Sequence Alignment

Algorithms in Computational BiologySpring 2006

Edited by Itai SharonMost slides have been created and edited by Nir Friedman, Dan Geiger, Shlomo Moran and Ydo Wexler

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Sequence Comparison

Much of bioinformatics involves sequences DNA sequences RNA sequences Protein sequences

We can think of these sequences as strings of letters DNA & RNA: alphabet ∑ of 4 letters Protein: alphabet ∑ of 20 letters

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Sequence comparison: Motivation

Finding similarity between sequences is important for many

biological questions.

Find homologous proteins Allows to predict structure and function

Locate similar subsequences in DNA e.g: allows to identify regulatory elements

Locate DNA sequences that might overlap Helps in sequence assembly

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Sequence Alignment

Input: two sequences over the same alphabet Output: an alignment of the two sequences

Two basic variants of sequence alignment: Global – all characters in both sequences participate

Needleman-Wunsch, 1970 Local – find related regions within sequences

Smith-Waterman, 1981

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Sequence Alignment - Example Input:

GCGCATGGATTGAGCGA and TGCGCCATTGATGACCA

Possible output:

-GCGC-ATGGATTGAGCGA

TGCGCCATTGAT-GACC-A

Three elements: Perfect matches Mismatches Insertions & deletions (indel)

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Scoring Function

Score each position independently: Match: +1 Mismatch: -1 Indel: -2

Score of an alignment is sum of position scores

Example:

-GCGC-ATGGATTGAGCGATGCGCCATTGAT-GACC-A

Score: (+1x13) + (-1x2) + (-2x4) = 3

------GCGCATGGATTGAGCGATGCGCC----ATTGATGACCA--

Score: (+1x5) + (-1x6) + (-2x11) = -23

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Sequence vs. Structure Similarity

Sequence 1 lcl|1A6M:_ MYOGLOBIN Length 151 (1..151)

Sequence 2 lcl|1JL7:A MONOMER HEMOGLOBIN COMPONENT III Length 147 (1..147)

Score = 31.6 bits (70), Expect = 10

Identities = 33/137 (24%), Positives = 55/137 (40%), Gaps = 17/137 (12%)

Query: 2 LSEGEWQLVLHVWAKVEA--DVAGHGQDILIRLFKSHPETLEKFDRFKHLKTEAEMKASE 59

LS + Q+V W + + AG G++ L + +HPE F +

Sbjct: 2 LSAAQRQVVASTWKDIAGADNGAGVGKECLSKFISAHPEMAAVFG--------FSGASDP 53

Query: 60 DLKKHGVTVLTALGAI---LKKKGHHEAELKPLAQSH---ATKHKIPIKYLEFISEAIIH 113

+ + G VL +G L +G AE+K + H KH I +Y E + +++

Sbjct: 54 GVAELGAKVLAQIGVAVSHLGDEGKMVAEMKAVGVRHKGYGNKH-IKAEYFEPLGASLLS 112

Query: 114 VLHSRHPGDFGADAQGA 130

+ R G A A+ A

Sbjct: 113 AMEHRIGGKMNAAAKDA 129

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Sequence vs. Structure Similarity

1A6M: Myoglobin

1JL7: Hemoglobin

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Global Alignment

Input: two sequences over the same alphabet Output: an alignment of the two sequences in

which all characters in both sequences participate

The Needleman-Wunsch algorithm finds an optimal global alignment between two sequences Uses a scoring function A dynamic programming algorithm

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The Needleman-Wunsch (NW) Algorithm

Suppose we have two sequences: s=s1…sn and t=t1…tm

Construct a matrix V[n+1, m+1] in which V(i, j) contains the score for the best alignment between s1…si and t1…tj. The grade for cell V(i, j) is:

V(i-1, j)+score(si, -)

V(I, j) = max V(i, j-1)+score(-, tj)

V(i-1, j-1)+score(si, tj)

V(n,m) is the score for the best alignment between s and t

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NW Algorithm – An Example

Alphabet: DNA, ∑ = {A,C,G,T}

Input: s = AAAC t = AGC

Scoring scheme: score(x, x) = 1 score(x,-) = -2 score(x, y) = -1

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NW Algorithm – An Example

A G C

A

A

A

C

-2 -4 -60

-2

-4

-6

-8

1 -1

-1 0

-3

-3

-2

-2

-5 -4

-1

-1

AG-C

AAAC

-AGC

AAAC

A-GC

AAAC

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NW – Time and Space Complexity

Time: Filling the matrix: Backtracing: Overall:

Space: Holding the matrix:

A G C

A

A

A

C

-2 -4 -60

-2

-4

-6

-8

1 -1

-1 0

-3

-3

-2

-2

-5 -4

-1

-1

O(n·m)

O(n·m)O(n+m)

O(n·m)

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NW – Space Complexity

In real-life applications, n and m can be very large

The space requirements of O(n·m) can be too demanding If n = m = 1000 we need O(1MB) space If n = m = 10000 we need O(100MB) space

We can afford to perform extra computation to save space Looping over million operations takes less than seconds on

modern workstations

Can we trade space with time?

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Why Do We Need So Much Space?

We can do the same computation in O(min(n,m)) space: Compute V(i, j) column by column, storing only two

columns in memory (or row by row if rows are shorter).

However… Trace back information requires

O(m·n) memory bytes. -2

-4

-6

-8

0

-1

-2

-4

-4

0

-1

-2

-6

-3

-1

-1

-2

-5

-3

1

GA C

A

A

A

C

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Space Efficient Version

Input: sequences s=s1…sn and t=t1…tm to be aligned.

Idea: perform divide and conquer find position (i, n/2) at which

some best alignment crosses a midpoint

Construct alignments A=s1…sn/2 vs. t=t1…ti and B=sn/2+1…sn vs. t=ti+1…tm

Return AB

s

t

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Finding a Midpoint

The score of the best alignment that goes through i

equals:score(s1…sn/2, t1…ti) + score(sn/2+1…sn, ti+1…tm)

Thus, we need to compute these two quantities for all values of i

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Finding a Midpoint

Define F(i, j) = score(s1…si, t1…ti)

B(i, j) = score(si+1…sn, tj+1…tm)

F(i, j) + B(i, j) = score of best alignment through (i, j)

Compute F(i, j) and B(i, j) in linear space complexity We compute F(i, j) in O(min(i, j)) We compute B(i, j) in exactly the same manner, going

“backward” from B(n,m)

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Time Complexity

Time to find a mid-point: c·n·m (c - a constant) Size of recursive sub-problems is (n/2,i) and (n/2,m-

i), hence:T(n,m) = c·n·m + T(n/2,i) + T(n/2,m-i)

Lemma: T(n, m) 2c·n·m

Proof:T(n,m) c·n·m + 2c(n/2)i + 2c(n/2)(m-i) = 2c·n·m.