An exact formula for the number of alignments between two DNA sequences.

Abstract

Given two DNA sequences, one is usually interested in measuring their similarity. For this purpose we have to consider different alignments of both the sequences. The number of alignments grows rapidly with the length of the sequences. In this short communication, we give an exact formula for the number of possible alignments using the theory of difference equations.

Full text

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An exact formula for the number of alignments
between two DNA sequences
Angela Torres,
Departamento de Psiquiatr´ ıa, Radiolog´ ıa y Salud P´ ublica
Facultad de Medicina
Universidad de Santiago de Compostela, Spain
Alberto Cabada
Departamento de An´ alisis Matem´ atico
Facultad de Matem´ aticas
Universidad de Santiago de Compostela, Spain
and
Juan J. Nieto∗
Departamento de An´ alisis Matem´ atico
Facultad de Matem´ aticas
Universidad de Santiago de Compostela, Spain.
Abstract: Given two DNA sequences, one is usually interested in measuring their similarity. To this
purpose we have to consider different alignments between both sequences. The number of alignments
growths very rapidly with the length of the sequences. In this note we give an exact formula for the
number of possible alignments using the theory of difference equations.
Key Words: DNA sequence, alignment, difference equation.
Proposed Running head: Number of alignments.
∗Corresponding author. Address: Departamento de An´ alisis Matem´ atico, Facultad de Matem´ aticas, Universidad
de Santiago de Compostela, 15782, Santiago de Compostela, Spain

