Short Communication

An Exact Formula for the Number of Alignments BetweenTwo DNA Sequences

ANGELA TORRESa, ALBERTO CABADAb and JUAN J. NIETOb,*

aDepartamento de Psiquiatrıa, Radiologıa y Salud Publica, Facultad de Medicina, Universidad de Santiago de Compostela, Santiago de Compostela,Spain; bDepartamento de Analisis Matematico, Facultad de Matematicas, Universidad de Santiago de Compostela, Santiago de Compostela, Spain

(Received 9 June 2003)

Given two DNA sequences, one is usually interested inmeasuring their similarity. For this purpose we have toconsider different alignments of both the sequences.The number of alignments grows rapidly with the lengthof the sequences. In this short communication, we givean exact formula for the number of possible alignmentsusing the theory of difference equations.

Keywords: DNA sequence; Alignment; Difference equation; Exactformula

INTRODUCTION

Mathematics, statistics and computer science play arelevant role in the study of genetic sequences and,generally, in molecular biology, bioinformaticsand computational biology (Paun et al., 1998; Tang,2000; Jamshidi et al., 2001; Lange, 2002; Morgenstern,2002; Percus, 2002; Nieto et al., 2003; Torres andNieto, 2003).

The comparison of genomic sequences (Forsteret al., 1999; Gusev et al., 1999; Li et al., 2001; Liben-Nowell, 2001; Jiang et al., 2002) is performed usingsequence alignment as a basic tool. It is obvious thatthe number of possible alignments grows rapidly.Here, we present an exact formula for the number ofalignments using the theory of difference equationsand a computer program to compute that number,and discuss some of its implications. Our methodgives an alternate way for computing the number ofpossible gapped alignments. For example, for twosequences of length 8 and 16, respectively,

the number of alignments is <40 £ 106. For twosequences of length greater than 107, such a numberis greater than 1080.

ALIGNMENTS

The information archive in each organism is thegenetic material, DNA or, in some viruses, RNA.DNA molecules are long chain molecules containinga message in a four-letter alphabet: A (adenine),G (guanine), C (cytosine) and T (thymine). In theRNA alphabet, T is replaced by U (uracil). Thegenetic information is encoded digitally as stringsover this four-letter alphabet. A DNA sequence isjust a string

x ¼ ðx1; x2; . . .; xnÞ

where xi, i ¼ 1; 2; . . .; n is one of the four nucleotidesA, G, C, T. We say that the sequence x is of length n.

Given another sequence

y ¼ ð y1; y2; . . .; ymÞ;

we wish to compare both sequences and measuretheir similarity and determine their residue–residuecorrespondences. To compare the nucleotides oramino acids that appear at corresponding positions intwo sequences, we must first assign those correspon-dences and a sequence alignment is just theidentification of residue–residue correspondence(Lesk, 2002). Any assignment of correspondencesthat preserves the order of the residues within the

ISSN 1042-5179 print/ISSN 1029-2365 online q 2003 Taylor & Francis Ltd

DOI: 10.1080/10425170310001617894

*Corresponding author. Tel.: þ34-981-563-100. Fax: þ34-981-597-054. E-mail: amnieto@usc.es

DNA Sequence, December 2003 Vol. 14 (6), pp. 427–430

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sequences is an alignment and we allow the

introduction of gaps. Thus, an alignment of sequences

x and y is an arrangement of x ¼ ðx1; x2; . . .; xnÞ and

y ¼ ðy1; y2; . . .; ynÞ and gaps in two rows, keeping the

ordering on the xi and 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 ¼ CGT and

y ¼ ACTT; an alignment is

Sequence alignments is one of the basic tools ofmolecular biology, computational biology andbioinformatics (Altschul et al., 2001; Lesk, 2002;Sadovsky, 2003; Torres and Nieto, 2003).

The number of alignments f(n,m) between two givensequences x ¼ ðx1; . . .; xnÞ and y ¼ ðy1; . . .; ymÞ is givenby the following recurrence relation (see Lange, 2002),

f ðn;mÞ ¼ f ðn21;mÞþ f ðn;m21Þþ f ðn21;m21Þ; ð1Þ

together with the initial conditions

f ð0;mÞ ¼ f ðn;0Þ ¼ 1; n;m[ {1;2;3; . . .}: ð2Þ

The initial conditions hold from the trivial fact thatthe only alignment is given by insert n blanks in thestring with none elements. The general recurrenceformula follows from the fact that every pair ofmodified sequences ends with one of the three pairs

xn

2

!;

2

ym

!or

xn

ym

!;

and the remainder of the alignment to the left of oneof these pairs constitutes an alignment between twoshorter sequences.

EXACT FORMULA

It is obvious that Eqs. (1) and (2) have a uniquesolution since Eq. (1) is an explicit recurrence relationwith given initial conditions (2). To solve suchproblem we shall use some results from the theory ofdifference equations (Kelley and Peterson, 2001).One of the advantages of our method is that it scalesbetter than direct evaluation of Eq. (1). Denote forn [ {1; 2; . . .}; k [ {0; 1; . . .}; the “falling factorialpower” as n0 ¼ 1 and, if k $ 1,

nk ¼ nðn 2 1Þ· · ·ðn 2 k þ 1Þ: ð3Þ

From the definition, one can verify the followingproperty

ðn þ 1Þk 2 nk ¼ knk21; k $ 0: ð4Þ

First, we assume that n $ m. We look for a solutiongiven by the following expression

f ðn;mÞ ¼Xmþ1

i¼0

AiðmÞni; ð5Þ

where

A0ðmÞ ; 1; for all m [ {0; . . .; n}: ð6Þ

Note that if the solution f is given by thisexpression then

f ðn þ 1;m þ 1Þ2 f ðn;m þ 1Þ ¼ f ðn þ 1;mÞ þ f ðn;mÞ

¼Xmþ1

i¼0

AiðmÞððn þ 1Þi þ niÞ:

Using property (4) we deduce that this lastexpression is equal toXmþ1

i¼1

ð2Ai21ðmÞ þ iAiðmÞÞni21 þ 2Amþ1ðmÞnmþ1

As consequence of these equalities, from property(4) and the initial conditions (2), we conclude that

f ðn;m þ 1Þ ¼ 1 þXmþ1

i¼1

ð2Ai21ðmÞ þ iAiðmÞÞni

i

þ 2Amþ1ðmÞnmþ2

m þ 2:

From this last expression and the equality (5)applied at the point ðn;m þ 1Þ; we have that thecoefficients Ai(m) satisfy the following recurrencesystem

Aiðm þ 1Þ ¼ AiðmÞ þ2

iAi21ðmÞ;

i [ {1; 2; . . .;m þ 1}; ð7Þ

and

Amþ2ðm þ 1Þ ¼2

m þ 2Amþ1ðmÞ: ð8Þ

Now, taking into account that 1 ¼ f ðn; 0Þ ¼1 þ A1ð0Þn; we arrive at A1ð0Þ ¼ 0; and as aconsequence of Eq. (8), we have that the followingproperty holds

Aiði 2 1Þ ¼ 0 for all i $ 1: ð9Þ

Now, we rewrite Eq. (7) with initial conditions (9)as

Aiðmþ 1Þ ¼ AiðmÞþAi21ðmÞ; m $ i2 1; Aiði2 1Þ ¼ 0:

In the sequel, we verify that the unique solution ofthis problem is given by

AiðmÞ ¼2i

ði!Þ2mi for all i [ {0; 1; . . .;m}:

For i ¼ 1; since A0ðmÞ ¼ 1 for all m $ 1 we havethat A1ðmÞ ¼ 2m ; 2m 1:

– C G T –

A C – T T

A. TORRES et al.428

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Assume that i $ 2. The initial condition followstrivially from the fact that ði 2 1Þi ¼ 0: Equation (7)follows directly from property (4).

When n # m, we look for a solution that satisfiesthe following expression

f ðn;mÞ ¼Xnþ1

i¼0

AiðnÞmi;

and conclude that f ðn;mÞ ¼ f ðm; nÞ:Thus, using that if 0 # k # n, then

nk ¼ k!n

k

!;

as a consequence of the development done in thissection, we arrive at the following result.

Conclusion: The unique solution of problems (1)and (2) is given by the following expression

f ðn;mÞ ¼Xmin {n;m}

k¼0

2km

k

!n

k

!: ð10Þ

We note that the formula allows for the situationwhere no residues are aligned between sequences,i.e. all the residues of one sequence are in the gaps ofthe other sequence. Of course, this is a non-biologicalresult and there are a huge number of suchalignments. In a forthcoming paper we will presentthe formula for the number of alignments with nooverlapping residues, the number of optimalalignments (with respect to an optimality criterion),and the expression for the number of multiplealignments. Also, we will discuss explicitly somescaling questions in the context of biologicalsequences.

CONSEQUENCES

Equation (10) permeates us to compute directly thevalue of the alignments between two strings of DNA.In the sequel we give some numerical values for suchcombinations

f ð2; 1Þ ¼ f ð1; 2Þ ¼ 5;

f ð4; 2Þ ¼ f ð2; 4Þ ¼ 41;

f ð8; 4Þ ¼ f ð4; 8Þ ¼ 3649;

f ð16; 8Þ ¼ f ð8; 16Þ ¼ 39490049:

Hence, we can see in this simple example that thenumber of alignments grows very fast. Equation (10)shows that the double sequence {f(n, m)} is mono-tone nondecreasing in both variables, i.e.f(n, m) # f(n, m þ 1) and f(n, m) # f(n þ 1, m), for alln, m [ N. In particular, f ðn;mÞ ¼ f ðm; nÞ $ f ðn; nÞ forall n # m and f ðn;mÞ ¼ f ðm; nÞ # f ðn; nÞ whenevern $ m.

From this fact, the behaviour of function f at thediagonal terms (n, n) plays an important role in thestudy of the considered functions. Clearly, { f (n, n)} isa nondecreasing monotone. And it grows very fast,as we can note in the following facts

f ð10; 10Þ ¼ 8097453;

f ð50; 50Þ < 15 £ 1036;

f ð100; 100Þ < 2 £ 1074;

Special mention of the inequalities

f ð106; 106Þ , 1080 , f ð107; 107Þ:

This last estimation is relevant since there are only1080 protons in our universe (see Nowak and May,2000, p. 84). A typical sequence contains about 200–500 base pairs (Lesk, 2002). Proteins are about200–400 amino acids long, and hence they are600–1200 letters long.

Using the Newton’s binomial formula andclassical inequalities, we arrive at the followingestimations from above and below on the growth off(n,n):

Xn

k¼0

2kn

k

!¼ 3n # f ðn; nÞ

#Xn

k¼0

ffiffiffi2

p k n

k

! ! Xn

k¼0

ffiffiffi2

p k n

k

! !

¼ ð1 þffiffiffi2

pÞ2n:

These estimations are shown in Fig. 1 when bothsequences are of the same length n.

Acknowledgements

Second and third author’s research partiallysupported by Ministerio de Ciencia y Tecnologıaand FEDER, project BFM2001-3884-C02-01, and byXunta de Galicia and FEDER, project PGIDIT02P-XIC20703PN. The authors thank the anonymousreviewers for their useful comments.

FIGURE 1 Upper and lower estimation of the growth of thenumber of alignments f (n, n).

NUMBER OF ALIGNMENTS 429

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