Z DNA are common in cells.
Special turns: 6 adenines in row produce 18o turn, alternating C-G and 5-methyl-C-G induces
left handed helical turns as in Z forms.
H-DNA: Unusual DNA structure found in polypyrimidine/ polypurine tracts incorporating a
mirror repeat (TGAGCGACT-TCAGCGGT) in structure, having interwinding of 3
strands to form triple helix, one strand contain either polypurines or polypyrimidines, two
strands contain only pyrimidines and 3rd one has purines.
Palindromes in DNA means inverted repeats of bases as
TTAGAATT AATCTTAA
AATCTTAA TTAGAATT
These are often the cause of cruciform structure of DNA, while self complementary
sequences in the same strand results in to hairpin structures.
Q. 58. Classical recombination is rare in RNA viruses. Why? What kind of recombination is
possible in RNA viruses?
Ans. Classical recombination is very rare in RNA viruses because there are probably no host
enzymes for RNA recombination.
Picorna viruses show a form of very low efficiency recombination. The mechanism is not
identical to the standard DNA mechanism, and is probably a “copy choice” kind of mechanism
in which the polymerase switches templates while copying the RNA. Recombination is also
common in the corona viruses – again the mechanism is different from the situation with DNA
and probably is a consequence of the unusual way in which RNA is synthesized in this virus. So
far, there is no evidence for recombination in the negative stranded RNA viruses giving rise to
viable viruses (In these viruses, the genomic RNA is packaged in nucleo-capsids and is not
readily available for base pairing).
87
Q. 59. What do you understand by complementation?
Ans. Complementation is interaction at a functional level but not at the nucleic acid level. For
example, if we take two mutants with a ts (temperature-sensitive) lesion in different genes,
neither can grow at a high (non-permissive) temperature. If we infect the same cell with both
mutants, each mutant can provide the missing function of the other and therefore they can
replicate (nevertheless, the progeny virions will still contain ts mutant genomes and be
temperature-sensitive). We can use complementation to group ts mutants, since ts mutants in
the same gene will usually not be able to complement each other. This is a basic tool in genetics
to determine if mutations are in the same or a different gene and to determine the minimum
number genes affecting a function.
Q. 60. What do you mean by phenotypic mixing, explain with examples?
Ans. If two different viruses infect a cell, progeny viruses may contain coat components derived
from both parents and so they will have coat properties of both parents. This is called
phenotypic mixing. It involves no alteration in genetic material, the progeny of such virions will
be determined by which parental genome is packaged and not by the nature of the envelope.
Phenotypic mixing may occur between related viruses, e.g. different members of the Picorna
virus family (Picornaviridae), or between genetically unrelated viruses, e.g. Rhabdo- and
Paramyxo- viruses. In the latter case the two viruses involved are usually enveloped since it
seems there are fewer restraints on packaging nucleo-capsids in other viruses’ envelopes than
on packaging nucleic acids in other viruses is icosahedral capsids.
Q. 61. In an experiment hundreds of new proline auxotrophs were isolated in E. coli and they all
fell into three complementary classes; what medium will you use to determine the order of the
biosynthetic steps affected in the pro mutants, using cross-feeding method and a few or one
mutant from each class?
Ans. Minimal medium without proline.
Q. 62. If in above cross feeding experiment growth pattern in minimal medium is as under what
might be the order of the gene products in the proline biosynthetic pathway?
Pro-117 & Pro-118= grow
Pro-117 & Pro-116= grow
Pro-118 & Pro-116= grow
Pro-117 & Pro-116= grow
88
Ans. As a mutant defective later in the pathway can cross-feed mutants defective at earlier steps in
the pathway. Thus the order is: 117 – 118 – 116.
Q. 63. How do cells survive to effect of restriction enzymes after DNA replication?
Ans. Immediately after DNA replication the restriction sites got methylated on one strand
(“hemimethylated”) and methylase rapidly modify the hemimethylated DNA, so it is not cut by
the restriction enzyme
Q. 64. The proBA genes are required for biosynthesis of proline. To isolate a Tn10 insertion near
the proBA genes, a strain with a nonsense mutation in the proB gene was transduced to TetR
with a P22 lysate grown on a random pool of Tn10 insertions in E. coli chromosome on brain
heart infusion agar (BHIA) with tetracycline. The TetR colonies from BHIA were then replica
plated onto minimal medium with or without proline. A few colonies could not grow on any of
the two medium while a few grew only on prolineless medium. What is your explanation for
these phenotypes?
Ans. The colonies those fail to grow on minimal medium with proline but grew on nutritionally
rich medium might appear due to Tn10 insertion in some other gene leading to creation of an
auxotrophic marker which could be a gene required for the biosynthesis of some other essential
metabolite. The colonies which grew on praline less medium could be either due to transduction
of proB+ with a linked Tn10 insertion, or due to the simultaneous acquisition of a random Tn10
insertion and an independent proB+ reversion.
Q. 65. phoP106 is a missense mutation in the Salmonella alkaline phosphatase gene that has a
dominant-negative phenotype. The strains having tandem duplication with one copy of the
phoP106 gene and one copy of the phoP+ gene are phenotypically PhoP-. What may be the
mechanism for the dominant-negative phenotype of the phoP106 mutant?
Ans. As phoP must function as a dimer, it is likely that the dominant-negative phenotype is due to
the formation of nonfunctional dimers.
Q. 66. Briefly describe the salient features of DNA replication in Salmonella, describing the
direction synthesis, properties of the two i.e., leading and lagging, strands, synthesis and joining
of Okazaki fragments and enzymes involved.
Ans. DNA replication by DNA pol III begins at the chromosomal origin (ori) and proceeds
bidirectionally (theta-replication) around the chromosome until the two replication forks meet
the ends. DNA replication is initiated by DNA primase, the DnaA protein, and specific DNA
89
sequences. Replication terminates on collision of the two DNA replication forks. Then
topoisomerase is required to separate the two strands of DNA. Although termination of
replication is facilitated by specific DNA sequences and a protein encoded by the tus gene, they
are not essential.
Q. 67. What are DNA polymerases? How they work?
Ans. Enzymes that replicate DNA using a DNA template are called DNA polymerases. However,
there are also enzymes that synthesize DNA using an RNA template (reverse transcriptases) and
even enzymes that make DNA without using a template (terminal transferases). Most organisms
have more than one type of DNA polymerase (for example, E. coli has five DNA polymerases),
but all work by the same basic rules.
1. Polymerization occurs only 5’ to 3’; 2. Polymerization requires a template to copy: the
complementary strand; 3. Polymerization requires 4 dNTPs: dATP, dGTP, dCTP, dTTP
(TTP is sometimes not designated with a ’d’ since there is no ribose containing equivalent);
4. Polymerization requires a pre-existing primer from which to extend. The primer is RNA in
most organisms, but it can be DNA in some organisms; very rarely the primer is a protein in
the case of certain viruses only.
Enzyme Templet Primer Other activities Other features
E. coli DNA pol I DNA DNA/RNA 3’-5’ exonuclease
and 5’-3’
exonuclease
Monomeric, add 16-20
dNTPs/second, can add about
3-200 dNTPS before
dissociation
E. coli DNA pol I
(Klenow fragment)
DNA DNA/RNA 3’-5’ exonuclease C-terminal fragment
E. coli DNA pol III DNA DNA/RNA 3’-5’ exonuclease
(on a separate
subunit)
multimeric structure
add 250-1000 dNTPs/second,
can add about >500 kb
dNTPS before dissociation
Taq pol DNA DNA/RNA extendase (adds
3’-A overhangs)
thermostable, used in PCR
Reverse transcriptase
DNA/RNA DNA/RNA (ribonuclease H) used to make cDNA
Terminal transferase None required DNA will synthesize
DNA in
nontemplated
reaction
Q. 68. During DNA replication RNA primers are needed, what is the probable length of those RNA
primers? Can you name some organisms which do not require RNA primers?
Ans. The length of RNA primers is usually only a few bases, as case of Basillus subtilis only 3-4
base length is enough. Many animal viruses, bacteriophage Φ29 and some plasmids don’t need
RNA for priming of replication.
90
Q. 69. In order to understand the biochemical mechanism of DNA replication, Kornberg purified
DNA polymerase I (Pol I) of E. coli that catalyzed DNA synthesis. However the same could not
catalyze DNA synthesis in vitro, thus it could not be proved an essential enzyme for DNA
synthesis in vivo. To test this question, De Lucia and Cairns isolated a temperature sensitive
mutant in the polA gene which encoded Pol I active at 30oC but inactive at 42 o
C. If Pol I is
essential for DNA synthesis in vivo, what you expect of polA mutant incubated at 42oC?
Ans. If the polA gene product is the primary DNA polymerase, the mutant bacteria would grow
normally at 30oC but not at 42 oC. However, DeLucia and Cairns’ isolated a polA mutant that
was not-conditional i.e., able to grow at both the temperatures of incubation, proving that the
polA gene is not essential under certain growth conditions and, thus, DNA Polymerase I is not
the primary DNA polymerase in E. coli.
Q. 70. Why mutants having defects in proofreading function of DNA polymerase III, grew to form
small, rough and deformed colonies even on on rich medium?
Ans. Bacteria defective in proofreading function of DNA polymerase III accumulate mutations in
high frequency Therefore; large number of “lethal mutations” that might arise during cell
division slows the growth of the colonies and deforms them. The mutation frequency is higher
on rich medium allowing rapid multiplication and resulting into more errors.
Q. 71. What are the transcription steps in E. coli, and the enzyme required in the process?
Ans. The first step is initiation of transcription, in which a promoter sequence is recognized by
RNA polymerase and a specific sigma factor. The “housekeeping” sigma factor of E. coli σ70
with RNA polymerase form holoenzyme (E σ 70). Holoenzyme binds to a set of sites upstream
of the start site of transcription. If the first nucleotide of the mRNA isis numbered +1, the
consensus sequence for E σ 70, TATAAT is located around -10 and TTGACA near -35. The
next step is chain elongation, RNA polymerase moves along the DNA in a 5’ to 3’ direction,
adding nucleotides to the 3’OH group of the growing RNA chain. The σ factor usually
dissociates from the RNA polymerase core soon after initiation of chain elongation. The
energy for elongation is derived from hydrolysis of NTPs. The final step is termination which
might be either ρ dependant or ρ independent.
Important enzyme required in transcription is RNA polymerase. RNA polymerase is a four
subunits structure containing two A, one B, and one B’-subunits. On association of σ factor
with the RNA polymerase RNA polymerase holoenzyme is formed.
Q. 72. What are the salient features of translation (protein synthesis) in E. coli?
91
Ans. In most of the bacteria including E. coli ribosome binds to a specific sequence (Shine-
Delgarno) on the mRNA. Shine-Delgarno sequence is complementary to a sequence at 3’ end of
the 16S rRNA. The complementation works to correctly align the ribosome to the start codon.
A typical Shine-Delgarno site (ribosome-binding site) has 3-9 purine bases and is located
around -3 to -12 nucleotides of the start codon. The start codon is usually AUG, but
occasionally GUG, UUG, or AUU is used. In bacteria, Archae, mitochondria and chloroplasts
the start codon is recognized by distinct tRNA and fMet-tRNA. Termination of translation
occurs at stop codons (UAG, amber; UAA, ochre; or UGA, opal) because there are no cognate
tRNAs to recognize these codons. However specific proteins called release factors recognize
the stop codons, bind thre andcatalyses hydrolysis of newly synthesized peptide chain leading
to its dissociation from ribosome and the mRNA.
Q. 73. Why transcription and translation are not coupled in eukaryotes?
Ans. Transcription and translation occur in different cellular compartments in Eukaryotes,
separated by a membrane barrier, transcription occurs inside the nucleus and translation occurs
in the cytoplasm.
Q. 74. What is CONSTIN?
Ans. Conjugal self transmissible integrating elements (CONSTIN) are the conjugative transposons
i.e., transferred through conjugal mode. These are self-transmissible to the recipient and
integrate into the host chromosome.
Q. 75. What is PCR, what for it is used and what is required for performing it?
Ans. Kary Mullis was awarded the Nobel Prize for discovering the process of PCR The polymerase
chain reaction (PCR) is a technique to amplify trace amounts of DNA. The technique is used to
amplify desired genomic DNA sequences with or without subsequent aim of cloning, for direct
sequencing of any DNA without cloning the sequence, for medico-legal diagnostics (i.e., DNA
fingerprinting),for creating mutations in desired DNA molecules and for analyzing of
prehistoric DNAs.. In each round of the PCR process DNA replication leads to its doubling and
the desired DNA sequences can be amplified millions of times in a few hours. To perform PCR
one needs:
1. DNA (templet)
2. Thermostable DNA polymerase (e.g., Taq polymerase)
3. Primers (oligonucleotides complementary to specific sequences in DNA)
4. dATP, dCTP, dGTP, dTTP
5. Thermocycler (PCR machine)
92
6. Buffer to perform the reaction
Q. 76. What are the limitations of PCR?
Ans. Important limitations are 1. Requirement of two primers, 2. Amplifies only sequence of
nucleotides between the two primers i.e., surrounding sequences are not amplified, 3. PCR
process of DNA replication is 100 to 10,000 fold error prone than in vivo DNA replication.
Q. 77. What are the different kinds of PCR, briefly describe?
Ans. On the basis of utility PCR might be diagnostic or analytical while on the technical basis
PCR might be of many different types a few are defined as under:
A. Allele-specific PCR: this method is used to identify a single base pair difference in DNA and is
based on single nucleotide polymorphisms (SNPs). For this prior knowledge of DNA sequence
is essential and the base pair with polymorphism is incorporated at 3’ end of a primer. PCR is
carried out under stringent conditions to avoid mis-priming. Amplification of the desired
product signals presence of the specific SNP in the test DNA.
B. MAMA-PCR : Mismatch amplification mutation assay (MAMA) developed to identify infrequent
mutation, uses primers designed to have a single mismatch with mutated allele and a double
mismatch with wild type thus permitting amplification of mutated but not the wild type DNA
templet.
C. AD-PCR (Arbitrarily primed PCR) or randomly like (UP-un-primed) PCR or RAPD-PCR: A
random amplified polymorphic DNA finger printing method was developed to differentiate
between different strains, serotypes, breeds, varieties etc.. In this method, a small
oligonucleotide is used as primer and annealing temperature is kept low to allow lot of non-
specific binding. This method has inherent variability even in the hands of same person
attempting at different time.
D. Multiplex PCR : (MP-PCR) This is similar to normal PCR but here more than one set of
primers are used to amplify specific product like a genus specific set and another group specific
set of primers. This type of PCR technique is quite common but one has to compromise to
sensitivity and specificity as has to select a PCR cycle for both set of primers.
E. Nested PCR: It is quite similar to multiplex PCR but here different primer sets are added at
different time. In first few PCR cycles are set according to one set of primer and then 2nd set of
primers is added and PCR cycle is set accordingly. It is quite more sensitive and specific in
comparison to MP-PCR.
F. ERIC-PCR (Enterobacterial repetitive intergenic consensus PCR): Recently a family of
repetitive elements called enterobacterial repetitive intergenic consensus (ERIC) or intergenic
93
repeat units (IRUs) has been defined using E. coli and Salmonella Typhimurium genome. These
126 bp elements contain a highly conserved central inverted repeats are located in extragenic
regions and are unrelated to REP consensus sequences. Inter ERIC distance and pattern is
specific for bacterial species and strains of different origin. This specificity can be employed to
differentiate between closely related different bacterial strains using single set of primers
commonly used ERIC primers e.g.,
ERIC-1 5’ ATGTAAGCTCCTGGGGATTCAC 3’
ERIC-2 5’ AAGTAAGTGACTGGGCTGAGCG 3’
G. REP-PCR: Similar to ERIC elements of enterobacteria and Alu repeats of mammalians
prokaryotic genomes also has some noncoding repetitive DNA. Repetitive extragenic
palindromic (REP) elements also known as palindromic units (PU) are 38 bp consensus
sequence containing six totally degenerate positions including a 5 bp variable loop between
each side of the conserved stem of the palindrome. Multiple functions have been thought of for
these elements including role in transcription termination, mRNA stability and chromosomal
domain organization in vivo. The distribution of REP is diverse in prokaryotic genomes and can
be examined with PCR or by slot blot hybridization with radiolabelled.Consensum probes.
H. Ligase chain reaction (LCR): A new DNA detection method that uses thermostable ligase to
discriminate exquisitely and amplify single base changes in the genes of interest. The enzyme
specifically links two adjacent oligo nucleotides when hybridized to a complementary target
only. Single base mismatch prevents ligation and amplification, thus this technique can
distinguish closely related but distinct strains, not possible to be differentiated by other
techniques. Although very useful technique and exploited a lot with strains of Neisserio,
Listeria, Erwinia and Chlamydia etc.
I. Assembly PCR: In the process synthesis of long DNA sequences are is targeted. In the method a
pool of oligonucleotides is made using PCR and oligos possess short some overlapping
segments. The oligos alternate between sense and antisence direction and the overlapping
sequences determine the order of PCR segments to be aligned to produce final long strand of
DNA.
J. Asymmetric PCR: It is used to selectively amplify one of the two strands of the target DNA.
The product finds use in synthesis of diagnostic probes where only one of the two
complementary strands is required. Here primer for the chosen strand is added in excess and in
due course of amplification limiting primer restricts the proliferation of non-selected strand.
Some extra cycles are needed for good yield of desired strand. The modification of this process
is known as Linear-after-the-exponential-PCR (LATE-PCR), in which limiting primer is
having higher melting temperature I than that of the excess primer.
94
K. Colony PCR: Used to screen bacterial colonies to find carriers of correct DNA vector
constructs. In the process bacterial colony is picked with topic and directly added to master-mix
or sterile water to use it as source of templet DNA. In the process either the first step at 95oC is
extended (when using standard polymerase) or a short denaturation step (100oC) is inserted in
to protocol (with special chimeric DNA polymerase).
L. Hot-start PCR: To reduce non-specific amplification (during initial stage of PCR) the reaction
mixes (without DNA polymerase is first brought to 95oC and then DNA polymerase is added in
the hot mixture. To overcome the problem, special DNA polymerases have been designed
(either bound with antibody or with covalently bound inhibitors which get dissociated or
inactivated at high temperature activation step.) which are inactive at ambient temperature and
get activated on start of elongation (hot) step only.
M. Inter-sequence specific (ISSR) PCR: This is used in DNA fingerprinting and targets some
simple sequences, often repeats, found in templet DNA. It produces a unique fingerprint of
amplified fragments like ERIC PCR.
N. Inverse PCR: It is used to amplify flanking regions of a known sequence (insert) or of a known
gene in different species. In the process a series of digestions and self ligations are attempted to
result in known sequence at either end of the unknown sequence.
O. Ligation mediated PCR: This method uses small DNA linkers ligated to DNA of interest and
multiple primers annealing to the DNA linkers. Linkers can be added after digestion of the
DNA with specific endonuclease as in amplified fragment length polymorphism (AFLP). This
technique has been used in DNA sequencing, genome walking and DNA foot printing.
P. Overlap extension PCR: It is a genetic engineering technique allowing the construction of
DNA sequence with an alteration inserted beyond the limits of the longest practical primer
length.
Q. Quantitative PCR (Q-PCR) or Real time PCR (RT-PCR, QRT-PCR or RTQ-PCR): To
quantify the amount of starting templet of DNA, cDNA or RNA this technique is used. In this
method primers tagged with dissociable dyes as Sybr Green or fluorophore containing DNA
probes (TaqMan) are incorporated into the reaction mix and amount of amplified product is
quantified in real time (while reaction is still going on).
R. Reverse transcription PCR (RT-PCR): In this process instead of DNA polymerase reaction is
started with reverse transcriptase to determine the sequence of RNA from RNA viruses or tissue
RNA or cellular RNA. In initial steps reverse transcriptase synthesises cDNA from RNA. Rapid
amplification of cDNA ends (RACE) with PCR is used to identify 5’ end of gene or to map
exons and introns in the gene.
95
S. Thermal asymmetric interlaced (Tail) PCR: It is used to identify unknown sequences flanking
to a known sequence. From the known sequence a nested pair of primers with differing
annealing temperature and a degenerate primer is added in order to amplify in the direction of
unknown sequence.
T. Touchdown PCR: In this variant of normal PCR, reduction of non-specific background
amplification is brought up by gradually decreasing the annealing temperature as the PCR
cycling proceeds. In start annealing temperature is set about 2-3oC higher then the Tm of
primers used while in the latter cycles it is 3-5oC lower than the primer Tm. The initial high
temperature give specificity and lower temperature in end cycles enhances efficiency of
amplification of specific product formed in initial cycles.
Q. 78.What is cDNA, and cDNA library, why we need it?
Ans. A DNA copy of mRNA is called cDNA. Cloned eukaryotic genes cannot be expressed
directly in prokaryotic organisms because of the introns and exones in eukaryotic genes.
Prokaryotes can not splice out introns. However, cloned cDNA can be expressed in bacteria
when a cDNA is coupled with bacterial promoter.
Purified mRNAs is converted to cDNAs using reverse transcriptase and oligo dT as a
primer. The oligo dT base pairs to the polyA tails of the mRNAs, and reverse transcriptase
synthesizes a complementary DNA (cDNA) strand using the RNA as a template. The
DNA/RNA duplexes are denatured with NaOH which also destroys the RNA. The single-
stranded DNA can be converted to double-stranded DNA using random hexamer primers
(synthetic DNA strands containing all possible combinations of six nucleotides) and DNA
polymerase. Thereafter, purified cDNA molecules can be cloned in desired vector(s) to create
a cDNA library. A cDNA encoding a particular protein can be identified by screening the
cDNA library with a “gene-specific” probe. Alternatively, the cDNA mixture can be amplified
by PCR using “gene-specific” primers.
Q. 79. How eukaryotic mRNA is purified?
Ans. Majority of eukaryotic mRNAs are polyadenylated i.e., having polyA tail which is added
(several hundred A nucleotides) after transcription to the 3’-ends of eukaryotic mRNAs. The
polyA tail is used as “handle” to purify the mRNA. Total RNA is extracted from the eukaryotic
cell and hybridized to oligo dT (short synthetic DNA made entirely of T nucleotides)
chemically immobilized on a support matrix (Sepharose beads). When RNA extract is passed
through the matrix, the polyA tails of the mRNAs base pair with the oligo dT, and “stick” the
mRNAs to the column while all other RNAs (rRNAs and tRNAs) elute out unbound. After
96
washing off the rRNAs and tRNAs from the matrix column, the purified mRNAs are extracted
by breaking the hydrogen bonding between the polyA and oligo dT.
Q. 80. What is shotgun sequencing?
Ans. It is the fastest method to sequence the entire genome of an organism. It involves sequencing
of all possible randomly cloned genomic DNA fragments and the method is called shotgun
sequencing.
Q. 81. What are the functions of DNA polymerase I of E. coli?
Ans. Pol-I of E. coli has a. 5’-to-3’ DNA polymerase activity, b. 3’-to-5’ exonuclease
(proofreading) activity, c. 5’-to-3’ exonuclease (Nick translation) activity, (i.e., simultaneous
polymerization in the 5’-to 3’ direction from a nick with concurrent removal of nucleotides
ahead by the 5’-to-3’ exonuclease activity) and it removes RNA primers during replication
Q. 82. What do you understand by templet strand and coding strand?
Ans. The terms template strand stands to the strand of DNA that serves as a template for the
synthesis of mRNA. The opposite strand which has base sequence directly corresponding to the
mRNA sequence is called the coding strand. It is mRNA-like strand because the sequence
corresponds to the codons that are translated into protein.
Q. 83. What do you mean by nick translation?
Ans. Nick translation means replacing the RNA primer with DNA, it is achieved by the coordinated
polymerase and 5’ exonuclease activities of DNA polymerase I. In Nick translation, in every
step the nick moves by one position to the left, as a UMP residue of RNA is replaced by a
dTMP residue of DNA with no change in the total number of nucleotides polymerized in the
process.
Q. 84. What do you understand by dye terminator method of DNA sequencing?
Ans. In Dye-terminator method of DNA sequencing the single reaction mixture contains all four
unlabeled dNTPs, all four fluorescently labeled ddNTPs (giving blue, red, yellow, and green
fluorescence), DNA polymerase, template (the DNA to be sequenced; double- or single-
stranded), and specific primer (oligonucleotide). The newly synthesized strands all start with
the same primer (and therefore have identical 5’-ends) and terminate with a specific ddNTP.
The DNA is resolved on a denaturing sequencing gel. Only “colored” DNA is detected. The
97
“color” (fluorescence) of each band identifies with ddNTP, which was incorporated to terminate
further DNA synthesis. The sequence is read 5’-to-3’ from the bottom-to-top of the gel.
Q. 85. What are integrons?
Ans. Integrons are mobile DNA elements with the ability to capture genes, notably those encoding
antibiotic resistance, by site-specific recombination. Integrons have an intergrase gene (int), a
nearby recombination site (attI), and a promoter (Pant). They are of at least three types based
upon the type their integrase gene. Class 1 integrons have a variable region bordered by 5’ and
3’ conserved regions. The 5’ region is made up of the int gene, ‘attI’, and the promoter ‘Pant’
which drives transcription of genes within the variable region. The 3’ region consists of an
ethidium bromide resistance locus (qacED1), a sulfonamide resistance gene (sulI), and an open
reading frame containing a gene of unknown function. In Class 2 integrons, integrase is located
within the 3’ conserved region. Class 3 integrons are still little understood.
Q. 86. How integrons differ from transposons?
Ans. Transposons have direct or indirect repeat sequences at their ends, but the regions flanking the
antibiotic resistance genes are not repeats in the integrons, and the integrons possess a site-
specific integrase gene of the same family as those found in phage but lacked genes associated
with transposition.
Q. 87. What are gene cassets?
Ans. Gene cassettes are the small DNA elements containing a few defined genes which can be
mobilized by integrons through site specific recombinations. A gene cassette may be either a
linear insert within integron or a transient small dsDNA circle carrying one and occasionally
two resistance genes and a recombination site (a 59-base element).
Q. 88. Name three important genetic elements associated with acquisition of antibiotic resistance
by bacteria.
Ans. There are integrons, plasmids and transposons carrying antibiotic resistance genes (R-factors).
Q. 89. How bacteria acquire drug resistance?
Ans. Bacteria can acquire drug resistance either through mutation (in environment may be having
antibiotic, but not necessary but presence of antibiotic helps in selection of resistant strains) or
acquire R-factors via horizontal transfer from related bacteria by transformation, transduction,
or conjugation. R factors may be on integrons, plasmids and transposons.
98
Q. 90 If a bacteriophage is added to a flask containing bacterial culture, the broth would become
clear, creating a scene as if the bacteriophage had killed all the bacteria. However, after a few
days, the medium once again becomes cloudy as the bacterial population rebounded – now
composed of virus-resistant bacteria. This happens even when all the bacteria in the flask are
the clonal offspring of a single bacterium. How the phenomenon can be explained
Ans. The explanation can be two fold, first, the small population of bacteria exposed to phage
might have developed immunity transferable to progeny, and the other reason might be the
existing resistance in a few mutants arising in the parent colony itself (as natural phenomenon)
and those mutants got selected on addition of phage. Origin of mutants in nature was proved by
fluctuation test conducted by Salvador Luria and Max Delbruck. They proved that mutations
occur even in absence of infection of bacteria and resistance is due to random, spontaneous
mutations that occurred sometime during the growth of the culture prior to exposure to phage.
Similar phenomenon can be used for emergence of antimicrobial drug-resistance microbes.
Q. 91. How bacteria can acquire new characteristics?
Ans. The bacteria acquire new characteristics either through adaptation, or through induced
mutations (under stressful environment), or through spontaneous mutation which can occur
even prior to exposure to the toxic conditions, yielding resistant progeny cells.
Q. 92. What do you understand with merozygote in bacteria?
Ans. During unidirectional transfer (which is the normal way of gene transfer in bacteria) of DNA
from a donor cell to a recipient cell. In the process the donor usually gives only a small part of
DNA to the recipient. Therefore certain segment of DNA become zygotic and bacteria may be
called as partial zygote or merozygote.
Q. 93. What do you understand with cut and paste mechanism of transposition?
Ans. Non-replicative transposition of some transposable elements is called “cut-and-paste”
mechanism. In the process, a transposase makes a double-stranded cut in the donor DNA at the
ends of the transposon and simultaneously makes a staggered cut in the recipient DNA. Each
end of the donor DNA is then joined to an overhanging end of the site of DNA. DNA
polymerase fills in the short, overhanging sequences, resulting in a short, direct repeat on each
side of the transposon insertion in the recipient DNA.
Q. 94. What is the headful packaging mechanism?
99
Ans. It can be explained with the example of P22 bacteriophage. First the circular P22 DNA is
replicated through theta-replication followed by circle replication generating a long concatemer
of linear, double-stranded DNA to be packaged into phage heads. Each phage head holds about
44 kb of double stranded linear DNA. The phage encoded endonuclease initially cuts the DNA
at a pac site. One phage head is filled with DNA, and then the DNA is cut by the endonuclease.
The next head is filled with DNA beginning with the remaining end. These reactions continue
progressively until 3-5 phage heads are filled with DNA. Phage genome is about 42 kb,
approximately 2 kb less than the length of one headful of DNA. Thus, the DNA packaged into
each phage head is terminally redundant and the ends are circularly permuted. This mechanism
is called the heedful packaging.
Q. 95. To act as a replicon what qualities must be present in the DNA molecule?
Ans. To act as a replicon each DNA molecule must have an origin of replication specific to the
organism. Other properties, which sometimes can be supplemented by the host, include genes
encoding specific proteins or RNAs necessary for the initiation of replication and/or regulation
of replication.
Q. 96. Explain salient features of DNA replication in E. coli.
Ans. DNA replication in E. coli is started by DNA pol III at the chromosomal origin of replication
(ori). The replication proceeds bidirectionally (theta-replication) along the chromosome til the
two replication forks meet at the terminus. For initiation of DNA replication, DNA primase, the
DnaA protein, and specific DNA sequences are the basic requirements. Replication terminates
on meeting of two DNA replication forks. Then a topoisomerase separates the two strands of
DNA. Termination is facilitated by specific DNA sequences and a protein encoded by the tus
gene, but they are optional for propagation of bacteria.
Q. 97. In order to understand the biochemical mechanism of DNA replication, Kornberg purified
an enzyme from E. coli (called DNA polymerase I or Pol I) that could catalyze DNA synthesis
although this enzyme could catalyze DNA synthesis in vitro, this does not prove that the
enzyme is essential for DNA synthesis in vivo. To test this question, De Lucia and Cairns
sought to isolate a temperature sensitive mutant in the polA gene which encodes Pol I. In such a
mutant, Pol I would be active at 300C but inactive at 420C. If Pol I was essential for DNA
synthesis in vivo, what would happen to a polA(Ts) mutant if it was shifted to 420C?
Ans. If the polA gene product was the primary DNA polymerase, the cells would grow normally at
30 C but when the cells were shifted to 420C DNA replication would stop and no further cell
division or growth could occur. [Note – DeLucia and Cairns’ isolated a polA mutant that was
100
not conditional, indicating that the polA gene is not essential under certain growth conditions
and, thus, DNA Polymerase I is NOT the primary DNA polymerase in E. coli.]
Q. 98. Mutants defective for the proofreading function of DNA polymerase III typically form
small, unhealthy looking colonies on rich medium. Why?
Ans. Such mutations in the dnaQ gene (called mutD) produce a “mutator” phenotype. Because they
are unable to proofread errors that occur during DNA replication, such strains accumulate
mutations at a high frequency. The resulting large number of “lethal mutations” that arise
during cell division slows the growth of the colonies. The mutation frequency is highest when
the cells are growing on rich medium which allows the cells to grow more rapidly than the
resulting errors can be corrected by other repair systems.
Q. 99. Briefly describe the salient features of transcription in E. coli, including: (a) the features of
the promoter required for the specific initiation of transcription, (b) the direction of
transcription relative to the 5’ and 3’ ends of the DNA and how elongation occurs, (c) the
typical transcription termination sites, (d) and the enzyme required.
Ans.
a. The promoter sequence is recognized by RNA polymerase + a specific sigma factor. The
“housekeeping” sigma factor in E. coli is called sigma 70 (s70). The RNA polymerase + s70
protein holoenzyme (Es70) binds to a set of sites centered around upstream of the start of
transcription. (The first nucleotide of the mRNA is called +1.) The consensus sequence for
Es70 is TATAAT (centered on -10) and TTGACA (centered around -35).
b. RNA polymerase moves along the DNA in a 5’ to 3’ direction, adding nucleotides to the
3’OH group of the growing RNA chain. The sigma-factor normally dissociates from the
RNA polymerase core soon after transcription is properly initiated and chain elongation has
begun. The energy for elongation comes from hydrolysis of NTPs.
c. Factor independent termination and Rho dependent termination
d. The enzyme required is RNA polymerase. RNA polymerase core is a large protein with four
subunits – two a subunits, one B-subunit, and one B’-subunit. When a sigma-factor
associates with the RNA polymerase core, it is called the RNA polymerase holoenzyme.
Q. 100. Briefly describe the ribosome binding site and start codon required for the specific
initiation of protein synthesis, and the codons and release factors that signal termination of
protein synthesis.
101
Ans. In bacteria the ribosome binds to a specific sequence on the mRNA called a ribosome-binding
(or Shine-Delgarno) site. This sequence is complementary to a sequence near the 3’ end of the
16S rRNA and serves to properly align the ribosome relative to the start codon. A typical
ribosome-binding site consists of 3-9 purine bases located about 3-12 nucleotides upstream of
the start codon. The start codon is usually AUG, but occasionally GUG, UUG, or AUU is used.
In bacteria, at least some Archaea, mitochondria, and chloroplasts the start codon is recognized
by a distinct tRNA, fMet-tRNA. Translation termination occurs at stop (or nonsense) codons –
UAG (amber), UAA (ochre), and UGA (opal). There are no cognate tRNAs that recognize these
codons, but they are recognized by specific proteins called release factors. When a release
factor binds to a stop codon, the ribosome catalyzes hydrolysis of the polypeptide chain. The
ribosome then dissociates from the mRNA.
Q. 101. Gene fusion or operons are useful, how?
Ans. Gene fusion or operons can be useful:
1. For in vivo cloning
2. To identify mutations in genes
3. Portable region of homology are used for constructing duplications e.g., Hfrs.
4. for purification of fusion proteins
5. To study cell localization
6. To study in vivo gene expression
7. They are used as linked marker with selectable phenotype
Q. 102. To utilize maltose as a carbon source E. coli need aus and malF gene products for
transport of maltose into the cell. The aus gene encodes a periplasmic protein while malF
gene encodes a cytoplasmic membrane protein. Tn-phoA insertion mutants were isolated in the
aus and malF genes in a host lacking the chromosomal phoA gene. A few Tn-phoA insertions
in aus and malF express alkaline phosphatase activity and some do not express alkaline
phosphatase activity. Reason why some aus::Tn-phoA insertion did not express alkaline
phosphatase activity and others do. In the same study could it be possible to find a colony with
malF::Tn-phoA insertions reacting to alkaline phosphatase antibodies but not expressing the
alkaline phosphatase activity, if yes, how?
Ans. The bacteria not expressing alkaline phosphatase activity despite of aus::Tn-phoA insertion
might be having insertions in the wrong orientation or in wrong reading frame. The bacteria
those react to alkaline phosphatase antibodies but not express the enzyme activity might
102
producing alkaline phosphatase either as hybrid protein (due to gene fusion) or might be
producing inactive protein due to incomplete transfer of the gene.
Q. 103. It is possible to produce gene fusions that not only express reporter genes but also express
a target gene if a suitable promoter is provided by a transposon. A decade ago, John Roth
constructed a tetracycline resistant derivative of Tn10 called T-pop in which tetracycline
induced formation of promoters pointing outward from both sides of the transposon. Do T-pop
insertions in lac operon may result in both a Lac- phenotype and regulation of lacZ expression
by tetracycline?
Ans. If T-pop got inserted between the promoter and beginning of the lacZ gene then only it is
possible. In this case, absence of tetracycline (which is inducer of the T-pop promoter) will
result in Lac-ve phenotype but in presence of tetracycline Lac+ve (lacZ expression) phenotype.
Q. 104. Fusions to the chloramphenicol transacetylase (cat) gene make the host cell resistant to
chloramphenicol. If you have a lacZ::cat gene fusion, can you select a mutant with disrupted
lacI repressor?
Ans. The transcription of the lacZ gene is repressed by the LacI repressor, a strain with the
lacZ::cat gene fusion can make very little chloramphenicol transacetylase in absence of lactose
or IPTG in medium thus will show sensitivity to chloramphenicol on medium lacking lactose or
IPTG. On the other hand mutants having defect in the lacI gene yield high-level expression of
cat gene and will be resistant to chloramphenicol even in absence of lactose or IPTG in
medium. Therefore we can select for mutation in lacI gene by simply by selecting
chloramphenicol resistant colonies on medium lacking lactose or IPTG.
Q. 105. Using reporter genes to facilitate purification of specific proteins a variety of reporter genes
kits are available, a kit for making fusions of the N-terminus of the maltose binding protein
from E. coli (aus) to the C-terminus of a target protein. The target gene is cloned into the
proper reading frame near the end of the aus gene on a multicopy plasmid. The MalE-fusion
protein is purified easily using affinity (for maltose) chromatography. From purified protein
MalE domain is removed by the use of highly specific protease. The question is why we use
lengthy cloning process for the purpose instead of transposition?
Ans. For the successful expression of desired protein the target gene needs to be cloned at the very
end of a reporter gene correct orientation and reading frame. Thus the cloning region is a
comparatively very small in comparison to the rest of the gene. Using targeted cloning chances
103
of success are quite high while in transposition the probability of finding a transposon insertion
precisely at the required position is very low, therefore the former procedure is preferred.
Q. 106. Lawes and Maloy wanted to map the temperature-sensitive lethal mutation clm-195(TS) in
Salmonella enterica serovar Typhimurium. Two-factor crosses were done with several linked
loci: an insertion mutation with a tetracycline resistant phenotype (zci::Tn10), the nit gene, and
the btuC gene. The results are shown in the table below. Calculate the linkage between each of
the genes in these two-factor crosses. In the results Tn10 indicates an insertion mutation with a
tetracycline resistant phenotype and genetic mapping of Tn10 is essentially similar to any other
gene.
Donor
Recipient Selected
phenotype
Recombinants Colonies
obtained
Clm+ 113 clm+ zci::Tn10 clm(T-sensitive) Tet-Resistant
Clm(Ts) 28
Nit+ 37 nit+ zci::Tn10 nit-9 Tet-Resistant
Nit- 203
Tet-sensitive 52 btuC11 btuC+ zci::Tn10 Btu-
Tet-Resistant 143
Ans. Frequency clm+ zci::Tn10 are coinherited = 113/(113 + 28) = 80%
Frequency nit+ zci::Tn10 are coinherited = 37/(37 + 203) = 15.4%
Frequency btuC zci+ (Tets) are coinherited = 52/(52 + 143) = 26.7%
Q. 107. Phage P22 HT was grown on the donor cells and using phage lysates, two-factor crosses
were done against Salmonella enterica serovar Typhimurium strains with mutations in the
metA, aceA, and iclR genes. If the results obtained are as shown in the table below, what may
be the probable genetic linkage. Calculate the cotransduction frequency and show the predicted
order of the metA, aceA, and iclR genes. Tn10 insertion mutation is identified by tetracycline
resistant phenotype.
Donor Recipient cells Selected
phenot
ype
Recombinants Number
obtained
metA::Tn10 iclR- metA+ iclR+ TetR IclR-
IclR+
60
40
aceA+ iclR- aceA- iclR+ Ace+ IclR-
IclR+
95
5
metA::Tn10 aceA+ metA+ aceA- TetR Ace-
Ace+
10
90
R stands fpor resistance and S for sensitive to tetracycline.
104
Ans. Frequencies of recombinants are as under:
TetR+IclR- recombinants =60/60+40=60%
Ace+IclR- recombinants =95/95+5=95%
TetRAcs+ =90/10+90=90%
The predicted order of genes is: metA-aceA-iclR
metA::Tn10 90 aceA 95 iclR
60
Q. 108. Design three-factor cross to confirm the order of clm, btuC, and zci::Tn10?
Ans. The experiment can be done with clm+ btuC17 zci+ donor and clm-160(TS) btuC+ zci::Tn10
recipient or any other similar combination of donar and recipient. Actually there need to be
difference in all the three “factors” in donor vs the recipient i.e., if the donor has a mutant allele
the recipient should have the wild type allele, otherwise recombinant classes will not be
recovered. In addition, one of the markers in the donor needs to be a selectable allele as here is
tetracycline resistance.
Q. 109. How complementation and recombination differ or similar to each other?
Ans.
Complementation Recombination
Only mixing of gene products to give a
phenotype
Only phenotype got changed not the genotype
There is no breakage/covalent rejoining of
DNA.
No mixing of gene products but a required product is
wholesome.
Genotype gets changed resulting to phenotypic
change.
It requires breakage/covalent rejoining of DNA.
Q. 110. During a study, cloning and sequencing of a gene encoding a multifunctional enzyme was
performed but to know further about the structure and function of the gene what could be the
strategy. Would the site-directed mutagenesis on the gene or other general method of
mutagenesis be the more useful?
Ans. If the gene showed substantial sequence similarity to other well understood genes, it will be
better to test some of the “computer” predicted site-directed mutations otherwise try some more
general faster method of mutagenesis.
105
Q. 111. If some one wanted to test the effect of multiple amino acid substitutions at a single site in
a protein, he/she might begin with site directed mutagenesis to convert that codon to amber
(UAG) but what to do further to solve the problem?
