The number of reduced alignments between two DNA sequences

Background In this study we consider DNA sequences as mathematical strings. Total and reduced alignments between two DNA sequences have been considered in the literature to measure their similarity. Results for explicit representations of some alignments have been already obtained. Results We present exact, explicit and computable formulas for the number of different possible alignments between two DNA sequences and a new formula for a class of reduced alignments. Conclusions A unified approach for a wide class of alignments between two DNA sequences has been provided. The formula is computable and, if complemented by software development, will provide a deeper insight into the theory of sequence alignment and give rise to new comparison methods. AMS Subject Classification Primary 92B05, 33C20, secondary 39A14, 65Q30


Background
Let us consider a DNA sequence as a mathematical string where x i ∈ {A, G, C, T} is one of the four nucleotides, i = 1, 2, . . . , n, i.e. A denotes adenine, C cytosine, G guanine and T thymine. In these conditions, the sequence x is of length n.
Our main goal is to compare the sequence x with another DNA sequence y = (y 1 , y 2 , . . . , y m ) , to measure the similarity between both strings and also to determine their residue-residue correspondences.
Sequence comparison and alignment is a central and crucial tool in molecular biology. For example, Pairwise Sequence Alignment is used to identify regions of similarity that may indicate functional, structural and/or evolutionary relationships between two biological sequences (protein or nucleic acid) [1].
To align the sequences CGT and ACTT, one can use EMBOSS Needle for nucleotide sequence [9] that creates an optimal global alignment of the two sequences using the Needleman-Wunsch algorithm to get Following Lesk [10], in order to compare the amino acids appearing at their corresponding positions in two sequences, theirs correspondences must be assigned and a sequence alignment is the identification of residueresidue correspondence. For some references on sequence alignment we refer the reader to [10][11][12][13][14][15][16].
To compare two sequences, there exist mainly three different possibilities leading to three different numbers of total alignments [10,11,13]: http://www.biomedcentral.com/1471-2105/15/94 1. The total number of alignments denoted by f (n, m) that was solved in [13]. 2. A gap in a sequence is followed by another gap in the other sequence as in Alignments 1 and 2 for the sequences x = CGT and y = ACTT (see Tables 1  and 2 below) Considering the two alignments as equivalents to the Alignment 3 (see Table 3) without gap in those positions, we have the number of reduced alignments denoted by h(n, m), and obviously h(n, m) < f (n, m). This case has been solved in [11], and we give here another representation in terms of hypergeometric series. 3. In the interesting case that the alignments 1 and 2 are equivalent, but different from alignment 3 we have a number or reduced alignments g(n, m) where h(n, m) < g(n, m) < f (n, m). This last case is new and we present an explicit formula for g.

Results and discussion Number of f (x, y) alignments
The total number of alignments f (x, y) satisfies the following recurrence relation [13] f The solution of the above partial difference equation is given by (see formula (10) in [13]) and the generating function [17,18] is Therefore the coefficients f (n, m) in the expansion are given in terms of a hypergeometric series by This relation seems to be new in this form. Here, the generalized hypergeometric series is defined as (see e.g. [19,Chapter 16]) It is assumed that b j = −k in order to avoid singularities in the denominators. If one of the parameters a j equals to a negative integer, then the sum becomes a terminating series.

Number of h(x, y) alignments
In this case, the recurrence relation for the h(n, m) coefficients is [11] h(n, with initial conditions h(n, 0) = h(0, m) = 1. Therefore, the generating function [17,18] is and the coefficients in the expansion h(n, m)x n y m are given by The above coefficients can be written in terms of (terminating) hypergeometric series as

Theorem 1. The coefficients α n,m in the expansion
α n,m x n y m (2) are explicitly given by where and [x] denotes the integer part of x.
Proof. If we expand, we have two summands to be computed, namely In order to compute the first sum (12) let us introduce Therefore, the summation to be done reads as where U, V, A and B must be computed in terms of the initial indices. The product of binomials can be simplified to Thus, and then Finally, the summation reads as A similar work with the second summand (13) leads to the final result.

Conclusions
A unified approach for a wide class of alignments between two DNA sequences has been provided. We conclude also that our approach gives an explicit formula filling a gap in the theory of sequence alignment. The formula is computable and, if complemented by software development, will provide a deeper insight into the theory of sequence alignment and give rise to new comparison methods. It may be used also, in the future, to get explicit formulas and compute the number of total, reduced, and effective alignments for multiple sequences.

Methods
We have performed a number of numerical computations to compare our formulae and Mathematica® [20] command Coefficient for the series expansion of (1), on a MacBook Pro featuring a 45 nm "Penryn" 2.66 GHz Intel "Core 2 Duo" processor (P8800), with two independent processor "cores" on a single silicon chip, 8 GB of 1066 MHz DDR3 SDRAM (PC3-8500). We would like to mention that our approach is amazingly fast, since e.g. g(100, 100) is computed by using Mathematica® in 0.125165 seconds by using the new formulas presented in this paper, while the use of Mathematica® command Coefficient needs 99.167659 seconds.