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INTRODUCTION
Mathematics, statistics, and compuster science play a central role in the study of genetic sequences
and, generally, in bioinformatics and computational biology (Jamshidi et al., 2001, Lange, 2002,
Morgenstern, 2001, Nieto et al, to appear, Paun et al, 1998, Percus, 2002, Tang, 2000, Torres and
Nieto, 2003).
It is very important to compare genomic sequences (Foster et al., 1999, Gusev et al., 1999,
Jiang et al., 2002, Li et al , 2001, Liben-Nowell, 2001). In this direction, sequence alignment is a
basic tool. It is almost obvious that the number of possible alignments growths very rapidly. Here,
we present an exact formula for the number of alignments using the theory of difference equations
and a computer program to compute that number, and discuss some of its implications.
ALIGMENTS
The information archive in each organism is the genetic material, DNA or, in some viruses, RNA.
DNA molecules are long chain molecules containing a message in a four-letter alphabet: A (adenine),
G (guanine), C (cytosine), and T (thymine). In the RNA alphabet, T is replaced by U (uracil). The
genetic information is encoded digitally as strings over this four-letter alphabet. A DNA sequence
is just a string
x = (x1,x2,...,xn)
where xi, i = 1,2,...,n is one of the four nucleotides A, G, C, T. We say that the sequence x is of
lenght n.
Given another sequence
y = (y1,y2,...,ym),
we wish to compare both sequences and measure their similarity and determine their residue-residue
correspondences. To compare the nucleotides or amino acids that appear at corresponding positions
in two sequences, we must first assign those correspondences and a sequence alignment is just the
identification of residue-residue correspondence (Lesk, 2002). Any assigment of correspondences that
preserves the order of the residues within the sequences is an alignment and we allow the introduction
of gaps. Thus, an alignment of sequences x and y is an arragement of x = (x1,x2,...,xn) and
y = (y1,y2,...,yn) and gaps in two rows, keeping the ordering on the xiand on the yi, but not
having any gap directly over another gap. This restriction implies that there are only a finite number
of possible alignments.
For example, given the sequences x = CGT and y = ACTT, an aligment is
-C
C
G
-
T
T
-
AT
Sequence alignments is one of the basic tools of molecular biology, computational biology and
bioinformatics (Altschul et al., 2001, Lesk, 2002, Sadovsky, 2003, Torres and Nieto, 2003).
The number of alignments f(n,m) between two given sequences x = (x1,...,xn) and y =
(y1,...,ym) is given by the following recurrence relation, see (Lange, 2002),
f(n,m) = f(n − 1,m) + f(n,m − 1) + f(n − 1,m − 1),
together with the initial conditions
(1)
f(0,m) = f(n,0) = 1,n, m ∈ {1,2,3,...}.
(2)
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The initial conditions hold from the trivial fact that the only alignement is given by insert n
blanks in the string with none elements. The general recurrence formula follows from the fact that
every pair of modified sequences ends with one of the three pairs
?
and the remainder of the alignement to the left of one of these pairs constitutes an alignement
between two shorter sequences.
xn
−
?
,
?
−
ym
?
, or
?
xn
ym
?
,
EXACT FORMULA
It is obvious that equation (1) – (2) has a unique solution since (1) is an explicit recurrence relation
with given initial conditions (2). To solve such problem we shall use some results from the theory
of difference equations (Kelley and Peterson, 2001). Denote for n ∈ {1,2,...}, k ∈ {0,1,...}, the
“falling factorial power” as n0= 1 and, if k ≥ 1,
nk= n(n − 1) ··· (n − k + 1).
From the definition, one can verify the following property
(3)
(n + 1)k− nk= knk−1,k ≥ 0.
(4)
First, we assume that n ≥ m. We look for a solution given by the following expression
f(n,m) =
m+1
?
i=0
Ai(m)ni,
(5)
where
A0(m) ≡ 1,
for all m ∈ {0,...,n}.
(6)
Note that if the solution f is given by this expression then
f(n + 1,m + 1) − f(n,m + 1) = f(n + 1,m) + f(n,m) =
m+1
?
i=0
Ai(m)((n + 1)i+ ni).
Using property (4) we deduce that this last expression is equals to
m+1
?
i=1
(2Ai−1(m) + iAi(m)) ni−1+ 2Am+1(m)nm+1.
As consequence of these equalities, from property (4) and the initial conditions (2), we conclude that
f(n,m + 1) = 1 +
m+1
?
i=1
(2Ai−1(m) + iAi(m))
ni
i
+ 2Am+1(m)nm+2
m + 2.
From this last expression and the equality (5) applied at the poit (n,m + 1), we have that the
coeficients Ai(m) verify the following recurrence system
Ai(m + 1) = Ai(m) +2
iAi−1(m),i ∈ {1,2,...,m + 1},
(7)
and
Am+2(m + 1) =
2
m + 2Am+1(m).
(8)
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Now, tacking into account that 1 = y(n,0) = 1 + A1(0)n, we arrive at A1(0) = 0, and as
consequence of equation (8), we have that the following property holds
Ai(i − 1) = 0for all i ≥ 1.
(9)
Now, we rewrite equation (7) with initial conditions (9) as
Ai(m + 1) = Ai(m) + Ai−1(m), m ≥ i − 1,
In the sequel, we verify that the unique solution of this problem is given by
Ai(i − 1) = 0.
Ai(m) =
2i
(i!)2mi
for all i ∈ {0,1,...,m}.
For i = 1, since A0(m) = 1 for all m ≥ 1 we have that A1(m) = 2m ≡ 2m1.
Assume that i ≥ 2. The initial condition follows trivially from the fact that (i − 1)i= 0.
Equation (7) follows directly from property (4).
When n ≤ m, we look for a solution that satisfies the following expression
n+1
?
and conclude that y(n,m) = y(m,n).
Thus, using that if 0 ≤ k ≤ n, then
f(n,m) =
i=0
Ai(n)mi,
nk= k!
?
n
k
?
,
as a consequence of the development done in this section, we arrive at the following result.
Conclusion The unique solution of problem (1) – (2) is given by the following expression
f(n,m) =
min {n,m}
?
k=0
2k
?
m
k
??
n
k
?
.
(10)
CONSEQUENCES
Equation (10) permeate us to compute directly the value of the alignements between two strings of
DNA. In the sequel we give some numerical values for such combinations
f(2,1)
f(4,2)
f(8,4)
f(16,8)
=
=
=
=
f(1,2)
f(2,4)
f(4,8)
f(8,16)
=
=
=
=
5,
41,
3649,
39490049.
As we can see with this simple example the number of alignements growths very fast. Equation
(10) says us that the double sequence {f(n,m)} is monotone nondecreasing in both variables, i.e,
f(n,m) ≤ f(n,m + 1) and f(n,m) ≤ f(n + 1,m), for all n, m ∈ N. In particular f(n,m) =
f(m,n) ≥ f(n,n) for all n ≤ m and f(n,m) = f(m,n) ≤ f(n,n) whenever n ≥ m.
From this fact, the behaivor of function f at the diagonal terms (n,n) plays an important
role in the study of the considered functions. Clearly {f(n,n)} is monotone nondecreasing. And it
is very fast, as we can note in the following facts
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f(10,10)
f(50,50)
f(100,100)
=
≈
≈
8097453,
15 × 1036,
2 × 1074,
Special mention to the inequalities
f(106,106) < 1080< f(107,107).
This last estimation is relevant since there are only 1080protons in our universe (see (Nowak and
May, 2000), page 84). A typical sequence contains about 200-500 base pairs (Lesk, 2002). Proteins
are about 200-400 amino acids long, and hence they are 600-1200 letters long.
Using the Newton’s binomial formula and classical inequalities, we arrive at the following
estimations from above and bellow on the growth of f(n,n).
?
k=0
n
?
k=0
2k
?
n
k
?
= 3n≤ f(n,n) ≤
n
?
√2
k?
n
k
?? ?
n
?
k=0
√2
k?
n
k
??
= (1 +√2)2n.
Such estimation is illustrated in the following figure:
Acknowledgments
First and second author’s research partially supported by Ministerio de Ciencia y Tecnolog´ ıa and
FEDER, project BFM2001 – 3884 – C02 – 01, and by Xunta de Galicia and FEDER, project
PGIDIT02PXIC20703PN.
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