Ans. On creating amber mutation at a particular site, function of multiple amino acids can be tested
at that site by moving the mutant into multiple isogenic strains that carrying different amber
suppressor mutations e.g., supE (insert Gln), supD (insert Ser). But efficiency of suppression is
important consideration before performing the experiment. If suppression is inefficient, the
failure to observe activity of the gene product might be simply due to insufficient amounts of
the gene product, not due to the particular amino acid substitution.
Q. 112. Some times IS1 insertion in the aus gene, in a specific orientation, produces polar effect
on the galT and galK genes, but in the opposite orientation the insertions are not polar in effect,
why?
Ans. The orientation specific polarity might be due to position of transcription termination signal
on one strand of the IS1 but not on the other strand. Thus, when inserted in an orientation such
that the RNA polymerase transcribes through the strand containing the terminator, transcription
will stop in the IS1 and thus prevents transcription of downstream genes in the same operon
such as galT and galK. While in the opposite orientation, transcription termination does not
occur. Translational termination signals in one orientation but not the other could have a similar
effect due to Rho-dependent transcription termination i.e., once translation stops, the resulting
un-translated (“naked”) mRNA produced allows Rho protein to bind and knock out the RNA
polymerase complex off the DNA. Under some rare exception when the insertion element is
located very close to the ribosome binding site of the downstream gene this effect may not be
observed. Polar effect may also occur due to transcriptional or translational termination signals
in both strands, but when one strand also contains an outward facing fortuitous promoter, the
ribosomes could hop back on to the mRNA soon after they fall off, so little naked RNA would
result and Rho dependent termination would not occur. The latter might have worked for
insertions in aus which are close to the ribosome binding site for galT so that Rho dependent
transcription termination doesn’t occur.
Q. 113. Describe translation in brief.
Ans. Translation is the RNA directed synthesis of polypeptides requiring all three classes of RNA.
The template for correct addition of individual amino acids is on the mRNA, the tRNAs carry
activated amino acids into the ribosome which is composed of rRNA and ribosomal proteins.
The ribosome is get hold on the mRNA for ensuring correct access of activated tRNAs and
106
containing the necessary enzymatic activities to catalyze peptide bond formation, the process
goes on codone by codone till the completion of peptide synthesis.
Q. 114. How can you say that genetic code is degenerate type?
Ans. There are 64 possible combinations of the 4 nucleotides code (triplets) just for 20 amino acids,
thus the code is degenerate, i.e., more than one codone for an amino acid.
Q. 115. What is the general structure of any tRNA?
Ans. On sequencing of more than 300 different tRNAs it has been deduced that tRNAs vary in
length from 60 – 95 nucleotides (i.e., 18 to 28 kD). However, the majority contain 76
nucleotides. The role of tRNAs in translation is to carry activated amino acids to the elongating
polypeptide chain. In general all tRNAs:
1. Exhibit a cloverleaf-like secondary structure.
2. Have a 5’-terminal –PO4 group.
3. Have a 7 bp stem that includes the 5’-terminal nucleotide and may contain non-Watson-Crick
base pairs, e.g. GU. This portion of the tRNA is called the acceptor since the amino acid is
carried by the tRNA while attached to the 3’-terminal –OH group.
4. Have a D loop and a TjC loop.
5. Have an anti-codon loop.
6. Terminate at the 3’-end with the sequence 5’-CCA-3’.
7. Contain 13 invariant positions and 8 semi-variant positions.
8. Contain numerous modified nucleotide bases.
Q. 116. What do you understand by Wobble hypothesis?
Ans. Out of the 64 triplet codons 3 are recognized as translational termination codons while the
remaining 61 codons might be considered as being recognized by individual tRNAs. Most cells
contain isoaccepting tRNAs, different tRNAs that are specific for the same amino acid,
however, many tRNAs bind to two or three codons specifying their cognate amino acids. As an
example yeast tRNA-phe has the anticodon 5’-GmAA-3’ and can recognize the codons 5’-
UUC-3’ and 5’-UUU-3’. It is, therefore, possible for non-Watson-Crick base pairing to occur at
the third codon position, i.e., the 3’ nucleotide of the mRNA codon and the 5’ nucleotide of the
tRNA anticodon. This phenomenon been termed as the wobble hypothesis.
Q. 117. Give the sequence of event in protein synthesis.
107
Ans. First the ribosome binds to the mRNA at specific site near the start codon to read it in the 5’ to
3’ direction and protein synthesis starts from the N-terminus to the C-terminus of the protein. In
active translation, synthesis may starts at multiple site along the mRNA strand i.e., it is
polyribosomic (multiple ribosomes or polysomes work simultaneously). However, chain
elongation occurs by sequential addition of amino acids to the C-terminal end of the ribosome
bound polypeptide.
Q. 118. Describe the initiation of translation?
Ans. Both prokaryotes and eukaryotes translation requires a specific initiator tRNA, tRNA-meti,
that is used to incorporate the initial methionine residue into all proteins. The initiation of
translation requires recognition of an AUG codon. In prokaryotes a specific tRNAmeti
[tRNAfmeti] is required to initiate translation. The methionine attached to this initiator tRNA is
formylated. Formylation requires N10-formy-THF and is carried out after the methionine is
attached to the tRNA. The fmet-tRNAfmeti still recognizes the same codon (start codon), AUG,
as regular tRNAmet. Although tRNAmeti is also required for initiation in eukaryotes too it is
not a formylated. In the polycistronic prokaryotic RNAs this AUG codon is located adjacent to
a Shine-Delgarno element in the mRNA. The Shine-Delgarno element is recognized by
complimentary sequences in the small subunit rRNA (16S in prokaryotes). In eukaryotes
initiator AUGs are generally, but not always, the first encountered by the ribosome. A specific
sequence (A/GCCA/GCCAUGA/G) surrounding the initiator AUG helps ribosomal
discrimination is present in most mRNAs.
During process of initiation of translation:
1. A ribosome must dissociate into its’ 40S and 60S subunits. The initiation factors eIF-1 and
eIF-3 bind to the 40S ribosomal subunit favoring anti-association to the 60S subunit. The
prevention of subunit re-association allows the pre-initiation complex to form.
2. A ternary complex (pre-initiation complex) consisting of the initiator, GTP, eIF-2 and the
40S subunit is formed. To this complex the mRNA binds. (eIF-2 is composed of three
subunits, a, b and g. The binary complex then binds to the activated initiator tRNA, met-
tRNAmet forming a ternary complex that then binds to the 40S subunit forming the 43S
pre-initiation complex. The pre-initiation complex is stabilized by the earlier association of
eIF-3 and eIF-1 to the 40S subunit.)
3. Then 60S subunit of ribosome associates with the pre-initiation complex to form the 80S
initiation complex.
In eukaryotes, mRNAs cap structure bonds with specific eIFs prior to association with the
pre-initiation complex. Cap binding is performed by the initiation factor eIF-4F which itself is a
108
complex of 3 proteins; eIF-4E, A and G. eIF-4E (24 kD) recognizes and binds to the cap
structure. The eIF-4A (46 kD) binds and hydrolyzes ATP besides having RNA helicase activity.
Helicase activity (unwinding) on mRNA is essential to straighten the mRNA strand and freeing
from secondary structure to allow access of the ribosomal subunits. The eIF-4G aids in binding
of the mRNA to the 43S pre-initiation complex.
After proper alignment of mRNA with the pre-initiation complex, the initiator met-tRNAmet
binds to the initiator AUG codon (facilitated by eIF-1) and the 60S subunit associates with the
complex facilitated by eIF-5, already bound to the preinitiation complex. Hydrolysis of the GTP
bound to eIF-2 act as energy source for the process of formation of the 80S initiation complex.
The GDP bound form of eIF-2 then binds to eIF-2B which stimulates the exchange of GTP for
GDP on eIF-2. After exchange of GTP, eIF-2B dissociates from eIF-2. This is termed the eIF-2
cycle which is the absolute necessity for eukaryotic translational initiation to occur. The GTP
exchange reaction can be affected by phosphorylation of the a-subunit of eIF-2. At this stage
the initiator met-tRNAmet is bound to the mRNA within a site of the ribosome termed the P-
site (peptide site). The other site within the ribosome to which incoming charged tRNAs bind is
termed the A-site (amino acid site).
Q. 119. Describe the chain elongation in translation process.
Ans. The process of chain elongation requires specific non-ribosomal proteins. In prokarotes these
are Efs and in eukaryotes eEFs. It is cyclic process i.e., at the end of one complete round of
amino acid addition the A site will be empty and ready to accept the incoming aminoacyl-tRNA
dictated by the next codon of the mRNA. This means that not only does the incoming amino
acid need to be attached to the peptide chain but the ribosome must move down the mRNA to
the next codon. Each incoming aminoacyl-tRNA is brought to the ribosome by an eEF-1a-GTP
complex. After deposition of the correct tRNA into the A site, GTP is hydrolyzed and the eEF-
1a-GDP complex dissociates. To keep the continuous source of energy for the process GDP get
exchanged for GTP by eEF-1bg similarly to the GTP exchange that occurs with eIF-2 catalyzed
by eIF-2B.
The peptide attached to the tRNA in the P site is transferred to the amino group at the
aminoacyl-tRNA in the A site. This reaction called transpeptidation is catalyzed by peptidyl
transferase. The elongated peptide now rest on a tRNA in the A site. The A site is freed in order
to accept the next aminoacyl-tRNA by the process of translocation i.e., moving the peptidyl-
tRNA from the A site to the P site, catalyzed by eEF-2 coupled to GTP hydrolysis. In the
process of translocation the ribosome is moved along the mRNA such that the next codon of the
109
mRNA resides under the A site. Following translocation eEF-2 is released from the ribosome.
The cycle keeps on moving time and again till the end of the process.
Q. 120. Describe translation termination in short.
Ans. Similar to initiation and elongation, translational termination requires specific protein factors
(releasing factors) i.e., RFs in prokaryotes (two, RF-1 and RF-2) and eRF in eukaryotes (one).
The termination signals are the same both in prokaryotes and eukaryotes, the termination
codons present in the mRNA. Of the three termination codons, UAA and UAG are recognized
by RF-1, whereas UAA and UGA are recognized by RF-2. All the three stop codons are
recognized by the single eRF. The eRF binds to the A site of the ribosome in conjunction with
GTP to stimulates the peptidyl transferase activity to transfer the peptidyl group to water
instead of an aminoacyl-tRNA. The resulting uncharged tRNA left in the P site is expelled with
concomitant hydrolysis of GTP. The inactive ribosome then releases its mRNA and the 80S
complex dissociates into the 40S and 60S subunits ready for another round of translation.
Q. 121. Name some important protein synthesis inhibitors.
Ans.
Mode of action Inhibitors
Inhibition prokaryotic peptidyl transferase Chloramphenicol
Inhibition of prokaryotic peptide chain initiation
mRNA misreading
Streptomycin
Neomycin
Inhibition of prokaryotic aminoacyl-tRNA binding to the ribosome small subunit Tetracycline
Inhibition of prokaryotic translocation through the ribosome large subunit Erythromycin
Prevention of dissociation of EF-G from the large subunit Fusidic acid
Mimics an aminoacyl-tRNA thus interferes with peptide transfer resulting in
premature termination in both prokaryotes and eukaryotes
Puromycin
Catalysis of ADP-ribosylation of and inactivation of eEF-2 Diphtheria toxin
Catalysis of cleavage of the eukaryotic large subunit rRNA Ricin
Inhibition of eukaryotic peptidyl transferase Cycloheximide
Q. 122. How transposons regulate the frequency of transposition? Do unregulated transposition
may be lethal to bacteria?
Ans. Unregulated transposition may be lethal to bacteria because it may disrupt essential genes, to
avoid this problem; most transposons carefully regulate the transposition frequency through
different mechanisms viz;
110
1. Regulated by Dam methylation: D-adenine methylase (encoded by dam) adds a methyl
group to the adenine residue in the sequence 5’ GATC 3’ (“dam sites”). Naturally the
sequence present in both strands of dsDNA is in methylated state. However, during
replication, initially only the parent DNA strand is methylated, i.e, the site is
hemimethylated for about 1-2 min, till the Dam adds a methyl group to the non-methylated
strand to restore the fully methylated state. The hemimethylated dam sites signals that DNA
replication has occurred which is recognized by specific DNA-binding proteins. These Dam
sites play an important role in transposition. Tn10 recognizes two dam sites for regulation of
transposition. One dam site overlaps the -10 region of the promoter for transposase (Pin).
When this dam site is fully methylated, RNA polymerase binds poorly to the promoter thus
poor transcription and limits expression of transposase. But in the hemimethylated state
RNA polymerase binds to the promoter more efficiently, resulting in to 10-fold increased
expression of transposase. Increased expression of transposase increases the transposition of
both Tn10 and IS10 (located at each end of Tn10.) Another dam site overlaps the
transposase binding site on the inside ends of IS10. In fully methylated transposase binds
poorly to this site thus limits transposition of IS10. However in hemimethylated state
transposase binds to this site more efficiently, resulting in 10-fold increase in transposition
of IS10.
2. The other mechanism is based on preferential action of transposase in cis. If concentration
of transposase high naturally rate of transposition is high, but even for multiple copy
transposons excessively high frequency of transposition is controlled through rapid removal
of transposase, which is very unstable and also diffuses through the cell to find the end of a
transposon, transposase probably binds to DNA immediately after it is made. Thus, the
efficiency of transposition decreases as the distance between the transposase gene and its
binding site increases. For the increase of distance by 50 Kb between the transposase gene
and the end of transposon (Tn10) the frequency of transposition decreases by 15-fold.
3. If transposase is in trans it got repressed. As in case of Tn10, translation of transposase is
repressed by an antisense RNA. The antisense transcript has a much longer half life then the
transposase transcript, such that under steady state conditions there are about 5 copies of the
antisense transcript in the cell and about 0.2 copies of the sense transcript. Therefore, the
antisense transcript regulates production of transposase in cell remains to remain in quite
low concentration. In addition, the antisense transcript represses expression of transposase
from any Tn10 entering the cell already having a copy of Tn10.
111
Q.123. What are the steps in in DNA replication performed by concurrent performance of several
proteins?
Ans. There are several steps in DNA replication which requires proteins (enzymes and non-
enzymes) for accomplishment of the process. They are:
1. Unwinding of Parent DNA with DNA Helicase.
2. Several DNA binding proteins binds to single stranded DNA and prevent rewinding and the
strands are held in position, facilitating binding of DNA polymerase to catalyze the
elongation of the leading and lagging strands.
3. DNA polymerase, besides elongation of strand also checks the accuracy (proof reading) of its
newly synthesized strand.
4. DNA primase units unite to form primosomes, which help to build the RNA primer for
replication of lagging strand (synthesis of Okazaki fragments).
5. Okazaki fragment is attached to the completed portion of the lagging strand through a
reaction catalyzed by DNA ligase.
Q. 124. What are the important characteristics to be considered while designing a PCR primer?
Ans. The good primer has following characteristics:
1. Length: Normally between 18-24 bases but for AP or RAPD it may be anywhere between 9-
18 bases and for amplifying a sequence for gene cloning long primers with 28-35 bases are
often preferred. Some people have used even 50 base long primers too. Short primers
increases propensity of false priming.
2. Meting Temperature I: Reasonable range for PCR is 56-63oC; it depends on GC% and
length of primer. GC contents should be about 50% or more. Hybridizing probes and
sequencing primers usually designed to have >50% GC content.
3. Degeneracy: Primers with least degeneracy are the most desired for specificity, it is
important at 3’ end as single mismatch can stop extension of the strand.
4. End stability: Primers are often designed to have stable 5’ and unstable 3’ ends. A stable 3’
end may bind to other complementary sites than the target leading to secondary band
formation in hanging 5’ end while too much instability at 3’ end may lead to false priming.
The stable 5’ end is called GC clamp which ensures adequate and efficient priming even at
lower annealing temperature.
5. Repeats and runs: Three or more dinucleotides repeat (as –GCGCGC--) should be least in
number for better stability of the primer.
6. Secondary structures: Due to complementarity of bases (even with only three homologous
bases) within the primer may lead to secondary structure formation particularly when
112
primers are long. Such structures may for hairpins. These may interfere in amplification if
available free energy in the system is low. Dimer formation is another problem, it occurs
when there is complementarity in two primers of a pair or of more pairs (in multiplex PCR).
It occurs more when homology is at 3’ end.
Q. 125. How you can quantities gene expression?
Ans. There are several methods including quantization of the gene product in the system,
quantifying the activity of the gene product, measuring the amount of mRNA amount of the
target gene. Quantification can be done with reverse transcription polymerization (RT-PCR) of
mRNA. Other method is real-time PCR for cDNA made from reverse transcription of mRNA.
Q. 126. How SYBR Green dye is utilized in real-time PCR to measure the amplification?
Ans. SYBR Green dye binds to minor groves of double helix of DNA, i.e., only double stranded
DNA bounds the dye. The dye yields fluorescence signals when bound and free dye is
minimally fluorescent. As the product form in PCR cycles the more and more dye bounds to
amplified DNA and produce fluorescence which can be quantified. The disadvantage of dye is
that it also bound to templet DNA, falsely primed product as well as to dimmers and may give
false reaction. Moreover, about 6% dye loss its activity during 30 PCR cycles i.e., as the
reaction progress its precision goes down.
Q. 127. How TaqMan probes works in real-time PCR?
Ans. TaqMan probes are dual labeled hydrolysis probes designed to exploit 5’ exonuclease activity
of Taq polymerase. TaqMan probe is a 18-20 bases long oligonucleotide able to anneal with
some region of the target sequence of DNA between two primers. It is having a reporter dye
attached to 5’ end and a quencher dye to its 3’ end. Both dyes till hydrolysis are in close
proximity and yield a little fluorescence. As the reaction progresses more and more product is
formed and i.e., more number of complementary sequences become available for annealing of
the probe thus, more and more probe get annealed to the amplified product. During the
extension process as the Taq polymerase reaches the region where probe is present it cleaves
the probe with exonuclease activity. As a result quenching and reporter dyes got separated and
freed into the reaction mixture and reporter dye start to fluorescent. With each new cycle the
fluorescent activity doubles and measured to quantify the amplification product.
Q. 128. What are common methods employed for delivery of transposons to bacteria, describe in
brief?
113
Ans. For experimental transposition in laboratory common methods are:
1. Phage transduction: During the packaging of a lytic phage DNA into capsid a few also got
packed with transposons containing DNA, those phages or those where transposon got
inserted into phage DNA, making phage incapable to lyse the cell. Such transposed phages
can be used for transposition of bacteria e.g., in λ phage Tn10 insertion [λ cI::Tn10] make it
crippled for lytic process because Tn10 insertion disrupts the cI gene. The phage can not
grow lytically in a supo recipient because the P gene product is required for phage
replication. A lysate of this phage prepared in an E. coli amber suppressor mutant. When E.
coli supo recipient is infected with the phage, and deletion is made on tetracycline plate
only those E. coli will be able to grow which posses tetracycline resistance (TetR) encoded
by the Tn10, i.e, TetR colonies are possible through transposition of Tn10 into the
chromosome. This type of transposition can also be achieved when phage of one species of
bacteria is used to infect other bacteria without further propagation e.g., P1 phage can infect
and replicate in E. coli, but can infect Myxococcus xanthus without any replication. A lysate
of P1 carrying a transposon (Tn5) grown in E. coli, can be used to infect M. xanthus with
selection for kanamycin resistant (KanR) colonies. KanR is encoded by the Tn5 thus KanR
colonies can appear only after transposition of Tn5 from phage onto the bacterial
chromosome.
2. Through plasmid delivery: Plasmid delivery systems take advantage of a transposon
insertion on a plasmid which is incapable of replication in the recipient bacteria i.e., the
transposon is carried on a suicide plasmid. The suicide plasmid can replicate in permissive
host and grown in required amount in that. The plasmid from permissive host can be
transferred to non-permissive host either through conjugation or through transformation
(chemical or electroporation) applying the required selection for an antibiotic resistance
encoded by the transposon.
3. Over expression of transposase: The frequency of transposition can be improved by
increasing the concentration of transposase in the cell or through increasing the stress of
selection. The transposase gene(s) can be cloned into a vector keeping expression under the
control of an easily regulated promoter (tac or ara). Introduction of transposon into a
bacterium producing high levels of transposase, the frequency of transposition will be high
even detected without direct selection.
Q. 129. What do you understand by transposon mediated DNA rearrangements?
Ans. In bacteria having multiple insertions of transposons in chromosome provides multiple
regions of homology which may lead to deletions and duplications through unequal crossing-
114
over which probably occurs between sister chromosomes soon after the DNA replication.
Consequences of promote unequal crossovers, half-crossover to generate duplications or
deletions.
Q. 130. What is a transposon pool? How it is made?
Ans. Transposon pool is a population of bacteria which have random transposon insertions at many
different sites in their genomes. In E. coli or Salmonella having about 3000 nonessential genes,
a transposon pool can be created through about 15,000 random insertions in most nonessential
genes. In such a pool ~ 1 in 3000 of bacterial cells contains a transposon insertion in any one of
the 3000 nonessential gene, and ~1 in 100 of those bacterium contains a transposon insertion
within 1 centisome (1 min or 1% of the chromosome length) of that particular gene. Transposon
pools made using any one of transposon delivery systems and phage lysis. To determine the
location of insertions in or near a gene, transposed bacteria are first selected onto a rich medium
containing an antibiotic resistance encoded by transposon. The selected colonies can then be
screened for insertion mutations in a gene or insertions near a gene using specific selection
criteria dependent on the target gene as you want to select a bacteria having insertion in aroA
gene then select the bacteria from the pool by plating it on growth medium lacking aromatic
amino acids.
Q. 131. What are important uses of transposons in molecular genetics?
Ans. Transposons in molecular genetics are used for:
1. Transposon can be to find role of nonessential genes through transposon insertions in the
genome because a transposon insertion in or near a gene usually causes complete loss of
function.
2, Transfer of desired mutation or character from one bacteria to other because it can be easily
selected as the phenotype of the insertion mutation is completely linked to antibiotic
resistance.
3. To determine the operon of a gene: Insertions in operons produces strong polar effect. Thus,
transposon insertions can be used to determine that which the genes are in the particular
operon.
4. For genetic mapping: Insertions near the gene but not in gene can be used for constructing
deletions therefore for genetic mapping. As the transposon insertions behave as point
mutations and can be utilized in fine-structure genetic mapping.
5. Transposon can be used for deletion or duplication mutants: As multiple insertions of
transposons provide a portable region of homology therefore, recombination between
115
transposons or crossing over during replication of chromosome might result in deletions or
duplications with defined endpoints, or to form cointegrates between different genetic
elements.
6. Specially designed transposons can be used for construction of operon(s) or gene fusions
tagged with reporter genes.
Q. 132. What are general properties of transposable elements?
Ans. Important properties of transposable elements are as under:
1. Move randomly – Although movement is decided by the presence of specific sequences in
the DNA but their random distribution make the transposition random. Transposable genetic
elements can move from any one DNA molecule other DNA molecule or from one to other
location on the same DNA molecule.
2. No self replication- Replication of transposable elements depends on replication process in
the cell mediated either through chromosomal, phage or plasmid replication. Exception for
the rule is some transposable phages.
3. Transposition is site-specific- Homology between current location and new location is not
required for transposition because this event is mediated through transposase coded by the
transposable genetic element. Transposase induce recombination of transposable element at
some specific sites (site-specific) and also known as result of illegitimate or nonhomologous
recombinations.
4. Transposition can be through duplication or deletion – Mostly transposition of transposable
element results in deletion of the element from the original site and insertion at a new site
but occasionally transposition event is accompanied by the duplication of the transposable
genetic element. That is one copy is retained at the original site and the other is transposed
to the new site (duplication).
Q. 133. What are the different kinds of transposable elements, how are they named, give their
importance, too.
Ans. Transposable elements are of two types namely insertion elements or insertion sequences and
transposons.
1. Insertion sequences (IS) – Insertion sequences carry no known genes except those required
for transposition and on both ends have repeat sequences. They are designated numerically
like IS1, IS2, IS3 and so on. Repeat sequences are needed for transposition. Is elements
have important role in i) Mutation (insertion into a bacterial gene may result in inactivation
of the gene), ii) Plasmid insertions into chromosomes (sites of plasmids insertion in
116
bacterial chromosome are often at or near IS elements in the chromosome), iii) Phase
Variation – (in Salmonella, two or more genes code for two or more antigenically different
flagellar antigens, the expression of these genes is regulated through insertion sequences by
the phenomenon of inversion of IS. In one orientation one of the genes is active while in the
other orientation the other flagellar gene is active. However phase variation in Neisseria is
due to transformation not due to IS inversion).
2. Transposons (Tn) – Transposons are transposable genetic elements carrying one or more
nonessential genes in addition to those required for transposition. They are designated as Tn
followed by an Arabic numeral. Their structure is similar to IS and nonessential genes are
located between the terminal repeats. In case of composite transposons terminal repeats are
insertion sequences, either the same or different at both the ends. Transposons can jump
from one DNA molecule to another thus those carrying antibiotic resistance genes leads to
development of plasmids bearing R factors for multiple drugs and thus evolution of multiple
drug resistance in bacteria. Transposons have been used in microbial genetics for muting
genes, finding nonessential genes and for chromosomal mapping.
Q. 134. Enumerate the molecular Koch’s postulates?
Ans. Molecular Koch’s postulates are used for determining the role of a particular gene in
virulence. These are:
1. The gene or its product should be found only in virulent strains.
2. The gene should be isolated and cloned.
3. Disruption of gene in virulent strain should diminish virulence.
4. The gene should be expressed by the pathogen during process of infection.
Q. 135. What modifications have been suggested in Koch’s molecular postulates to make them
simpler?
Ans. Only three Molecular postulates have been suggested to be required to prove the role of a
pathogen in pathogenesis. They don’t include need of isolation in pure culture of the pathogen.
They are:
1. There must be PCR amplification of 16s rRNA gene of suspected pathogen from from the
lesion.
2. Isolation and sequencing of rRNA gene for proving its identity with that of the pathogen.
3. Loss of amplifiable rRNA after successful treatment to cure the disease.
117
Q. 136. Even after knowing the full genome of bacteria or of any other organism why is it still
important to study their physiology? Name the molecular technique used to understand the
physiology of bacteria?
Ans. Study of physiology is still important because as it is known that more than a third of genes
are still unrecognized and tentative identification of gene remains questionable until
physiologically observed. Many genes for known proteins are still not recognizable in database
and to understand interaction of many gene products no other technology is available.
Moreover, very closely related gene sequences may code for totally unrelated products in
function. Microarrays are the most common and advanced technique to understand the
physiologic role of different genes. In microarrays, all the possible genes of an organism are
spotted at a chip and then mRNA of the organism expressed under different growth and
physiological conditions are used to identify the genes having the defined specific role.
Q. 137. Name the molecular methods used for determining virulence factors and their genes?
Discuss their limitations.
Ans. The common techniques are:
1. Cloning: suspected genes are cloned in non-pathogenic E. coli looked for their effect on
virulence. All genes can not be expressed in E. coli, or may not elaborate the effect due to
failure to secretion or activation of the product.
2. Transposon mutagenesis: Transposon with a selectable marker is introduced to generate
mutant banks of the pathogen or in a required specific gene. Every selectable colony has a
mutation. However insertion of transposon may have polar effect on downstream genes and
mutations can be created in genes not required for growth of the bacterium on media used
for selection.
3. Transcriptional fusion (operon fusion): In this technique reporter gene (e.g. lacZ, uidA, cat)
is fused with promoter operator of the virulence genes. The reporter protein will be
expressed only under conditions required for expression of the virulence gene. It is a
cumbersome technique.
4. IVET (in vivo expression technology): This technique was tried for identification of genes
of Salmonella Typhimurium responsible for its ability to cause typhoid like disease. To find
the promoters important for expression of virulence genes in vivo a promoter less purine
synthesis gene (purA) is introduced at random in chromosome of a strain from which purine
synthesis gene has been deleted. For this, chromosome of target bacterium is first digested
with RE into 4-6 kbp fragments and these fragments are cloned upstream to a promoter less
purA fused with promoter less lacZY into a plasmid and then plasmids are introduced into
118
Salmonella (purA mutant) which can not maintain the plasmid (defective to replicate in
Salmonella, suicide), thus to survive in host the bacterium must incorporate the plasmid in
its genome (at the site of homology). With this technique only house keeping type genes
could be identified.
5. Signature tag mutagenesis (STM): This is the in vivo version of transposon mutagenesis.
The transposons used in this technique carry a specific sequence and only one transposon
enters at a time in a bacterium i.e. only mutate one gene at a time. The selected pool of
mutants is inoculated into mice and then bacteria are re-isolated from the mouse spleen or
other organ to check the missing mutant. The missing mutant might carry mutation in genes
responsible for survival or pathogenesis in vivo.
Q. 138. Now when full genome of many organisms are known, does there is still requirement for
cloning the genes?
Ans. Yes, the genome sequence is just the basis for further studies as there are still many genes
with no known function, or for which functions have wrongly been assigned. There are genes
having multiple functions, which can be visualized and explored on suitable cloning.
Q. 139. What are the inducers for transcription of heat shock response genes?
Ans. It is misnomer to say heat shock response it is actually stress response and stress factors as
high temperature, accumulation of secretory proteins, carbon or phosphorus starvation,
exposure to heavy metals, ethanol, UV light, inhibition of topoisomerase or phage infection.
Q. 140. How many types of genes are known to be associated with heat shock response in E. coli?
Ans. There are atleast 18 known genes related to heat shock in E. coli, including the Chaperonins
dnaK, dnaJ, groEL, groEs, grpE, proteases clpP, lon & ftsH.
Q. 141. What are CIRCE and their role?
Ans. Controlling inverted repeats of chaperone expression (CIRCE) is the inverted repeats found
near to promoter of groE genes in more than 40 Bacteria. Its consensus sequence is
TTAGCACTC-N9-GAGTGCTAA. It is responsible for negative regulation of heat shock
proteins. Rhe repressor which binds to CIRCE is product of hcrA gene and requires GroE
binding for its activity. This system is absent in E. coli, which has a similar DnaK, presence of
denatured proteins titrate away the repressor in E. coli.
Q. 142. What is intein?
119
Ans. In most of the eukaryotes and prokaryotes, protein splicing occurs when a gene contain the
carboxy and amino termini of a protein separated by a region of DNA coding for an embedded
protein. The embedded protein in called intein, it has catalytic activity and act both as an
exonuclease and a protein splicing agent.
Q. 143. What is inversion system how it help the pathogenic bacteria?
Ans. Bacteria posses a variety of invertible DNA sequences and their orientation is controlled by
DNA invertase through site-specific recombination between ends defined DNA segments.
Normally two genes in opposite orientation are there in the same segment and at any given
time, only one gene orients in proximity of a promoter. One gene is expressed other is
automatically shut off. Common examples are flagellar antigens of Salmonella, fimbriae of E.
coli and tail fibers of P1 and Mu phages.
This system helps the pathogens to overcome the host defense.
Q. 144. What do you understand by his251, HISG or HisG, hisGDCBHAFI, ∆hisG251, ∆(hisgD),
∆(hisD-rfb), hisG::Tn10, hisD::mudJ, Φ(hisD-lacZ), hisG(Oc) and hisD(Op).
Ans. His251= allel number 251 having mutation in his gene.
HISG or HisG= hisG gene product or phenotype or HisG enzyme.
hisGDCBHAFI= genes of histidine biosynthesis operon showing sequence of different genes.
∆hisG251= deletion in hisG gene the Arabic numeral indicate the mutant number.
∆(hisGD)= deletion encompassing two or more genes.
∆(hisD-rfb)=deletion from hisD extending upto rfb; deletion encompassing more than one
operon.
hisG::Tn10= Tn10 insertion in hisG gene.
hisD::MudJ= MudJ insertion in hisG gene.
Φ(hisD-lacZ) =fusion of hisD and lacZ operons.
hisG(Oc) = suppression of hisG gene through ochre nonsense mutation.
hisD(Op) = suppression of hisG gene through opal nonsense mutation.
Q. 145. What do you understand by minutes in physical map of genome of bacteria?
Ans. It takes about 100 minutes for an E. coli Hfr strain to transfer its entire chromosome to a
recipient cell and length of DNA passed in one minute is taken as one centisome it is roughly
equals to 1% of genome or referred as 1 minute.
Q. 146. How can you use triphenyl tetrazolium chloride (TTC) for selection of mutants?
120
Ans. TTC is used to select the carbohydrate fermenting mutants, those ferment carbohydrate
produce acid and in presence of acid TTC is reduced to colourless substance, thus the
fermenters produce colourless colonies while non-fermenters reduce TTC into red TTC
formazone and colonies are red in colour.
Q. 147. How will you differentiate between a lysogen and pseudo-lysogen?
Ans. EBU (Evans blue uranine) agar can be used for the purpose on which pseudo-lysogen (phage
infected cells) form dark blue colonies due to internalization of blue dye while lysogen and
nonlysogen form light coloured colonies.
Another method is through growing the test bacteria on Green plates. This contains LB base with
0.67% glucose, alizarin yellow GG and methyl blue. Pseudolysogen form dark coloured
colonies. This media is often toxic for Salmonella and colonies grow very small.
Q. 148. How can you select a tetracycline sensitive colony?
Ans. For the purpose Bochner Maloy medium is used, this medium contains chlortetracycline,
fusaric acid and ZnCl2 in LB agar. Chlortetracycline is added before autoclaving to inactivate
the drug (the inactivated drug maintains the inducer function for resistant bacteria) and fusaric
acid is added after autoclaving. Fresh medium strongly inhibit growth of tetracycline resistant
bacteria.
Q. 149. What are runaway replication vectors?
Ans. These are the vectors, which are designed to get maximum production of a particular product
or cloned DNA from relatively small amount of culture. The vectors in general have some Ts
mutation in copy number controlling genes, which are inactivated on growing at a particular
temperature and plasmid starts to replicate in uncontrollable manner at the expanse of host life
and leads to production of maximum product/DNA.
Q. 150. What are the expression vectors, how they differ from cloning vectors?
Ans. Expression and cloning vectors are not much different except the presence of a suitable
promoter and a ribososme binding site just upstream to the cloning site in expression vector
which is not necessary to be present in cloning vectors.
Q. 151. Give some examples of good promoters used in cloning vector.
Ans. Phage T5 promoter is very competent to compete out the E. coli promoters, the tac promoter
combined with -35 region of the trp promoter or -10 region of lac promoter has high efficiency
of expression.
121
Q. 152. What are the different techniques for assembly of a contiguous DNA sequences?
Ans. There are three approaches viz., shotgun approach, clone contig approach, and directed
shotgun approach.
Q. 153. Give flow diagram for shotgun techniques for assembly of a contiguous DNA sequences.
What are its advantages and disadvantages?
Ans. 1. Extract DNA, 2. Sonicate DNA to obtain DNA fragments, 3. Electrophorase on agarose to
collect fragments of 1.6 to 2 kb, 4. Prepare a cone library, 5. Obtain end sequences of DNA
inserts, 6. Prepare DNA contigs.
Advantages: Genome can be sequenced rapidly, can be done even if nothing is known about
genome.
Weaknesses: complexity and data analysis to identify overlaps. Applicable for short genomes of up
to 5 Mb only.
Q. 154. Give flow diagram for cone contig techniques for assembly of a contiguous DNA
sequences. What are its advantages and disadvantages?
Ans. This technique is based on the assumption that upto 5 Mb size shotgun method is the best and
this technique exploit the RE enzymes to create smaller segments of large DNA molecule. The
individual clone is checked and interpreted by looking for some features as STSs (sequence
tagged site mapping) and SSLPs (simple sequence length polymorphism).
Q. 155. What are the cloning vectors for large sequences?
Ans. If the size of cloned DNA is more than a few kb bacterial plasmids yield unstable cones
therefore are of little use in cloning larger segments. Bacteriophages came next to clone up to
125 kb sequences. To overcome this problem cosmids (about same limit as the phages) are
developed these are hybrids of lamda phage and plasmids. But limit of size of the head of the
plasmid to fill the DNA was a big hindrance to overcome yeast artificial and bacterial artificial
chromosomes (YACs and BACs) are developed to clone > 100 kb sequences. YAKs are known
to have highest capacity to carry cloned DNA (600 to 1400 kb) without having any stability
problem. BACs can be used to clone more than 300 kb. P1 derived artificial chromosomes
(PACs) combine the features of P1 phage and BACs and can handle clones up to 300 kb.
Fosmids (origin of replication from F plasmid and a lamda cos site) are similar to cosmids but
with lower copy number and with higher stability.
Q. 156. What is chromosomal walking?
122
Ans. To arrange the contigs of a chromosome an overlapping series of cloned DNA is build and we
start with any one clone and find out other with overlapping sequence, once it is found 3rd clone
is searched to have overlapping sequence to 2nd and in this way are cones are arranged to
assemble the chromosome. This technique is called chromosomal walking.
Q. 157. How can you find a gene in a sequenced genome?
Ans.
1. By analysis to search for Open reading frames (ORF), the method is less successful for
eukaryotic genes due to presence of introns.
2. Through searching for expressed sequences i.e., mRNA. cDNA sequencing can be used to
map the genes in a sequence.
3. By homology analysis and comparison of related genome.
4. By determining the effect of some kind of inactivation at a particular sequence (transferring
some similar DNA to chromosome by homologous recombination).
5. Over expression in a multicopy vector.
Q. 158. What are the tools used in searching of ORFs in a sequence?
Ans. Locating start and termination codones, codon bias, exon-intron boundaries and upstream
control sequences (as most vertebrate genes have a CpG islands upstream of many genes).
Q. 159. What are different methods to determine the function of a gene?
Ans. There are two basic methods with many modifications:
1. Computer analysis of gene function: the function is attributed as per homology of structure/
sequence reported in other organisms, basis is that related genes have similar sequences and
new gene can be discovered by virtue of similarity to an equivalent gene in related
organism. There may be two types of homologies. Orthology; presence of homologous gene
in different organisms, b. paralogy: homologous genes present in the same organism.
2. Experimental analysis: This group of method involves gene inactivation by mutagenesis,
gene expression in cloning vector, directed mutagenesis, IVET, STM.
Q. 160. What do you mean by Exon-intron boundaries?
Ans. These are some sequences which are used in searching introns and exons in a sequence. The
sequence of upstream exon-intron boundary is often 5’-AG GTAAGT-3’ and down stream
boundary is less defined one 5’-PyPyPyPyPyPyNCAG -3. Arrow shows the precise boundary
point.
123
Q. 161. What do you mean by codon bias?
Ans. It means that different codons for an aminoacid are not used at the same frequency, e.g. in
human genome out of 4 codons for alanine (GCA, GCC, GCT or GCG), GCG is more
commonly used. The study of codon bias helps in identification of ORFs.
Q. 162. What may be the functions of lot many unassigned genes identified on sequencing and
computer analysis of sequences but no known or assigned functions?
Ans. The possible functions for unassigned genes may be, DNA synthesis and cell cycle, RNA
synthesis and processing, protein synthesis, stress responses, cell wall synthesis and
morphogenesis, transport of biomolecules within molecules, lipid, energy and carbohydrate
metabolism, development, DNA repair and recombination, meiosis and mitosis, chromosomal
structure, cell architecture, secretion and protein trafficking.
Q. 163. What are the prerequisites for in vitro genetic engineering?
Ans. Availability of the versatile and specialized cloning vectors and an efficient transformation
system are the prerequisites for in vitro genetic engineering.
Q. 164. What are the major tools in genetic engineering?
Ans. These are 1. RE enzymes, 2. Cloning and expression vectors, 3. Transformation system, 4.
Transposable elements and 5. Many advanced methods for in vivo genetic engineering.
Q. 165. What are different kinds of vectors used in genetic engineering?
Ans. Conjugative, conjugative helper, mobilizable, replicative and narrow host range vectors.
Q. 166. Enumerate some phenotypic variations among bacteria without any genotypic change
appearing heritable.
Ans. One good example is protoplasmic form of bacteria originated after lysozyme treatment;
removal of lysozyme always is not associated with reversion of normal morphology (bacillary
form). Depending on conditions a protoplasmic form can be propagated for generations
together.
Q. 167. Enlist different kinds of transposable elements and their properties.
124
Ans.
Transposable
elements
Size
(kb
)
Terminal
repea
ts
(bp)
Target size Phenotype
conferred
Origin
Insertion sequences (IS)
IS1 0.768 18/23 9 None G –ve bacteria
IS2 1.327 32/41 5 None G –ve bacteria
IS3 1.3 32/38 ¾ None G –ve bacteria
Is4 1.426 16/18 11/12 None G –ve bacteria
IS10 1.329 17/22 9 None G –ve bacteria
IS50 1.531 12/18 9 None G –ve bacteria
IS903 1.050 18/18 9 None G –ve bacteria
IS256 1.350 Not
known
Not detected None G +ve bacteria
Composite transposons
Tn5 5.7 IS50(IR) 9 Km G –ve bacteria
Tn9 2.5 IS1(DR) 9 Cm G –ve bacteria
Tn10 9.3 IS10(IR) 9 Tc G –ve bacteria
Tn903 3.1 IS903(IR) 9 Km G –ve bacteria
Tn1681 2.1 IS1(IR) 9 Heat stable toxin –ST G –ve bacteria
Tn4001 4.7 Is256(IR) Not detected Gm Tb Km G +ve bacteria
Complex Transposons, Tn1 Family
Tn1, Tn3 4.957 38/38 5 Ap G –ve bacteria
Tn21 19.5 35/38 5 Hg Sm/Sp Su G –ve bacteria
Tn 501 8.2 35/38 5 Hg G –ve bacteria
Tn 1721 11.4 35/38 5 Tc G –ve bacteria
Tn951 16.5 5 Lac G –ve bacteria
Tn551 5.3 35/35 5 Ery G +ve bacteria
Tn 1000 (γδ) 5.8 36/37 5 None G –ve bacteria
Others
Tn7 14 22/28 5 Tp Sm/Sp G –ve bacteria
Tn554 6.2 None 0 Sp Ery G +ve bacteria
Tn916 15 20/26 10 Tc, conjugal transfer G –ve bacteria
Tn4291 Not det. Not
detected
Not detected Mec G +ve bacteria
DR, direct repeats; IR, inverted repeats; Ap, Ampicillin; Cm, chloramphenicol; Ery, erythromycin; Gm, gentamicin;
Hg, mercuric ions; Km, kanamycin; Mec, methicillin; Sm, streptomycin; sp, spectinomycin; Su, suphonamide;
Lac, lactose metabolism; Tb, tobramycin; Tc, tetracycline; Tp, trimethoprim; Sm/Sp, resistant to both Sm and Sp.
Q. 168. What are the differences between complex and composite transposons, give examples?
Ans. Composite transposons are made of simple units i.e., An internal (central) sequence carrying
functions specific to the transposon but without transposition function, the central region is
flanked by long direct or inverted sequence repeats (1-2 kb) e.g. Tn5 and Tn10, while, complex
transposons do not follow the modular structure of transposon and both functions (specific to
transposon and transposition) are in central region and are flaked by short inverted repeats (~40
bp) e.g. Tn1 and Tn3.
Q. 169. What are the difference between Class I, Class II and Class III transposable elements?
125
Ans. Class I: These are the complex transposons or IS elements which have short terminal repeats
e.g. Tn1, Tn3, IS1, Class II: These are Composite transposons having long terminal repeats e.g.
Tn5, Tn10, Class III: Transposing bacteriophages are grouped in this class e.g. Mu and D108.
Q. 170. What are the different methods of acquisition on new genes among microbes?
Ans. Transformation, Transduction, Conjugation and cell fusion are the four common methods of
acquisition of new genes by bacteria and other microbes. These methods are used by both G-ve
and G+ve bacteria but not necessarily by all strains.
Q. 171. What the methods by which new genes acquired could be maintained in the host?
Ans. New information acquired can be maintained in two ways: 1. either the new DNA is
incorporated into the existing replicons (chromosome or plasmid) or 2. the new DNA maintains
itself as an independent replicon.
Q. 172. What are the different methods of incorporating new information into replicon of a
microbe?
Ans. Site specific recombinations, insertions/ transpositions with transposons or integrons, or
integration of some plasmids as F plasmids to form Hfr and F’ strains are the few methods of
incorporation of new DNA into replicon.
Q. 173. How rifampicin, streptomycin, kanamycin, chloramphenicol and puramycin interfere with
bacterial propagation?
Ans. Rifampicin binds to RNA polymerase to inhibit RNA synthesis, streptomycin and kanamycin
binds to 30s ribosome at the same site at which mRNA binds, chloramphenicol binds with 50s
ribosome at the site which binds with tRNA while puramycin interfere with maturation of
polypeptide chain, i.e., causes premature termination.
Q. 174. What is leader sequence?
Ans. It is a sequence of untranslated mRNA on 5’ end before a region encoding first polypeptide, it
contains regulatory sequences.
Q. 175. What is spacer sequence on mRNA?
Ans. In polycistronic mRNA there are often untranslated sequences about tens of bases long,
which separate the coding sequences corresponding to different polypeptide chain.
126
Q. 176. What are different functions of DNA and RNA binding proteins, name with examples?
Ans. The different functions of DNA and RNA bindind proteins are as follows:
1. Packaging of DNA: Similar to histones of eukaryotic DNA Bacteria requires Bacterial
Nucleoid proteins.
2. DNA recombination; RecA
3. DNA repair: DNA glycosylases, nucleases
4. DNA replication: Origin recognition proteins, DNA polymerase, DNA ligases, ssDNA
binding proteins, DNA topoisomerases.
5. Transcription initiation: Similar to eukaryotic TATA-binding proteins bacteria also have
some proteins which bind to σ subunit of RNA polymerase
6. RNA synthesis: RNA polymerases
7. Transcription regulation: Repressors or transcription factors
8. DNA degradation: Restriction enzymes
9. Intron splicing: snRNP proteins
10. mRNA polyadenylation: CPSF, CstF
11. mRNA editing: Adenosine deaminases
12. rRNA and tRNA processing: Ribonucleases
13. Translation: Amino acyl tRNA synthetases, translation factors
14. RNA turnover: Ribonucleases
15. Ribosome structure: Ribosomal proteins
Q. 177. What are different classes of coding and noncoding RNA?
Ans. Coding RNA are of only one kind i.e., mRNA and constitute about less than 4% of total RNA
in cells while noncoding RNA are of two main types viz., rRNA (most abundant type making
upto 80% of RNA in cells and are the major component of ribosomes) and tRNA (are about 61
types to transfer 20 amino acids). Other noncoding RNA are small nuclear RNA (snRNA) or U-
RNA (are rich in uridine nucleotides), transfer messenger RNA (tmRNA, which attaches to
mRNA, label wrongly synthesized peptides for immediate degradation) and small cytoplasmic
RNA (scRNA).
Q. 178. What is pre-RNA and what modifications change it to RNA?
Ans. Pre-RNA or precursor RNA is the RNA molecule which after different modification forms
different types of RNA molecules. The modifications that changes pre-RNA are
a. End modification: A single unusual nucleotide called cap is attached to 5’ end and a poly(A)
tail is attached to 3’ end.
127
b. Splicing: It is to make mRNA from un-spliced heterogeneous nuclear RNA (hnRNA) by
removing introns (rare in bacteria).
c. Cutting: it is to form rRNA and tRNA form pre-rRNA and pre-tRNA (made of more than
one unit of RNA), respectively.
d. Chemical modification: These are made to make mRNA, tRNA and rRNA through addition
of specific chemical groups to specific nucleotides. This processing of mRNA is called
‘editing’.
Q. 179. What is alternative splicing?
Ans. By the process of alternating splicing a single pre-mRNA molecule can be used to make
different mRNA molecules by the different combination of exons, it is common in eukaryotes
and Archaea.
Q. 180. What are the different kinds of chemical modification occurring with rRNA and tRNA
nucleotides?
Ans. Different types of modifications in nucleotides of pre-tRNA and pre-rRNA are:
a. Methylation: addition of one or more –CH3 group to the base or sugar, as of guanosine to
give 7-methylguanosine.
b. Deamination: Removal of amino group from the base, generation of Inosine from
guanosine.
c. Sulfur substitution: replacement of oxygen by sulfur to give rise to 4-thiouridine.
d. Base isomerization: Changing position of atom in the ring component of bas, as uridine give
rise to pseudouridine.
e. Double bond saturation: Converting double bond in uridine to give rise dihydrouridine.
f. Nucleotide replacement: As replacement of bases with queosine.
Q. 181. What are the functions of different RNA processing events?
Ans. mRNA ends are modified by 5’-end capping for transport of mRNA and initiation of
translation, 3’ polyadenylation also helps in translation initiation and mRNA turn over. Splicing
removes introns. The rRNA is modified to increase their range of activity. tRNA are modified
to recognize two or more codones by wobble tRNA, and for their recognition by aminoacyl-
tRNA synthetase. The mRNA is edited for using one mRNA for coding two or more proteins or
to place termination codon (in mitochondrial mRNA) or to convert abbreviated transcripts into
functional RNAs.
128
Q. 182. What are the pseudo genes and what is their role?
Ans. A pseudo gene is a nonfunctional gene which has become defunct, non-readable due to some
mutation. These are of two types, conventional pseudo genes (a functional gene inactivated due
to accumulation of deleterious mutations) and processed pseudo genes (arises by reverse
transcription of mRNA of a functional gene followed by integration of cDNA back into
chromosome). Processed pseudo genes, as they are made from mRNA, lack the transcription
unit, i.e., lack promoter however those derived from genes transcribed by RNA polymerase III
can be transcribed latter on also due to presence of internal promoters e.g., Alu elements.
Another important difference between two types of pseudo genes is absence of introns in
processed pseudo genes.
Q. 183. What are the mechanism of mRNA degradation in bacteria and yeasts?
Ans. In bacteria, mRNA is degraded by a multienzyme complex, degradosome, which has atleast
an endonuclease to cut the mRNA from inside and an exonuclease to chew it up from ends.
In yeasts, mRNA is degraded by two mechanisms:
a. De-adenylation pathway: it is triggered by removal of poly(A) tail either by exonuclease or
by loss of polyadenylate binding protein which stabilizes the tail, it is followed by
decapping from 5’ end and then are rapidly digested by 5’ end chewer exonucleases.
b. De-adenylation independent pathway: it is induced by presence of termination codone in
mRNA at wrong place to inhibit its translation, cap cleavage and exonuclease degradation
occurs even when poly(A) tail is intact. It is mostly to degrade mRNA which has got altered
due to mutation or incorrect splicing.
Q. 184. What is Pan-editing and what role it plays?
Ans. Pan editing involves extensive insertion of nucleotides into abraded RNA to make it a
functional RNA, it is common in mitochondria of trypanosomes having crypto genes (genes
lacking many nucleotides present in mature mRNA) which encode for pre-RNAs. These pre
RNAs are usually processed by insertion of many U nucleotides at the positions defined by
short guide RNAs.
Besides, insertional editing is also seen in some viral genes (Paramyxo virus P gene),
insertion of Gs at specific positions in mRNA, however here guide RNAs are not needed and
insertions are made by RNA polymerases during synthesis itself.
It is also seen with other mitochondrial genes as five genes of human mitochondria
produces mRNA without any termination codone and ending with U or UA, polyadenylation
covert them to UAA to create termination codone.
129
Q. 185. Describe the clover leaf structure of tRNA?
Ans. Following are the features of clover leaf structure of tRNA (most of tRNA can make clover
leaf structure except a few of vertebrate mitochondria, human tRNASer has no D arm):
a. Acceptor arm: Formed by 7 base pairs between 5’ and 3’ ends of the molecule. The
aminoacid attaches to A of invariant CCA terminal sequence at extreme 3’ end of tRNA.
b. D arm: Named due to presence of dihydrouridine, a modified base.
c. Anticodone arm: Contain triplet of nucleotide called anticodone, it base pairs with mRNA.
d. V loop: It contains 3-5 nucleotides in class 1 and 13-21 nucleotides in class 2 tRNAs.
e. TψC Arm: it contains sequence thymidine-pseudouridine-cytosine.
Q. 186. What are the main features of aminoacyl-tRNA synthetase?
Ans. They are two types their characters are:
Characteristic features Class I enzymes Class II enzymes
Active site of enzyme Parallel β-sheet Anti-Parallel β-sheet
Interact with tRNA at Minor grove of acceptor arm Major grove of acceptor arm
Orientation of bound tRNA V lop faces away from enzyme V loop faces enzyme
Aminoacid attachment With 2’-OH group of terminal
nucleotide of tRNA
With 3’-OH group of terminal
nucleotide of tRNA
Q. 187. What do you understand with ‘wobble’?
Ans. Non-standard base pairs can form between the third nucleotide of the codone and the first
nucleotide of the anticodone. This is called wobble. As uridine (U) can base pair with G at third
nucleotide of a codone and Inosine (I) can base pair with A, C, or U.
Q. 188. What is the importance of Kozak consensus sequence?
Ans. Kozak consensus sequence 5’-ACCAUGG-3’ contains initiation codone AUG and helps the
geneticists in determining the ORF in most of the fragmented genes found in eukaryotes. It is
often required for proper expression of cloned gene, i.e., protein expression in prokaryotic
vectors.
Q. 189. What do you mean by Pharming? Name the common plants used in biopharming.
Ans. The word ‘Pharming’ has come from ‘Farming’ and ‘Pharmaceuticals’ which means farming
the pharmaceutically important chemicals through genetically modified plants and farm
animals. The process of pharming in animals and plants is also known as gene pharming and
biopharming, respectively.
130
Common plants used in biopharming are genetically modified tobacco, potato, barley, rice,
corn, banana, alfalfa, safflower, soybean, sugarcane, tomato, spinach, lettuce etc. However,
corn is the most commonly used crop.
Q. 190. What are Plantibodies?
Ans. Plantibodies are the human antibodies produced in plants.
Q. 191. Give some examples of gene pharming and biopharming. What are the major issues with
the products of pharming?
Ans. The different examples of pharming are listed below. The major issue involved are technical
(can be overcome in due course of time with technological advancement), ethical and social
(arising mainly with the use of different drugs and even organs for transplantation grown in
animals, rouge proteins like those causing mad-cow disease), oral tolerance to different antigens
may lead to unknown problems. Gene flow from GM crops to farmed crops and wild flora and
GM animals to other animals may complicate the evolution and change the environment.
Gene Pharming Products Produced in
Haemoglobin Transgenic pigs (New Jersey)
Factor IX (a blood clotting protein Transgenic sheep (Scotland)
Malaria antigen for immunization Transgenic goat
Human Calcitonin Transgenic rabbit
Human albumin, collagen I and II Transgenic cows (USA)
Tissue plasminogen activator (tPA) Transgenic pig, sheep, goats
Antithrombin III (AT III) Transgenic goats
Antibodies to harness cancer Cow
Insulin Cow
Growth hormone Cow
Bio-Pharming Products Produced in
(Plant made pharmaceuticals, PMPs)
Sperm antibodies to be used in contraceptive jellies Contraceptive corn
Hepatitis B proteins and malaria proteins GM corn
Measles proteins Rice
Corona virus Tomatoes
Rabies vaccine Spinach
Cytokines and chemokines Different plants
131
Q. 192. Caenorhabditis elegans has emerged as a good model for multi-cellular eukaryotic
development t and for many other studies, why?
Ans. Sydney Branner in 1960 started the use of this microscopic nematode as model for many
experiments due to following qualities: Can be grown in laboratory, has short life cycle, short
generation time, transparent at all stages of cycle and permit internal examination without
killing the worm, every cell division is precisely charted and mapped, complete connectivity of
302 cells making its nervous system is known. Its genome is just 100 Mb with known sequence.
Q. 193. What are the specific properties of Archaeal DNA polymerase?
Ans. Archaeal DNA polymerase is special in the sense that it has DNA synthesis activity similar to
eukaryotic DNA polymerase subunit δ, whereas it’s proof reading action is similar to ε unit of
E. coli DNA polymerase III.
Q. 194. What do you mean by Holliday model, what for it is used?
Ans. Holliday model refers to general or homologous recombination between two homologous or
partly homologous dsDNA molecules with identical or nearly identical sequences; or a single
molecule having two separate regions of homology. In this process heteroduplexes are formed
for exchange of polynucleotides between two regions of homology and gaps are then sealed by
DNA ligase, three dimensional chi forms produces different results on cuts in different planes.
Q. 195. What may be role of junk DNA or noncoding DNA present in most of the cells on earth?
Ans. Possible functions of junk DNA may be; a. Genome organization, b. broad ranging control
functions, however nothing is known about its exact role.
Q. 196. Why so much junk DNA is present in most of the cells?
Ans. There may be two reasons; a. either this so called junk DNA has some important role not
known to biologists or b. it is a parasitic DNA (selfish DNA) maintained just because of no
selection pressure to get rid off it.
Q. 197. Name a few nonstandard amino acids present in cells.
Ans. Besides 20 standard amino acids, about 300 more amino acids have been identified in cells, a
few more important amino acids in cells including, a. 4-hydroxyproline (a proline derivative)
found in plant cells and collagen of connective tissues, N-methyl-lysine (a derivative of
lysine)- found in myosin of muscles, γ-carboxyglutamate found in prothrombin and other
Ca++ binding proteins, desmosine- a derivative from four residues of lysine, it is found in
132
elastin and selenocystine found in ausation peroxidase, contains selenium in place of
oxygen in serine. Citrulline is a key intermediate along with ornithine in biosynthesis of
arginine and in urea cycle.
Q. 198. A DNA molecules yields 36% A, 23 % T, 20% C and 21 % G, what conclusion you can
draw from it?
Ans. Neither A and T nor G and C matches in concentration indicating that there is no base pairing
and DNA is ssDNA, might be of some bacteriophage DNA.
Q. 199. Name the component proteins of E. coli replication fork and their functions.
Ans. The important proteins of replication fork and their functions are:
Proteins M Wt.
kDa Subunits Functions
SSB 75.6 4 Bind to ssDNA
Protein I (Dna T protein) 66 3 Primosome constituent
Protein n 28 2 Primosome assembly and function
Protein n’ 76 1 Primosome constituent
Protein n” 17 1 Primosome constituent
Dna C protein 29 1 Primosome constituent
Dna B protein (Helicase) 300 6 DNA unwinding, Primosome constituent
Dna G protein (primase) 60 1 RNA primer synthesis, Primosome constituent
DNA polymerase III 900 2 X 10 Progressive chain elongation
DNA polymerase I 103 1 Filling of gaps and primer excision
DNA ligase 74 1 Ligation of Okazki fragments
DNA topoisomerase II
(gyrase)
400 4 Supercoiling
Rep (helicase) 65 1 Unwinding
DNA helicase II 75 1 Unwinding
DNA topoisomerase I 100 4 Relaxing negative supercoils
Q. 200. What are ribozymes how they loss their activity or can be inhibited?
Ans. Ribozymes are the RNA molecules which catalyzes the biotransformation reactions. Some
examples of ribozymes are self splicing group I introns, peptidyl transferase and Rnase P.
Q. 201. How one can determine the diverse groups of pathogen without using cultivation
techniques? Or what are the methods which can be used to detect uncultivable pathogens?
Ans. In past decade, culture-independent molecular methods of microbial identification and
characterization have been developed and applied in the context of microbial ecology. Several
of these techniques involve the use of rRNA gene sequences as tools for species identification
by means of phylogenetic sequence analysis. rRNA genes can be amplified by PCR directly
133
from mixed-community DNA preparations (e.g., representative of both host and resident
microbes in the case of clinical samples) and cloned, and individual clones can be sequenced.
The occurrence of a particular rRNA gene sequence in a clone library indicates that the
organism that carries this sequence is present in the sampled community. On the basis of rRNA
sequence comparisons, species-specific DNA or RNA hybridization probes can subsequently be
designed and used to enumerate the various types of organisms present in a sample. Several
laboratories have used rRNA-based molecular techniques to identify and characterize human
pathogens and commensals.
Q. 202. Give salient features of rRNA analysis to detect uncultivable pathogens.
Ans. The technique involves three steps viz., PCR, cloning, and sequence analysis. Small-subunit
(SSU) rRNA genes are amplified from DNA samples by PCR with one of two sets of primers
(i) bacterial genes are amplified with the primers 8F (5’-AGAGTTTGATCCTG GCT CAG) and
805R (5’-GACTACCAGGGTATCTAAT), and (ii) all SSU rRNA genes in a sample are
surveyed with the universal primers 515F (5’-GTGCCAGCMGCCGCGGTAA) and 1391R (5’-
GACGGGCGGTGWGTRCA). DNA fragments are excised from agarose gels and purified. A
portion of each PCR product is cloned into suitable vector (TA cloning kit). For each clone
library is constructed, transformants are grown overnight at 37°C in a 96-well culture plate.
Screen for positive transformants. Aliquot of culture supernatant is used as a template in PCR
with vector-specific primers. Part of PCR product is separated by agarose gel electrophoresis in
order to screen for cloned inserts. The remainder of the PCR product is digested with 2 U each
of MspI and HinPI restriction enzymes for 3 h at 37°C. The restriction fragments are separated
by gel electrophoresis. Restriction fragment length polymorphism (RFLP) types can be sorted
by visual inspection or with computer aided programmes. Representative RFLP types are
sequenced. The majority of both strands of DNA are sequenced in all instances. Initial microbial
species identifications are made by a batch BlastN search. SSU rRNA gene sequences are
aligned to an existing database of rRNA gene sequences using the computer application.
Phylogenetic analysis can be done for phylogenetic-tree estimations. The percent sequence
identities are uncorrected.
Q. 203. What are different methods used for tree construction in bacterial genetics?
Ans. Two common methods used for tree construction in bacterial genetics are distance
measurement and evolutionary parsimony. Besides, use of genome signature and protein
signature are important methods for drawing tree structure.
134
Q. 204. What are the advantages of using rRNA gene analysis for determining phylogeny?
Ans. The 16S rRNA gene, encoding for one of the three RNA molecules of the ribosome, is the
most widely used DNA sequence for the detection and phylogenetic analysis of
microorganisms. The major advantages of the 16S rRNA gene are:
• It is generally highly conserved throughout the living world. Universal PCR primer sets
exist which can amplify the 16S rRNA gene from the overwhelming majority of Bacteria,
Archaea and Eukaryotes, respectively.
• It contains more and less conserved regions allowing for the design probes specific for
higher and lower taxons.
• There is an apparent lack of lateral transfer of the rRNA genes.
• During evolution, the 16S rRNA gene is believed to have changed at a fairly constant rate.
Thus it can be considered as an evolutionary clock with each nucleotide difference
translating to an evolutionary time unit.
Thus, the ~1500 bp sequence, amplified by universal primers carries enough information to
predict the phylogeny of the host organism with high precision. An extensive and rapidly
growing database exists for this gene. Some important databases are ARB, RDP and NCBI.
The intergenic sequence (IGS) between the 16S / 18S (for prokaryotes) and the 23S / 28S
(for eukaryotes) rRNA genes is highly variable thus can be used for strain differentiation.
Conserved regions of the 5’ end of the 16S rRNA and of the 3’ end of the 23S rRNA genes
enable the design of universal primers for the IGS region.
Q. 205. What are the genes other than rRNA which are common and conserved in bacteria? Give
some consensus primers used for such genes.
Ans. The following genes are common to all bacteria and are more or less highly conserved. Thus,
they offer an alternative to the use of the 16S rRNA gene in analyzing the total bacterial
diversity. They are present in only a single copy, providing a significant advantage over 16s
rRNA which may be present in up to 15 copies (i.e. in Clostridium paradoxum).
1. rpoB gene (1300) (Dahllof et al., 2000, Drancourt et al., 2004, Mollet et al., 1997) –
encoding the RNA polymerase beta subunit. Resolution goes below the species level.
2. gyrB gene (2200) (Fukushima et al., 2002) encoding the DNA gyrase (topoisomerase II) B
subunit. There is a curated gyrB database maintained at the Marine Biotechnology Institute
in Japan. Resolution goes below the species level.
3. gyrA gene (550) (Brisse and Verhoef, 2001) encoding the DNA gyrase (topoisomerase II) A
subunit. Resolution goes below the species level.
135
4. tmRNA (300) (Zwieb et al., 2003), also called 10Sa RNA is a catalytic RNA species
employed in a trans-translation process to add a C-terminal peptide tag to the incomplete
protein product of a broken mRNA. There is a curated tmRNA database maintained at the
University of Texas Health Science Center. Reliable resolution at the species level; in many
cases, even below that. The tmRNA is a naturally amplified within cells to 100-10.000
molecules per cell, enabling its application in FISH as well.
5. recA (1600) (Lloyd and Sharp, 1993) encoding RecA, a small protein (352 amino acids in
Escherichia coli), implicated in homologous DNA recombination, SOS induction, and DNA
damage-induced mutagenesis. Resolution goes below the species level.
6. EF-Tu (tuf) gene (300) (Ludwig et al., 1993, Baldauf et al., 1996) – encoding the bacterial
elongation factor EF-Tu. Resolution goes below the species level.
7. groEL (cnp60, hsp60) gene (1800) (Wong et al., 2002) – encoding the heat shock protein
GroEL/Hsp60. High resolution (below species level).
8. atpD gene (150) (Ludwig et al., 1993) – encoding beta-subunit of bacterial F1F0 type ATP-
synthases. Its resolution power is similar to that of the 16S rRNA gene.
9. ompA gene (350) (Wertz et al., 2003) – encoding OmpA (outer membrane protein A). High
resolution (below species level).
10. gapA gene (150) (Wertz et al., 2003) – encoding the ausationydes-phsophate
dehydrogenase. High resolution (below species level).
11. pgi gene (350) (Wertz et al., 2003) – encoding the glucose-6-phosphate isomerase. High
resolution (below species level).
12. Santos and Ochman have published consensus PCR primer pairs for fusA, gyrB, ileS, lepA,
leuS, pyrG, recA, recG, rplB and rpoB.
Other (often referred to as functional, i.e. encoding for related enzymes carrying out a
defined function) genes meeting the above conditions at least for a narrower group of
microorganisms, can also be used in molecular microbial ecology. The application of functional
genes narrows down the analysis to a functionally (sometimes also phylogenetically) defined
group of microbes. The main advantage of this approach is, that it enables the detection and
analysis of microbial groups with no cultivated members. Examples are:
1. pmoA/amoA (1700) (Bourne et al., 2001; Radajewski et al., 2000) – The pmoA gene
(encoding for the active site subunit of the particulate methane monooxygenase) of
methanotrophic bacteria and the evolutionarily related amoA gene (encoding for ammonia
monooxygenase) of aerobic ammonia ausation is widely used for the analysis of
methanotrophic and/or nitrifying bacterial communities. Conserved primer pairs that enable
136
the parallel amplification of nearly all pmoA/amoA and related genes from environmental
samples exist. Their use has led to the discovery of functional groups of bacteria with novel
pmoA/amoA related genes. Members of these groups have not yet been cultivated. They
may play a role in the oxidation of atmospheric methane.
2. mmoX (90) (Auman et al., 2000) – Encoding for one of the proteins of the soluble methane
monooxygenase (present in some of the methanotrophs only). This gene has been used to
complement the results of pmoA/amoA analyses of environmental samples.
3. mxaF (180) (McDonald and Murrell, 1997) – The mxaF gene encodes for the large subunit
of the methanol dehydrogenase of aerobic methanol ausation bacteria. It has been used for
the analysis of methanol ausation communities as well as for confirming the results of
pmoA/amoA sequence analyses on methanotrophs.
4. nifH (2000) (Lovell et al., 2001;Widmer et al., 1999) – The nifH gene encodes the
nitrogenase iron protein. It was used for the analysis of nitrogen fixing bacterial populations
in various environments.
5. nirS (400) (Priemé et al., 2002; Rösch et al., 2002) – The nirS gene encodes the cytochrome
cd1-containing nitrite reductase of denitrifiers, enabling the analysis of this functional
group.
6. nirK (240) (Priemé et al., 2002; Rösch et al., 2002; Avrahami et al., 2002) – The nirK gene
encodes the Cu-containing nitrite reductase of denitrifiers, enabling the analysis of this
functional group.
7. norB (90) (Ren et al., 2000) – The norB gene encodes for the nitric oxide reductase.
Bacterial populations able to reduce nitric oxide can be analysed.
8. mcrA (350) (Lueders et al., 2001) – Encoding for the methanogen-specific methyl-
coenzyme M reductase a-subunit, this gene is used for the analysis of methanogen
communities. Recently, a novel Archaeal lineage was revealed by analysis of mcrA
sequences from rice field soil.
9. rbcL (14.000) (Wyman et al., 2000) – The rbcL gene encodes the RuBisCo (ribulose
bisphosphate carboxylase/oxidase) large subunit, present in many autotrophic bacteria. It
was used to follow, and also to predict, the changes in phytoplankton communities,
including phytoplankton booms. Most of the entries in GenBank are from chloroplasts of
different plant hosts.
Primers Sequence (5’-3’) Direction Taerget gene
AGPROTM1 GCTGGSGGGAKTTGAACC rev tmRNA
SGPROTM1 GGGGCTGATTCTGGATTCG fw tmRNA
fusAF CATCGGCATCATGGCNCAYATHGA fw fusA
fusAR CAGCATCGGCTGCAYNCCYTTRTT rev fusA
137
gapAfw AGAACATCATCCCGTCCTCTACC fw gapA
gapArev CCAGAACTTTGTTGGAGTAACC rev gapA
groELfw GACGCTCGYGTRAAAATGCTSC fw groEL
groELrev GCAGTGCAACTTTGATACCCACG rev groEL
gyrAfw ATGAGCGACCTTGCGAGAGAAAT fw gyrA
gyrArev CTCGTCACGCAGCGCGCTGATGCC rev gyrA
gyrBBAUP2 GCGGAAGCGGCCNGSNATGTA fw gyrB
gyrBBNDN1 CCGTCCACGTCGGCRTCNGYCAT rev gyrB
ileSBCUP1 GCCCGGCTGGGAYWSNCAYGG fw ileS
ileSBKDN1 TGGAGCCGGAGTCGAWCCANMMNTC rev ileS
lepABAUP1 CATCGCCCACATCGAYCAYGGNAA fw lepA
lepABAUP2 TGCATCATCGCCCACRTNGAYCAYGG fw lepA
lepABIDN1 CATGTGCAGCAGGCCNARRAANCC rev lepA
leuSBKDN1 GGGGCAGCCCCARWANCKYT rev leuS
leuSF GAGACCGTGCTGGCCAYGARSARRT fw leuS
ompAfw AAAGCTCAGGGCCTTCA fw ompA
ompArev GCGGCTGAGTTACAACGTCTTT rev ompA
pgi_fw GAGAAAAACCTGCCTGTACTGCTGGC fw pgi
pgi_rev CGCGCCACGCTTTATAGCGGTTAAT rev pgi
pyrGBAUP1 GGCGTGGTGTCCTCCNTNGGNAARGG fw pyrG
pyrGBDDN2 GGAAGGGCAGGCACTCNATRTCNCCNA rev pyrG
recABDUP1 CCCGAGTCCTCCGGNAARACNAC fw recA
recABGDN2 CGTTGCCGCCGGKNGTNRYYTC rev recA
recABHDN1 GAAGGGTGGGGCCANYTTRTTYTT rev recA
recAF TNGARATHTAYGGICCIGARTC fw recA
recAR ACNACYTTNACICGIGTYTCRCT rev recA
recGBHUP2 GGGCGACGTGGGCDSNGGNAARAC fw recG
recGBMDN1 GGGTCCGGGGGATNGGNGTNGC rev recG
RL1 GATGATATCGATCAYCTDGG fw rpoB
RL2 TTCVGGCGTTTCAATNGGAC rev rpoB
rplBBDUP1 CAAGGTGGAGCGCATCSANTAYGAYCC fw rplB
rplBBHDN1 GCCGCCGCCGWDNGGRTGRTC rev rplB
rplBR CGCCGCCGCCGWRNGGRTGRTC rev rplB
rpoBBDUP1 GGGCACCTTCATCATCAAYGGNDBNGA fw rpoB
rpoBBDUP4 CATGGGCGACATCCCNHWNATNAC fw rpoB
rpoBBJDN2 CCGATGTTCGGGCCYTCNGGNGTYT rev rpoB
rpoBBJDN3 GATGTTCGGGCCCTCNGGNGTYTC rev rpoB
UP1 GAAGTCATCATGACCGTTCTGCAYGCNGGNGGNAARTTYGA fw gyrB
UP2
AGCAGGGTACGGATGTGCGAGCCRTCNACRTCNGCRTCNGT
CAT rev gyrB
Q. 206. Give consensus sequences for rRNA gene analysis of Archaea and prokaryotes or for both
(universal).
Ans. Some commonly used sequences are:
Primers Sequence (5’-3’) direction Used for
f27 AGAGTTTGATCMTGGCTCAG fw eubacterial
8f AGAGTTTGATCCTGGCTCA fw eubacterial
1492rpl GGTTACCTTGTTACGACTT rev eubacterial
r1492 TACGGYTACCTTGTTACGACTT rev eubacterial
A8F TCCGGTTGATCCTGCCGG fw Archaeal + eukaryotic
E528f CGGTAATTCCAGCTCC fw eukaryotic
U514f GTGCCAGCMGCCGCGG fw universal
U1492r ACCTTGTTACGACTT rev universal
38r CCGGGTTTCCCCATTCGG rev universal – 23S/IGS
72f TGCGGCTGGATCTCCTT fw universal – 16S/IGS
G1 GAAGTCGTAACAAGG fw IGS
L1 CAAGGCATCCACCGT rev IGS
138
Q. 207. What are different mechanisms of action of some common antibiotic resistance in bacteria?
Ans.
Antibiotics Mechanism of action Mechanism of resistance by bacteria
Amino-glycosides
(Streptomycin, kanamycin,
gentamicin, spectinomycin)
Bind to ribosome to inhibit
protein synthesis
Alteration in target sites in ribosomes as
mutation in ribosomal protein S12 leads to
resistance to spectinomycin,
Plasmid or transposon encoded resistance is due
to enzymes viz., amino-glycoside-O-
nucleotidyl transferase, amino-glycoside-N-
acetyltranferase, and amino-glycoside-O-
phosphotransferase
Ampicillin Β-lactam antibiotics inhibit
to penicillin binding
proteins required for
peptidoglycan synthesis
Mutational alteration in penicillin binding
proteins, chromosomally encoded Β-lactamase
(Pseudomonas aeruginosa and Klebsiella
pneumoniae)
Lack of permeability to Β-lactams.
Plasmid and transposon mediated Β-lactamases.
Β-lactamase is a periplasmic enzyme prone for
leakage.
Chloramphenicol Diffuses into bacterial cells
to inhibit peptidyl
transferase activity.
Mutational alteration in outer cell membrane,
Plasmid and transposon mediated resistance is
due to production of chloramphenicol acetyl
transferase.
Tetracycline Diffuses into bacterial cells
to inhibit binding of
aminoacyl-tRNA to
ribosome a site.
No mutational resistance but Plasmid and
transposon mediated resistance is due to
production of membrane bound carrier protein
(tetA permease) that catalyses efflux of
tetracycline.
Q. 208. How can you select a tets bacterium from tetr culture?
Ans. Tetr cells are sensitive to lipophilc chelators like fusaric acid in presence of Zn++ which
provide chance to select Tets colonies of bacteria.
Q. 209. What is resolution power of gels, how concentration of Agarose and acrylamide affect the
resolution?
Ans. Resolution power is the ability of a gel to differentiate between two or more DNA molecules
of different length. It is affected by the concentration of agarose as follows;
Concentration % Range of linear dsDNA fragments to be
resolved
Agarose 0.5
0.8
1.5
3.0
2-20 kb
0.6-15 kb
0.2 – 4 kb
0.08-1.5 kb
Arcylamide concentration
5
12
20
80-500 bp
40-200 bp
10-100 bp
139
Q. 210. Name some signals for virulence regulatory mechanisms of bacteria.
Ans.
Signal Potential mechanism
Carbon limitation Activate transcription by directly interacting with RNA polymerase,
respond to cAMP levels
Iron starvation Represses transcription by interfering with RNA polymerase binding (Fur)
Supercoiling Respond to temperature or osmotic stress (HNS)
Temperature, ionic
conditions, pH, nicotinic
acid
Activate two component system, transmembrane sensor-mediated
phosphorylation of regulator activates virulence gene expression by
binding upstream promoters (BvgAS in Bordetella and PhoPQ in
Salmonella)
Temperature, pH, osmolarity,
amino acids
ToxR transmembrane transcriptional activator stabilized in active
conformation by ToxS (ToxRS in Vibrio)
Autoinducer Homoserine –lactone molecules detected by membrane sensors activate
gene expression in response to bacterial population density.
Gene rearrangement Leads to variation in expression locus, which may results to production of
no product (antigenic variation in Neisseria)
DNA inversion Leads to phase variation in gene expression (H1/H2 antigens of
Salmonella)
Replication slippage Leads to altered reading frame, sometimes causing frame shift mutations
and loss of protein expression (P protein of Neisseria)
Q. 211. What for drop dialysis, spin column purification are performed?
Ans. Drop dialysis and spin columns are used to remove impurities from DNA.
Drop dialysis can be done as follows:
1. Pour about 10 ml of TE in sterile Petri dish and place with sterile forceps a 25 mm
Millipore filter nylon membrane with shiny side up in it.
2. After 2-3 min, till the membrane is wet, place 20-100 µl of DNA in the centre of filter and
cover the Petri dish carefully without disturbing any thing.
3. Carefully remove the DNA from the membrane after 1-2 hr of incubation at room
temperature.
Spin column is prepared with Sephadex G50 as follows:
1. Hydrate 10 g resin in 100 ml TE and autoclave for 15 min, cool to RT and allow to settle,
decant supernatant, and add fresh TE and store at 4oC till use.
2. Take a 500 µl microfuge tube and make hole in bottom with 23 gauge needle, put some
glass bead (212-300 µm) of glass wool in tube. Swirls the slurry of matrix and transfer 600
µl to the tube and make the column to flow freely.
3. Place the column in 1.5 ml tube and centrifuge at 3000 rpm for 2 min and then transfer the
column to fresh tube, apply the 25-30 µl of DNA to be purified and spin as above for 2 min
to recover all the volume loaded. It will contain pure DNA.
140
Q. 212. N15 labeled DNA of T4 phage is mixed with an equal amount of T4 DNA of normal
density. The solution is then heated to denature and allowed to renature, how many species of
DNA molecules can be detected with CsCl2 density gradient and in what proportion?
Ans. Three types of DNA species ratio of 1 N14 N15:2 N14 N15: 1 N15 N15, because both type of
DNA will hybridize to reanneal randomly.
Q. 213. What evidence supports the notion that the loops of a bacterial chromosome are in some
way isolated from one another?
Ans. A single break in chromosomal DNA leads to loss of supercoiling in a part of DNA but rest of
the molecule still remains in the supercoiled state.
Q. 214. What for these refers; Gly-tRNAgly, GlytRNAleu, tRNAgly?
Ans. Gly-tRNAgly, it referes to acylated tRNA for glycine with glycine or charged; tRNA tRNAgly,
it refers to nonacylated or uncharged tRNA for glycine, GlytRNAleu, refers to mischarged
tRNA, that carrying wrong amino acid.
Q. 215. What are advantages of classical genetic approach in understanding microbial genetics?
Ans. On using classical genetic approach 1. Mutants can be isolated and characterized without any
a prior understanding of the molecular basis of the function, 2. Number genes involved in
control of a fuction can be predicted, 3. Interacting (physically or functionally) genes can be
identified.
Q. 216. What are limitations of PCR?
Ans. A prior knowledge of the sequences of the primer annealing sites must be known to amplify a
DNA sequence of any length. Another problem is the length of amplicon, regions >5 kb are
difficult to amplify.
Q. 217. Compare the transcription and translation processes to differentiate between bacteria and
archaea.
Ans. Transcription and translation in archaea are more similar to these processes in eukaryotes than
in bacteria. Archaean RNA polymerase and ribosomes are similar to those eukaryotes. Archaea
only have one type of RNA polymerase in contrast to many types in eukaryotes but its structure
and function in transcription is similar to that of the eukaryotic RNA polymerase with similar
transcription factors directing the binding of the RNA polymerase to a promoter of gene.
However, several other archaean transcription factors are similar to those of bacteria. Post-
141
transcriptional process of archaea is simpler in eukaryotes. Most most archaean genes, similar
to those of bacteria, lack introns but there are many introns in their tRNA and rRNA genes nd a
few other protein encoding genes.
Q. 218. What is LCR and what for it is used? How it differ from PCR.
Ans. Ligase chain reaction (LCR) is a technique adapted and applied to rapid discriminatory
detection alleles. LCR uses a thermostable ligase and can discriminate DNA sequences
differing in only a single base pair. For LCR we need four oligonucleotides (two pairs of
oligonucleotides or primers) to amplify the target sequence. One pair of oligonucleotides is
complementary to one strand of the target DNA sequence and the second pair is complementary
to the first pair. Two oligonucleotide primers of each pair are hybridized to denatured target
DNA so that the 3'-end of one primer is next to the 5'-end of the other primer. DNA ligase links
the primers covalently if the nucleotides at the junction are perfectly base-paired to the target.
Thus mutation of a single-nucleotide (deletion or substitution or insertion) is detected.
LCR differs from PCR as it uses the probe molecule rather than producing amplicon
through polymerization of nucleotides. Two probes are used per each DNA strand and are
ligated together to form a single probe. LCR uses both a DNA polymerase enzyme and a DNA
ligase enzyme to drive the reaction. Like PCR, LCR is also conduced on a thermal cycler to
drive the run and each cycle results in a doubling of the target nucleic acid molecule. LCR
usually have more specificity than PCR.
Q. 219. Write a note on amplification inhibiting oligonucleotides.
Ans. Some oligonucleotides from double-stranded to single stranded or in some cases vice versa in
a temperature-dependent manner act as inhibitors of amplification. Reversible inhibitory
property might be due to change in conformation of at least a portion of the primer or oligo.
They can be used as inhibitors in a hot-start PCR, to inhibit activity of the thermostable DNA
polymerase below a desired activation temperature (Tact). The inhibitor is thermally inactivated
above Tact and liberates the polymerase activity to initiate DNA amplification process. The
oligos have 6 segments linked in tandem in the order of: 1. flap; 2. 5'-Exo oligo (5-50 bases
long); 3. Exo Loop; 4. TLO (template-like oligonucleotide; 5. Pol Loop; and 6. PLO (primer-
like oligonucleotide, 8-50 bases long), in the 5' to 3' direction. These are conjugated with Taq
polymerase. An inhibitor oligo (Ni17-3) has been patented (Mao & Xin, USA), it is 5'-
T*G*G*GATATCCCTTTTCTTTCATTCTTACATATGTAAGAATGAAAGAAAA-3', is an
example of inhibitor oligo.
142
Q. 220. How the Streptomyces chromosomal replication differs from the linear replicons?
Ans. Most of the bacteria have circular chromosome with a origin of replication from where DNA
replication starts but like the φ29 phage of bacillus and adenovirus Streptomyces have linear
replicons. Similar to φ29 phage and adenovirus, Streptomyces linear plasmids and
chromosomes have covalently attached protein at their 5’ DNA ends. However, unlike
adenovirus and φ 29 and several other linear replicons having origin of replication near
terminii, Streptomyces linear DNAs (chromosome and plasmid) replicate bidirectionally from a
site located near the center of the molecule, generating 3’ leading-strand overhangs at the
telomeres. Incomplete 5’ ends of the lagging strands produced by the joining of Okazaki
fragments are then completed also, the telomeric 3’ overhangs are also repaired to produce full-
length duplex DNA molecules. Streptomyces chromosomes and its plasmids have similar
termini. Streptomyces telomeres have some similarities and differences with the telomeres of
eukaryotic chromosomes as 3’ overhangs of both types of replicons contain multiple short
repeats but Streptomyces telomeres contain inverted repeats which are absent in eukaryote
telomeres (which have long series of tandem direct repeats). Moreover, the proteins that bind to
eukaryote telomeres are not covalently attached to the terminus as in Streptomyces replicon.
143
Chapter 5. Bacteriophages
A bacteriophage (derived from Bacteria +phagein “to eat”) is virus which infects bacteria.
They are the most common organisms on Earth. Typical structure of bacteriophages means an outer
capsid enclosing genetic material which can be ssRNA, dsRNA, ssDNA, dsDNA ranging from 5 to
500kb of nucleotides in circular or linear arrangement. Their size ranges between 20 to 200 nm.
Phages are ubiquitous and can be found in all reservoirs populated by bacteria including soil, water,
air and the intestines of animals and human beings. Sea water contains up to 9×108 phages/ml and
up to 70% of marine bacteria may be having infected of phages. Phages have also been used as an
alternative to antibiotics in the former and are seen as a possible therapy against multi drug resistant
strains of many bacteria. For therapy temperate phages are useless because they may not kill
bacteria through lysogeny and only lytic phages are used for therapeutic purpose. FDA approved
(2006) using bacteriophages on cheese to kill the Listeria monocytogenes giving them GRAS status
(Generally Recognized as Safe). Phages have potential for counteracting bio-weapons and toxins of
anthrax and botulism. They can be used as spray in horticulture for protecting plants and vegetable
produce from decay and the spread of bacterial disease. Other possible applications for
bacteriophages are as a biocide for surface decontamination in hospitals, medical devices, uniforms,
curtains, even sutures for surgery. Success in veterinary treatment of pet dogs with otitis and cows
with mastitis has been reported. Bacteriophages are not always harmful to bacteria in some cases
phages in form of prophages may provide benefits to the host bacterium while they are dormant by
conferring new functions to the bacterial chromosome. This phenomenon is called as lysogenic
conversion for example, through lysogenic conversion a harmless strain of Vibrio cholerae become
a highly virulent one causing fatal cholera.
Q. 1. How phages can be classified?
Ans. Bacteriophages are viruses and are classified as per their morphology and genome structure into six
groups A to F (Bradley’s scheme).
Examples with their characteristics Type Characteristics
Name Head Tail Host
T2 100x70 110x25 E. coli
PST 126x95 115x25 Pasteurella
PBS1 120 200x22 Bacillus
A dsDNA, Elongated or regular hexagonal
head, long tail with contractile sheath
and tail fibres
P1 90 220x20 Shigella
T1 60 140x10 Escherichia
λ 54 140x7 Escherichia
B3 55 163x8 Pseudomonas
B dsDNA, hexagonal head, long tail without
contractile sheath but tail fibres are
present
MSP8 70x55 150x10 Streptomyces
T3 55 15x8 Escherichia
P22 60 7x7 Salmonella
Φ29 42x32 13x6 Bacillus
C dsDNA, Elongated hexagonal or hexagonal
head, short non-contractile tail with tail
fibers
AA-1 65 18x13 Acinetobacter
144
ΦX174 25 Escherichia
S13 26 Escherichia
D ssDNA, head with large capsomeres no tail
M20 30 Escherichia
MS2 25 Escherichia
µ2 24 Escherichia
Φcb23 22 Caulobacter
E ssRNA, head with small capsomeres no tail
7s 25 Pseudomonas
E2 Three pieces od dsRNA, a lipid containing
enveloped head with small capsomeres
Φ6 Pseudomonas
syringae
M13 800x8 Escherichia
fd 700x5 Escherichia
F Flexible filaments containing ssDNA
f1 850x5 Escherichia
G dsDNA , enveloped, no rigid head Phages of Thermoproteus tenax
According to ICTV, phages have been grouped into order Caudovirales consisting 13 families, 11
of DNA phages and 2 of RNA phages.
DNA/
RNA
Family Characteristics Examples and Host
dsDNA Plasmaviridae Enveloped, pleomorphic, released by budding, circular
supercoiled dsDNA, lipids but no capsid
Acholeplasma laidlawii
Rudiviridae Non-enveloped, lemon-shaped, helicle or rods, can exist
as plasmid/ integrated in chromosome, UV induces
release, no lysis, linear dsDNA
Sulfobus shibatae B12,
infect archae bacteria
Fuselloviridae Enveloped, lemon-shaped, circular, supercoiled dsDNA,
with short tail fibers attached to one pole. 60 nm in
diameter and 100 nm in length, pleomorphic, no capsid
Infect Archaea and
Sulphobacteria
Lipothrixviridae Rod shaped or filamentous, enveloped, lysogenize the
host, linear dsDNA
Thermoproteus tenax
Tectiviridae Icosahedral double capsid, non-enveloped, grow in
presence of R-plasmid as sex pili act as receptors for
these, linear dsDNA, tail produced for injecting DNA
PRD1 of many bacteria
Corticoviridae Lipid containing, non-enveloped, lytic infaction, circular
supercoiled dsDNA
R plasmid of
Alteromonas
Myoviridae Non-enveloped with contractile tail with isometric head T even phages of E. coli
Siphoviridae Non-enveloped with long non-contractile tail, lysogenic,
isometric head
λ phage of E. coli
Podoviridae Non-enveloped with short non-contractile tail, isometric
head
T7 phages of E. coli
dsRNA Cystoviridae Lipid containing, dodecahedral, Enveloped, segmented
genome, lytic infection, absorb on side of pili
Φ6 of Pseudomonas spp.
ssDNA Microviridae Isometric (icosahedral), non-enveloped, absorbed on pili,
extruding, circular ssDNA
Φx174 of E. coli
Inoviridae Non-enveloped, rod shaped (helicle symmetry) , released
through extrusion, circular ssDNA
fd phages of E. coli
ssRNA Leviviridae
(+ strand)
Non-segmented, non-enveloped, isometric, absorbed on
side of pili, lytic infection, quasi-icosahedral, linear
ssRNA
MS2 group Coliphages
Q. 2. What are different classes of ssRNA phages?
Ans. On the basis of immunological cross reactivity of the coat protein, buoyant density of virion
and ratio of Adenine to uracil and the amino acids which do not occur in coat protein ssRNA
phages have been divided into 4 groups. Each group produces specifc enzyme to replicate its
145
RNA which do not replicate other RNA viruses. Groups are; I (f2, MS2, R17); II (GA); III (QB)
and IV (SP, F1).
Q. 3. What do you mean by Twort-d’ Herelle phenomenon?
Ans. Twort-d’ Herelle phenomenon is nothing but action of lytic phages observed independently by
Twort (1915) on Staphylococcus colonies and by d’Herelle (1917) on Shigella colonies.
Q. 4. What are factors which lead to non-lysis of bacteria on infection with lytic Bacteriophages?
Ans. There may be many reasons important ones are: 1. loss of surface receptors as a result of
mutation, 2. presence of restriction endonucleases to which phage DNA may be susceptible, 3.
bacteria might have gone lysogenized with same or related bacteria, 4. bacteria might been
grown in conditions different than required to produce receptors, 5. phages may be having
foreign DNA (as used for transduction) not able to cause lysis. When there is lysogeny, immune
cell forms dark centred (due to growth of resistant colonies) plaques.
Q. 5. What the different receptors for T4, Φ6, ΦX174, R17, QB, K20, χ1 and PBS1 on surface of
host bacteria?
Ans. T4 Diglucosyl residue of LPS or OmpC protein
Φ6, ΦX174, R17, QB F pilus
K20 OmpF and LPS
χ1, PBS1 Flagellum
Q. 6. Name some cofactors required for adsorption of T2, λ, T4 and T6 phages onto their hosts.
Ans. Monovalent cations for T2, Mg++ for λ, Ca++ for T4, L-tryptophan for T6.
Q. 7. Do RNA phages replicate in the manner same to retroviruses of eukaryotes?
Ans. Except ΦR73, a ssRNA phage of E.coli, most of the RNA phages do not opt the mechanism
similar to retroviruses (i.e. they do not integrate into bacterial chromosome using reverse
transcriptase), rather they multiply by using RNA replicase which binds to coat cistron where 3’
end of RNA also anchor to start RNA replication
Q. 8. Lysogenic or phage conversion is important in certain bacteria to make them pathogenic, give
some examples with the virulence factor conveyed by the phage.
Ans. Few examples are:
1. Diphtheria toxin of Corynebacterium diphtheriae coded by genes on prophage β.
146
2. Botulinum toxin type C & D of Clostridium botulinum coded by genes on prophages.
3. Cholera toxin gene of Vibrio cholerae is on ΦCTX phage.
4. Serum resistance in E. coli is coded by λ phage.
5. Shock resistance in E. coli is entails pIV (a filamentous phage) infection.
6. Verotoxin of E. coli, STX phenotype in entails presence of temperate phages.
Q. 9. Name different types of transductions and their use and how they differ?
Ans. There are two types of transductions a. generalized transduction, b. specialized transduction.
Generalized transduction is often used earlier for gene mapping while specialized transduction
is used to transfer specific genes from one bacterium to another.
Generalized transduction: transducing particles are made of totally chromosomal DNA. One
allele is replaced in recipient by the transduced allele.
Specialized transduction: Transducible genes lye near to integration site of prophage DNA
and transduced particles are hybrids of phage and chromosomal DNA. Transduced DNA is
added into bacterial chromosome.
Q. 10. What is high frequency transduction?
Ans. When an integrated transducing phage is induced to enter lytic cycle it produces large number
of transducing particles in lysate and when this lysate is used to infect a sensitive culture
transduction frequency increases by about 1000 fold and it is called high frequency
transduction.
Q. 11. What is the specificity of rough specific phages?
Ans. Rough specific phage recognizes the core antigen of LPS of members of Enterobacteriaceae
as its receptor and thus is a wide host range phage.
Q. 12. What do you understand by multiplicity reactivation of bacteriophages?
Ans. If double stranded DNA phages are inactivated using ultraviolet irradiation, we often see
reactivation if we infect cells with the inactivated virus at a very high multiplicity of infection
(MOI) – this is because inactivated viruses cooperate in some way. Probably complementation
allows phages to grow initially, as genes inactivated in one virion may still be active in one of
the others. As the number of genomes present increases due to replication, recombination can
occur, resulting in new genotypes, and sometimes regenerating the wild type phages.
147
Q. 13. What is MOI, how it is determined?
Ans. The multiplicity of infection (abbreviated MOI) is the average number of phage per
bacterium. The MOI is determined by simply dividing the number of phage added (ml added x
PFU/ml) by the number of bacteria added (ml added x cells/ml). The average number of phage
per bacterium in the population could be 0.1, 1, 2, 10, etc, depending upon how you set up the
experiment.
Q. 14. What is the meaning of defective interfering particles of bacteriophages?
Ans. The replication of the helper phage may be less effective than if the defective virus (particle)
was not there. This is because the defective particle is competing with the helper for the
functions that the helper provides. This phenomenon is known as interference, and defective
particles which cause this phenomenon are known as “defective interfering” (DI) particles. Not
all defective viruses interfere, but many do. Note that it is possible that defective interfering
particles could modulate natural infections.
Q. 15. When the viruses are called pseudo types?
Ans. We can pseudo types where virus capsid (coat) of a entirely different virus contain nucleic
acid of another virus, e.g., a retrovirus nucleocapsid in a rhabdovirus envelope. This kind of
phenotypic mixing is sometimes referred to as pseudo type (pseudovirion) formation. The
pseudotype described above will show the adsorption-penetration-surface antigenicity
characteristics of the rhabdovirus and will then, upon infection, behave as a retrovirus and
produce progeny retroviruses. This results in pseudo types having an altered host range/tissue
tropism on a temporary basis
Q. 16. Give steps of propagation of bacteriophages following lytic cycle.
Ans. The steps are: Adsorption, Penetration, Un-coating of phage DNA (start of Eclipse phase),
Synthesis of phage DNA and phage proteins, Assembly/ maturation, Release of phage particles.
Q. 17. What for structural and non-structural terms stands?
Ans. All proteins in a mature virus (bacteriophage) particle are said to be structural proteins – even
if they make no contribution to the morphology or rigidity of the virion – non-structural proteins
are those viral proteins found in the cell but not packaged into the virion.
Q. 18. What are the methods for determining the concentration of bacteriophages, what are their
limitations?
148
Ans. Some methods e.g., electron-microscopy enable every virion to be counted but are not
informative about infectivity. Other methods e.g. immunoassays are a less sensitive measure of
how much virus is present, but again are not informative about infectivity. The best method-
plaque assay, measure the number of infectious virus particles.
Q. 19. What do you understand with plus and minus strand RNA viruses?
Ans. Plus stranded viruses, the virion (genomic) RNA is the same sense as mRNA and so functions
as mRNA. This mRNA can be translated immediately upon infection of the host cell.
Minus or negative stranded viruses, the virion RNA is negative sense (complementary to
mRNA) and must therefore be copied into the complementary plus-sense mRNA before
proteins can be made. Thus, besides needing to code for an RNA-dependent RNA-polymerase,
these viruses also need to package it in the virion so that they can make mRNAs upon infecting
the cell.
Q. 20. What do you mean by a phage plaque, how it is formed?
Ans. Plaques are clear zones formed in a lawn of cells due to lysis by phage. At a low multiplicity
of infection (MOI) a cell is infected with a single phage and lysed, releasing progeny phage
which can diffuse to neighboring cells and infect them, lysing these cells then infecting the
neighboring cells and lysing them, etc, ultimately resulting in a circular area of cell lysis in a
turbid lawn of cells.
Q. 21. What are the factors which decide that lytic cycle should go for a lysogenic phage?
Ans. Lytic growth is favored when cells are growing rapidly and the MOI is low.
Q. 22. What are the factors affecting plaque morphology?
Ans. The morphology of the plaque depends upon the phage, the host, and the growth conditions.
Usually phage infection is studied in a layer of soft agar (or “top agar”) which allows the phage
to diffuse rapidly. The size of the plaque is proportional to the efficiency of adsorption, the
length of the latent period, and the burst size of the phage. A diversity of plaque sizes can result
if the phage infects cells at different times during the bacterial growth phase: phage that adsorb
early make larger plaques than those that adsorb later. To avoid this, phage are often pre-
adsorbed to cells under conditions that do not allow phage growth (e.g. low temperature) then
shifted to permissive conditions to induce all of the phage to develop at the same time. A clear
plaque will be formed if the host is completely susceptible to the phage. (Often a clear plaque
will be slightly turbid at the edge because the cells at the edge of the plaque are not yet fully
149
lysed.) If the host is partially resistant to the phage then the plaque may be uniformly turbid (for
example, if 10% of the cells survive phage infection).
Q. 23. Why do plaque assays accurately measure the titer of temperate phage?
Ans. If a phage lysogenized a host cell immediately upon infection, it would never form a plaque.
Instead, when temperate phage infects a population of exponentially growing cells, each phage
produces a plaque with “bulls-eye” plaque morphology, a turbid center surrounded by a ring of
clearing. This characteristic plaque morphology is due to the role of the MOI and cell
physiology on the lysis-lysogeny decision. Lytic growth is favored when cells are growing
rapidly and the MOI is low. Lysogeny is favored when cells are growing slowly and the MOI is
high. This is why temperate phage typically has plaques with turbid centers. When the cells are
initially infected with phage the ratio of phage to cells is usually about 1 phage per 106 cells.
Initially the nutrients are plentiful so the bacteria grow rapidly and, since the MOI is low, the
phage grows lytically. After several lytic cycles the local MOI increases and most of the cells
are lysed, producing a plaque in the lawn of cells. As the cell lawn becomes saturated, the rate
of cell growth slows down and, since lysis requires rapid metabolism, the plaque stops
increasing in size. However, any lysogens that formed in the center of the plaque are immune to
lysis and can continue to grow since they do not have to compete with nearby cells for nutrients.
Thus, lysogens begin to grow in the center of the plaque, giving the plaque a turbid, bulls-eye
appearance.
Q. 24. What were the major problems with phage therapy in past?
Ans. Major Problems which puzzled the people included:
1. Paucity of understanding of the heterogeneity and ecology of both the phages and the
bacteria involved.
2. Failure to select phages of high virulence against the target bacteria before using them in
patients.
3. Use of single phages in infections which involved mixtures of different bacteria.
4. Emergence of resistant bacterial strains. This can occur by selection of resistant mutants (a
frequent occurrence if only one phage strain is used against a particular bacterium) or by
lysogenization (if temperate phages are used, as discussed below).
5. Failure to appropriately characterize or titer phage preparations, some of which were totally
inactive.
6. Failure to neutralize gastric pH prior to oral phage administration.
7. Inactivation of phages by both specific and nonspecific factors in body fluids.
150
8. Liberation of endotoxins as a consequence of widespread lysis of bacteria within the body
(which physicians call the Herxheimer reaction). This can lead to toxic shock, and is a potential
problem with chemical antibiotics as well.
9. Lack of availability or reliability of bacterial laboratories for carefully identifying the
pathogens involved, necessitated by the relative specificity of phage therapy.
Q. 25. What are the potential advantages of phage therapy? Name the diseases which can be treated
with phage therapy.
Ans. Potential advantages of phage therapy are:
1. They are both self-replicating and self-limiting, since they will multiply only as long as
sensitive bacteria are present and then are gradually eliminated from the individual and the
environment.
2. They can be targeted far more specifically than can most antibiotics to the specific problem
bacteria, causing much less damage to the normal microbial balance in the body. The bacterial
imbalance or “dysbiosis” caused by treatment with many antibiotics can lead to serious
secondary infections involving relatively resistant bacteria, often extending hospitalization time,
expense and mortality. Particular resultant problems include infection by Pseudomonas, which
are especially difficult to treat, and Clostridium difficile, cause of serious diarrhea and
membranous colitis.
3. Phages can often be targeted to receptors on the bacterial surface which are involved in
pathogenesis, so that any resistant mutants are attenuated in virulence.
4. Few side effects have been reported for phage therapy.
5. Phage therapy would be particularly useful for people with allergies to antibiotics.
6. Appropriately selected phages can easily be used prophylactically to help prevent bacterial
isease in people or animals at times of exposure, or to sanitize hospitals and help protect against
hospital-acquired (nosocomial) infections.
7. Especially for external applications, phages can be prepared fairly inexpensively and locally,
facilitating their potential applications to underserved populations.
8. Phage can be used either independently or in conjunction with other antibiotics to help reduce
the development of bacterial resistance.
The common diseases where phage therapy has been ausat effective in treatment
of human infections including staphylococcal skin disease, skin infections caused by
Pseudomonas, Klebsiella, Proteus and E. coli, P. aeruginosa infections in cystic fibrosis
patients, neonatal sepsis and surgical wound infections. Similarly, bacteriophages have also
been used to treat animal disease as mastitis.
151
Q. 26. What is the difference between the mechanism of generalized transduction by phages P22
and P1?
Ans. Both phages use a headful packaging mechanism. P22 occasionally initiates packaging of
chromosomal DNA from a pseudo-pac site. In contrast, P1 does not recognize pseudo-pac sites
on the chromosome. P1 seems to begin packaging from random double-stranded breaks on the
chromosome that are generated during phage infection.
Q. 27. Wild-type P22 cannot transduce plasmid pBR322 but P22 HT can transduce pBR322. Based
upon the way these two phage package DNA, propose an explanation for this result.
Ans. P22 only transduce DNA with pac or good pseudo-Pac sites while P22 HT recognizes and
packages DNA fragments relatively nonspecifically. Thus, the results suggest that pBR322 does
not have a pseudo-pac site so it cannot be transduced by P22, but because P22 HT can package
the DNA “nonspecifically”, pBR322 can be transduced by P22 HT.
Q. 28. What will you do to avoid obtaining lysogens that are resistant to further infection with P1
phage? What problems may arise and what steps are recommended to avoid the problem?
Ans. To avoid obtaining lysogens that are resistant to further infection, a virulent mutant of phage
P1 is often used for generalized transduction. However, this poses a second problem because
any cells that are co-infected with both a transducing particle and a phage particle will lyse,
preventing the isolation of transductants. One way to avoid lysis by the virulent phage is to mix
a solution of Ca+2 and Mg+2, recipient cells, and the phage lysate at a low multiplicity of
infection (MOI), incubate for 15-20 min, then add sufficient citrate to chelate the divalent
cations and plate the mixture on selective medium. Another way to limit lysis by the virulent
phage is to UV irradiate the lysate before mixing the phage with recipient cells.
Q. 29. The F-plasmid is about 100 Kb. Because P22 HT only packages about 45 Kb of DNA, it
cannot package the entire F plasmid. However, when a Tetracycline sensitive (TetS) recipient is
infected with P22 HT grown on a strain with a F-plasmid marked with Tetracycline resistance
(TetR), it is possible to obtain rare TetR transductants. Suggest a likely explanation for this
result.
Ans. Transduction of plasmids requires packaging of a linear concatemer of the plasmid DNA and
re-circularization of the plasmid DNA in the host cell. Thus, the simplest explanation is that the
TetR colonies arise by transduction of rare, spontaneous deletions of the F plasmid. The
deletion mutants would have to reduce the size of the F plasmid to less than 45 Kb (allowing
152
sufficient terminal redundancy for re-circularization by homologous recombination in the
recipient), without removing the TetR marker or the genes required for vegetative replication.
Q. 30. You isolated a phage from sewage that grows on E. coli. The phage acts as a generalized
transducing phage on some strains of E. coli but it acts as a specialized transducing phage on
other strains of E. coli. A. How could you distinguish generalized transduction from specialized
transduction using simple genetic tests? [Indicate any donor or recipient strains you would use
and how you would do the experiment.] b. Suggest an explanation for these results.
Ans. A. In generalized transduction – many different chromosomal markers will be transduced so
you can try transducing several different auxotrophic recipients to prototrophy with the phage.
In specialized transduction – only markers adjacent to the integrated phage will be transduced,
sometimes the resulting transductants can be detected by cross-streaking after transduction.
b. The phage is probably unable to integrate into the chromosome of strains where it does
generalized transduction. This may be due to (i) lack of a host protein that is required (e.g. IHF)
or (ii) lack of an att site or other homology required for recombination (possibly a cryptic
prophage).
Q. 31. P1 transduction was done to determine the order of the fadD, gap, and pabB genes relative to
each other. The donor was fadD gap+ pabB and the recipient was fadD+ gap pabB+. Gap+
transductants were selected. The coinheritance of the non-selected markers is shown in the table
below. Based upon these results, what is the predicted order of the genes?
fadD pabB Number of colonies
+ + 98
- - 54
+ - 19
- + 9
Ans. The rarest class of recombinants was Gap+ FadD- PabB+. The recipient marker inherited in
this strain is pabB+, suggesting that this is the middle marker. The three-factor cross data also
allows you to calculate the percent coinheritance between the selected marker and each of the
unselected markers. When selecting for Gap+ the total number of colonies obtained was 180.
The coinheritance of PabB- was 54 + 19 = 73 thus 73/180 = 41%. The coinheritance of FadD-
was 54 + 9 = 63 thus 63/180 = 35%. Thus, the order of the genes is predicted to be: fadD pabB
gap.
Q. 32. The general transducing phage P1 was grown on a wild-type bacterial strain (cys+ pyr+
trp+). The resulting phages were mixed with bacteria which require cysteine, uracil, and
153
tryptophan (Cys- Pyr- Trp-). Recombinants were selected on medium containing uracil and
tryptophan. The results are below.
Phage and cells added Number of colonies
Phage only control (108 phage) 0
Cells only control (108 cells) 4
Phage (108) + Cells (108) 200
a. Why were there no colonies on plates with phage only? B.What was the likely source of the
four colonies on the plate with recipient cells only? C.What is the predicted genotype of these
four colonies?
Ans. A. This is a control to prove that the phage were not contaminated with donor cells. If there
are no bacteria contaminating the phage, no colonies will be obtained. B. The selection was for
Cys+ colonies. Given the number of cells plated, the four colonies that arose in the absence of
phage were probably due to spontaneous revertants. C. It is unlikely that the other mutations
also reverted, so the phenotype is most likely: cys+ pyr- trp-.
Q. 33. When a generalized transducing particle enters a recipient cell, the DNA can recombine with
the recipient chromosome. Alternatively, sometimes a phage protein binds to the ends of the
transducing DNA, causing it to circularize and protecting it from nucleases. When this happens
the transducing DNA is not a substrate for the RecBCD recombination pathway and thus
remains in the cytoplasm as an “abortive transductant”. Abortive transductants typically form
tiny colonies that never grow to full size and usually fail to form colonies when picked and re-
streaked on a fresh plate. For example, when a Salmonella del(proBA) recipient is transduced
with phage P22 grown on a prototrophic strain of Salmonella selecting for growth on minimal
medium without proline, some large colonies appear and some tiny colonies appear. The large
colonies reproduce when re-streaked on the same medium, but the tiny colonies fail to
reproduce when re-streaked. Suggest an explanation for these results.
Ans. The large colonies are “true transductants” due to recombination between the pro+ alleles on
the linear transducing DNA and the chromosomal DNA, resulting in repair of the auxotrophic
mutation. The tiny colonies are due to abortive transduction. The DNA on an abortive
transducing particle can be transcribed and translated, allowing complementation of the
chromosomal mutation. However, because an abortive transducing particle does not have an
origin of replication it cannot be replicated. Every time a cell divides only one of the two
daughter cells will get a copy of the abortive transducing particle that complements the
chromosomal auxotrophy. The other daughter cell will retain a proportion of the complementing
proteins made before cell division, but it will only be able to continue growing until the gene
products are degraded or too dilute to satisfy the auxotrophic requirement. Hence, at each cell
154
division only one of the two cells will be able to continue to divide and produce daughter cells,
so instead of the cells reproduces exponentially (i.e. no. cells = 2[no. generations]) the abortive
transductant only reproduces geometrically (i.e. no. cells = 1 + [2 x no. generations]) and most
of the cells in the colony cannot reproduce when re-streaked.
Q. 34. What is the phenotype of a lambda int mutant?
Ans. An int mutant would be unable to integrate or excise but would be fully proficient for lytic
growth.
Q. 35. Which of the following genes are required for efficient lysogenization following infection of
a sensitive host by phage lambda: int, xis, N, cII, Q?
Ans. Int, N, and cII are required for efficient lysogeny, xis is only required for excision. Q is
required for expression of late genes during lysis.(Note that N is also required for expression of
lysis genes.)
Q. 36. What is the phenotype of a lambda xis mutant?
Ans. An xis mutant would be able to integrate, but unable to excise and would remain fully
proficient for lytic growth.
Q. 37. How could you isolate lambda int or xis mutants?
Ans. There are a variety of creative tricks for isolating int and xis mutants. However, the key
concept for each trick is the inability to integrate or, more often, the inability to excise. One of
the most commonly used approaches is to begin with a lambda lysogen inserted in the gal
operon. Because the insertion prevents the synthesis of gene products required for the
catabolism of galactose, this strain forms white colonies on MacConkey-Galactose plates.
However, precise excision of the prophage yields Red colonies on MacConkey-Galactose
plates. Thus, any int or xis mutants will be unable to form red colonies.
Q. 38. A lysogen of the double temperature sensitive mutant lambda cI(Ts) cII(Ts) was isolated at
30 C. What would the phenotype be if the lysogen was shifted to 42 C? Which mutation is
epistatic?
Ans. When shifted to 42 C the cI repressor would be inactivated, the lysogen would be induced, the
phage would enter the lytic cycle, and the cells would lyse. The cII protein is only required for
establishment of lysogeny, not for maintence of lysogeny or for lytic growth, so the cI mutation
would be epistatic to the cII mutation.
155
Q. 39. Most E. coli recA mutants do not prevent lambda lysogeny. However, a unique recA mutant
was isolated that permits lysogeny of lambda at 30 C but not at 42 C. What is the probable
mechanism for this observation?
Ans. When cells encounter DNA damage (e.g. by UV irradiation or treatment with mitomycin C)
wild-type RecA becomes activated to a form called RecA*. RecA* interacts with the lambda cI
repressor and induces autoproteolysis of the repressor. The resulting proteolytic fragments are
not functional for repression. Hence, one explanation for the new recA missense mutant is that
the mutation does not affect the conformation of RecA at 30 C, but it causes the RecA protein to
assume the activated RecA* conformation at 42 C. Thus, at 30 C lambda would make cI and be
able to lysogenize the mutant, but at 42 C the RecA* conformation would stimulate
autoproteolysis of cI, preventing lysogeny.
Q. 40. After hydroxylamine mutagenesis of a wild-type lambda lysate, it is possible to isolate rare
lysogens in the above recA mutant strain at 42 C. What is the probable explanation for this
result? How could you test your hypothesis? Would you expect to obtain these lambda mutants
using proflavin as a mutagen? Why?
Ans. Missense mutations in lambda cI that prevent the interaction of the repressor protein with
RecA. If wild-type cells were lysogenized with mutant phage with this phenotype, the resulting
lysogens would not be induced by UV light or mitomycin C. This phenotype would be most
likely to result from a missense mutation, so it would be unlikely that a frame shift mutagen like
proflavine could produce this type of mutant.
Q. 41.The lysis-lysogeny decision of P22 is very similar to phage lambda. A S. Typhimurium (P22
sieA) lysogen is wild-type for immunity and sensitive to super infection with other P22 phage.
When infected with wild-type P22, no plaques arise because the phages are homoimmune. Clear
plaque mutants of phage P22 fall into two categories: (i) when these mutants infect the lysogen
no plaques arise; (ii) when these mutants infect the lysogen, clear plaques are produced. What is
a likely explanation for each of these classes of clear plaque mutants?
Ans. Class (i) mutants are probably due to mutations in the proteins required for establishment and
maintenance of lysogeny (c1, c2, c3). These mutations are complemented by the repressor
protein made in the lysogen. Class (ii) mutants are probably due to mutations in the repressor
binding site. These mutations are dominant. Because the repressor made by the lysogen cannot
bind to these sites, the incoming phage will always enter the lytic pathway.
156
Q. 42. You isolated a new phage that forms turbid plaques on Klebsiella aerogenes. Describe a
simple test to determine whether the turbid plaques are due to formation of lysogens.
Ans. The turbidity in a turbid plaque arises from the growth of surviving bacteria within the zone of
lysis. These surviving bacteria can be isolated by picking from the plaque and streaking for
isolation on an agar plate. Isolated colonies are restreaked out several times to purify them away
from contaminating phage. The surviving bacteria may be phage resistant for one of two
reasons. Either they are lysogens and prevent superinfecting phage from lysing due to
repression of the lytic cycle of the entering phage by repressor produced by the resident
prophage (a phenomenon called “superinfection immunity”). Alternatively they are rare
receptor minus mutants to which the phage cannot adsorb. A simple to test determine if the
isolated colonies are phage resistant is to streak a colony at 90¡ across a plate previously
inoculated with a bead of the Boneyard phage. Sensitive bacteria will lyse upon meeting the
phage whereas resistant bacteria will continue to grow beyond the point of contact with the
phage. However, this test will not distinguish between a receptor mutant and a lysogen. To do
that, isolated colonies must be grown up in broth, induced via UV or some other DNA
damaging agent and then allowed to grow overnight. The next day the supernatant is tested for
plaque forming units on a lawn of phage sensitive K. aerogenes. Receptor mutants will not
release phage whereas the lysogens will.
Q. 43. When an Hfr carrying lambda prophage transfers its chromosome into a F- cell that does not
contain lambda, the prophage is induced and the F- cell is killed. Why?
Ans. The lambda prophage in the Hfr donor expresses the lambda cI protein from PRM. The
concentration of cI protein in the cell is maintained at a level sufficient to repress the lytic
functions of the phage. However, there is no cI protein in the F- recipient, so when the lambda
DNA enters the F- cell RNA polymerase can bind to PR and PL and express the genes required
for lysis. Because each recipient only gets one copy of the lambda DNA, the concentration of
cIII protein will be low, favoring lytic growth.
Q. 44. If a lambda cI mutant infected an hfl mutant of E. coli would you expect the phage to grow
lytically or to form a lysogen?
Ans. These two mutations have opposite effects – a lambda cI mutant will grow lytically (because
it lacks the repressor), but lambda forms lysogens at a high frequency on a hfl mutant host
(because cII accumulates at high levels). When combined, the lambda cI mutant phenotype will
be epistatic to the hfl mutant phenotype (because accumulation of cII does not affect the
lysis/lysogeny decision in the absence of cI) – that is, the phage will grow lytically.
157
Q. 45. What is transduction? Describe its types and significance in bacteria.
Ans. Transduction is the transfer of genetic information from a donor to a recipient by way of a
bacteriophage. The phage coat protects the DNA in the environment so that transduction, unlike
transformation, is not affected by nucleases in the environment. Not all phages can mediate
transduction. In most cases gene transfer is between members of the same bacterial species.
However, if a particular phage has a wide host range then transfer between species can occur.
The ability of a phage to mediated transduction is related to the life cycle of the phage.
Types of Transduction
a. Generalized Transduction – Generalized transduction is transduction in which potentially
any bacterial gene from the donor can be transferred to the recipient. Phages that mediate
generalized transduction generally breakdown host DNA into smaller pieces and package their
DNA into the phage particle by a “head-full” mechanism. Occasionally one of the pieces of host
DNA is randomly packaged into a phage coat. Thus, any donor gene can be potentially
transferred but only enough DNA as can fit into a phage head can be transferred. If a recipient
cell is infected by a phage that contains donor DNA, donor DNA enters the recipient. In the
recipient a generalized recombination event can occur which substitutes the donor DNA and
recipient DNA
b. Specialized transduction – Specialized transduction is transduction in which only certain
donor genes can be transferred to the recipient. Different phages may transfer different genes
but an individual phage can only transfer certain genes. Specialized transduction is mediated by
lysogenic or temperate phage and the genes that get transferred will depend on where the
prophage has inserted in the chromosome. During excision of the prophage, occasionally an
error occurs where some of the host DNA is excised with the phage DNA. Only host DNA on
either side of where the prophage has inserted can be transferred (i.e. specialized transduction).
After replication and release of phage and infection of a recipient, lysogenization of recipient
can occur resulting in the stable transfer of donor genes. The recipient will now have two copies
of the gene(s) that were transferred. Legitimate recombination between the donor and recipient
genes is also possible.
Significance – Lysogenic (phage) conversion occurs in nature and is the source of virulent
strains of bacteria.
Q. 46. What is the correlation between transduction frequency and distance between two genes
(markers)? What are limitations of the method?
158
Ans. For measuring distance between two given markers Wu (1966) gave a formula,
F= (1-d/L) 3
Where f= % transduction, L= length of transducing DNA in kb, d= physical distance between
two markers in kb.
Saunderson and Roth modified Wu formula in 1988 as follows;
100/f= 1 +[(L-m)3 – (L-m-d)3]/ (L-m-n-d)3
Where m is excess size of the selected donor marker in kb, n is the excess size of an unselected
donor marker in kb, f= % transduction, L= length of transducing DNA in kb, d= physical
distance between two markers in kb.
Limitations are: 1. only closely linked genes can be mapped.
2. For localization of newly discovered genes hfr strains are required.
3. Recombination hot spots (chi sequences) can influence the genetic distance.
4. Packaging sites are not purely random and are restricted to certain specific regions and may
interfere in mapping.
Q. 47. Explain specific transduction with an example.
Ans. The attachment site for phage Phi80 is between the opp operon and the trp operon on the E.
coli chromosome. Aberrant excision of the Phi80 prophage integrated at this site can yield
specialized transducing phage carrying either the opp genes or the trp genes. During process
low (LFT) and high (HFT) frequency opp transducing lysates can be produced.
Because lambda is packaged by a headful mechanism, specialized transducing phage that
carry a small region of the bacterial chromosome lack a corresponding amount of phage DNA.
The phage DNA that is missing corresponds to the opposite side I prophage genome. Phage that
carry the gal genes located on the left side of the prophage lack the phage genes required for
lytic functions (e.g. the tail genes) that are located on the right side of the prophage, and phage
that carry the bio genes located on the right side of the prophage lack the phage genes required
for lysogeny functions (e.g. the int gene) that are located on the right side of the prophage.
Specialized transducing phage that carry genes located on the left side of the prophage (e.g.
lambda dgal) are proficient for lysogeny but deficient for lysis. These phages require a helper
phage to lyse a recipient cell (e.g. a wild-type lambda phage that provides the missing functions
in trans). The “d” written in front of gal indicates that the phage is defective for lytic growth.
Specialized transducing phage that carry genes located on the right side of the prophage
(e.g. lambda bio) are proficient for lysis but deficient for lysogeny. These phage can infect a
recipient cell and generate a lysate, but require a helper phage to form lysogens a recipient cell
159
(e.g. a wild-type lambda phage that provides the missing functions in trans). (Because they have
all the functions required for lytic growth, there is no “d” written in front of bio.)
These are examples of specialized transducing phage resulting from integration of phage
lambda at its primary attachment site (attB) on the E. coli chromosome.
However, specialized transducing phage can be produced by analogous mechanisms by
phage integrated at other sites as well (for example, lambda integrated at secondary attachment
sites or other temperate phage integrated at their particular attachment sites.
Q. 48. How can you determine presence of phages in a culture of bacteria?
Ans. There are a few methods which can either determine or indicate presence of Bacteriophages in
a culture, these are;
1. Electron microscopy.
2. Plaque assay.
3. Serologically (antibodies against tail and head can be used).
4. Molecular analysis (restriction endonucleases cut the chromosome at certain places the
change can be visualized after RE analysis, PCR method can be used to detect the foreign
DNA in a specific gene).
5. Genetic analysis.
Q. 49. What do you mean by phenotypic mixing in Bacteriophages?
Ans. On infection of bacteria with two types of T-even phages (T2 and T4) at the same time,
infected bacteria may produce phage particles having head of T2 and genome of T4 or vice-
versa, this is called phenotypic mixing.
Q. 50. How the phage antibodies can be used for determining the phage types, do these antibodies
can neutralize the phage particle as in case of eukaryotic viruses?
Ans. The antibodies to head are used for deciding the group and against tail for type of phage. The
antibodies to head can agglutinate the phages but could not neutralize their activity/ infectivity.
Antibodies to head also cross react with heads of similar phages e.g. all T-even phages cross
react but no cross reaction with other groups.
Antibodies to tail result into loss of infectivity of phage but can not prevent adsorption of
phage particle to bacterial cells. Tail antibodies are more specific than head antibodies and show
no or very little cross reaction.
Q. 51. Give general composition of T4 bacteriophage of E. coli.
160
Ans. It has a typical bacteriophage structure with oblong head (78x111 nm), contractile tail (113 x
20 nm). Besides protein (gp= gene protein) subunits head contains linear dsDNA molecule
(1.21 x 106 Daltons or 1.51 x 105 bp), tail sheath contains certain polyamines (putrescine,
spermadine and cadaverine), DNA, Ca++ and ATP while base plate contain
dihydroptteroylhexaglutamate.
Q. 52. During one step growth experiment what is synchronization and how it is achieved?
Ans. The success of one step growth infection depends on synchronization because as per
requirement all bacteria should be same stage of growth and infected at the same time if with
the phage particles. It can be brought about by a. restricting the attachment of phage particles to
bacterial cells for a short time or b. first treating the bacterial cells with potassium cyanide (a
reversible metabolic poison) and then adding the virion to the culture. In latter cases phages
undergo early stage of life cycle but macromolecules could not be synthsized to proceed further
after a fixed period unattached virion particles are either neutralized or washed off. Washed
bacterial cells restore their metabolism, but all starting from the same point that is in
synchronized manner.
Q. 53. How can you determine the burst size of phage?
Ans. It is determined any of the three methods i.e., 1. One-step growth experiment to determine the
mean burst size by finding the ratio between number of plaques during plateau phase and latent
phase or 2. With the single burst experiment, in which bacterial culture is infected with low
MOI (<1) and then small aliquots (1 or 2 infected cells per sample) are take at different time to
determine the plaque numbers. The aliquots are diluted and bacteria lysed there releases phages
but these phages do not infect the bacteria due to low density and the suspension of phages can
be used to determine the pfu of phages. At any time phage can not infect (rarely infect) bacteria
if the culture has bacteria or phage density below 106/ ml due very low probability of collision
of two, 3. Premature lysis experiments, samples removed from one-step growth experiment are
lysed and lysates are determined for pfu. Immature lysis leading to lysis from without does not
yield infective phages.
Q. 54. How can you induce premature lysis in bacteria infected with phages?
Ans. There are many different methods of lysis of bacteria but those lyse the bacteria without
hurting the phage morphology is used for this purpose as. A. Shaking the bacterial cultures with
chloroform or by super infecting the bacteria with MOI (~100) as for recovering the T4 phage
161
particles from infected bacteria one can super infect culture with T6 phage to induce lysis
without yielding any T6 particles.
Q. 55. Of the following phages which is odd one out and why? Φ29, λ, P22, P2, P4, P1, Mu.
Ans. Φ29, all others are temperate phages while it is a lytic one.
Q. 56. Give some examples of temperate phages with their hosts and other characteristics.
Ans. Λ, P22, P2, P4, P1 and Mu are some important temperate phages.
Phage Host Related Dimensions Topology and morphology
λ
P22
P2
P4
P1
Mu
E. coli
Salmonella
E. coli, Shigella,
Serratia
E. coli
Φ80, 82, 424
L, Lp7, SF6
PK, 186, W Φ
P7
D108
H 60 nm,T 150x17
H 65 nm, T 18 nm
H 57 nm,T 140x17
H 46 nm,T 133x17
H 87 nm,T 226x18
H 56 nm,T 130x18
Icosahedral, noncontractle tail, cohesive ends
Icosahedral, 6 short tails, terminally
redundant
Icosahedral, contractle tail, cohesive ends
Icosahedral, contractle tail, cohesive ends
Icosahedral, cont. tail, terminally redundant
Icosahedral, cont. tail, ends from host DNA
Q. 57. What are two mechanism of insertion of prophage DNA into host chromosome?
Ans. On the basis of mechanism of insertion of prophage DNA into host chromosome temperate
phages are grouped into two types, viz., 1. Those insert their DNA as a result of single
recombination event between circular phage DNA and circular host DNA e.g. λ , P2 phage, 2.
Insertion as transposon through transposition as Mu.
Q. 58. Name some specialized transducing phages and generalized transducing phages, what are
their important characteristics?
Ans. Specialized transducing particles are always produced by integrative temperate phages e.g. λ,
Φ80, P1, P22.
Generalized transduction particles are the progeny of either virulent or temperate phages,
but bacteriophage DNA neither integrate into bacterial chromosome, nor disintegrate the host
DNA e.g. λ, P1, P22.
Q. 59. What are challenge phages, how they have been used in study of bacterial genetics?
Ans. These are derivatives of P22 with kanamycin resistant (Knr) disruption in mnt gene, an
arc(Am) mutation and substitution of any desired DNA binding site for Omnt. Thus binding of a
protein to the substituted Omnt site controls the decision between lysis and lysogeny for the
challenge phage. If no protein is bound to, ant will be expressed and host cells will lyse but in
case of protein binding, ant expression will be repressed yielding Knr lysogen.
These are used to:
162
1. For in vivo assay to determine the affinity of a DNA binding protein for a specific DNA
sequence.
2. Strong selection of mutant containing mutation in a DNA binding site that identify
nucleotide which directly interact to a DNA binding protein.
3. a strong selection for mutation in a DNA binding protein that identify the amino acid
involved in recognition of a DNA binding site.
Q. 60. How P22 modulates the super infection immunity of a lysogen?
Ans. P22 has gene systems products to grant superinfection immunity to host (lysogen),
a. The al locus of P22 modifies O-antigen of host to reduce adsorption of super infecting
phage
b. The sieA system prevents super infection by both P22 and related phages.
c. The sieB system prevents super infection by heteroimmune phage but not by P22
derivatives.
Q. 61. Why wild type T4 phage often produces small plaques (1 mm dia) while T4 mutant can
produce plaques of >2 mm in size? Explain Lysis inhibition phenomenon.
Ans. When MOI is more lysis is delayed and growth of uninfected bacteria catches up with phage
multiplication and exhaust the nutrients in media to inhibit expansion of plaques. T4 mutants
are not inhibited by high MOI and lead to formation of larger plaques.
Q. 62. T4 DNA polymerase can not distinguish between dCTP and HMC then how it prevent dCTP
to be incorporated into it DNA to protect itself T4 nuclease?
Ans. T4 encodes for an enzyme called dCTPase which degrades both dCDP and dCTP to dCMP
and make it non available for incorporation in phage DNA and side by side inhibiting
replication of bacterial DNA. Another enzyme dCMP ausation convert dCMP to dUMP
which on acquiring methyl group become dTMP thus disturbing the T:C ratio turning things in
favour of phage..
Q. 63. What mechanism help T4 DNA containing hydroxymethyl cytosine protect it from E. coli
endonucleases degrading DNA containing HMC?
Ans. Glycosylation of HMC protects T4 DNA from E. coli endonucleases activity.
Q. 64. Give a few examples where bacteriophages play some role in causation of systemic diseases.
Where they are mostly found?
163
Ans. Bacteriophages play crucial roles in the pathogenesis of many systemic diseases as botulism,
salmonellosis, dysentery, diphtheria, haemorrhagic enterocolitis by EHEC and endodontic
infection by phage carrying enterococci in human beings. Phages commonly inhabit the oral
cavity, both as free virions and as prophage in lysogenic bacterial strains, in water, swage and
almost everywhere bacteria are there.
Q. 65. Describe role of bacteriophage in failure of live Vibrio cholerae vaccines developed earlier
after deletion of ctx gene?
Ans. Earlier vaccines tested against V. cholerae having chromosomal deletion of the ctx gene
encoding cholera toxin to make the bacterium avirulent. However, the vaccine strain could
readily acquire a functional copy of the cholera toxin gene on infection with CTX phage,
causing reversion of an avirulent vaccine strain to make it a fully virulent strain.
Q. 66. What are common characters of all virulence genes transferable through bacteriophages?
Ans. Most of the bacteriophage-encoded virulence factors are bacterial toxins and proteins that
alter antigenicity. Several bacteriophage-encoded proteins such as superantigens their effectors
translocated by type III secretion system.
Q. 67. Enlist a few phages of different bacteria carrying virulence genes.
Ans.
Bacterial host Bacteriophage Virulence genes Function
Clostridium
botulinum
Clostridial phages bontC
bontD
Botulinum toxin type C (BoNT/C)
Botulinum toxin type D (BoNT/D)
Corynebacterium
diphtheriae
Β-phage, ω-phage tox Diphtheria toxin
Escherichia coli 933J, H19B, 933W
λ-phage
stx-1, stx-2, lom, bor
sltx
Shigatoxin 1, 2, OMP, serum resistance
Shiga-like toxin
Pseudomonas
aeruginosa
φCTX ctx Cytotoxin
Salmonella enterica SopEφ
Gifsy-2
sopE
sodC-I
SPI-I effector protein
Periplasmic superoxide dismutase
Shigella flexnerii Sf-6
Sfv, Sfx, Sf-II
oac
gtr
O-antigen acetylase
Glucosyl transferase
Staphylococcus
aureus
Φ13 sac
entA
Staphylokinase
Enterotoxin A
Streptococcus
pyogenes
T12, CS112 speA, speC Erythrogenic toxin
Vibrio cholerae CTXφ ctxAB,
ace, zot
Colera toxin
Enterotoxins
Q. 68. Name a bacteriophage which is known to encode a gene product which reduces virulence of
the bacterium.
164
Ans. Gifsy-2 phage of Salmonella enterica serovar Typhimurium although contributes significantly
towards virulence but it also carries a gene grvA, coding an antivirulence factor.
Q. 69. Enlist mechanisms of selecting gene-prophage associations and beneficiaries in association?
Ans.
Mechanism
Number
Selecting mechanism for gene-prophage
associations
Benefiting
entities
1 Gene survival via greater mobility Gene
2 Genetic hitchhiking on more-fit bacterial lineages Gene
3 Gene escape from immune surveillance Gene
4 Gene extra-bacterial survival Gene
5 Faster gene evolution Gene
6 Epistasis linking VF and phage genes Bacteria
7 Dissemination of an effective toxin dose Bacteria
8 Lysogen allelopathy Bacteria
9 Direct enhancement of phage fitness Phage
10 Indirect enhancement of phage fitness Phage
Q. 70. How autochthonous organisms become harmful to host when host catch infection with a
pathogenic bacteria?
Ans. Autochthonous or commensal bacteria such as nontoxigenic Corynebacterium diphtheriae,
Streptococcus pyogenes, Staphylococcus aureus, and E. coli are common members of the
normal human and animal microflora. But autochthonous organisms undergo in situ phage
conversion if the host catch infection with just a few toxinogenic (phage mediated or plasmid
mediated) organisms.
Q. 71. Write a short note on lysogen allelopathy.
Ans. The process in which bacteria kill or inhibit growth of competitor bacteria in the environment
use the use of its phages it is called lysogen allelopathy. In lysogen allelopathy, phage particles
released from induced bacterial lysogens can block the growth of competing bacteria by
infecting and subsequently killing those bacteria. By releasing phage just as toxins are released
from induced lysogens, the remaining (uninduced) lysogens may gain advantage over
competing bacteria through acquisition of nutrients and niche or space for growth and
reproduction.
Q. 72. How bacteriocin mediated killing is different from lysogen allelopathy?
Ans. Bacteriocins, and some defective prophages, can also bring a similar competition-minimizing
result as lysogen allelopathy but, unlike intact phages, they are not equipped with disseminating
machinery for delivery of effective lethal toxin dose. Besides, allelopathic effects can also work
as a defensive tool against phage-attacked bacterial. Another important phenomenon is the
165
survivors of competitors get converted to the similar type as the alleopathic bacteria. Lysogen
allelopathy can be used in treatment of bacterial infections as a novel approach.
Q. 73. Classify phages (Viruses) of algae with examples?
Ans. Unicellular eukaryotic green algae show a unique viral ecology; in contrast to tailed viruses of
bacteria and archaea their viruses tend to be large non-tailed dsDNA viruses. But, other kind of
viruses have also been detected in higher algae. Algal viruses can be classified in two groups:
1. Large dsDNA viruses (Phycodnaviridae). Large icosahedral particles 130-190 nm diameter,
large dsDNA genomes 160-380 kb. Viral capsid assembly and DNA packaging occurs in the
cytoplasm. Paramecium bursaria chorella virus (PBCV-1) infect eukaryotic unicellular-like
green algae (Chlorella algae are small, spherical, unicellular, nonmotile, asexual reproducing
green algae) while Ectocarpus virus (EsV) and Feldmannnia virus (FsV) infect marine brown
alga
2. Small viruses. Heterosigma akashiwo nuclear inclusion virus (HaNIV) and Heterosigma
akashiwo RNA virus (HaRNAV) infect the algae Heterosigma akashiwo (a toxic bloom-
forming microalgae). HaRNAV 25 nm diameter, polyhedral symmetry and HaNIV 30 nm
diameter, replicates to high copy number (105) before host lysis, nonenveloped and hexagonal
implying icosahedral symmetry. Virus forms a crystalline array located in different parts of cell.
Chara algae virus: They also infect higher green algae in oceans.
Q. 74. Give some examples of protozoan viruses.
Ans. 1. Virus-like particles (VLPs) of Polysporoplasma mugilis (a parasite in a marine fish, Liza
aurata). VLP are 18–20 nm in diameter related to the Picornaviridae and densely packed inside
membraned vacuoles.
2. Leishmania (causative agent of leishmaniasis in human beings) RNA virus: L. guyanensis
and L. braziliensis are infected with persistent, single-segmented, non-enveloped dsRNA
viruses termed LRV1 while L. major harbors a similar virus, LRV2-1.
3. Eimeria nieschulzi associated with RNA virus: The indirect evidences suggested that this
parasite of gastrointestinal tract also possess a RNA virus.
4. dsDNA virus in cytoplasm of Cryptosporidium parvum: It is a Phycodna-like virus of
protozoan parasite, these types of viruses are rare in animalia.
5. Extrachromosomal dsRNAs virus of C. parvum: Neither their mechanism of spread is
elucidated nor their structure is known. It has been classified into a new family Partitiviridae.
Its replications is predicted to be similar to satellite viruses of the family Totiviridae i.e., head-
166
full replication mechanism. Their dsRNA genome is enclosed in two separately capsid
segments. Virion RNA polymerase complexes from Cryptosporidium parvum, has been
identified.
Q. 75. What is LUCA and what is its significance?
Ans. It is thought that the first cellular life form evolved on earth about 4.5 billions years ago is the
common progenitor to all life forms it is designated as the Last Universal Common Ancestor
(LUCA). Sequence analysis of the major domains of extant life forms (bacteria, archaea and
cyanobacteria) suggested that about 360 genes are common to all life forms which might have
descended from LUCA. Archaea and Bacteria are thought to have diverged early from LUCA,
~4 billion years ago then emerged the Cyanobacteria _~2.6 billion years ago). Although all
three kinds of prokaryotes are much much different in structure, metabolism and reproduction,
all of these prokaryotes have the tailed phage, which appears to have evolved prior to the
divergence of the three forms.
Q. 76. What is the major difference in viruses of filamentous fungi and Mycoplasma?
Ans. Most filamentous fungi get infected with dsRNA virus whereas Mycoplasma has infection
with ssDNA viruses.
Q. 78. Wrie a note on archaeal viruses.
Ans. Although Archaea have phages as frequently as bacteria, only few has been described. The
described archaeal viruses (phages) constitute about less than 1% of the total viruses
enumerated at the International Committee on the Taxonomy of Viruses. Limited studies
indicated that archaeal viruses might be as diverse as of bacteria and eukarya. Typical phage of
archaea does not differ from bacterial phage as it infect the archaeal halophiles on anchoring
with its tail. However, there are many other and diverse groups of phages morphologies
including filaments, lemons and corndogs, it mamy spherical or, an icosahedrally symmetric
virus. In some phages the protruding turrets (with unknown function but may be helping in
attaching with host) extending up to 13 nm above the capsid surface at the 12 five-fold
symmetrical positions of the icosahedron. Sulfolobus turreted icosahedral virus is a phage of
hyperthermophilic archaeon Sulfolobus sulfataricus (growing at temperatures above 80°C and a
pH as low as 2. It has 17,663-bp circular dsDNA genome, with 36 ORFs, of which proteins for
four are known with their functions, two from other Sulfolobus viruses one from Sulfolobus
genome, the 4th one a gene encoding a 37-kDa major capsid subunit. There is a lot of scope in
proteomics and genomics of archeal phages.
167
Most of the archaeal phages have been reported from Sulfolobales of Crenarchaeota
kingdom of Archaea. They are unique both in their genome sequence and morphology and
grouped in three families viz., Rudiviridae, Lipothrixviridae, Rudiviridae virions are stiff 23nm
× 800nm to 900nm helical rods with linear dsDNA genomes. Fuselloviridae, and Guttaviridae.
Fuselloviridae members are 60 to 90 nm spindle shaped with ~15.5 kb circular dsDNA
containing about 35 ORFs. Lipothrixviridae members are ~ 24×1,980nm flexible rods with
putative attachment fibers at both ends containing linear dsDNA genome of ~ 42 kb in length
with ~74 ORFs with an example of Sulfolobus islandicus filamentous virus (SIFV). A forth
group of virus is isolated from Acidianus, (Acidianus filamentous virus 1; AFV1), it is an
enveloped 900nm × 24nm filamentous virus with claw-like structures at each end. Its genome is
of ~21kb dsDNA with 40 ORFs.
Q. 79. Do VLPs of yeasts can be considered phage-like organisms, give an example?
Ans. Virus-like particles (VLPs) isolated Wickerhamia fluorescens, a yeast strain is certainly a
phage like particle with 42nm diameter with a single 4.6 kbp dsRNA molecule. It encodes for a
capsid protein of 74.5 kDa.
Q. 80. Write a short note on UOPs.
Ans. The most common unicellular organism parasites (UOPs) are no doubt Bacteriophages but
they also include: 1. the cyanophages of cyanobacteria, 2. the actinophages of Actinomycetes ,
3. the mycoviruses of yeasts, 4. the phycoviruses of unicellular, eucaryotic algae, 5.
Bdellovibrio spp. feeding on gram negative bacteria, 6. Legionella spp. parasitizing amoebas, 7.
Vampirococcus/ vampirovibrio parasitizing and predating on Chromatium bacteria, 8.
Virioplankton of planktonic environments
Q. 81. Where we see the role of UOPs?
Ans. The UOPs have multifaced role in environmental ecology as: 1. Bacteriophages are used as
indicators of fecal pollution; 2. Bacteriophages have been used as models for studies of viral
disinfection; 3. Through generalized transduction phages are instrumental in horizontal transfer
of genetic material between hosts; 4. Lysogenic bacteriophages are often associated with phase
variation, and communication of several virulence factors in bacteria; 5. Some UOPs become
etiological agents for zoonotic diseases being carried on their unicellular hosts or unicellular
reservoirs; 6. UOPs may be employed as anti-microbial therapeutics e.g., phage therapy.
168
Chapter 6. Plasmids and Conjugation
A plasmid can be defined as a circular, double-stranded unit of DNA that replicates within a cell
independently of the chromosomal DNA. Plasmids are common in bacteria and are frequently used
in genetic engineering research to transfer genes between cells. Plasmids have also been identified
in a few eukaryotes. Plasmids can be classified into two groups on the basis of the number of genes
and functions they carry; a. The large plasmids (~100 kb) having code for approximately 100
genes, these are usually low copy number and their replication is precisely coordinated with the cell
division cycle. B. Smaller plasmid (~6 to10kb) coding for 6 to10 genes and are multi-copy (10–20
copies per cell) plasmids. Cryptic plasmids found in some bacterial species, they have no known
functions. The host range of a particular plasmid is usually limited to closely related bacteria;
however, some have broader host range. Although most of the bacterial plasmids confer one or the
other quality and specified functions, they are not essential for cell growth. Plasmids conferring
antibacterial resistance are called R plasmids and they are of medical importance.
Q. 1. What is the difference between a plasmid and episome?
Ans. Plasmids are extrachromosomal genetic elements capable of autonomous replication while an
episome is a plasmid that has to integrate into the bacterial chromosome for propagation.
Q. 2. How plasmids are classified?
Ans. Classification of Plasmids is based on their transferability and their effect on host:
1. Classification based on Transfer properties
a. Conjugative plasmids - Conjugative plasmids are those that mediated conjugation. These
plasmids are usually large and have all the genes necessary for autonomous replication and for
transfer of DNA to a recipient (e.g. genes for sex pilus).
b. Nonconjugative plasmids - Nonconjugative plasmids are those that cannot mediate conjugation.
They are usually smaller than conjugative plasmids and they lack one or more of the genes
needed for transfer of DNA. These can be of two type; mobilizable and non-transmissible.
c. Non-transmissible plasmids: These plasmids can neither be transferred through conjugation nor
can be mobilized by other plasmids, e.g. virulence plasmids.
d. Mobilizable plasmids: A nonconjugative plasmid can be transferred by conjugation if the cell
also harbors a conjugative plasmid e.g. col plasmids.
169
2. Classification based on phenotypic effects
a. Fertility plasmid (F factor)
b. Bacteriocinogenic plasmids - These plasmids have genes which code for substances that kill
other bacteria. These substances are called bacteriocins or colicins.
c. Resistance plasmids R factors - carry antibiotic resistance genes e.g. Rbk of E. coli.
d. Killer plasmids or Col plasmids of E. coli.
e. Suicide vectors, plasmids containing sacB gene used for transfer of mutated genes.
f. Cloning vectors, puc, pmBr etc.
g. Virulence plasmid, responsible for pathogenicity as Ti of Agrobacterium tumefaciens for causing
gall disease on dicotyledonous plants.
h. Degradative plasmid; code for enzymes for metabolism of unusual compounds, e.g., TOL of
Pseudomonas putida.
Q. 3. What is the general composition of R plasmids?
Ans. R plasmids are conjugative plasmids in which the genes for replication and transfer are
located on one part of the R factor and the resistance genes are located on another part.
RTF (Resistance Transfer Factor) - carries the transfer genes.
R determinant - carries the resistance genes. The resistance genes are often parts of transposons.
Q. 4. How different antibiotic resistance genes act in bacteria?
Ans. General Mode of action of resistance genes is:
a) Modification (detoxification) of antibiotic - e.g. ß-lactamase
b) Alteration of target site - e.g. Streptomycin resistance
c) Alteration of uptake - Tetracycline resistance
d) Replacement of sensitive pathway - e.g., new folic acid pathway for resistance to sulfa drugs
Q. 5. What is conjugation and what are the mating types in bacteria?
Ans. It is transfer of linear and ordered transfer of DNA from a donor to a recipient by direct
physical contact between the cells. In bacteria there are two mating types a donor (male) and a
recipient (female) and the direction of transfer of genetic material is one way; DNA is
transferred from a donor to a recipient.
Mating types in bacteria
a. Donor - The ability of a bacterium to be a donor is a consequence of the presence in the cell of an
extra piece of DNA called the F factor or fertility factor or sex factor. The F factor is a circular
piece of DNA that can replicate autonomously in the cell; it is an independent replicon.
170
Extrachromosomal pieces of DNA that can replicate autonomously are given the general name
of plasmids. The F factor has genes on it that are needed for its replication and for its ability to
transfer DNA to a recipient. One of the things the F factor codes for is the ability to produce a
sex pilus (F pilus) on the surface of the bacterium. This pilus is important in the conjugation
process. The F factor is not the only plasmid that can mediated conjugation but it is generally
used as the model.
b. Recipient - The ability to act as a recipient is a consequence of the lack of the F factor.
Q. 6. What are different stages in conjugative transfer of plasmid DNA? How can you classify
plasmids on the basis of these characteristics?
Ans. There are four stages:
1. Formation of donor-recipient pairs (effective contact)
2. Mobilization, i.e., preparation for DNA transfer
3. DNA transfer
4. Formation of functional replicating plasmid in recipient.
All plasmids do not undertake all four processes and can be divided into four as:
a. Non-transmissible plasmids can not start processes of effective contact and DNA transfer.
b. Conjugative plasmids: carry genes for process of effective contact but may not prepare its
DNA for transfer.
c. Mobilizable plasmids: Can prepare its DNA for transfer but can not initiate effective contact
e.g. ColE1
d. Self-transmissible plasmids: Have both conjugative and mobilizable properties, F1, Vir
plasmid of Salmonella.
Q. 7. What is plasmid donation?
Ans. When a non-conjugative plasmid is transferred through effective contact conferred by a
conjugative plasmid, process is called Donation.
Q. 8. What is conduction?
Ans. For donation mobilization functions are plasmid specific and all non-mobilizable plasmids can
not be transferred just by complementation by any conjugative plasmid, however in some cases
mobilization can occur through conduction. In the process of conduction, non-mobilizable
plasmid can recombine to a self transmissible plasmid and is carried along with.
Q. 9. What are the different physiological sates of F factor in donor cells, and they differ?
171
Ans. Physiological states of the F factor
a. Autonomous (F+) - In this state the F factor carries only those genes necessary for its replication
and for DNA transfer. There are no chromosomal genes associated with the F factor in F+
strains. In crosses of the type F+ X F- the F- becomes F+ while F+ remains F+. Thus, the F
factor is infectious. In addition, there is only low level transfer of chromosomal genes.
b. Integrated (Hfr) - In this state the F factor has integrated into the bacterial chromosome via a
recombination event. In crosses of the type Hfr X F- the F- rarely becomes Hfr and Hfr remains
Hfr. In addition, there is a high frequency of transfer of donor chromosomal genes.
c. Autonomous with chromosomal genes (F') - In this state the F factor is autonomous but it now
carries some chromosomal genes. F' factors are produced by excision of the F factor from an
Hfr. Occasionally, when the F factor is excising from the Hfr chromosome, donor genes on
either side of the F factor can be excised with the F factor generating an F'. F' factors are named
depending on the chromosomal genes that they carry. In crosses of the type F' X F- the F-
becomes F' while F' remains F'. In addition there is high frequency of transfer of those
chromosomal genes on the F' and low frequency transfer of other donor chromosomal genes.
Q. 10. Describe the mechanism of conjugation under influence of different kinds of F factors.
Ans.
a. Mechanism of F+ X F- crosses
i) Pair formation - The tip of the sex pilus comes in contact with the recipient and a conjugation
bridge is formed between the two cells. It is through this bridge that the DNA will pass from the
donor to the recipient. Thus, the DNA is protected from environmental nucleases. The mating
pairs can be separated by shear forces and conjugation can be interrupted. Consequently, the
mating pairs remain associated for only a short time.
ii) DNA transfer - The plasmid DNA is nicked at a specific site called the origin of transfer and is
replicated by a rolling circle mechanism. A single strand of DNA passes through the
conjugation bridge and enters the recipient where the second strand is replicated.
iii) This process explains the characteristics of F+ X F- crosses. The recipient becomes F+, the
donor remains F+ and there is low frequency of transfer of donor chromosomal genes. Indeed,
there is no transfer of donor chromosomal genes. In practice however, there is a low level of
transfer of donor chromosomal genes in such crosses.
Mating of F+ and F- Bacterial
b. Mechanism of Hfr x F- crosses
i) Pair Formation
172
ii) DNA transfer - The DNA is nicked at the origin of transfer and is replicated by a rolling circle
mechanism. But the DNA that is transferred first is the chromosome. Depending upon where in
the chromosome the F factor has integrated and in what orientation, different chromosomal
genes will be transferred at different times. However, the relative order and distances of the
genes will always remain the same. Only when the entire chromosome is transferred will the F
factor be transferred. Since shearing forces separate the mating pairs it is rare that the entire
chromosome will be transferred. Thus, the recipient does not receive the F factor in a Hfr X F-
cross.
iii) Legitimate recombination - Recombination between the transferred DNA and the chromosome
results in the exchange of genetic material between the donor and recipient.
iv) This mechanism explains the characteristics of Hfr X F- crosses. The recipient remains F-, the
donor remains Hfr and there is a high frequency of transfer of donor chromosomal genes.
c. Mechanism of F" x F- crosses
i) Pair formation
ii) DNA transfer - This process is similar to F+ X F- crosses. However, since the F' has some
chromosomal genes on it these will also be transferred.
iii) Homologous recombination is not necessary although it may occur.
iv) This mechanism explains the characteristics of F' X F- crosses. The F- becomes F', the F'
remains F' and there is high frequency transfer of donor genes on the F' but low frequency
transfer of other donor chromosomal genes.
Q. 11. Why does the transfer of genes from an Hfr require a functional recA gene in the recipient,
but transfer of genes on a F' does not?
Ans. Transfer of genes from an Hfr into an F- recipient cell requires the recA gene product because
the incoming genes need to be inherited by homologous recombination. This two cross-over
event is mediated by the recA gene. Since the incoming genes are on an unstable linear
fragment of DNA, if the genes aren't recombined into the chromosome by RecA, then the DNA
will be degraded by exonucleases. Transfer of an F' can occur in a recA mutant because the F'
does not need to recombine with the recipient chromosome in order to be stably inherited. The
final step in the transfer of an F' is a site-specific recombination event between the first and
second copy of oriT to be transferred into the recipient cell. This single cross-over event is
mediated by one of the tra genes encoded on the F' plasmid, and is independent of recA. This
site-specific recombination event occurs at the conjugal bridge between the two cells, and
results in a new F' plasmid in the recipient cell that may complement mutant genes on the
recipient chromosome, if the wild-type copies are present on the F'.
173
Q. 12. Name some G+ve bacteria which undergo conjugation?
Ans. Streptomyces spp., Streptococcus spp. and Clostridium spp.
Q. 13. How plasmids get integrated into bacterial chromosome?
Ans. A plasmid can integrate into bacterial chromosome either through transposition due to
presence of transposons or IS elements in replicon or through recombination at specific sites as
some Is elements are common in bacterial chromosome and plasmid replicon.
Q. 14. What do you understand by plasmid primers?
Ans. The F plasmid or any plasmid that can integrate into chromosome and excised imprecisely
leaving a few components of plasmid in bacterial DNA; the leftover fragments of plasmid DNA
in bacterial chromosome are called plasmid primers.
Q. 15. What is the importance of conjugation in bacteria?
Ans. Among the Gram negative bacteria this is the major way that bacterial genes are transferred.
Transfer can occur between different species of bacteria. Transfer of multiple antibiotic
resistance by conjugation has become a major problem in the treatment of certain bacterial
diseases. Since the recipient cell becomes a donor after transfer of a plasmid it is easy to see
why an antibiotic resistance gene carried on a plasmid can quickly convert a sensitive
population of cells to a resistant one.
Gram positive bacteria also have plasmids that carry multiple antibiotic resistance genes, in some
cases these plasmids are transferred by conjugation while in others they are transferred by
transduction. The mechanism of conjugation in Gram + bacteria is different than that for Gram -
. In Gram + bacteria the donor makes an adhesive material which causes aggregation with the
recipient and the DNA is transferred.
16. Q. What is difference between F’ excision and lamda (λ ) excision in E. coli?
Ans. Λ DNA is of viral origin and its excision affect the viability (cause lysis) of the host cell while
excision of F’ has no effect on host cell viability. F’ DNA has no restriction over size of DNA
excised whereas it is strict for λ DNA.
Q. 17. What is the difference between Type IA F’ plasmid, Type IB F’ plasmid and Type II F’
plasmid?
174
Ans. These are different types of F’ plasmids depending on the site of Bacterial DNA segments
they carry: Type IA F’ plasmid- carries bacterial DNA sequences located near the origin of the
ancestral Hfr,
Type IB F’ plasmid- carries bacterial DNA located near the termini of the ancestral Hfr,
Type II- carries both bacterial DNA sequences located near termini and centre of Hfr and are
the most common types.
Q. 18. Name some chromosome mobilizing plasmids of bacteria.
Ans.
Bacteria Chromosome mobilizing
plasmids
Bacteria Chromosome
mobilizing
plasmids
Salmonella F E. coli F, F’
Pseudomonas FP2, FP39, FP110, R91-5, R65 Enterococcus pAD 1
Streptomyces SCP1, SC2 Agrobacterium Ti
Q. 19. How conjugation system of Agrobacterium is different than other bacteria?
Ans. Conjugation in Agrobacterium helps in transfer of bacterial DNA to a plant cell DNA, not
from bacteria to bacteria as occurs in most of the other bacteria.
Q. 20. Name the bacteriocins produced by different bacteria.
Ans.
Bacteria Bacteriocin Bacteria Bacteriocin Bacteria Bacteriocin
E. coli Colicins B. subtilis Subtilisins H. influenzae Inflenzacins
P. pyogenes Pyocins Halobacterium Halocins Klebsiella Klebocins
Q. 21. Write a short note on plasmid R100.
Ans. It was first isolated from Shigella flexneri in Tokyo. It is self transmissible drug resistance
plasmid producing F pili. It belongs to IncFII group and can coexist with F plasmid. It is
transferable among members of Enterobacteriaceae but not to Pseudomonas. It confers
resistance to streptomycin (100 mg/ml), chloramphenicol (200 mg/ml), tetracycline (125
mg/ml) and sulphonamide (200 mg/ml)
Q. 22. How copy number of a plasmid in bacteria is regulated.
Ans. The copy-number of a plasmid in bacteria cells is determined by regulating the initiation of
plasmid replication, which is controlled by regulating a. the amount of available primer for the
initiation of DNA replication, b. amount or function of essential replication proteins. The major
mechanisms used to control initiation of plasmid replication are:
175
1. Regulation by antisense RNA;
2. Regulation by binding of replication proteins to repeated 18-22 bp sites called iterons.
Q. 23. Name the functions and receptors of different colicins.
Ans.
Colicin
(MW)
Cross
resistance
group
Nature of
plasmid
Mode of action Receptors
A B Non-conjugative Pore formation Vitamin B12 transport
receptors
B (89 kDa) A Conjugative Pore formation Ferric enterobactin
receptor
D (89 kDa) B Conjugative Inhibition of protein
synthesis
Ferric enterobactin
receptor
E1 (56
kDa)
6646bp
A Non-conjugative Pore formation Vitamin B12 transport
receptors
E2 (62
kDa)
A Non-conjugative DNase (single or double
strand breaks)
Vitamin B12 transport
receptors
E3 (60
kDa)
A Non-conjugative RNase, cleaves 16s rRNA
(cleaves 49 bases from
3’end)
Vitamin B12 transport
receptors
Ia/Ib(80 kDa) B Conjugative Pore formation Chelated iron transport
K (43 kDa) A Non-conjugative Pore formation Nucleoside porins
M (27 kDa) B Inhibition of murein and
LPS synthesis
Ferrochrome receptor
V B Conjugative
Q. 24. In case of two unequal sized or equal sized chromosomes, what criteria is used to define
both DNA molecules as bonafide chromosomes instead of one as chromosome and other as a
megaplasmid or big plasmid?
Ans. The simple criterion is that plasmid genes are not essential under all growth conditions, that is,
plasmids are dispensable under certain growth conditions while chromosomes carry genes
essential for growth of the organism under all kinds of growth conditions. Therefore, if both
DNA molecules carry essential genes for growth then both DNA molecules are considered
chromosomes.
Q. 25. What are eukaryotic plasmids?
Ans. The plasmids found in eukaryotes are called eukaryotic plasmid e.g. 2µ circle of yeasts. It is a
6.3 kb circular, extra-chromosomal element in the nucleus of majority of Saccharomyces
cerevisiae strains. It is multi-copy (50 to 100) plasmid coated with Nucleosomes, similar to host
chromosome. Replication initiation is synchronous and dependent of host replication. Origin of
replication is called an autonomous replication sequence.
176
Q. 26. In 2µ circle plasmid of Saccharomyces cerevisiae replication is initiated by host replication
initiation enzymes i.e., replication is initiated only once per cell cycle thus only two plasmid
molecules would be produced. Then, how could the 2µ circle forms 50 to 100 copies per cell?
Ans. Instead of bidirectional replication 2µ circle also undergoes rolling circle replication similar to
prokaryotic plasmids. The 2µ circle has a site-directed inversion mechanism as it has two copies
of a 599 bp inverted repeat sequence the "flip" sites, and encodes a site-directed recombinase
(FLP) the "flip" protein which promotes recombination between these repeats. Recombination
between the flip sequences inverts the adjacent regions of the plasmid. This iinversion permits
the plasmid to switch from bidirectional-replication to rolling circle replication, producing
multiple copies of the plasmid per cell cycle. The origin of replication is located very close to
one of the flip sites, so one replication fork will pass through the adjacent flip soon after
replication has begun, but the second replication fork has to travel half way around the plasmid
before it passes through the other flip site. After one of the bidirectional DNA replication forks
has passed the first flip site but before the replicating fork reaches the second flip site,
recombination inverts the intervening sequence. After this inversion, both of the DNA
replication forks will be moving in the same direction producing long concatemers (A
concatemer is a long continuous DNA molecule that contains multiple copies of the same DNA
sequences linked in series). Concatemers are converted to monomeric plasmids by site-specific
recombination. Replication is terminated only when a second inversion takes place at the flip
sites, causing the replication forks to collide.
Q. 27. Describe the regulation of copy number in 2µ circle of yeast? How it differs with bacterial
plasmids?
Ans. It is regulated through three plasmid encoded proteins (REP1, REP2, and D) which repress
expression of FLP protein. Repressor proteins’ concentration is proportional to copy number of
the 2µ circle. In case, plasmid copy number is high, expression of FLP is repressed, but when
the plasmid copy number is low, expression of FLP is induced. In case of bacterial plasmids the
control of plasmid copy number is usually regulated by modulating the initiation of DNA
replication, in yeast, 2µ circle controls its copy number by regulating a protein which affects
amplification of the plasmid. Similar mechanism for controlling plasmid copy number is used
by an algal (Chlamydomomas) plasmid.
Q. 28. What are the characteristics of plasmid replication?
Ans. Plasmid replication is autonomous, independent of cell division and chromosomal replication.
Replication can occur at ant time in the host cell cycle i.e., the timing of plasmid replication is
177
nearly a random event. Copy number is regulated by factors encoded by plasmid and host and
modulated by growth conditions. Copy number mutants are common in occurrence. Cloning or
insertion of some foreign DNA may interfere with copy number and replication of plasmid.
Similar to chromosomal DNA, some plasmids also require synthesis of protein as a part of
initiation factor therefore protein synthesis inhibitors as chloramphenicol can also inhibit
plasmid replication while in case of some plasmids (ColE3) addition of chloramphenicol has
opposite effect. For primer synthesis plasmids use host RNA polymerase. Almost every
plasmid, big or small needs one or more host factors for initiation of replication of their DNA.
Instead of double stranded a single strand origin of replication is essential for replication of
plasmid.
Q. 29. What are different modes of plasmid replication?
Ans. Bacterial plasmids mainly replicate through three modes:
a. Theta (θ) replication: It is similar to replication of bacterial chromosome as in F plasmid.
The replicating DNA looks like theta (θ). It is usually due to bidirectional replication but
some plasmids (R plasmids, R100 and ColE1) can undergo θ replication with unidirectional
replication.
b. Sigma (σ) replication: It is quite similar to rolling circle replication and generates unit length
instead of concatemers, single stranded circle e.g., pC194, ΦX174 and some small R
plasmids of Staphylococcus aureus and Bacillus cereus. Initiation of replication is not
required time and again and it continues (though under control) as in bacteriophages.
c. Rolling circle replication: Common in cryptic plasmids and in eukaryotic plasmid. Initiation
is required a fresh after synthesis of one concatemer. Replication origin is a single strand
origin.
Q. 30. What might be the role of sex pheromones in bacteria?
Ans. In Gram negative bacteria conjugation is mediated through sex pili, however in Gram positive
conjugation is mediated through sex pheromones excreted by recipient cells. Pheromones
induce aggregation of donor bacteria around the recipient bacteria. After acquisition of plasmid,
recipient stops to excrete pheromone. Plasmid containing donors responds differently to
different pheromones produced by recipients. A single recipient may produce many kinds of
pheromones.
Q. 31. What is the nature of pheromone produced by Enterococcus faecalis?
178
Ans. Short polypeptide (HOOC-Leu-Phe-Ser-Leu-Val-Leu-Ala-Gly-NH2) acting as pheromone is
produced by plasmid negative Enterococcus faecalis to attract male cells for conjugation,
Q. 32. How conjugation occurs in Staphylococcus aureus?
Ans. Conjugation in S. aureus is phage mediated in nature.
Q. 33. Do all plasmids contain circular DNA, if no, give examples?
Ans. No, some plasmids as in Streptomyces spp. and Rhodococcus spp. have linear DNA.
Q. 34. What is curing and curing agent?
Ans. Curing means freeing of host cells from plasmids. A curing agent is one which promotes the
bacteria to loss their plasmids and strain without plasmid in then called ‘cured’.
Q. 35. What are different methods of curing of bacteria?
Ans. Bacteria can be cured i.e., made free of from plasmid by using one or more of the following
methods.
1. Using acridine dyes (proflavine and acridine orange): Acridine dyes interfere with replication
of plasmids without affecting replication of chromosomal DNA, thus producing plasmid
free cells. Plasmid free cells can be selected using suitable screening method. Effect of dye
is on curing is plasmid specific, host and dye concentration dependent. Besides pH (~7.6) of
the growth medium also affect the curing action of the dye.
2. Elevating growth temperature: Only temperature sensitive plasmids can be cured by this
technique, as a few plasmid of E. coli can be removed by growing E. coli at 42-45oC
instead of 37oC.
3. Detergents: Some detergents may also help to washout plasmid when used in low
concentration in growth medium.
4. Incorporation of antibiotics: Some antibiotics which do not inhibit cell growth may inhibit
plasmid replication as chloramphenicol.
5. Using plasmids of same incompatibility group: A plasmid containing bacteria can be
transformed with temperature sensitive (Ts) plasmid containing some selectable marker and
of same incompatibility as one already present in the bacteria and selected using the
selectable marker. The Ts plasmid is lost on growing bacteria at elevated temperature.
Q. 36. How plasmids are inherited by the progeny of bacteria?
179
Ans. Plasmid are inherited by progeny of bacteria either through plasmid partition system for low
copy number plasmids or through random distribution of plasmids in daughter cells in case of
high copy number plasmids, or through post-segregation killing.
Q. 37. What do you understand by plasmid partition in bacteria?
Ans. Plasmid partition means plasmid distribution among bacterial progeny cells in the end of cell
division of host. This mechanism is usually practiced for low copy number plasmids. the
process involves plasmid (par functions) as well as host specified control.
Q. 38. What is post-segregation killing?
Ans. In case of low copy number plasmids, plasmid free cells are killed after segregation of two
daughter cells; this function is brought about by susceptibility of the plasmid less cells to toxin
produced by plasmid containing cells which have antidote mechanism to protect themselves. In
some cases as in case of R1 plasmid the control over toxin is brought at the level of production
rather then through inactivation with antidote. The control locus, sok, overlap the toxin gene,
hok, when plasmid is there both genes are transcribed but mRNA having some complementary
region anneal together and ribosomal binding is inhibited therefore, no toxin is produced. In
case the plasmid is lost hok is transcribed but sok in not and toxin produced kills the cell.
Q. 39. For transfer of plasmid from one cell to another (through conjugation) what are the three
functions required to be present in plasmid.
Ans. The three universal functions required for transfer of plasmid are; mob, tra and oriT. These
must be present for conjugal transfer of plasmid; mob, tra functions may be complemented
through a helper plasmid also but oriT is must be of specific origin.
Q. 40. Some times bacteria become resistant to certain antibiotics without the import of any foreign
gene, why?
Ans. It happens due to some mutation in gene producing the receptor for a particular antibiotic.
These types of mutants can easily be selected on antibiotic plates, e.g., selection of Nalr strains
used in laboratories for genetic manipulations. Acquisition of resistance through mutation is not
much important clinically because of greater role of plasmids and transposons, however in case
of Mycobacterium tuberculosis this system of mutation is very important in generation of drug
resistant strains. In this bacteria plasmids and transposons play little role.
180
Q. 41. Name the plasmid incompatibility groups with varying type of pili phenotype and how these
pili are important for Bacteriophages?
Ans. There are several incompatibility groups (shown in table) of the bacterial plasmids. Two
plasmids of same group can not remain in the same cell of the host. Different plasmids produce
different types of pili which may be expressed either constitutively (C) or repressed (R) and act
as receptor for phages. Pili often act as receptor or ligands for phages. Some examples are as
under.
Receptors for
phages
Inc.
group
Plasmid
name
Pili
morphology
Serolog
ical
type On side On tip
Expression Mating type
B R724,
TP113
Thin flexible I1 C or R Universal
C Thick flexible C C-1 C or R Surface preferred
D Thick flexible D fd C Surface preferred
FI F Thick flexible F R17 fd C or R Universal
FII R1 Thick flexible F R17 fd R Universal
H R27,
TP124,
TP116
Thick flexible H R Universal
I1 Colb-P9,
R114, R64
Thin flexible I1 If1 R Universal
I2 R621a,
TP114
Thin flexible I1 If1 R Universal
Iγ Thin flexible Iα If1 R Universal
J Thick flexible C C -1 R Universal
K Thin flexible Iα R Universal
M Rigid M X C Surface obligatory
N N3, R15 Rigid N Ike,
PP4
C Surface obligatory
P RP4, R702 Rigid P PRR1 PR4, X C Surface obligatory
T Thick flexible T T C Surface preferred
U Rigid U X C Surface obligatory
V Thick flexible Nd R Universal
W Sa, R388 Rigid W PR4, X C Surface obligatory
X Thick flexible X X c Surface preferred
Q. 42. What is the difference between donation and conduction of plasmids?
Ans. In donation, a mobilizable plasmid is transferred in presence of conjugative plasmids or self
transmissible plasmids and its frequency is high. In conduction, a self transmissible plasmid
carries along a non-mobilizable plasmid through fusion or recombination of two plasmids, in
conduction chromosomal DNA may also get transferred at some times.
Q. 43. How can you detect presence of a plasmid in bacterium?
Ans. It can be done in two ways, a. by genetic methods (conjugation etc.) and by b. physical
methods (isolation of plasmid).
Q. 44. How can you increase plasmid yield of a bacterial culture? Why the process does not lead to
multiplication of host DNA?
181
Ans. It can be done by adding protein synthesis inhibitors as methyl-tryptophan or
chloramphenicol, because inhibition of protein synthesis inhibit replication of host DNA
because the process need new protein for each initiation of replication while many plasmid use
stabilized proteins for initiation of replication and keep on replicating.
Q. 45. F plasmid spread rapidly in bacterial population while self transmissible R plasmids spread
at slow pace, why?
Ans. Although not a rule, most bacteria carrying R plasmid rapidly loss their fertility factor (F),
other keeps on possessing it. At any given time only about 0.02% population carrying R
plasmid is competent donor and it can only be increased by adding fresh recipient cells to
disturb the equilibrium. R-ve cells which have just received the plasmid remain competent to
donate the plasmid for a short till the ‘fertility inhibitor’ is synthesized through the transcription
and translation of fin repressor gene. fin repressor gene has got inactivated in F plasmid due to
insertion of IS3 in finO gene; therefore it remains always a competent donor.
Q. 46. What are different common genes in F and R plasmids and their functions?
Ans. Some common genes and their functions in F plasmids are:
Genes Functions
traA pilin protein
traB,C,E,F,G,H,K,L,Q,U,V,W assembling pili
traG, N stabilize conjugal pairs
traS,T inhibiting mating between two F+ bacteria
traY nicking at oriT
oriT initiate rolling circle replication
incB,C,E decides incompatibility
finO,P inhibition of fertility inhibition, fertility repressor
Q. 47. What is the difference between oriT and oriV?
Ans. oriT is the origin for transfer of DNA during conjugation while oriV is the replication origin
in circular DNA of plasmid.
Q. 48. What is a killer plasmid?
Ans. It is plasmid that codes for certain toxins which kills other bacteria e.g. plasmids carrying
bacteriocin genes as Col plasmids of E. coli.
Q. 49. Write short note on plasmids of Streptomyces.
Ans. There are two principal conjugative plasmids in S. coelicolor, SCP1, and SCP2. SCP1 is a 350
kb, linear plasmid with 80 kb terminal inverted repeats and protein bounded to 5’ end, it codes
182
for antibiotic methylenomycin. SCP2 is a 31 kb cryptic plasmid. Both the plasmids are of low
copy number. Integration of SCP1 requires IS466 located near main plasmid DNA and right side
inverted repeat and another copy near agarase gene. Besides, seven more plasmids which can
mobilize the chromosome have also been reported with varying copy number in S. lividAns.
Presence of plasmid in Streptomyces is often associated with pock formation because presence
of plasmid in Streptomyces is often associated with inhibition or delayed formation of mycelia
and spores. Besides, one phage of Streptomyces ΦSF1 act as plasmid in prophage form, another
is SLP1 which often remain integrated in S. coelicolor.
Q. 50. What do you understand by slow stop and rapid stop mutation?
Ans. The temperature sensitive mutations affecting DNA synthesis are of two type; a. those
mutations which stop formation of new replication fork at elevated temperature but already
working forks complete their work are known as Slow Stop mutations, b. when all forks stop
to synthesize DNA on elevation of temperature and no new fork is formed mutations are named
Rapid Stop mutations.
Q. 51. What are the different genes and their functions in tra region of F plasmid?
Ans. There are two types of genes related to transfer region, viz., translated and non-translated
genes.
A. Translated genes:
1. Transfer genes
a. Pilin synthesis genes [traA (prepropilin), traQ (prepropilin hydrolase), traX (pilin N-
acetylase)]
b. Pilus assembly genes [traB, traC, traE, traF, traG (N-terminus), traH, traK, traL, traU,
traV, traW, trbC]
c. Mating pair stabilization [traG (C-terminus), traN]
d. DNA transactions [traD (DNA transport protein), vI (helicase I), traY (oriT-nuclease)]
2. Transfer related genes
a. Surface exclusion [traS, traT]
b. Regulation [finP (plasmid specific antisense RNA), traJ (positive regulator of traY), traM
(binds to oriT, allows tra expression), traY (allows expression traM)]
B. Non-translated genes
1. Transfer genes (DNA transaction, oriT)
2. Transfer related genes
a. Transcriptional promoters [traIp, traJp, traMp, traTp, traYp]
b. Repressor binding site [traJo]
183
Q. 52. Name some common broad host range plasmids and related vectors.
Ans. Though there are many plasmids and vectors with broad host range used in different
laboratories, a few common ones are:
Plasmid Inc.group Phenotypic characters
F IncF1 Tra+ (self transmissible)
RK2, RP4,RP1,R68, Incp1 Ar, Kr, Tr, Tra+
pRK290 Incp1 Mob+ (mobilizable), Tcr
pLAFR sries Incp1 Tr, Mob+,multiple site cloning cosmid, lacZα for pLAFR3
pHH1JI Incp1 Gr,Sr, low level Smr, Tra+
pR68.45 Incp1 Ar, Knr, Tr, Tandem duplication of IS21, form Hfr
pME487 Incp1 Ts (temperature sensitive) pR68.45
pRK2013 Incp1/ColE1 Kr, Tra+, mobilizable replicon
RSF1010 IncQ Smr, Sr, Mob+
pMMB66EH/HE IncQ Ar, tac promoter expression vector
R6K IncX Ar, Sr
pGP704 IncX OriR6K cis region, MobRP4, Ar, multiple cloning site, suicide
vector requiring R6Kπ
pBR322 pMB1 Ar, Tr, Mob+
pBR327 pMB1 Ar, Tr, Mob-
pUC series pMB1 Ar, lacZα, Mob-
pACYC177 P15A Ar, Kr
pACYC184 P15A Tr, Chr
Q. 53. What do you understand by integrative suppression?
Ans. Integration of F plasmid into certain bacterial mutants defective in DNA replication may lead
to normal replication of bacteria, this phenomenon of suppression of mutation is called
integrative suppression. It is because of the fact that replication of F does not require dnaA gene
product however uses all other proteins of bacteria, i.e., if an E. coli has dnaA(Ts) it will not
replicate at 42oC but F keeps on replicating. If the F integrates into bacterial chromosome then
E. coli also go on replicating but now replication origin is not of E. coli but oriV of F plasmid.
Q. 54. Name few characters transferred through plasmids.
Ans. The important characters of bacteria which are usually plasmid encoded are: 1. Antibiotic
resistance (R factors), 2. Heavy metal resistance, 3. degradation of unusual substrates
(hydrocarbon degradation) for growth, 4. Restriction/modification enzymes to protect DNA and
to degrade unprotected DNA, 5. Bacteriocins, 6. Toxins (on virulence plasmids) as Staph
aureus virulence factors: coagulase, hemolysin, enterotoxin and virulence factors of pathogenic
E. coli strains as hemolysin and enterotoxin, 7. fertility factors.
Q. 55. What is the suitable size of cloning vector (plasmid)? Do copy number is also important?
Ans. Plasmid size varies from 1-1000 kb but plasmids of <10 kb are ideal for cloning a gene or
sequence. Yes, high copy number plasmids are always preferred for cloning purpose.
184
Q. 56. What is the advantage of cloning in M13?
Ans. Genes can be obtained in single strand form. This is very useful when sequencing a gene.
Q. 57. What do you understand from 2µ plasmid?
Ans. This is circular plasmid of yeast, 2µ indicates about its ring size which is equal to 2two
microns.
Q. 58. What do you mean by Mobilome?
Ans. The term mobilome include all mobile genetic elements within a genome or intergenome,
including IS elements, transposons (in eukaryotes and prokaryotes), prophages and plasmids (in
prokaryotes).
Q. 59. Although plasmids can replicate in living cells of prokaryotes, Archaea and eukaryotes, are
not considered a form of life like viruses. Why?
Ans. Similar to viruses, plasmids are not considered a form of "life" because plasmids are "naked"
DNA and do not encode genes necessary to encase the genetic material for transfer to a new
host. Plasmids requires direct, mechanical transfer by conjugation or an intentional uptake as in
transformation. Transformation with plasmids neither parasitic nor symbiotic in nature.
Plasmids provide a mechanism for horizontal gene transfer among microbes and often provide a
selective advantage under a given environment e.g., antibiotic drug resistance, virulence
factors which provide a selective environmental niche. On the otherhand viruses mostly aim to
destroy host cell to get release so that they can further infect other hosts or host cells.
Q. 60. What do you mean by gene therapy? What are the factors affecting its success?
Ans. It is the modern method of treating the genetic and some acquired diseases through
supplementation of the required gene through transforming the target cells. The success of gene
therapy is decided by the efficiency of insertion of therapeutic gene at the desired target site in
chromosome of the targetted cells without causing cell injury, oncogenesis or an adverse
immune response.
Q. 61. What is ZFN and its use?
Ans. ZFN known as Zinc Finger Nuclease, it offer a way to cause a site-specific double strand
break in DNA genome and initiate homologous recombination. ZFN is thought to be a tool in
targeted gene correction. Plasmids encoding ZFN can be used to deliver a therapeutic gene to a
185
target chromosomal site. This approach to gene therapy has been considered more safer viral-
based therapeutic genes delivery system.
Q. 62. What is addiction system in bacteria?
Ans. Several low copy number plasmids encode for addiction system also named as
postsegregational killing system (PSK) e.g., hok/sok system of plasmid R1 of E. coli. In this
system plasmid encode for production of a long-lived poison and short live antidote. Progeny
cells having a copy of the plasmid survive because they can produce antidote while plasmid-
less cells die or their growth is supressed due to presence of poison in the inherited cytoplasm.
Q. 63. Name important yeast cloning vectors.
Ans. There are two important yeast cloning vectors namely, yeast integrative plasmid (YIp), and
yeast replicative plasmid (YRp). Yip vector integrates into the host chromosome for survival
and replication. While YRp transport a sequence of chromosomal DNA including an origin of
replication but these are less stable and often lost during budding of yeast.
Q. 64. During isolation of plasmid DNA, it may appear in different conformations what are they?
Ans. Plasmid DNA may appear in one of the following five onformations (listed in order of
electrophoretic mobility from slowest to fastest:
1. Single stranded "Nicked Open-Circular" DNA.
2. Fully intact with both strands, but been enzymatically relaxed and called Relaxed
Circular DNA. It has no coiling.
3. Linear DNA with free ends, either because of cut in both of the strands or because the
DNA was linear in vivo.
4. Covalently Closed-Circular (super coiled) DNA, fully intact with no cut and with built
ina twists.
5. Denatured supercoiled DNA. It is like supercoiled DNA, but has some unpaired regions
making it a little loosely packed often results from excessive alkalinity during plasmid
extraction.
Q. 65. Wrie a short note on bacterial sex pheromones.
Ans. Plasmid-free enterococci and other gram positive bacteria secrete a variety of short peptides
known as sex pheromones. They attract specific donor (plasmid carrying) cells. The
pheromones are hydrophobic octa-peptides or hepta-peptides from processing of the signal
sequences of lipoprotein precursors. Genes for pheromones are on bacterial chromosome. C-
186
terminal hepta- or octa-peptide of the signal sequence often forms the pheromone. In response
of presence of pheromone aggregation of donor and recipient (plasmid-free) cells is formed
leading to mating. In a broth medium pheromone induces donor cells to synthesize a plasmid-
encoded aggregation substance (AS). The As is a surface protein that binds to recipients and
initiates the contact necessary for transfer of plasmid DNA but As is not needed for conjugation
on solid medium. AS has has also been shown as a virulence factor. On acquiring a copy of
plasmid pheromone secretion activity is stopped and recipient statrt to act as donor. Pheromone
shutdown mechanism partly depends on plasmid-encoded peptide which acts as a competitive
inhibitor of pheromone. Precursors of the inhibitor peptides also resemble with unattached
signal sequences and are also the C-terminal hepta- or octa-peptide residues correspond to the
inhibitor.
Q. 66. Name some antibiotics which can be used for plasmid curing with examples.
Ans. Several quinolones group antibiotics (ciprofloxacin, nalidixic acid, norfloxacin, oxolinic acid,
pefloxacin, etc) and novobiocin in subinhibitory concentrations can eliminate plasmid from
enterobacteria, particularly R plasmids (R446b, R386, R386, pIP24, S-a) and a few virulence
plasmid (pWR105, pWR24 and pWR110), but at a low frequency.
Q. 67. Give some examples of plasmids of Archaea.
Ans. Several small plasmids have been commonly reported from Archaea and they are widely
spread among different genera and species of Archaea e.g., Pyrococcus abyssi and
Thermococcus sp. have been shown to possess single or multiple plasmids. Each of the three
strain of Pyrococcus abyssi possessed one plasmid of 3.5 kb (pGN27), 16.8 kb (pGN23) and
5.3 kb (pGN31) size, respectively, whilst the three other strains had two plasmids of 3.5 kb
(pSN559, PSIM560, pSN690) and 24 kb (pLN559, pLN560, pLIM690), respectively.
Q. 68. Write a short note on linear plasmids.
Ans. Linear plasmids are common in eukaryotic microbes. These are contemporary manifestations
of ancient viruses adjusted to mitochondria or to the cytoplasm. In both the cases, infectious
viral functions do not exist but display minimized gene equipment and an archetypical mode of
replication. Linear plasmids of filamentous fungi are selfish DNA elements routinely residing
in the mitochondria, where they coevoluted with their hosts. Besides the archetypical viral B-
type DNA polymerase, they exclusively encode a viral RNA polymerase. Rarely they affect the
host, some times, symptoms redolent of a molecular disease, as the accumulation of defective
mitochondria and early onset of senescence, manifest in plasmid-harbouring strains. Almost all
187
yeast linear plasmids known so far resides in cytoplasm thus have more complex enzyme
repertoire of viral origin for autonomous extranuclear and extramitochondrial replication and
transcription. They encode for a helicase, ssDNA binding proteins, and a capping enzyme
besides those present in similar elements in mitochondria. A few cytoplasmic linear plasmids
encode protein toxins (tRNase, whereas the other clearly involves a DNA-damaging mode of
action) which benefits the respective host in competing with other yeasts and impart their host a
killer phenotype.
All linear plasmids have a terminal structure called invertron. On the basis of invertrons,
there are two types of linear plasmids, hairpin plasmids having covalently closed ends and
plasmids with proteins bound on their 5′ end. Hairpin plasmids are common Borrelia, a human-
pathogen, cryptic hairpin elements are present in mitochondria of Rhizoctonia solani, a plant
pathogenic fungus. Plasmids having proteins attached on 5′ terminii are more common in nature
and often encode (in actinomycetes) for synthesis of antibiotics, degradation of xenobiotics and
heavy-metal resistance, and for growth on hydrogen as the sole energy source. Mjority of linear
plasmids have no define role in eukaryotes, thus are cryptic.
Q. 69. Write a shortnote on plasmids of Borrelia sp. and their probable origin.
Ans. Spirochetes Borrelia is unique among prokaryotes as they possess both linear double-stranded
DNA plasmids with covalently closed ends and circular plasmids. Linear plasmids have been
found in all species in size ranging 15 to 200 kb. Although linear plasmids are often cryptic but
in B. burgdorferi they encode for major surface proteins through ospA and ospB genes and are
responsible for antigenic variation in B. hermsii. Some people hypothesized that the linear
plasmids are actually minichromosomes because the finding indicates that the Borrelia
chromosome migrates in pulsedfield electrophoresis gels as a linear molecule of 950 kb similar
to the linear plasmids. Moreover, telomeric DNAs from the linear plasmids of B. burgdorferi
and B. hermsii when compared with telomeric sequences of other linear double-stranded DNA
replicons, sequence similarities were noted with poxviruses and particularly with the iridovirus
agent of African swine fever. The fact that African swine fever virus and a Borrelia sp. share
the same tick vector suggests that linear plasmids of Borrelia might have originated through a
horizontal genetic transfer across different kingdoms.
188
Chapter 7. Mutation and Mutagenesis
A randomly occurring change in genetic code of an organism is mutation. If the change is
somatic, non-inheritable it is not called mutation or just referred as somatic mutation. Somatic
mutation may sometimes be transferable to progeny produced by vegetative propagation. It may
results as a copying errors during cell division or through lesions caused by exposure to mutagens
or might occur as self induced. Lethal mutations do not accumulate but other (non-lethal or neutral)
keeps on accumulating to make a pool of genetic variation. Mutations are the evolutionary tools as
during natural selection non-useful mutations are discarded for more favorable mutations. Those
mutations do not influence the fitness of an individual to survive are called Neutral. DNA repair
mechanism of each and every individual keeps on mending day to day changes occurring in
genome before they become permanent still the system flaws and mutations occur. Each and every
individual of the same species has slight differences and can be recognized individually it is due to
difference in their physical make up mediated through their genetic code. This form of subtle
variations in DNA is called polymorphism. These variations some times lead to better health and
reproduction some times to diseases and weakness. This normal occurrence of variations in genes is
responsible for the existence of beauty in the nature.
Q. 1. What is mutation and wild type phenotype?
Ans. Mutation is any permanent change in the sequence of bases of DNA, irrespective of
phenotypic change. Due to mutations genes can exist in number of forms either with same or
variable activity, all different forms of a gene are called alleles. The form of gene in which it
existed in first bacterium isolate from nature is called the wild type alleles and all others are
called mutant alleles. Process of induction is called mutagenesis and can be achieved by one of
the several ways as chemical mutagenesis. The agents used for induction of mutation are called
mutagens. Common mutagens: Ethyl methane sulphonate (CH3CH2O. SO2.CH3), N-methyl-N-
nitro-N-nitrosoguanidine), irradiation (UV, γ).
1. Mostly mutations are the basis of evolution and provide raw material for development of
new strains, races etc.
2. Most mutations are random and non-adaptive in nature
Functional form of a gene is also some times referred as wild type genotype
Q. 2. How can you determine that the mutagen is acting in desired manner and is able to induce
mutations?
189
Ans. A satisfactory mutagen on treatment of large population of cells will generate about 5 to 10%
auxotrophic cells that will be unable to grow on defined minimal medium as M9. If you are
unable to isolate auxotroph in mutagenesis experiment than probably the mutagen is inactive.
Q. 3. Classify mutations?
Ans. According to origin: Spontaneous and induced.
Spontaneous mutations originate due to:
1. Errors occurring during replication (due to tautomerism in nucleotide bases, corrected by
proofreading DNA polymerase or by mismatch repair mechanism)
2. Spontaneous alteration in a nucleotide ( alteration in 5-methylcytosine, about 5% cytosine
bases in bacteria and viruses are present as methylated, an occasional loss of amino group from
cytosine or MeC changes it to uracil which then pairs with adenine rather than guanine, and GC
pair changes in to AT pair)
According to site: Random and site directed.
According to effect of mutation: Conditional and non-conditional, silent, leaky.
Conditional:
1. Temperature sensitive- product of gene is sensitive to temperature for its activity.
2. Suppressor sensitive- Difference in expression is due to presence of other gene(s) product
made by suppressor genes , it happens usually when mutation induce frame shift leading to
defective, shortened protein and are named as Am (UAG, amber mutation), Oc (UAA,
ochre mutation) or Op (UGA, opal mutations).
Non-conditional: Expressed all growth conditions.
Silent: When an amino acid is substituted (as result of point mutation or else) without having
detectible change in phenotype.
Or
A base change in a codon (usually at 3rd position) without having any change in amino acid of
coded protein.
Leaky: Activity of coded protein is slightly affected as in case of genes coding for enzymes
required for synthesis of essential substances. The bacterium can grow when essential substance
is present in medium, when absent growth is slow or absent. These can be brought about by:
1. Deletion of 3n bases, causing absence of 1 or more amino acid from completed protein.
2. Deletion or insertion leading to shift in reading frame so that all the codons under go
change.
3. Due to a chain termination mutation, a stop codon appears and truncated protein is formed.
190
According to number of changes:
1. Point mutation (involving only one base pair, further divided in to two: base pair
substitution and frame shift mutations).
2. Double mutation: If two changes are present either in single or two genes.
3. Deletion: By removal of all or a part of a gene.
4. Deletion-substitution: Replacement of all or a part of a gene.
5. Insertion: Genetic material is added without any removal of genetic material.
Q. 4. What do you mean by fluctuation test, what does it proved?
Ans. The test used by Luria and Delbruck, 1943) to prove spontaneous and random nature of
mutations which were thought to be brought under selection pressure. They showed that it is
population size rather than selection pressure which determine the number of mutants in any
population.
Q. 5. What are the uses of mutations?
Ans. Technique of mutation is used for;
1. To elucidated gene function.
2. To elucidate metabolic pathways.
3. To understand metabolic regulation.
4. To correlated between in vitro and in vivo functions.
5. To determine site of action of external stimuli/ agents.
6. To find out relationship between apparently unrelated systems
7. To elucidate interaction of one or more proteins.
8. To attenuate the strains for vaccine production etc.
Q. 6. What are different methods for isolation of mutants?
Ans. The different methods for isolation of mutants are a. Screening using master plate for replica
plating; b. Enrichment, non-growing cells (auxotrophic mutants) are enriched using radio active
metabolic precursors having β-disintegration and short half life (3H, 35S, 32P) or penicillin in
growth medium and, c. Selection.
Q. 7. What is ‘radioactive suicide’?
Ans. Radioactive suicide phenomenon is commonly exploited for enrichment of non-growing cells
(auxotrophic mutants) in a population by incorporation of radio active metabolic precursors
having β-disintegration and short half life (3H, 35S, 32P) in the growth medium, actively growing
191
cells incorporate radioactive element in its cell mass leading to concentration of radioactive
compounds in cells and eventually death of the cells.
Q. 8. What are different methods to isolate ts-mutants?
Ans. To isolate ts-mutants one uses same method as for other mutants, however permissive
conditions for growth is a lower temperature i.e., 30oC and selection or screening or enrichment
is done at 37oC.
Q. 9. What do you mean by cs-mutants (cold sensitive) mutants how they are isolated?
Ans. Cs-mutants are cold sensitive mutants of E. coli, wild type strain can grow at 20oC but cs-
mutant can not however it grows normally at ≥37oC. These are isolated similar to ts-mutants but
here permissive temperature for growth is kept ≥37oC while for enrichment or screening you
need to incubate at 20oC.
Q. 10. What may be different types of changes in bacterial cells when we select an antibiotic
resistant strain?
Ans. The common mechanism through antibiotic resistance or resistance to other substances is
noticed in bacteria might be due to:
1. Mutant cells may produce an enzyme that inactivate or hydrolyse the antibiotic or inhibitor.
2. Change in ligands on cells to bind the inhibitor, e.g., change in S12 ribosomal protein
prevent binding of streptomycin and mutants are resistant to the drug.
3. Mutant might have developed phenomenon to throw out the inhibitor from cells.
Tetracycline resistance may be either due to prevention of entry or active efflux of the
antibiotic.
4. Change in regulation of operation of genes as Tryptophan regulon change may lead to
resistance to methyltryptophan.
5. Mutant may produce excess amount of the target protein or other substance limiting the
effect of inhibitor.
Q. 11. What are different methods of genetic analysis of mutants?
Ans. The three common methods used for analysis of mutants are;
1. Genetic recombination to elucidate position of one gene in relation to other by two or three
factor crosses.
2. Complementation studies are useful in determining number of genes and their role in
regulation of a phenotype.
192
3. Deletion mapping for determining physical distance between two or more genes.
Q. 12. Although RNA viruses are prone to mutations, some RNA viruses have no or very few
variants. Why?
Ans. Some RNA viruses are remarkably invariant in nature. Probably these viruses have the same
high mutation rate as other RNA viruses, but are so precisely adapted for transmission and
replication that fairly minor changes result in failure to compete successfully with parental
(wild-type, wt) virus.
Q. 13. What do you understand by hot mutants, attenuated mutants and ts mutants?
Ans. Hot mutants grow better at elevated temperatures than the wild type strain. They may be more
virulent since host fever may have little effect on the mutants but may slow down wild type.
Attenuated mutants are those which cause much milder symptoms (or no symptoms)
compared to the parental- these are said to be attenuated. These have a potential role in vaccine
development and they also are useful tools in determining why the parental virus is harmful
Temperature sensitive (ts) mutants grow better at low temperature e.g. 31 oC but not
at e.g. 39 oC, wt grows at 31 and 39 degrees C. The reason for this is often that
the altered protein cannot maintain a functional conformation at the elevated temperature.
Q. 14. Dr. Mohanish isolated a strain of Salmonella Typhimurium that could grow as well as the
wild-type parent on rich medium, but when transferred from rich medium to minimal medium
the strain a very long lag phase compared to the wild-type parent. How could you determine if
the growth that occurred after the long lag phase was (i) due to a second mutation that permitted
growth in minimal medium or (ii) simply due to a requirement for longer time for adaptation to
the minimal medium?
Ans. A simple approach is to take an aliquot from the culture that grew in minimal medium after
the long lag phase and to subculture the bacteria in rich medium. After the culture has grown in
rich medium, subculture into minimal medium. If the original growth in minimal medium was
due to a mutation, the strain should begin growing soon after subculture. In contrast, if the
original growth in minimal medium was due to adaptation, it should require the strain just as
long to adapt upon subculture from rich to minimal medium as it took following the first
subculture.
193
Q. 15. Four independent mutations were obtained that affect the synthesis or assembly of fimbriae.
The properties of the mutations are described in the table below (where indicates that fimbriae
were produced and indicates that no fimbriae were produced).
(a) Based upon the above results, indicate whether each of the single mutations is a temperature
sensitive (Ts) or cold sensitive (Cs) mutation.
(b) Based upon the above results, what is the predicted order of the mutant gene products
in the pathway of fimbriae synthesis and assembly?
Growth temperature
Mutation
30°C 42°C to 42°C to 30°C
fim + + + +
fim (null allele) - - - -
fim-1 - +
fim-2 + -
fim-3 - +
fim-4 + -
fim-1 fim-2 - - + -
fim-1 fim-3 - - - -
fim-1 fim-4 - - - +
Ans.
(a) fim-1 and fim-3 are Cs mutations; fim-2 and fim-4 are Ts mutations.
(b) It is not possible to infer the order of fim-1 vs fim-3 because the double mutant is defective
under all growth conditions.
The results suggest that fim-2 acts before fim-1 because the product if only made if the cells are
first grown at 30°C where fim-2 is active, then shifted to 42°C where fim-1 is active – hence
fim-1 must convert an intermediate made by fim-2 into the product. The results suggest that fim-
1 acts before fim-4 because the product if only made if the cells are first grown at 42°C where
fim-1 is active, then shifted to 30°C where fim-4 is active – hence fim-4 must convert an
intermediate made by fim-1 into the product. Thus, the gene products probably act in the
pathway in the following order: fim-2 fim-1 fim-4
Q. 16. An amber suppressor mutation is needed for some useful bacterial genetic tricks. How could
you select for a suppressor mutation in Salmonella enteritidis?
Ans: Begin with a strain with amber mutations in two different genes and select for the
simultaneous repair of both mutant phenotypes. There are many ways you could do this, one
way is to begin with a S. enteritidis strain that has two amber mutations: one mutation
inactivates a gene required for resistance to the antibiotic tetracycline, and the other mutation
results in auxotrophy for tryptophan. By plating this strain on minimal medium containing
tryptophan and tetracycline, you can select for repair of both mutations. Reversion of two
194
separate mutations is very rare – for example, if the reversion frequency of the tet (Am)
mutation is 10-6 and the reversion of the trp (Am) mutation is 10-7, then the probability of the
two events occurring at the same time is the multiple of these numbers or 10-6 x 10-7 = 10-13, a
frequency too rare to readily detect. Therefore, simultaneous repair of both mutations is much
more likely to occur by another mutation that can suppress both amber mutations (i.e. an amber
suppressor mutant) designated “sup”.
Q. 17. Succinate dehydrogenase (sdh) catalyzes the conversion of succinate to fumerate under
aerobic conditions. Fumerate reductase (fdr) catalyzes the conversion of fumerate to succinate
under anaerobic conditions. Mutants defective for succinate dehydrogenase are unable to use
succinate as a carbon source. It is possible to isolate suppressors of a sdh mutant that allow
growth on succinate as a carbon source aerobically. List 6 different types of suppressors that
might be obtained and describe how each suppressor would restore the sdh+ phenotype.
Ans: Some examples include the following:
(i) Informational suppressor -- e.g. a missense suppressor tRNA which would occasionally
misread the sdh mRNA and produce a functional protein. [Note that if you begin with one
mutation, it is unlikely that you will find multiple classes of tRNA suppressors.]
(ii) An allele specific, intragenic interaction suppressor with a mutation at a second site in the
sdh gene that restores the structure and function of the Sdh protein.
(iii) A non-allele specific, intragenic suppressor with a mutation at a second site in the sdh gene
that increases the stability or activity of the Sdh protein.
(iv) An allele specific, interaction suppressor in another protein that interacts with the Sdh
protein that restores the structure and function of the Sdh protein.
(v) A bypass suppressor -- e.g. a mutation that allows the fdh gene product to catalyze the
synthesis of succinate under aerobic conditions. [Such mutations were actually found using
this approach.]
(vi) An over expression suppressor that increases the amount of the mutant Sdh protein to
sufficient levels that it allows cell growth.
(vii) A true revertant that restores the wild-type sdh DNA sequence.
Q. 18. What is polarity?
Ans. Polarity is failure to express downstream genes in an operon due to a mutation located
upstream. For example, a mutation in the lacZ gene which prevents expression of the lacY and
lacA genes.
195
Q. 19. What is the most likely cause of each of the two classes of mutations?
Ans. The first class of mutation is probably a promoter mutation that prevents RNA polymerase
binding and thus prevents expression of the entire lac operon. The second class of mutation
includes mutations in the lacZ gene that result in premature translation termination -- this may
be due to a nonsense mutation, a frameshift mutation, or an out-of-frame deletion.
Q. 20. How would you determine if a mutation was a nonsense mutation without directly
sequencing the DNA?
Ans: Testing for suppression with a collection of nonsense suppressors (i.e. suppressors caused by
mutations in tRNA genes that result in insertion of different amino acids instead of translation
termination).
Q. 21. Why are no nonsense mutations found among the first class of mutations?
Ans. The first class of mutations affects a DNA-binding site for RNA polymerase. Because this
DNA sequence is not transcribed or translated, any mutation in the sequence will not result in
translation termination and thus is not a nonsense mutation.
Q. 22. What causes the polarity of the two different classes of mutations?
Ans. The first class of mutations results in polarity because they directly prevent transcription of
the lac operon. The second class of mutations results in polarity because they cause translation
termination which, because Rho binds to the un-translated sequences, results in
transcription termination.
Q. 23. Many antibiotics act by binding to bacterial ribosomes and inhibiting translation. Resistance
to such antibiotics is often due to mutations that produce specific nucleotide changes in one of
the ribosomal subunits which prevent the antibiotic from binding to the ribosome. In such cases,
the antibiotic resistant allele is often recessive to the antibiotic sensitive allele. Why?
Ans. If the antibiotic stops the ribosomes in their tracks, it will prevent subsequent ribosomes from
transcribing that message. If cells have both resistant and sensitive ribosomes, both the resistant
and sensitive ribosomes will bind to the mRNA -- however, the sensitive ribosomes will
become stuck and block resistant ribosomes from translating the message. Thus, the sensitive
phenotype is dominant to the resistant phenotype.
Q. 24. How deletion and insertion mutations can occur? Describe the mechanism.
196
Ans. Deletion and insertion mutations can be formed in two ways: deletions or insertions of short
regions can occur by strand slippage, and deletions or insertions of longer regions can occur via
homologous recombination.
Strand slippage occurs by mispairing of the template strand and the newly synthesized
strand during DNA replication. If the newly synthesized strand denatures from the template
during DNA synthesis and if it is complementary to different stretches of the template strand, it
will occasionally pair with the wrong sequence. Such mispairing occurs in runs of a single
nucleotide or in short, directly repeated sequences. If the template strand loops out, then a
deletion will result. In contrast, if the newly synthesized strand loops out, then a deletion will
result.
In addition to strand slippage, deletions or insertions can be formed by intra-molecular
recombination between direct repeats of homologous DNA. Recombination can also result in an
insertion mutation, duplicating the material between a pair of direct repeats. One way that this
can happen is by recombination between sister chromosomes at the DNA replication fork. The
probability of forming a deletion or insertion depends on the length of DNA homology between
the direct repeats. Recombination between longer repeats will occur more often than
recombination between shorter repeats. RecA dependant homologous recombination requires at
least 25 bp of perfect homology. However, recombination even between very short repeats (<10
bp) is a major source of deletions. Thus, direct repeats are "hot spots" for deletion formation.
Q. 25. Describe frame shift mutations in brief.
Ans. Frame-shift mutations occur by strand slippage and mispairing during DNA replication.
Mispairing usually occurs within runs of a particular nucleotide. Following strand slippage, the
denatured strand can reanneal with one or more base pairs bulged out and the downstream
nucleotides correctly base paired. If this mispaired strand if elongated by DNA polymerase, the
mutation can be "fixed" in the new sequence. When this double-stranded DNA is subsequently
replicated it will produce one daughter with the frame-shift mutation and one daughter with the
wild-type sequence. This process can introduce either plus (the addition of 1 or 2 bp) or minus
(the loss of 1 or 2 bp) frame-shift mutations.
Q. 26. How can you move mutations from a cloned gene onto the chromosome? Explain with
examples.
Ans. By the use of conditional replicons ("suicide plasmids") to contruct genetic duplications and
null alleles. Plasmids that are conditional for their replication can beused to create defined
duplications within a target genome. In such instances a specific DNA fragment is inserted into
197
a plasmid which is then introduced into a recipient strain and placed under conditions where the
plasmid cannot replicate. Since the plasmid cannot replicate, selection for some property of the
plasmid, such as an antibiotic resistance marker, results in isolates that have integrated the
plasmid into the host chromosome via homology between the cloned fragment and the
corresponding region of the recipient chromosome. (This system can also be used to obtain gene
disruptions by cloning an internal portion of a gene sequence into such a suicide plasmid, thus
generating two incomplete gene copies upon integration of the plasmid into the chromosome.)
Plasmid vectors used as the backbone for such insertional mutagenesis have several properties
desirable for this type of gene inactivation. (1) The plasmid must be conditional for replication
to allow selection for integration into the chromosome. This can be achieved by using a plasmid
that is able to replicate autonomously only in permissive hosts or by using conditional replicons
(e.g. a plasmid that is temperature sensitive for replication). (2) The plasmid must carry a
selectable marker (e.g. antibiotic resistance). (3) Ideally, the plasmid should be transferable to a
variety of other bacteria. Plasmids that can be transferred by conjugation are preferable for
situations in which other means of transfer such as transformation or electroporation are not
efficient. (4) It is convenient if the plasmid has an array of unique cloning sites. The plasmid
pGP704 is an example of such a system. Plasmid pGP704 is a derivative of pBR322 that is
AmpR but has a deletion of the pBR322 origin of replication (oriE1) but that carries, instead, a
cloned fragment containing the origin of replication of plasmid R6K. The R6K origin of
replication (oriR6K) requires for its function a protein called pi, encoded by the pir gene. In E.
coli the pi protein can be supplied in trans by a prophage (lambda pir) that carries a cloned copy
of the pir gene. The plasmid also contains a 1.9-kb BamHI fragment encoding the mob region
of RP4. Thus, pGP704 can be mobilized into recipient strains by transfer functions provided by
a derivative of RP4 integrated in the chromosome of E. coli strain SM10. However, once
transferred it is unable to replicate in recipients that lack the pi protein.
Insertion mutations in a chromosomal gene are isolated by first subcloning DNA fragments
into pGP704, then mating the plasmid clone into a recipient which lacks the pir gene. Because
these plasmids cannot replicate in the recipient, the cloned gene can be incorporated into the
recipient genome by homologous recombination involving two crossovers between the the gene
present on the plasmid and the corresponding gene in the chromosome. The procedure requires
that the mutation has an easily selectable phenotype (for example, if the cloned copy of the gene
is disrupted with a KanR insertion). This method can also be used for integrating gene fusions
on plasmids into the chromosome to study expression of the gene fusion in single copy and its
wild-type context.
198
1. Use of suicide vectors for allelic exchange of nonselectable mutations. Because double-
crossover events that incorporate a gene from a plasmid into the chromosome are rare, it is not
feasible to simply screen for such events if the cloned gene cannot be directly selected. In such
cases, a two-step procedure is used instead. First, the entire plasmid is integrated into the
chromosome by a single-crossover between the homologous genes, producing a chromosomal
duplication.
Second, the chromosomal duplication is segregrated by homologous recombination between
the flanking direct repeats, ultimately leaving one copy of the gene on the chromosome -- either
the wild-type copy or the mutant copy. Because the direct repeats are often short, the desired
recombination event may be very rare. The ApR phenotype of the plasmid provides a direct
selection for integration of the plasmid into the chromosome. Once the plasmid is integrated, it
is possible to simply screen for segregrants by testing for the loss of the plasmid ApR marker.
However, because the segregrants are rare, often a way of selecting for the loss of the integrated
plasmid from the chromosome is needed. This can be accomplished by including a counter-
selectable marker on the plasmid. One way of selecting against the integrated plasmid is to use a
plasmid that also carries the sacB gene from Bacillus subtilis. Expression of the sacB gene is
toxic for gram-negative bacteria when grown in the presence of 5% sucrose, providing a direct
selection for loss of the plasmid. The resulting segregrants SucroseR colonies are screened for
the simultaneous loss of ApR to ensure that that the sucrose-resistant phenotype is due to loss of
the integrated plasmid.
Many modifications on this basic theme have been used in a variety of bacteria. Two other
slightly different approaches are worth describing. (1) The inability of plasmids with a colE1
origin to replicate in some bacterial species has been used for allelic exchange in Pseudomonas
aeruginosa. (2) Another approach takes advantage of the recessive nature of streptomycin
resistance conferred by chromosomal rpsL mutations. Plasmid pRTP1 is a colE1 replicon
carrying the wild type rpsL gene in addition to AmpR. The colE1 origin does not support
replication in a variety of bacterial species (ie. Bordetella pertussis), and thus acts as a suicide
vector when delivered from E. coli into such strains. Integration of the plasmid carrying the
wild type rpsL gene into a StrR recipient results in a merodiploid strain that is StrS. Subsequent
selection for StrR results in colonies that have lost the vector (like the segregration of sacB
vectors described above).
2. By using incompatible plasmids. The incompatibility of certain plasmids has been widely used
to select for marker exchange in a variety of gram-negative species. This procedure requires
three steps. (1) The mutant DNA fragment is cloned into a broad host range plasmid -- for
example, a KanR insertion mutation in a gene may be cloned into a TetR pLAFR derivative
199
maintained in an E. coli host with the necessary mobilization functions. (The pLAFR plasmids
require helper functions for their mobilization). (2) The KanR TetR pLAFR plasmid is then
mated into the recipient strain by selecting for TetR (with an appropriate counterselection
against the donor cells). (3) Recipients carrying this pLAFR plasmid are then mated with a
strain carrying a second IncP1 plasmid with a different selectable marker. Plasmid pPH1JI
which confers resistance to gentamicin (Gen) is often used for this step (sometimes called "the
kickout step"). Since two IncP1 plasmids cannot coexist in the same cell, selection for the GenR
from the pPH1JI plasmid results in loss of the KanR TetR pLAFR plasmid. However, if both
GenR and KanR are selected, exconjugants arise that have lost the original TetR pLAFR
plasmid, and the KanR marker has integrated into the chromosome via the homology
surrounding the insertion. Loss of the pLAFR vector can be confirmed by scoring for TcS.
Plasmid pPH1JI is somewhat unstable at high temperature, such that it can be cured from the
final strain at a frequency of between 5 and 20% by simply growing overnight at 42 degrees in
the absence of antibiotic selection.
After putative mutants have been isolated, the presence of the chromosomal insertion should
be verified. If the mutation is an insertion or deletion that results in a significant size difference,
the allelic exchange event can confirmed by Southern hybridizations or colony PCR. If the
mutation is a base substitution, the colony PCR product can be directly sequenced or confirmed
by RPLP analysis if the mutation alters the restriction digest pattern of the amplified fragment.
Q. 27. What are counter selectable markers, why they are used, give some examples?
Ans. Counter selectable markers are those coded by counter-selectable genes which under
appropriate growth conditions promote the death of the microorganisms harboring it.
Consequently, counter-selectable markers have been used for the positive selection of mutants
that have undergone defined genetic alterations leading to the loss of the marker. In different
studies, applications such as the construction of mutants, the isolation of insertion sequence (IS)
elements, and the curing of plasmids have been described. The most-used counter selectable
markers are the genes that confer sucrose, streptomycin, or fusaric acid sensitivity. They have
been used to construct mutants or vaccine strains in Mycobacterium tuberculosis, Helicobacter
pylori, Bordetella pertussis, and many other bacteria. The 3 most common systems are:
The fusaric acid sensitivity system. The first counter selectable marker described was tetAR, a
gene coding for tetracycline resistance or sensitivity. Expression of tetAR results in alteration
of the host cell membrane which interferes with tetracycline permeation and thereby renders
the cell resistant to tetracycline. These alterations render the bacteria hypersensitive to
lipophilic chelating agents such as fusaric or quinalic acids. It is therefore possible to select
200
clones which have lost the tetAR gene by virtue of their resistance to fusaric acid. This system
is, to the best of our knowledge, effective only in Escherichia coli, in which the threshold of
fusaric acid sensitivity is strongly dependent on the host strain. As a consequence, this marker
was mostly used for abrogating tetracycline resistance in E. coli that had been previously
mutagenized with a transposon harboring a tetracycline cassette. One problem encountered
with this system is the generation of out-of-frame deletions with polar effects on downstream
genes, which obscure the interpretation of the phenotype.
The streptomycin sensitivity system. The streptomycin sensitivity system takes advantage of
the fact that the S12 ribosomal protein is the target of streptomycin, a widely used antibiotic.
Mutations in the rpsL gene which encodes this protein are responsible for resistance to high
concentrations of streptomycin. However, resistance is recessive in a merodiploid strain. When
both wild-type and mutant alleles of rpsL are expressed in the same strain, the strain is sensitive
to streptomycin, possibly because of a general inhibition of translation by the wild-type
ribosome. Consequently, it is possible to select mutants that have lost the wild-type allele
encoding streptomycin sensitivity by plating the culture on streptomycin. Positive selection
based on streptomycin sensitivity functions in different bacteria, such as E. coli, B. pertussis, or
Mycobacterium smegmatis. A prerequisite for this technology to be effective is a streptomycin-
resistant strain. Therefore, the use of rpsL as a counterselectable marker leads to mutants
resistant to streptomycin, an antibiotic which is used in some chemotherapeutic regimens. This
is a severe limitation, particularly in the construction of attenuated strains meant to be used as
live vaccine candidates.
The sucrose sensitivity system. The Bacillus subtilis sacB gene coding for levansucrase is
perhaps the most popular counterselectable marker. In its natural gram-positive environment its
expression is harmless to the bacterium. However, cloning of sacB in E. coli and other gram-
negative bacteria leads to the death of the transformed bacteria when they are plated in the
presence of sucrose. The mechanism of this toxicity is not completely understood, but it has
been proposed that the accumulation of levans (high molecular-weight fructose polymers
synthesized by the levansucrase) in the periplasm of gram-negative bacteria might be toxic.
The effect of the sacB gene in the presence of sucrose has been demonstrated in most of the
gram-negative bacteria in which it has been tested, including Erwinia chrysanthemi, Legionella
pneumophila, Anabaena sp., Yersinia sp., Rhizobium sp., Xanthomonas sp., Pseudomonas
aeruginosa, Klebsiella pneumoniae, H. pylori, and Agrobacterium tumefaciens. The same
mechanism of toxicity has been invoked for Corynebacterium and Mycobacterium sp., the only
gram-positive bacteria for which sacB was shown to be functional as a counter selectable
marker.
201
Q. 28. Describe some immunological method(s) to identify mis-sense mutations.
Ans. Temperature sensitive mutants are nearly always due to missense mutations in a structural
gene for a protein. With this exception, it is usually difficult to determine if a mutation which
causes complete loss of an enzymatic activity is due to a missense mutation or some other type
of mutation. One way to identify missense mutations affecting a particular gene is to test for the
presence of protein using a specific antibody that recognizes that protein. A positive antibody
reaction is described as immunological cross-reacting material or CRM for short. Testing for
CRM is not trivial. It requires antibody to the wild-type protein. To make polyclonal antibody
against a particular protein, the protein must first be purified. The purified protein is then
injected into animals (e.g. rabbits) to stimulate the production of antibodies that recognize the
protein. A good antibody response typically requires a booster injection also. The process
usually takes many weeks before an adequate level of antibodies has been developed. Serum is
then collected from the animals and assayed for specific antibodies to the wild-type protein in
vitro. Once the specific antiserum has been produced, it can be used to test for the presence of
the protein in cell extracts. Even though the protein may be inactive due to a missense mutation,
if the protein is made by the mutant the extract will have cross-reacting material (CRM+).
However, if the mutant does not make the protein or if the protein is rapidly degraded, then
cross-reacting material will not be detected in the extract (CRM-). The tests for CRM can be
done in microtiter dishes with 96 wells, allowing many mutants to be tested at one time. Thus,
although this process of obtaining the antibody is rather slow, once the specific antibody is
available, it is possible to rapidly test for mutants that make the inactive protein using simple
immunological tests.
Q. 29. What do you mean by localized mutagenesis?
Ans. A small DNA fragment can be randomly mutagenized in vitro by exposing it to a chemical
mutagen. The DNA fragment can then be moved into an appropriate recipient cell by taking
advantage of a linked selectable genetic marker. This approach allows localized random
mutagenesis of a small, specific region of DNA without producing secondary mutations
elsewhere on the chromosome. Such localized mutagenesis is especially useful for obtaining
rare point mutations in or near a gene of interest (for example, temperature-sensitive mutations
in a gene, mutations in the promoter or operator of a gene, or mutations that affect amino acids
at the active site of an enzyme).
Q. 30. Name a suitable chemical for localized mutagenesis.
202
Ans. Hydroxylamine is a very useful mutagen for localized mutagenesis. Unlike many mutagens,
hydroxylamine can mutagenize DNA packaged inside of phage heads. This allows the
mutagenesis of transducing particles in vitro.
Q. 31. Write down the steps in localized mutagenesis using phage transduction.
Ans. Localized mutagenesis of transducing DNA with hydroxylamine involves several steps. (1) A
phage lysate containing transducing particles is mutagenized with hydroxylamine in vitro. The
extent of mutagenesis of the phage particles can be monitored indirectly by following decrease
in phage titer (or "killing") due to mutations in essential phage genes, or directly by following
the increase in clear plaque phage mutants in the lysate.
(2) Recipient cells are then transduced with the mutagenized lysate, selecting for a specific
marker carried by the transducing particle. When the transducing fragment is recombined onto
the chromosome, only the small, localized region carried on that mutagenized transducing
fragment is inherited.
(3) The resulting transductants are then screened for point mutations that affect the linked gene
(that is, the "unselected marker). The frequency of such mutations will depend upon how
heavily the transducing lysate was mutagenized and how closely linked the selected and
unselected markers are.
Q. 32. Do all mutations are random and non-induced type?
Ans. No, in 1988 Cairns and colleagues renewed the controversy about whether some mutations are
induced by the selective. In contrast to the experiments done by Luria and Delbruck where any
T1s mutants rapidly died on the selection plates, Cairns and colleagues used a "non-lethal
selection" E. coli Lac- mutants were plated on medium with lactose as a carbon source and the
number of Lac+ revertants were determined over a period of several weeks. Under these
conditions the Lac- cells cannot form visible colonies, but the cells do not die. They observed
that although some Lac+ revertants appear within a few days as expected for pre-existing
revertants, Lac+ revertants continued to appear over time suggesting that starvation of the Lac-
mutants stimulated mutation to Lac+. Hence, they dubbed this process "adaptive mutagenesis".
Starvation (and possibly other stressful conditions) may stimulate increased mutagenesis
(possibly by several different mechanisms), and only those mutations that allow the cells to
respond to the environmental stress accumulate in the resulting population.
Q. 33. What is mutation rate, how it is calculated, how it differs from mutation frequency?
203
Ans. The mutation rate is the number of mutations per cell division. Instead of the mutation rate,
often the "mutant frequency" is reported. The mutant frequency is simply the ratio of mutants
divided by the total number of bacteria in the population. Because the cell population is so
large, the number of cell divisions is approximately equal to the number of cells in the
population (N). The mutant frequency is easy to measure, but is less reliable than the mutation
rate because it may show large fluctuations depending upon when the first mutation appeared in
the population.
Q. 34. What is the typical mutation rate in E. coli, is it same for all genes?
Ans. For E. coli typical mutation rate works out to be between 10-6 and 10-7 mutations per gene per
generation. No, there are certain "hot spots" or "cold spots" for spontaneous mutations. A "hot
spot" is a site that has a higher rate of mutations than predicted from a normal distribution, and a
"cold spot" is a site with a lower rate of mutations than predicted from a normal distribution.
Higher eukaryotes too, have the same rate of spontaneous mutation.
Q. 35. Why mutations are too commonly observed in bacteria and what is major consequence of
mutations?
Ans. In bacterial populations mutations are constantly arising due to errors made during replication.
If there is any selective advantage for a particular mutation (e.g. antibiotic resistance), the
mutant will quickly become the major component of the population due to the rapid growth rate
of bacteria. In addition, since bacteria are haploid organisms, even mutations that might
normally be recessive will be expressed. Thus, mutations in bacterial populations can pose a
problem in the treatment of bacterial infections. Not only are mutations a problem, bacteria
have mechanisms by which genes can be transferred to other bacteria. Thus, a mutation arising
in one cell can be passed on to other cells.
Q. 36. To determine the frequency of StrR mutants a fluctuation test was done with 50 tubes each
containing 108 cells and 42 of the tubes contained no mutants. Use the Luria-Delbruck
calculation to determine the mutation rate to StrR.
Ans. First calculate the average number of hits per cell, h = -ln (42/50) = -ln(0.84) = 0.17
Then divide the average number of hits per cell by the number of cells in the population a
= h / N = 0.17 / 108 = 1.7 x 10-9
Q. 37. To determine the frequency of putP mutants a fluctuation test was done using 20 tubes with
a final concentration of 107 bacteria each. From each tube 0.1 ml of culture was plated on
204
medium that selects for putP mutants. Seventeen of the tubes yielded putP mutants but 3 of the
tubes yielded no mutants. Based upon these results, use the Luria-Delbruck calculation to
determine the mutation rate to putP-.
Ans. First calculate the average number of hits per cell, h = -ln (3/20) = 1.9
Then divide the average number of hits per cell by the number of cells in the population
a = h / N = 1.9 / 107 = 1.9 x 10-7
Q. 38. Suggest two reasons why the rate of mutation to StrR is so much less than the rate of
mutation to Pro-.
Ans. One reasonable explanation is that any mutation that disrupts any of the proline biosynthetic
genes would result in a Pro- phenotype, but only very specific base substitutions in ribosomal
genes result in streptomycin resistance (i.e., Str resistance is a smaller target size for mutations)
and this is the actual reason. A second potential reason could be that there are redundant genes
that encode the wild-type Str sensitive phenotype and the Str resistant mutant phenotype is
recessive to the wild-type.
Q. 39. What is a mutator phenotype and mutator gene? Give examples.
Ans. Inactivation of any of the gene products responsible for mechanisms that limit errors
following DNA replication results in an increased spontaneous mutation rate is called a
"mutator" phenotype and the defective gene responsible for mutator phenotype is called mutator
gene. A few examples of mutations that yield a mutator phenotype are as follows:
Repair system Mutations Increase in
mutation
frequency
Major types of mutation
Mismatch repair mutHSL 102-103 Transition, Frame shifts
Proofreading mutDc 103-104 Transitions, Transversions, Frame shifts
G-A mismatch repair mutT 103-104 AT to CG Transversions
G-A mismatch repair mutY > 102 GC to TA transversions
Q. 40. What are the two major types of mutations?
Ans. 1. Base substitution leading to missense or silent mutations.
2. Insertion and deletions leading to frameshift or chain termination mutations.
Q. 41. Write down the characteristics of the different types of mutations?
205
Ans.
Mutation
Type
Missense Nonsense Frameshift Deletion Insertion
Phenotype Silent, leaky,
null
Null (rarely
silent and
leaky)
Null (rarely silent
and leaky)
Null (rarely
silent and
leaky)
Null (rarely
silent and
leaky)
Reversions True reversion
Missense
suppression
Intragenic
suppression
True reversion
Nonsense
suppression
Intragenic
suppression
True reversion
Frameshift
suppression
Intragenic
suppression
True reversion
Bypass
suppression
No-True
reversion
Bypass
suppression
Change in
protein
Amino acid
substitution
Premature
translation
termination,
truncated
protein
Multiple amino acid
substitution,
Premature
translation
termination
Addition of
amino acids,
Premature
transcriptional
or translation
termination
Loss of amino
acids or
change in
down stream
amino acids
Q. 42. List different mutagens and their mechanism of action.
Ans.
Mutagen Mode of action
X-rays (5 nm wavelength) Causes single and double strand breaks
UV rays (254 nm) Form pyrimidine dimers between adjacent pyrimidine bases
Nitrous acid (HNO2) Deamination and intra-strand cross-linking
Hydroxylamine (NH2OH) Hydoxylate cytosine
N-methyl-N’-nitro-N-nitrosoguanidine
(MNNG)
Leads to production of 6 methyl guanine at replication fork and give
rise to transition
Ethyl methane sulfonate
(CH3SO3C2H5) (EMS)
Alkylate purines to cause transition
Methyl methane sulfonate
(CH3SO3CH3) (MMS)
Alkylate purines to cause transition
2-aminopurine (base analog of
adenine)
May replace adenine through hydrogen bonding with cytosine
5-bromouracil (base analog of
thymine)
Replaces thymine as it hydrogen bonds with guanine
Acridine orange (intercalating dye) Causes frame shift mutation
ICR191 (nitrogen mustard)
(intercalating)
Causes frame shift mutation
Mitomycin C and dimethyl psoralen Crosslink two DNA strands, psoralen get activated by 360 nm light
Transposons Insert themselves randomly causing to deletion and inversions
Mutator mutations Mutations affecting DNA replication as mismatch repair system (mutS,
H, U), 3’to 5’ exonuclease activity of DNA polymerase III (dnaQ)
Artificial creation by mutagenic
techniques e.g. Site directed
mutagenesis
Mutations are generated by using mutated primers in PCR and then
PCR products are cloned.
Q. 43. If another Aro- mutant was isolated that behaved the same as the aroB mutant in cross
feeding experiments, describe a genetic test you could do to determine if both mutations affect
the same enzymatic step. [No DNA sequencing or enzyme assays allowed.]
Ans. Note that if the mutations behave the same in cross feeding experiments, doing additional
cross feeding experiments would not be a very good approach. Assuming that both mutations
were recessive, and then the best test would be an in vivo complementation test with two copies
206
of the genes, one copy with a mutation in aroB and one copy with a mutation in the new Aro-
mutant. This test must come with controls for each mutation against itself.
Q. 44. What results would you would obtain if the two mutations affected different enzymatic steps
are grown together.
Ans. If the two mutations affected two different steps, then the complementation test would allow
growth without aromatic amino acids.
Q. 45. Given many mutations in a gene, how could you identify deletion mutations?
Ans: Deletion mutations typically do not revert, so failure to revert is a hint that a mutation may be
a deletion. Failure to produce wild-type recombinants with several other mutations that can
recombine with each other is stronger evidence that a mutation is a deletion.
Q. 46. Given a collection of mutations that affect a gene, including several deletion mutations, how
would you construct a deletion map?
Ans. Recombination tests with pair-wise combinations of each of the mutations failed to produce
wild-type recombinants indicates that two mutations affect the same base pair. This could be
due to two point mutations that affect the base pair, a point mutation and a deletion that affect
an overlapping DNA sequence, or two deletion mutations that affect an overlapping DNA
sequence. However, if two mutations that recombine with each other both fail to recombine
with a third mutation, the third mutation is likely to be a deletion that removes the DNA
corresponding to the mutant base pairs in each of the other two mutations.
Q. 47. Why are there so many mutations in some deletion intervals and not in others?
Ans. There are several reasons why the number of mutations that map within a deletion will not be
uniform. The larger a deletion interval the more targets there are per round of mutagenesis. The
nature of the mutagen used will affect the spectrum of mutations recovered; each mutagen will
may "hot spots" for mutagenesis which may not be evenly distributed throughout the gene. Not
all residues of the gene product are essential for proper structure and/or function. The codons
corresponding to the critical residues will be the only mutable sites recoverable as missense
mutants. Such critical residues are often unevenly distributed within the gene.
Q. 48. What are the main drawbacks in using NTG as mutagen?
Ans. N-methyl-N’-nitro-N-nitroguanidine (NTG) is a powerful mutagen which alkylates the bases
inducing mutagenesis by mispairing and it is independent of error prone repair system. Its main
207
drawback is it tends to cause multiple clustered mutations (acting on replication fork) thus
complicating subsequent genetic analysis.
Q. 49. How can you induce mutagenesis in plasmids? Give protocol for one method.
Ans. Hydroxylamine has commonly been used for mutagenesis of plasmids, besides one can also
use transposon mutagenesis in the same way as used for bacteria.
For hydroxylamine induced mutagenesis following steps are followed:
1. Five µl of purified plasmid DNA is mixed with 30 µl distilled water, 20 µl of phosphate EDTA
buffer (0.5 M potassium phosphate buffer pH 6 and 5 mM EDTA) and 45 µl of 7 M freshly
prepared hydroxylamine (350 mg hydroxylamine is dissolved in 560 µl of 4 M NaOH and then
volume is made to 5 ml).
2. Incubate overnight at 37oC, drop dialyze the DNA to remove hydroxylamine and transform
appropriate recipient.
Q. 50. What do you mean by drop dialysis of plasmid DNA?
Ans. Drop dialysis is done to remove unwanted materials from DNA, it is done as follows:
1. Take 10 ml of TE buffer in a petriplate, pick up a sterile 25 mm membrane (pore size 0.22 µm)
and float it in petridish over TE keeping shiny side up and allow it to be completely wet and
carefully place 20-100 µl of DNA in the centre of the filter. Keep it undisturbed for 2 hrs at RT.
Carefully take up the DNA with micropipette.
Q. 51. How the site directed mutations are introduced in a specific gene?
Ans. Site directed mutagenesis can be achieved by exploiting in vitro synthesis of mutated DNA
and then introducing the cloned mutated sequence into the target bacterium. We can deliberately
incorporate a mismatch into primer or we can design a primer with deletion or can use different
method to introduce deletions in specific sequence as by cutting with selected RE.
Q. 52. What is flip flop mechanism; give some examples from bacteria and bacteriophages?
Ans. By site specific recombination mechanism specific sequence of DNA get inverted to turn off
one gene while turn on the other alternative gene it is called flip flop mechanism. e.g.,
production of alternative tail fibres by Mu and P1 phages, production of two or more than two
sets of flagellins by Salmonella, antigenically different type I Fimbriae of E. coli and Morexella
bovis.
Q. 53. Explain the phase variation of flagellar antigens of Salmonella enterica.
208
Ans. Flagellar antigens are different in different phases of growth and bacteria achieve this ability
through flip-flop mechanism of gene control to off or on the required gene controlling specific
flagellar protein. Two phases of flagella of Salmonella are encoded by two genes, H1 and H2.
Acell can alternate between by expression of these two genes by inversion of a short sequence.
In response to some external stimuli rh1 is activated, its product act as suppressor to H1 and in
same operon a gene hin (similar to gin of Mu phage and cin of P1 phage) product induces
inversion of sequence through site specific recombination to turn off H1 and turn on H2 or vice-
versa.
Q. 54. What is negative selection in mutagenesis experiments?
Ans. When we select for loss of character rather than for acquisition of new character in
mutagenesis it is called negative selection. For examples during signature tag mutagenesis with
the help of tagged transposon selection is made for the identification of loss of virulence, or
using replica plating selection for not growing auxotrophic mutants, or enrichment and then
selection of auxotrophic mutants.
Q. 55. What is hypermutation, how you can achieve it?
Ans. Under normal conditions some mutations occur in each and every individual but due to some
known and unknown causes the frequency increases to the illogical limits. It usually occurs
when damage to genome is beyond normal repair system and SOS system is activated as result
of rescue operation called by the protection system which reconstructs the genome.
Q. 56. What are the different DNA repair systems in cells and their components?
Ans. Most cells have five kinds of DNA repair systems, viz.,
a. Direct repair system: Act directly on damaged nucleotide, replaces it with normal one. (e.g.,
DNA photolyases, O6-methylguanine-DNA methyltransferase)
b. Base excision repair: Remove damaged base of nucleotide creating AP site which is then
mended by DNA polymerase. (e.g. DNA glycosylases, AP endonucleases, DNA polymerase I
and DNA ligase).
c. Nucleotide excision repair: It directly remove damaged nucleotide replacing with a normal one.
(e.g. ABC exonucleases, DNA polymerase I, DNA ligase)
d. Mismatch repair: Corrects errors in replication by excising a stretch of ssDNA containing
offending DNA and then repair the gap. (e.g., Dam methylase, MutH, MutL, MutS proteins,
DNA helicase II, SSB, DNA polymerase III, exonuclease I, DNA ligase)
e. Recombination: Repair double stranded break. (e.g. RecABC, ligase etc.)
209
Q. 57. What do you understand with short patch and long patch repair system in E. coli?
Ans. These are nucleotide excision repair systems of E. coli. In short patch repair process up to 12
nucleotide base length can be excised at one time through UvrABC endonuclease or also called
exonuclease (excinuclease). Long patch repair system needs Uvr proteins and can excise up to 2
kb long damaged patch of DNA.
Q. 58. What do you understand by dark repair system?
Ans. DNA damaged by UV radiation gets easily repaired in error free method by photolyase
activation which repairs the dimers in presence of light but even when there is no light repair
occurs in error free way by short patch repair or nucleotide excision repair method which called
dark repair. In short patch repair process up to 12 nucleotide base length (20 bp in eukaryotes)
can be excised at one time through UvrABC endonuclease or also called exonuclease. This
method of repair is more effective against the pyrimidine-(6-4)-pyrimidone structure. It is
brought about by excinuclease
Q. 59. What are the different genes and their products of SOS system of E. coli?
Ans. lexA Encode for repressor of SOS genes as method of autoregulation.
polB (dinA) encode for polymerization unit of DNA polymerase II.
uvrA, uvrB, uvrC encode ABC exonuclease
umuC, umuD encode Protein required for error prone repair
sulA encode a cell division inhibitor protein
recA encode RecA protein for error prone repair
There are other genes too with unknown function which get activated during SOS response
e.g., dinB, dinD and dinF and the genes which are known function in DNA metabolism but
without any repair role e.g. ssb (encoding for single strand binding proteins, SSB), uvrD
(encoding for DNA helicase II and himA (encoding a unit of integration host factor required
during recombination, replication and transposition), dnaG, encode for DNA primase to trigger
new round of replication.
Q. 60. What is the difference between eukaryotic and prokaryotic long patch repair systems?
Ans. In prokaryotes as E. coli Long patch repair system needs Uvr proteins and can excise up to 2
kb long damaged patch of DNA while in eukaryotes replacement involve only 24-29
nucleotides of DNA.
Q. 61. Mitomycin C and dimethyl psoralen crosslink two DNA strands to induce mutations;
however, psoralen is used widely by most researchers, why?
210
Ans. Dimethyl psoralen get activated by 360 nm light to produce cross linking in two DNA strands
therefore it gives better control over timing of cross linking event to induce mutations.
Q. 62. What are the two types of structures formed on exposure of DNA to UV rays?
Ans. Two types of dimers formed during UV induced DNA damage are cyclobutane dimers and
pyrimidine-(6-4)-pyrimidones.
Q. 63. According to time of damage (with respect to replication) to DNA, how many types of
damages are recorded? What are their characteristics?
Ans. These can be of two types viz., extrareplicational and intrareplicational damage.
Intrareplicational damage occurs due to mistakes made by DNA polymerases without any
alteration in DNA duplex structure, although it is a heteroduplex. If the damage is not repaired
next cycle of replication may produce two types of homoduplexes, one normal and other
mutant.
Extrareplicational damage is caused by number of chemical and physical mutagens.
Q. 64. Haemophilus influenzae and Streptococcus pneuminae are resistant to mutagenic effect of
UV rays, why? What methods can be used to create mutation in these pathogens?
Ans. Both of the bacteria are natural mutant to umuC/umuD genes therefore defective in
production of Protein required for error prone repair in SOS repair mechanism, thus they are
repaired either through error free photolyase or dark repair mechanism, therefore UV either kills
the bacteria or it may not induce error prone SOS mechanism. The mutgens which are
independent of SOS as nitrosoguanidine and EMS (ethyl methylsufonate) directly creating
mispairing lesions can be used for mutagenesis in Haemophilus inflenzae and Streptococcus
pneuminae.
Q. 65. What are the factors influencing the severity of nonsense polarity?
Ans. The important factors are:
1. Time RNA polymerase spends at pause site before starting transcription spontaneously.
2. Efficiency of Rho protein action.
3. Distance between nonsense codone and the pause site.
4. Distance between the pause site and beginning of the next gene.
Q. 66. What are different kinds of bacteria used in Ames test?
211
Ans. There are Salmonella and E. coli systems of Ames mutagenicity tests, in the former one
reversion of histidine mutant of S. Typhimurium is used while in latter E. coli mutants for
tryptophan synthesis is observed for reversion on exposure to a mutagen.
Q. 67. What is the sensitivity of Salmonella and E. coli systems of Ames mutagenicity tests?
Ans. Salmonella system is about 84% sensitive, i.e., it finds 84% of the known mutagens while E.
coli system is about 91% sensitive.
Q. 68. In Ames test for mutagenicity assay, why liver microsomal fraction is included?
Ans. It is because some chemicals are not mutagenic themselves but converted to mutagen by liver
enzymes.
Q. 69. If a cell contains 3000 genes and the average mutation rate for each gene is 1 x 10-6 per
generation, what is the average number of mutations per cell per generation?
Ans. 3000 x 10-6= 0.003 mutations/ cell/ generation.
Q. 70. In an experiment to measure mutation rate is locus conferring resistance to T1 phage,
fluctuation test was done. Five of the 12 small independent cultures contained no phage
resistant mutant, the average number of cells per culture were 5 x 107, calculate the rate of
mutation.
Ans. µ= -2.303 log (5/12)/ 5 x 107= 1.8 x 10-8
Q. 71. In a fluctuation test for measuring mutation in genes F and H, for gene F, 22 of the 40
cultures had no mutants, with N=5.6 x 107, for gene H, 15 of the 37 cultures had no mutants
with N=5.6 x 107, calculate the mutation frequencies for the two genes.
Ans. F= -ln(22/40)/ (5.6 x 107)= 1.1 x 10-8
H= -ln(15/37)/ (5.6 x 107)= 1.8 x 10-8
Q. 72. What are the characteristics of nonsense suppressors?
Ans. The nonsense suppressor mutants have following important features:
1. A particular suppressor mutant can not suppress all UAG chain terminating mutations
because some times the inserted amino acid fails to produce the active protein.
2. Often suppression is incomplete because specific activity of protein may be less than wild
type protein.
212
3. Suppressor mutant will survive only if there are 2 or more tRNA genes for one codone, if
there is only one and that is mutated to become suppressor than the there will be no tRNA
to read the normal codone which than starts to work as stop codone.
Q. 73. If there is a UAG suppressor gene, then how the bacteria survive because many genes have
UAG as stop codone and then those proteins will not be terminated and elongate to unusual
length?
Ans. Even if there is UAG suppressor mutation bacteria can survive because:
1. Protein factors active in translation termination respond to chain termination even when
there is a suppressor tRNA and hardly in 10 % cases termination mutation is suppressed,
i.e., 90 % proteins will be of normal type.
2. Normal chain termination often uses pairs of distinct termination codons such as the
sequence UAG-UAA, thus the existence of a UAG suppressor will not prevent usual
termination
Q. 74. What are different methods to isolate auxotrophic mutants?
Ans. Common methods to isolate auxotrophic mutants are a. enrichment using the chemicals
(penicillin, ampicillin, cycloserine, radioactive bases) which selectively kill the growing cells,
b. by replica plating on normally enriched medium and defined medium not containing the
nutrient for which you are searching auxotrophy, c. Both methods can be used in succession.
Q. 75. What is positive selection; differentiate it from negative selection with example?
Ans. Positive selection is defined as the detection and selection of mutant cells and rejection of
unmutated cells. If we want to select nalidixic acid resistant mutants, bacteria is grown in on
plates containing media with nalidixic acid so that only resistant colonies could grow. In
negative selection, bacteria which cannot perform a desired function is selected for example if
we want mutants which cannot grow without one or more extra growth factors or are deficient
in a certain metabolic pathway than to identify such mutants replica plating is used for negative
selection. In the process two identical prints of a set of colonies are made on plates with and
without the required nutrient. The colonies which do not grow on the plate lacking the nutrient
can then be selected from the identical (replica) plate, which contains the nutrient.
Q. 76. What is the difference between mutagenesis and carcinogenesis?
Ans. Mutagenesis leads to mutation i.e., any change in genome of an organism it may be beneficial,
harmful or silent while for carcinogenesis more than one mutation is necessary. A series of
213
mutations in certain classes of genes, usually required for normal cell life, get mutated and
transform the cell into a cancer cell. The mutations in genes which play vital roles in cell
division, apoptosis or cell death, and DNA repair may lead to process of carcinogenesis.
Q. 77. What do you understand by PCR mutagenesis?
Ans. PCR mutagenesis is a method for site directed mutagenesis. Through this method you can
generate mutations by base substitutions, insertions, and deletions in a double-stranded plasmid
without subcloning into a M13-based bacteriophage (for ssDNA rescue). In the process a
plasmid vector is used as the template and two primers containing the desired mutation
(substitution, deletion or addition, preferably towards 5’ end) with each complementary to the
opposite strands of the vector. The PCR product can replace the homologous region in plasmid.
The template (wild type) plasmid (dam methylated in nature) is removed by digestion with
DpnI (specific for methylated DNA). You can recover intact mutated plasmid after the PCR
reaction
Q. 78. How spontaneous mutations are caused?
Ans. Spontaneous mutations may be due to: If any base got changed through repositioning of a
hydrogen atom, changing the hydrogen bonding pattern of the affected base (incorrect) which
pairs with different than normal base during replication. The process is called tautomarism, or
through loss of a purine base (A or G) to form an apurinic site (AP site) i.e. depurination, or
through deamination or replacement of amine group by keto group as cytosine to uracil, adenine
to hypoxanthine, or transition i.e., a purine changes to another purine, or a pyrimidine to a
pyrimidine or transversion i.e., a purine becomes a pyrimidine, or vice versa. However, reasons
of occurring such changes are not exactly known but environmental changes have important
role in all these.
Q. 79. What major difference you see in heredity of mutations in bacteria and animals and sexually
propagating plants?
Ans. All mutations in bacteria are hereditary i.e., acquired by parents are passed to the progeny
while mutations in higher, sexually reproducing, organisms can be either germline or somatic
mutations. Germline mutations, in cells dedicated to reproduction, can be passed on to
descendants. The somatic mutations, arise in somatic cells, i.e., those cells not dedicated
reproduction, are not transmitted to progeny.
Q. 80. Define amorphic mutations, antimorphic mutations, neomorphic mutations, lethal mutations
and back mutation.
214
Ans. Amorphic mutations: Those lead to complete loss of gene function.
Antimorphic mutations: mutations resulting in an altered gene product which is antagonistic
in action to the normal allele.
Neomorphic mutations: Resulting in to gain of new or abnormal function.
Letal mutations: Lead to death of the organism.
Back mutation: also known as reversion means restoration of the original gene sequence and
thus the original function.
Q. 81. What do you understand by mutator strain?
Ans. Mutator strains are the variants which have naturally high efficiency of mutation. You can
select a mutator strain of bacteria by repeatedly exposing it to antibiotics, and selecting resistant
mutants. The selected bacteria might have a much higher mutation rate than the original
population.
Q. 82. What is advantage of using mutator strain to create random libraries?
Ans. The advantage of using mutator strains to create random libraries is in the the accessibility of
the procedure and minimal need of specialized equipment or cloning techniques. One can
potentially create a large random library in two or three days, with little effort and prior
experience in recombinant DNA techniques.
Q. 83. Why the bacterial strains become mutator strain?
Ans. The strains which have defect in their DNA repair (proof reading and editing) mechanism or
rearrangement of DNA during and after replication become mutator strains which are prone for
mutations and in presence of any mutagen/ carcinogen they show high frequency (usually 105
times more than the parent strain.
Q. 84. Mutator strains of E. coli and Salmonella enterica often have mutated mutD, mutS and mutT
genes, how these mutations affect the replication?
Ans. Defect in mutD hinders the 3’-5’ exonuclease activity of DNA polymerase III and prevent
correction of the incorrectly incorporated bases leading to transition of bases and account for
~85% of total mutations. Defect in mutS gene hinder with mismatch repair activity leading to
mutations due to transition and transversions. Defects in mutt prevents degradation of 8-
oxodGTP in mismatch involving A:G resulting into AAT-CG transversions.
Q. 85. Do mutator strains also occur in eukaryotes? If yes, why they become mutator?
215
Ans. Yes, several strains of Saccharomyces spp. and other species are mutator strains. They have
mutation either in urf (a sequence called ‘unassigned reading frame’).
Q. 86. What is mitochondrial mutator strain?
Ans. Some strains of S. cervisiae having mutated mitochondrial var-l gene coding for a
mitochondrial ribosomal protein can easily be induced for high frequency of mutations. Most of
the mutations are point mutations or small deletions. Mitochondrial mutator strains of yeast
may lead to deletion in chromosomal genes but mitochondrial protein synthesis genes are not
affected. It also indicates that mitochondrial genes can also affect the chromosomal genes. In S.
pombe, three mutator strains (ana r-8, ana r-14, and diu r-301) were shown to produce
respiratory deficient mutants at different rates. The frequency of respiratory deficient mutants in
a culture could be increased by adding carcinogen/ mutagens as ethidium bromide in growth
medium.
Q. 87. What are hypermutator strains, where can you find them and detect them? What may be
their probable utility?
Ans. Hypermutators strains of bacteria those display strong mutator (hypermutator) activity with
comparatively high frequency of mutation to readily adjust to host environment they are of
emmence public health concern. About 1-20% of clinical isolates are classified as hypermutator
strains. Hypermutator clinical strains have enhanced the risk of emergence of antibiotic
resistance leading increased probability of treatment failure. Easiest method to detect the
presence of hypermutators in pathogenic bacterial population is disk diffusion method of
antimicrobial sensitivity testing where they form hazy zones around or appearanc of squatter
colonies in clear zone around some antibiotics including phosphomycin, rifampicin and
spectinomycin.
Hypermutator strains are used for creating random mutant libraries and to enhance the
recovery of rare resistance mutations, e.g., for elucidation of modified drug targets of bacterial
cell. Hypermutator strains can be used to find out new drugs to combat infection in more
effective manner.
Q. 88. How environment affect rate of mutation?
Ans. Impetus for mutation comes from environment and it is the need of every organism is to
adjust to everchanging environment which make them to mutate over the time and evolve best
fit strains. Even in hyper mutator strains the frequency of mutation is highly dependent on the
environment.
216
Chapter 8. Transformation Genetics
In molecular genetics, transformation means inheritable genetic change in a cell or an organism
resulting from the acquisition, genomic incorporation, and expression of foreign gene(s) or just a
part of DNA. Different modes of transfer or acquisition of genetic material are defined by different
names as introduction of DNA by viruses is named Transduction, through direct cell to cell contact
between bacteria is called conjugation while transformation of eukaryotic cells in tissue cultures is
defined as transfection. RNA may also be transferred into cells using similar methods, normally to
produce non-heritable changes, sometimes used in medicine but it is not considered a true
transformation.
Q. 1. What is transformation? Name the factors affecting it and steps involved in it. What is the
significance of transformation in bacterial genetics?
Ans. Transformation is gene transfer resulting from the uptake by a recipient cell of naked DNA
from a donor cell. Certain bacteria (e.g. Bacillus, Haemophilus, Neisseria, Pneumococcus) can
take up DNA from the environment and the DNA that is taken up can be incorporated into the
recipient's chromosome.
Factors affecting transformation
a. DNA size and state - Double stranded DNA of at least 5 X 105 daltons works best. Thus,
transformation is sensitive to nucleases in the environment.
b. Competence of the recipient - Some bacteria are able to take up DNA naturally. However,
these bacteria only take up DNA a particular time in their growth cycle when they produce a
specific protein called a competence factor. At this stage the bacteria are said to be
competent. Other bacteria are not able to take up DNA naturally. However, in these bacteria
competence can be induced in vitro by treatment with chemicals (e.g. CaCl2).
Steps in transformation
a. Uptake of DNA - Uptake of DNA by Gram+ and Gram- bacteria differs. In Gram + bacteria
the DNA is taken up as a single stranded molecule and the complementary strand is made in
the recipient. In contrast, Gram- bacteria take up double stranded DNA.
b. Legitimate/Homologous/General Recombination - After the donor DNA is taken up, a
reciprocal recombination event occurs between the chromosome and the donor DNA. This
recombination requires homology between the donor DNA and the chromosome and results
in the substitution of DNA between the recipient and the donor as illustrated in
Significance - Transformation occurs in nature and it can lead to increased virulence. In addition
transformation is widely used in recombinant DNA technology.
217
Q. 2. Name the bacteria which have natural genetic transformation system (natural competence).
Ans. Natural transformation occurs in genera Acinetobactercalcoaceticus, Achromobacter,
Azatobacter vinelandii, Bacillus subtilis, Butyrivibrio, Camplylobacter, Clostridium,
Haemophilus inflenzae, Micrococcus, Mycobacterium, Neisseria, Pasteurella novicida, some
Pseudomonas spp., Rhizobium spp., Streptococcus pneumoniae, Streptomyces and
Synechococcus spp., Xanthomonas phaseoli etc.
Q. 3. Name a few bacteria which can be made competent or may have artificial competence.
Ans. E. coli, Pseudomonas spp. Salmonella spp. and Staphylococcus spp.
Q. 4. Do bacteria can take up DNA from different sources with same efficiency?
Ans. Not always, e.g. H. influenzae can take DNA from related bacteria but not from others
because it recognise on or more sequence for binding sites (8-12 bp), however no such difference
has been not in Streptococcus and Bacillus.
Q. 5. How natural competence is induced in different bacteria?
Ans. Except Neisseria gonorrhoeae most bacteria can regulate their stage of competence, it is
mostly shift down i.e. transfer of bacteria from a nutrient rich to nutrient deficient environment.
Some common methods are:
Bacteria Mode of induction of competence
B. subtilis shift from 42oC to 37oC, only 15% cells are competent at any time
Staph. aureus Bacteriophage infection
Strept. pneumoniae During exponential phase, upto 100% cells may be competent under suitable
conditions
Hemophilus influenzae Nutritional shift down, any substance which cause inhibition of nucleic acid
synthesis without affecting protein synthesis
Q. 6. What is the similarity in transformation efficiency of Bacillus and E. coli?
Ans. Concentration of poly-β-hydroxybutyric acid concentration in cell membrane correlates well
with transformation efficiency in both.
Q. 7. What is the dissimilarity in transformation efficiency of Bacillus and E. coli?
Ans. In E. coli and other bacteria having artificial competence transformation method is best
successful with plasmid and phage DNA rather than chromosomal/ or linear DNA while in
Bacillus and other naturally competent organisms linear DNA and chromosomal DNA can be
transferred successfully.
218
Q. 8. How artificially competent cells of E. coli and other pathogens can be made targets for
transformation with linear and chromosomal DNA?
Ans. By making them deficient in exonuclease V encoded by recBCD gene in E. coli and similar
genes in other microbes and mutating a locus called sbcB.
Q. 9. What is the role of Ca++ in transformation?
Ans. It may activate the biochemical pathway for synthesis of poly-β-hydroxybutyric acid in
membranes and making the cells competent for transformation.
Q. 10. What is the difference between transfection and transformation, do we can induce
transfection in E. coli?
Ans. In transformation source DNA may be either plasmid or another bacterial DNA while in
transfection it is bacteriophage (viral) DNA. In transfection success is measured by formation of
number of plaques (no growth) while in transformation the number of bacterial colonies on
selective media are the measure of success. Yes, in E. coli we can perform transfection with
number of coli phages.
Q. 11. How transformation DNA is processed in naturally competent bacteria?
Ans. Donors’ dsDNA bound to competent cells (through receptors) and is attacked by nucleases to
yield shorter dsDNA fragments (15-30 kb in case of Bacillus subtilis, 8-9 kb for S. pneumoniae
and ~18 kb for H. influenzae) that penetrate into cells. Linear DNA can either enter ssDNA
(Bacillus, Streptococcus) or dsDNA (Haemophilus). DNA goes into cytoplasm either as
vesicular (G-ve bacteria, Haemophilus) or non-vesicular (G+ve bacteria) transformasome.
Thereafter DNA is incorporated in bacterial DNA by recombination.
Q. 12. Besides artificial and natural competence, how DNA can be made to enter the bacterial
cells?
Ans. Another popular method is electroporation or electrotransformation carried out under
influence of transient high voltage shock (about 2500 volts) to the bacteria under controlled salt
conditions using specific equipment.
Q. 13. What are the advantages of electroporation over chemical transformation?
Ans. In electroporation
1. Preparation of cells is simple and quick, only washing of the cells with low ionic solution is
needed.
219
2. Transformation efficiency is often higher (~109-1010/µg of DNA) than in chemical
transformation (~108/µg of DNA).
3. Highly purified DNA is not required for electroporation.
4. Size of transforming DNA (2-44 kb) affects little the transformation.
5. A wide variety of bacteria can be transformed without much change in protocol.
Q. 14. What is transduction and how many types is it?
Ans. Bacteriophage mediated gene transfer between bacteria is known as transduction, it is of two
type; a. generalized transduction (any part of bacterial genome can be transferred), b. specialized
transduction (only special sequences are transferred).
Q. 15. What are the factors which act against electrotransformation (electroporation)?
Ans. Presence of S layer on cell surface and tetrad mode of cell growth.
Q. 16. What is genetic transfusion? Give some examples.
Ans. Gene transfer by means of cell fusion (protoplast fusion), in this method diploid bacteria can
be produced and this transfer can take place between bacteria of same spp. or genus as well as
between bacteria of different genera of families. Diploid bacteria usually segregate to produce
haploid offsprings and may result into extensive recombination; however there are reports of
formation of stable, non-complementing diploid Bacillus subtilis cells. Protoplast fusion has
been reported in Actinoplanes, Brevibacterium, Bacillus, Mycobacterium, Providencia,
Staphylococcus and Streptomyces.
Q. 17. What are the different methods of bacterial transformation?
Ans. There two different methods of transformation, natural transformation and electroporation. In
Natural method, bacterial cells can be made competent viz., naturally competent, induced
competence.
Q. 18. How can you concentrate competent cells from a given culture of Bacillus subtilis known to
contain competent cells?
Ans. Competent cells from a given culture of Bacillus subtilis can be concentrate by gradient
centrifugation as competent cells are more buoyant than noncompetent cells. They are
nonreplicating, have single genome and small size and contain 3-4 times more poly-β-
hydroxybutyric acid (PHB).
220
Q. 19. How can you identify a competent Streptococcus pneumoniae cell from noncompetent cell?
Ans. A competent Streptococcus pneumoniae cell posses a new protein antigen (product of coma &
comD 10 kDa) on their surface which can be identified by immunological methods.
Q. 20. What are different methods for producing competent cells?
Ans. There are two common methods for production of chemically competent cells from different
bacteria which yield a transformation efficiency of about 2 × 108 transformants per µg of DNA;
1. Hypotonic Ca++ shock method; an early exponential phase culture is centrifuged and
resuspended in cold hypotonic CaCl2 solution, DNA is added to these cells in form of Ca-
DNA complex that easily get adsorbed to the cell surface. Then cells are briefly heat
shocked for transport of DNA into cells.
2. By using PEG and DMSO; Polyethylene glycol-DNA complex easily adsorb onto cells,
dimethylsulfoxide facilitate the DNA entry.
Q. 21. What is natural competence? How it differs in ability of acquisition of DNA from artificial
competence?
Ans. About 1% of all bacteria species are naturally capable of taking up extraneous DNA under
laboratory conditions without any aid like chemical or electroporating methods; this property is
called natural competence. Many bacteria might be acquiring DNA in their natural environments
for evolution of competent generation to fight the competition in nature. Such naturally
competent bacteria carry sets of genes specifying the cause of the machinery for bringing DNA
across the cell's membrane into the cell-sap.
The naturally competent cells get easily transformed with linear DNA while artificially
transformed cells need circular DNA in form of plasmid DNA for efficient transformation.
Q. 22. What are the common tools used for transformation of plants?
Ans. In plants transformation can be performed through different mechanisms, common methods
used are:
1. Agrobacterium mediated transformation: The easiest and most simple plant transformation
techniques uses Agrobacterium as transforming tool. The bacterium invades the injured or
exposed plant cells. The Agrobacterium contains a cloning vector a plasmid carrying
desired gene. Some cells get transformed by the bacterium, which inserts its DNA along
with plasmid into the cell. Placed on selectable rooting and shooting media, the plants will
re-grow. Some plants species can be transformed very easily while others don’t respond at
all.
221
2. Particle bombardment: Micro particles of gold or tungsten coated with desired genes are
shot with strong force into young plant cells in cell culture or plant embryos. Some genetic
material might persist in plant cells to transform them which can be grown in to full
transformed plants with the help of growth hormones. Along with plant cells plant plastids
can also be transformed with this method. The transformation efficiency is low but almost
all plants can be transformed with this method.
3. Electroporation: Under the influence of millisecond high voltage electric shock make
transient holes in cell membranes. The extraneous DNA present can got entry in that short
duration to yield transformation in the way similar to earlier two methods.
4. Transduction: The desired genes can be packaged into a suitable plant virus and the
modified virus is used to infect the plant. The desired genetic material (DNA) can insert into
the plant chromosomes to produce transformation of some cells. The limitation is that single
stranded RNA structure of most of the plant viruses.
5. Transfection: The most plant viruses consist of single stranded RNA which replicates in the
cytoplasm of infected cell and if the RNA of the infecting virus carries some external gene
it can also express. The results is not a real transformation because the desired genetic
material never reaches the nucleus of the plant cell and do not integrate into the host
genome. Thus progeny is non transfected but as most of the plants can be propagated
through vegetative methods the transfected line can be maintained.
Q. 23. What is artificial competence and what for it is use; how it is created?
Ans. Induction permeability in cells for DNA in laboratory is called as artificial competence. This
method is used for transformation cells in laboratory. Several methods as chilling of cells in the
presence of Ca2+ make the cells permeable to plasmid vectors. This method works well for
circular plasmid but not for linear molecules viz., fragments of chromosomal DNA as lineal
DNA is probably degraded by exonucleases of the cell. Electroporation is another method to
make holes in bacterial (and other) cells, by briefly shocking them with a high (10-20kV/cm)
voltage. Plasmid DNA can enter the cell through these holes. This method is amenable to use
with large plasmid DNA, the holes get repaired soon through natural membrane-repair
mechanisms of the cells.
Q. 24. What do you understand by DNA transformation?
Ans. Creating changes in DNA to alter its expression thus the physiology is called DNA
transformation. The term originated from work of Stanley Cohen and Herbert Boyer where they
used cut and paste DNA technique to create the custom-made Escherichia coli containing
222
recombined or recombinant DNA. The recombinant DNA was of a plasmid and bacteria were
able to take plasmid and expression of a foreign DNA sequence cloned in plasmid caused
transformation of E. coli.
Q. 25. Give protocol for making competent yeast cells for transformation and further steps in
transformation of yeasts.
Ans. Take overnight grown culture, dilute (1:100) in fresh yeast extract-peptone-adenine-dextrose
(YPAD) medium or other suitable medium (YPD); incubate for 6 h (~107 cells/ml). Centrifuge
15 ml broth culture at ~1000g for 5 min. Discard supernatant, resuspend cell pellet in 1 ml
distilled sterile water. Transfer in a microfuge vial, spin at top speed for 5 min, discard
supernatant, repeat the process thrice with H2O and twice with 1 ml LiOAc Mix. Resuspend cell
pellet in 70 ul LiOAc Mix and add 10 ul boiled salmon sperm DNA and make it 100 ul with
H2O. Make aliquot in 50 ul volumes. 10 ml of LiOAc Mix: 8.88 ml H2O+ 1 ml of 1 M LiOAc
(Lithium Acetate) + 100 ul 1 M Tris-Cl, pH 7.5 + 20 ul 0.5 M EDTA
YPAD: 10g yeast extract, 20g bacto peptone, 10mL of 0.1% adenine-hemisulfate dissolved in
1L of H20; adjust pH to 7 with 1N NaOH, sterilize through autoclaving and then add 50mL
sterile 40% glucose per liter.
YPD medium: Yeast extract 10g, peptone 20g in 1L of H20; adjust pH to 7 with 1N NaOH,
sterilize through autoclaving and then add 50mL sterile 40% glucose per liter.
YPG medium: Yeast extract 10g, peptone 20g, glycerol 30 ml, in 1L of H20; adjust pH to 7
with 1N NaOH, sterilize through autoclaving.
Alternate protocol for making competent yeast cells: Grow cells in YPAD to an OD600 0.6 to 1.0
(~0.6 - 1 × 107 cfu/ml). Centrifuge 10ml culture and collect pellet of the cells. Wash the cells
with 0.5 ml of mix containing 1.0 M. sorbitol, 10 mM Bicine-NaOH (pH 8.35), 3% ethylene
glycol and 5% DMSO. Resuspend cells in 200µl of the same solution with which cells are
washed. Competent cells for transformations are ready which can be freezed (slowly) over
liquid nitrogen and stored at -70oC until needed.
To 50 ul of competent cells (from first protocol or 20 ul from second protocol) add 10 ul of
PCR product or plasmid containing the desired DNA sequence, vortex briefly. Add 6 volume
(360ul) of PEG Mix and incubate 30°C or at room temperature for 30 min. Then add 45 ul of
223
DMSO (~10%). Incubate for 15 min at 42° C if incubated earlier at 37oC or at 37oC if incubated
earlier at room temperature. Spin the contents for 3-5 minutes at 2000 rpm in microcentrifuge.
Resuspend cells with 200 ul of YPAD and plate onto suitable growth medium to select the
desired trait for which transformation is attempted through negative (replica plating) or positive
selection (on selective media).
10 ml of PEG Mix: 8.88 ml of 45% PEG (Poly Ethylene Glycol)+ 1 ml of 1 M LiOAc + 100 ul
1 M Tris-Cl, pH 7.5 + 20 ul 0.5 M EDTA
Q. 26. Name a few commonly used E. coli strains in transformation experiments and genetic
engineering.
Ans. Several E. coli strains such as HB101, DH5α, GM2929, XL1-Blue, TG1, BL21, and JM109
can be used in genetic engineering and transformation work.
Q. 27. Briefly describe the method to prepare competent cells for electoroporation i.e., electroshock
Competent Cells of E. coli.
Ans. To prepare competent cells for electroporation inoculate a 25ml SOB-Mg medium with single
colony of E. coli and incubate at 37oC overnight. Take 7.5 ml of broth culture and inoculate into
750ml of SOB-Mg media in a 2L flask, incubate at 37oC on shaking platform (50 rpm) till you
get an OD550 or 600 equal to 0.75 units (~3-6x108 cfu/ml). Centrifuge the whole culture at 3000 g
for15 min at 40C. Discard supernatant and resuspend cell pellet in a volume of chilled 10%
redistilled glycerol in ultrapure water equal to original and centrifuge again, and discard
supernatant, repeat the washing step twice and then resuspend the cell pellet in 4 ml of chilled
10% glycerol. Aliquot the cell suspension in 20 µl volume and freeze on dry ice/ethanol bath,
and store at -80C till needed.
Q. 28. How we can increase efficiency of transformation in E. coli using competent cells made
through chemical method?
Ans. Escherichia coli cells become 4 to 6 times more transformable and 20 to 30 times more
competent after 24 h incubation in cold calcium chloride than using immediately after calcium
chloride treatment.
Q. 29. Briefly describe method for getting competent E. coli using CaCl2 method for transformation
using a plasmid Vector System.
Ans. First step in transformation is preparation of competent cells of E. coli, for which following
procedure can be followed:
224
1. Grow overnight E. coli DH5α in LB medium from single colony.
2. Transfer 0.5 ml of broth culture in to fresh 10 ml sterile LB medium and grow for about 4
h on shaking platform to have ~109 cfu/ml.
3. Transfer the growth into a precooled centrifuge tube and cool it on ice for 10 min.
4. Pellet the cells at ~3000g for 10 min at 4oC. Discard supernatant and resuspend the cell
pellet in 10 ml prechilled 0.1M CaCl2 by vortaxing.
5. Repeat step 4 to collect the pellet and resuspend that in 1 ml of prechilled 0.1M CaCl2 by
vortaxing.
6. Aliquot it in 200 µl volumes for transformation, competent cells are ready which can be
used within 24 h is stored chilled.
7. For transformation, take 200 µl of competent cells in a microfuge tube (transformation
tube) and add 10 µl of the desired vector containing the cloned gene or sequence and
incubate on ice for 5 min.
8. Transfer the microfuge tube to water bath at 42oC and hold for about 90 seconds then
immediately return the transformation tube to ice.
9. Add 800 µl of prewarmed SOC medium into transformation tube and incubate at 37oC for 1
h.
10. Spin the tube to pellet the cells and remove about 800 µl of supernatant and then vortex to
resuspend cells in remaining medium. Pour the contents of transformation onto plate of
suitable selective medium depending on the indicator system used in cloning vector and E.
coli. Incubate overnight at 37oC and next day pick up the transformed colonies.
Q. 30. Write a short note on eukaryotic cell transformation.
Ans. Transformation in sexually breeding taxa is of two type, in somatic and in germ line cells.
Somatic transformation is often a transient transformation which may loss the introduced DNA
and is of little use in generating transgenic plants/ animals. Germ line transformation involves
chromosomal integration of the introduced functional DNA sequence in germ line cells so that
the integrated DNA sequences are passed on to the progeny of the transgenic organisms.
Chromosomal integration includes incorporation into chromosomes of the organelles including
mitochondria and plastids as their chromosomes display maternal inheritance. Now transformed
somatic cells can be grown in to fully fertile organism giving rise to transgenic organisms.
Besides virus and bacterial suicide vectors, polyethylene glycol (PEG) induced and
electroporation are commonly used techniques for transformation of eukaryotic cells. In plants
certain certain cell wall-degrading polysaccharidase enzymes are used to make competent cells
for getting better transformation efficiency.
225
Chapter 9. Bioinformatics in Microbial Genetics
Information chaos in biology brought in order with bioinformatics
Bioinformatics is the science of application of information technology in the field of molecular
biology. The term bioinformatics was coined by Pauline Hogeweg (1979), it entails the creation
and advancement of databases, algorithms, computational and statistical techniques, and theory to
solve formal and practical problems arising from the handling and analysis of biological data. It
encompasses mathematical and computing approaches used to gather understanding of biological
processes. It includes mapping and analyzing nucleic acid and protein sequences, aligning different
nucleic acid and protein sequences to compare them and creating and viewing 3-D models of
protein structures and construction of phylogenetic trees. In brief it deals with sequence analysis,
genome annotation, computational evolutionary biology, biodiversity, analysis of gene and protein
expression, gene regulation, prediction of protein structure, mutations in cancer, comparative
genomics and proteomics, modeling biological system, high-throughput image analysis, protein-
protein docking, software and tools used for bioinformatics and web services in bioinformatics.
Q. 1. How it is determined that the DNA sequence you recently determined is of some
significance?
Ans. After determining some DNA sequence it is always important to detect that which protein it
code or it code for something or not at all. Though may be a small but it is not easy to infer it at
first or second look unless we use some tool of bioinformatics. The tool(s) rapidly determine
the protein coding regions in a sequence using program based on statistical differences between
coding and non-coding sequences often called GeneMarks. Thereafter more tools help you tell
either similar gene(s) have been analysed earlier or not. If yes, what functions it may have. For
using such program you need some basic knowledge of their mode of action and also how to
input your information to get further knowledge.
Q. 2. Which are the major techniques which brought impetus for evolution of Bioinformatics?
Ans. The techniques which are behind development of bioinformatics include applied mathematics,
informatics, statistics, computer science, artificial intelligence, molecular modeling and
simulation, computational biochemistry, genomics, proteomics, cytometery, and biophysics etc.
Q. 3. What type of work is carried out under bioinformatics?
Ans. Bioinformatics deals mainly with genomics and proteomics specifically sequence alignment,
sequence analysis (genome adnotation), gene prediction, genome assembly, protein structure
226
analysis, prediction and analysis of gene expression and protein-protein interactions modeling.
Besides, computational evolutionary biology, analysis of mutations and carcinogenesis,
comparative genomics, modeling biological systems, high throughput image analysis is other
emerging fields of bioinformatics.
Q. 4. What are the major uses of genetics in modern medicine?
Ans. Bioinformatics plays important role in molecular medicine in designing gene therapy, finding
more drug targets, personalized medicine, preventive medicine, nanomolecule designing for
targeted organs and cells as for cancer therapy, designing of vaccines, vaccine vectors and other
genome applications in biotechnology, antibiotic resistance and evolutionary studies.
Q. 5. What are the major tools used to understand host-parasite interaction?
Ans. Bioinformatics, microarrays and proteomics are important modern tools for investigating host
and pathogen interactions.
Q. 6. What are the important software tools in bioinformatics?
Ans. Bioinformatics uses several software tools, BLAST, an algorithm for searching large
databases of protein or DNA sequences, is the most commonly known. There are many online
bioinformatic meta search engines for sequence profiling tools, to fetch information from other
databases. SOAP-based bioinformatics web services are new generation tools in form of
integrated platform, where an application running on one computer in a laboratory uses
algorithms, data and computing resources on servers located at distant locations.
Q. 7. What are different common sequence search and sequence alignment tools of bioinformatics
available for microbial geneticists?
Ans. For sequence search and sequence alignment, particularly to find out significance of the
sequence you have identified or you are interested, different available tools are:
1. NCBI BLAST (National Centre for Biotechnology Information, Basic Local Alignment
Search Tool) available at www.ncbi.nlm.nih.gov/blast: it can be used for amino acid and
nucleic acid sequences for proteins and DNA, respectively. It indicates statistical
significance of matches found in database inferring functional as well as evolutionary
relationship between sequences of interest and finds the gene family to which the sequence
belonged (http://blast.ncbi.nlm.nih.gov/Blast.cgi).
227
2. FASTA: It also works both for protein and DNA sequence queries as the BLAST service.
Its services include FASTA, FASTX/FASTY, TFASTX/TFASTY, FASTF/TFASTF,
FASTS/TFASTS, LALIGN/PLALIGN, PRESS, GREASE AND GARNIER/CHOFAS
available at http://ftp/viginia.edu/pub/fasta.
3. EMBOS (European Molecular Biology and Bioinformatics Search Site): Can be used to
find out homology between different DNA sequences and express results in form identity
percentage and strength of homology (www.ebi.ac.uk/emboss/align;
www.ebi.ac.uk/Tools/emboss, emboss.sourceforge.net).
4. Codon W1.3: The software package (www.mobiol.ox.ac.uk/cu) estimates different
nucleotide composition along with their codon position.
5. ALIGN Query: This program compares different DNA or protein sequences
(http://www.igh.cnrs.fr/bin/align-guess.cgi).
6. Pairwise fast local alignment for gigabase (PRFLAG): It brings local alignment for two
different DNA sequences giving different compositional analysis at a very fast speed
(http://bioinformatics.itri.org.tw/prflag/prflag.php).
7. EMB BLAST: Permits one to specifically search databases
(http://www.ch.embnet.org/software/aBLAST.html).
8. WU BLAST: It is a powerful software package for gene and protein identification
(http://blast.wustl.edu/blast/README.html).
9. FASTA33: Provides sequence similarity and homology searching against nucleotide and
protein databases using the Fasta3 programs
(http://www.ebi.ac.uk/Tools/fasta33/index.html).
Q. 8. How the phylogeny trees are constructed and what are techniques used?
Ans. Phylogenetic trees using input sequences are constructed using computational phylogenetic
technique such as Distance-matrix methods (neighbor-joining, UPGMA) which determine
genetic distance through multiple sequence alignment. Some sequence alignment methods as
ClustalW creates trees using the simpler algorithms based on distance. Maximum parsimony
method can also be used for making phylogenetic trees by implying an implicit model of
evolution (i.e. parsimony). In advanced methods optimality criterion of maximum likelihood
within Bayesian Framework is used to create an explicit model of evolution to phylogenetic
tree estimation. Now a days, many of these techniques as NP-hard, heuristic search and
optimization is are used in combination with tree-scoring functions to identify a reasonably
good tree that fits the sequence data. Mathematical methods as T-theory can also be used to
construct phylogeny trees. Efficiency of Tree-building methods can be assessed by their speed,
228
ability to use diverse kind of data, consistency, robustness and ability to find and reports its
inability to use the data.
Q. 9. What are different Tools available for finding out phylogeny or for phylogenetic analysis,
give a brief description?
Ans. To determine phylogenetic relationship of an organism using the DNA sequence different
bioinformatics tools are used, important ones are:
1. MEGA (Molecular Evolutionary Genetic Analysis): Uses parsimony, distance matrix and
likelihood methods for analyzing protein or DNA sequences and measures distances to form
bootstrapping and consensus tree (http://www.megasoftware.net).
2. PHYLIP 3.6: PHYLIP is a free package of programs for inferring phylogenies. It is
distributed as source code, documentation files, and a number of different types of
executables. This package of programmes contains several (34) phylogeny tree drawing
programmes (http://evolution.genetics.washington.edu/phylip.html;
http://bioportal.bic.nus.edu.sg/phylip/index.html; http://cmgm.stanford.edu/phylip/).
3. NJplot (Neighbor joining plot): It find out different phylogenies displaying branch length,
boot strap information, swap branches and change in root position and saves plots as post
script files (http://pbil.univ-lyon1.fr/software/njplot.html).
4. Phylogenetic tree: With this tool the source sequence multiple alignments could be formed
from low and upper case letters. Upper case letters will show that there is a homology
between correspondent fragments in alignment
(http://www.genebee.msu.su/services/phtree_reduced.html).
5. FastDNAml: Used to construct phylogenetic tree of DNA sequences using maximum
likelihood to explore a very few of alternative tree topologies relative to typical pasimony
program (http://bioinformatics.oxfordjournals.org/cgi). The program can be run on a wide
variety of computers ranging from UNIX workstations to massively parallel systems. It is
available from the Ribosomal Database Project (RDP) by anonymous FTP. It is based on
version 3.3 of Felsenstein's dnaml program having several enhancements (algorithmic
changes) to improve performance with minimal memory usage, making it feasible to
construct very large trees. Trees containing 40–100 taxa can easily be generated. With this
method phylogenetic estimates are possible even when hundreds of sequences exist.
6. HENNIG 86: It is a fast parsimony program which is used to branch and bound search for
most parsimonious trees and interactive tree agreement. It can handle up to 180 taxa and
999 characters (http://www.cladistics.org/education/hennig86.html). Perhaps one of the
229
most frustrating things for first-time users of Hennig86 is the way it treats anything
numbered. Everything begins with zero (0) not one (1).
7. Nexus: This is amalgamation of hundred of methods and technique to construct and
compare phylogenetic trees. This program is used to search for interactive tree
rearrangement in a given phylogenic prediction tree file
(http://bioinfo.unice.fr/biodiv/Tree_editors.html).
Q. 10. Name some online Phylogeny software.
Ans.
1. Hypergeny: visualise large phylogenies with this hyperbolic tree browser
(http://bioinformatics.psb.ugent.be/hypergeny/home.php).
2. iTOL - interactive Tree Of Life annotate trees with various types of data and export to
various graphical formats; scriptable through a batch interface (http://itol.embl.de/).
3. Phylodendron: different tree styles, branch styles and output graphical formats
(http://iubio.bio.indiana.edu/treeapp/treeprint-form.html)
4. PhyloWidget: It views, edits and publishes phylogenetic trees online; interfaces with
databases (http://www.phylowidget.org/).
Q. 11. Name some phylogenetic tree applications available offline.
Ans.
1. Dendroscope: It is an interactive viewer for large phylogenetic trees and networks all
(http://www-ab.informatik.uni-tuebingen.de/software/dendroscope/welcome.html).
2. FigTree: It is a modern treeviewer with coloring and collapsing all trees
(http://tree.bio.ed.ac.uk/software/figtree/).
3. TreeDyn: it is very powerful open-source software for tree manipulation and annotation
allowing incorporation of Meta information (http://www.treedyn.org/).
4. TreeView: A classic treeviewing software that is very highly cited
(http://taxonomy.zoology.gla.ac.uk/rod/treeview.html).
Q. 12. List the common phylognetic tree programmes.
Ans. The important ones are as under:
1. BEAST (A. J. Drummond, A. Rambaut)
2. Bosque Integrated graphical software (S. Ramirez, E. Rodriguez)
3. ClustalW (Thompson et al.)
4. fastDNAml (G.J. Olsen)
5. Geneious (A. J. Drummond, M.Suchard, V.Lefort et al.)
230
6. HyPhy (S.L. Kosakovsky Pond, S.D.W. Frost, S.V. Muse)
7. IQPNNI (L.S. Vinh, A. von Haeseler)
8. MEGA (Molecular Evolutionary Genetics Analysis) (Tamura K, Dudley J, Nei M & Kumar
S).
9. Mesquite (W. P. Maddison and D. R. Maddison)
10. MOLPHY (J. Adachi and M. Hasegawa)
11. MrBayes (J. Huelsenbeck, et al.)
12. Network (A. Roehl)
13. Nona (P. Goloboff)
14. PAML (Phylogenetic analysis by maximum likelihood) (Z. Yang)
15. PAUP (Phylogenetic analysis using parsimony) (D. Swofford)
16. PHYLIP (J Felsentein)
17. PhyloQuart (V. Berry).
18. QuickTree (K. Howe, A. Bateman, R. Durbin)
19. RAxML-HPC (A. Stamatakis)
20. SEMPHY (M. Ninio, E. Privman, T. Pupko, N. Friedman)
21. SplitsTree (D.H. Huson and D. Bryant)
22. TNT (P. Goloboff et al.)
23. TreeGen (J. Hein)
24. Treefinder (G. Jobb)
25. TREE-PUZZLE (H.A. Schmidt, K. Strimmer, A. von Haeseler)
26. T-REX (V. Makarenkov, et al.)
27. Winclada GUI and tree editor (requires Nona) (K. Nixon)
28. Xrate (I. Holmes)
Q. 13. What are different multipurpose analysis tools available for geneticists and people of
bioinformatics?
Ans. The most useful ones are:
1. The Sequence manipulation Suite: It is comprehensive sequence analysis package of
some web-based programs for analyzing and formatting DNA and protein sequences. The
output of each program is a set of HTML commands, which is rendered by your web
browser as a standard web page. You can print and save the results, and you can edit them
using an HTML editor or a text editor. (http://www.bioinformatics.org/SMS/).
2. EMBOSS: A package of free software tools for sequence analysis
(http://bioinfo.pbi.nrc.ca:8090/EMBOSS/index.html).
231
3. ExPaSy Proteomics Tools: Contains number of web based useful tools for protein analysis
(http://www.expasy.ch/tools/).
4. DeCypher: An easy access to a variety of nucleic acid and protein analysis packages
(http://decypher2.stanford.edu/index_by_algo.htm).
Q. 14. Give web addresses of some sequence alignment tools.
Ans.
1. BLAST-2: It produces the alignment of two given sequences using BLAST engine for local
alignment (http://www.ncbi.nlm.nih.gov/blast/bl2seq/bl2.html)
2. Pairwise FLAG: To performs local alignment for two different DNA sequences
(http://bioinformatics.itri.org.tw/prflag/prflag.php).
3. SCAN2: It permits alignment of genome length DNAs
(http://linux1.softberry.com/berry.phtml?topic=scanh&prg=SCAN2)
4. Advanced Pip Maker: It aligns two DNA sequences (http://pipmaker.bx.psu.edu/cgi-
bin/pipmaker?advanced).
5. CoreGenes: It helps in determining the core set of genes in a maximum of five small
genomes (http://pumpkins.ib3.gmu.edu:8080/CoreGenes/)
6. ALIGN Query: Compares any two sequences
7. LALIGN: It detects multiple matching subsegments in two sequences
(http://www.ch.embnet.org/software/LALIGN_form.html)
8. ClustalW: It is used for multiple Sequence Alignment
(http://www.ebi.ac.uk/Tools/clustalw2/index.html)
9. DbClustal: It aligns sequences from a BlastP database and perform search with one query
sequence (http://igbmc.u-strasbg.fr:8080/DbClustal/dbclustal.html)
10. MultAlin: It is also used multiple sequence alignment
11. (http://prodes.toulouse.inra.fr/multalin/multalin.html)
12. AliBee Multiple Alignment: Besides alignment we can also use this for phylogenetic
analysis of the data (http://www.genebee.msu.su/services/malign_reduced.html)
13. DiAlign: This constructs pairwise and multiple alignments by comparing whole segments of
the sequences, sequence of up to one MB and a maximum of 100 sequences can be loaded
at a time.( http://bibiserv.techfak.uni-bielefeld.de/dialign/submission.html).
14. BOXSHADE: This program accepts a wide variety of file formats and allows the requester
considerable flexibility in defining the output appearance (http://mobyle.pasteur.fr/cgi-
bin/portal.py?form=boxshade).
232
15. CINEMA: It is a colour INteractive Editor for Multiple Alignments for DNA and proteins
(http://www.bioinf.manchester.ac.uk/dbbrowser/CINEMA2.1/).
16. Multiple Align Show: This accepts a group of aligned sequences (in FASTA or GDE
format) and formats the alignment as per given specifications
(http://www.bioinformatics.org/SMS/multi_align.html).
Q. 15. Give names of important tools used to detect Transcription elements in a given sequence.
Ans.
1. Cister: It is a Cis-element Cluster Finder (http://zlab.bu.edu/~mfrith/cister.shtml).
2. CONPRO: This program is used as a consensus promotor predictor
(http://stl.bioinformatics.med.umich.edu/conpro/).
3. Dragon: Is a Promotor finder
(http://sdmc.krdl.org.sg:8080/promoter/promoter1_3/DPFV13.htm).
4. MatInspector: It is used for detection of transciption factor binding sites
(http://www.helmholtz-muenchen.de/en/ieg/group-ag-biodv/matinspector.html).
5. Promotor Scan: This program predicts Promoter regions based on scoring homologies with
putative eukaryotic Pol II promoter sequences. It is done using the PROSCAN Version 1.7
suite of programs developed by Dr. Dan Prestridge, (http://www-
bimas.cit.nih.gov/molbio/proscan/).
6. SignalScan: It finds and list homologies of published signal sequences with the input DNA
sequence. Analysis is done using the SIGSCAN Version 4.05 suite of programs developed
by Dr. Dan Prestridge (http://www-bimas.cit.nih.gov/molbio/signal/).
7. TATAprediction: It predicts TATA signal in eukaryotic genes
(http://l25.itba.mi.cnr.it/~webgene/wwwHC_tata.html).
8. TESS: Used for search of different Transcription element. ESS is a web tool for predicting
transcription factor binding sites in DNA sequences. It can identify binding sites using site
or consensus strings and positional weight matrices from the TRANSFAC, JASPAR, IMD,
and our CBIL-GibbsMat database. (http://www.cbil.upenn.edu/cgi-bin/tess/tess).
9. Tfsearch: It predicts Transcription Factor Binding Sites in the given sequence
(http://www.cbrc.jp/research/db/TFSEARCH.html).
10. Promotor Inspector: Used for prediction of promoter regions in mammalian genomic
sequences (http://genomatix.gsf.de/products/index_promi.html).
11. Promotor Prediction: It is a Neural network promoter prediction system, used for any given
sequence (http://www.fruitfly.org/seq_tools/promoter.html).
233
Q. 16. If you want to explore for the presence of some gene (gene prediction) in a DNA sequence
you recently found, what will be your resource tools?
Ans. Either one or more of the following can be used to precisely predict the genes in a given
sequence.
1. ORF finder: The ORF Finder (Open Reading Frame Finder) is a graphical analysis tool
which finds all open reading frames of a selectable minimum size in a user's sequence or in
a sequence already in the database. This tool identifies all open reading frames using the
standard or alternative genetic codes. The deduced amino acid sequence can be saved in
various formats and searched against the sequence database using the WWW BLAST
server. The ORF Finder should be helpful in preparing complete and accurate sequence
submissions. It is also packaged with the Sequin sequence submission software.
(http://www.ncbi.nlm.nih.gov/gorf/gorf.html)
2. FGENEH: It helps in finding splice sites, protein coding exons and Gene models
construction, promotor and poly-A search (http://dot.imgen.bcm.tmc.edu:9331/gene-
finder/gf.html)
3. Gene Finder: The program predicts protein coding exons in genomic DNA
(http://argon.cshl.org/genefinder/).
4. GeneID: Used for gene identification and structure prediction
(http://apolo.imim.es/geneid.html)
5. GeneMark: It works on inhomogeneous Markov model approach combined with training
datasets to predict genes (http://genemark.biology.gatech.edu/genemark/)
6. GeneParser2: It is used for identification of protein coding regions in a DNA sequence
(http://beagle.colorado.edu/~eesnyder/GeneParser.html).
7. Generation: Simple program for microbial gene prediction
(http://compbio.ornl.gov/generation/index.shtml)
8. Genie: This Gene finder is based on hidden Markov models in a sequence of nucleotides
(http://www.fruitfly.org/seq_tools/genie.html).
9. GENLANG: a Linguistics based method to find genes
(http://www.cbil.upenn.edu/genlang/genlang_home.html).
10. GenScan: Good for identification of gene structures in genomic DNA
(http://genes.mit.edu/GENSCAN.html).
11. GenViewer: A program for predicting and analysis of protien-coding gene structures in a
sequence (http://www.itb.cnr.it/sun/webgene//).
12. Glimmer: A John Hopkins system for finding genes in microbial DNA
(http://www.cs.jhu.edu)
234
13. Grail: A DNA sequence analysis tool (http://compbio.ornl.gov/gallery.html).
14. HMMGene: A tool for prediction of vertebrate and C. elegans genes
(http://www.cbs.dtu.dk/services/HMMgene/).
15. NetGene2: A server which helps in Neural network predictions of splice sites
(http://www.cbs.dtu.dk/services/NetGene2/).
16. Procrustus: Gene recognition via spliced alignment (http://hto-
13.usc.edu/software/procrustes/).
17. SliceSite Prediction: A server helps in Splice site prediction by neural network
(http://www.fruitfly.org/seq_tools/splice.html).
18. Wise2: Intelligent algorithm for DNA searches. The Wise2 form compares a protein
sequence to a genomic DNA sequence, allowing for introns and frameshifting errors. The
model parameters which are included are 1. Human gene parameters, 2. Local start/end in
the protein, 3. 6:23 Algorithm (http://www.ebi.ac.uk/Tools/Wise2/index.html).
19. WebGene: It is an integrated computing system for protein-coding gene prediction.
Encompasses several tools for prediction and analysis of protein coding gene structures
(http://www.itb.cnr.it/sun/webgene//).
20. Xgrail: The server helps to find exons and other features.
(http://www.hgmp.embnet.org/Registered/Option/xgrail.html)
Q. 17. What are the tools available for determining gene expression during micro array analysis?
Ans. For analysis of gene expressions following tools are readily available online.
1. BRB ArrayTools: It is an integrated package for the visualization and statistical analysis of
DNA microarray gene expression data. It was developed for analysis of microarray data and
development of improved methods for the design and analysis of microarray based
experiments. The input data is in Excel spreadsheets describing the expression values and a
spreadsheet providing user-specified phenotypes for the samples arrayed. The analytic and
visualization tools are integrated into Excel as an add-in. The analytic and visualization
tools themselves are developed in the powerful R statistical system, in C and Fortran
programs and in Java applications. Visual Basic in the applications integrates the
components and hides the complexity of the analytic methods from the user. The system
incorporates a variety of powerful analytic and visualization tools developed specifically for
microarray data analysis (http://linus.nci.nih.gov/BRB-ArrayTools.html0.
2. Eisen Lab Tools: It requires registration or licence to use the software. It performs multiple
functions including ScanAlyze, Cluster, TreeView (http://rana.lbl.gov/EisenSoftware.htm).
235
3. TIGR ArrayViewer: This program is designed to facilitate the presentation and analysis of
microarray expression data (http://www.jcvi.org/cms/research/software/).
4. XCluster: A clustering tool from Stanford University is useful in studying gene expression.
Most of the files that are output by the clustering program are readable by TreeView (.cdt,
.gtr, .atr), which allows you to browse clusters.
(http://fafner.stanford.edu/~sherlock/cluster.html).
5. ClustArray: Uses clusters expression data and produces an annotated graphical visualization
of the clusters. It allows further analysis of individual genes with Ann-spec for discovery of
promoter elements, and Probeviz for design of cDNA arrays of selected genes.
(http://www.cbs.dtu.dk/services/DNAarray/).
Q. 18. Name some software which can be used for primer designing?
Ans. Important online helps in Primer designing are:
1. Primer3: It is a very powerful online PCR primer design program
(http://biotools.umassmed.edu/bioapps/primer3_www.cgi)
2. Gnefisher: Using assumption that genes with related function from different organisms
show high sequence similarity it can design even degenerate primers from sequences of
homologues genes. (http://bibiserv.techfak.uni-bielefeld.de/genefisher2/).
3. iCODEHOP: It helps in generating degenerate PCR primers from protein multiple sequence
alignments (https://icodehop.cphi.washington.edu/i-codehop-context/Welcome).
4. PRIDE and GenomePRIDE: Both are primer design programs especially developed for the
large scale design of walking primers or PCR primers, respectively. GenomePRIDE is also
able to design oligos (50-70mers) used for DNA chips.
Q. 19. Give web address for some tools for protein modification research?
Ans. A few important ones are hereunder, however there are many and address can be found on
(http://bomi.ou.edu/russell/mol-biol.html).
1. NetOGlyc 3.1: The NetOglyc server produces neural network predictions of mucin type
GalNAc O-glycosylation sites in mammalian proteins.
2. Netphos 2.0: This (http://www.cbs.dtu.dk/services/NetPhos/) server produces neural
network predictions for serine, threonine and tyrosine phosphorylation sites in eukaryotic
proteins. For kinase specific phosphorylation predictions resources are available at:
(http://www.cbs.dtu.dk/services/NetPhosK/ )
3. big_Pi Predictor: This GPI anchor prediction helps in designing different modicications in
proteins (http://mendel.imp.univie.ac.at/gpi/gpi_server.html).
236
4. FindMod: Predict potential protein post-translational modifications (PTM) and find
potential single amino acid substitutions in peptides. is a tool that can predict potential
protein post-translational modifications (PTM) and find potential single amino acid
substitutions in peptides. The experimentally measured peptide masses are compared with
the theoretical peptides calculated from a specified Swiss-Prot/TrEMBL entry or from a
user-entered sequence, and mass differences are used to better characterize the protein of
interest (http://ca.expasy.org/tools/findmod/).
5. GlycoMod: This free software predicts the possible oligosaccharide structures that occur on
proteins from their experimentally determined masses
(http://ca.expasy.org/tools/glycomod/).
Q. 20. Name some methods for developing transgenic plants.
Ans. Development of transgenic plants started in 1970s with the soil bacterium Agrobacterium
tumefaciens containing a plasmid, part of plasmid get transferred to competent plant cells. Host
range limitations of Agrobacterium-mediated gene transfer led to discovery of alternate
methods viz., direct gene transfer to protoplasts, pollen transformation, pollen tube pathway,
electroporation, microlaser, liposome-fusion and liposome-injection, macro-injections, tissue-
electroporation, silicon carbide fiber-mediated transformation, micro-injection, site-directed
recombination and microtargeting.
Q. 21. Which proteomics approach is used to find the cause of a certain disease supposed to be due
to some mutation or defect in gene?
Ans. The best approach will be determining the environmental factor(s) influencing the expression
of the gene of interest.
Q. 22. What for physical maps from genomic clones are used?
Ans. Physical maps are the tools which depict the actual size and distance between sequences of
DNA shown to be genetically linked.
Q. 23. How shotgun cloning differs from the clone-by-clone method?
Ans. No genetic or physical maps of the genome are needed to begin shotgun cloning which is
required for using clone-by-clone method to sequence the entire genome.
Q. 24. What is the relation of gene density and chromosome number with complexity of an
organism?
237
Ans. As the complexity of life increases gene density decreases and chromosome number increases.
In less complex organisms, such as one-cell prokaryotes and eukaryotes genomes are highly
gene-dense with very few or no introns and hardly any wasted space, occupied by non-useful
sequences.
Q. 25. The expansion of genome in eukaryotes is often associated with gene duplication and
occurrence of multiple alleles, what may be the advantage of this apparent waste?
Ans. In simplest form, it is just keeping a backup of genetic information, in case one gene copy
become nonfunctional or redundant, a backup is available for normal life requirements.
Q. 26. There are only 300 major genes for antibody formation while thousands of different
antibodies form in body, how?
Ans. It is due to recombination, deletions, and random assortment of DNA segments of antibody
genes.
Q. 27. What do you mean by e value?
Ans. It stands for “expectation value” lower the e value more significant is the match. It determines
the statistical significance of a match to predict that either a match has taken by chance or not.
It is used in BLAST analysis as an important parameter to avoid the random hits during
database search. Lower e value signifies stronger alignment.
Q. 28. How can you perform Homology Modeling when your target identity with template to
about 30%?
Ans. You can opt for threading or you break your sequence into some pieces and then search
template for all the fragments separately or go for comparative modeling because for Homology
Modeling there should be >30% identity with the templet.
Q. 29. How the sensitivity and selectivity of Blast is determined?
Ans. Suppose during Blast search it returns 100 hits. Of these, 25 were false positives and we
knew that there were 150 sequences in the database which should have returned a hit with our
sequence. Then determine the ratio of true positive for sensitivity and ratio of false positive for
specificity. Number true positives (TP) must have been 100-25 = 75, i.e., search program found
only 75 out of the 150, i.e. false negative (FN) were 75, then
Sensitivity = TP/(FN+TP) = 75/(75+75) = 0.5 or 50%
Selectivity or Specificity= FN/(TP+FN) = 75/(75+75) = 0.5 or 50%
