pairwise alignmentstaffweb.ncnu.edu.tw/shieng/pairwise_alignment.pdf · 2003. 9. 25. · pairwise...

Pairwise Alignment

Guan-Shieng Huang

shieng@ncnu.edu.tw

Dept. of CSIE, NCNU

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Approach

1. Problem definition

2. Computational method (algorithms)

3. Complexity and performance

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Motivations

• Reconstructing long sequences of DNA formoverlapping sequence fragments

• Determining physical and genetic maps fromprobe data under various experimentprotocols

• Database searching

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• Comparing two of more sequences forsimilarities

• Protein structure prediction (building profiles)• Comparing the same gene sequenced by two

different labs

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Similarity & Difference

1. Common Ancestor Assumption

2. Mutation:(a) substitution (transition, transversion)(b) deletion(c) insertion

We use indel to refer to deletion or insertion.

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What is the difference between acctga andagcta?

acctgaagctgaagct - a

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Key Issues

1. notion of similarity/difference

2. the scoring system used to rank alignments

3. the algorithm used to find optimal scoringalignment

4. the statistical method used to evaluate thesignificance of an alignment score

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Edit Distance

Measure similarity by

1. substitution: −1

2. indel: −2

3. match: +1

a c c t g aa g c t - a1 -1 1 1 -2 1 = 1

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a c c t g aa - g c t a1 -2 -1 -1 -1 1 = −3

a c c t g a- a g c t a

-2 -1 -1 -1 -1 1 = −5

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x: x1x2x3 . . . xm

y: y1y2y3 . . . yn

Alphabet:• Σ = {A,G,C, T} for DNA sequence

• Σ = {A,G,C, U} for RNA sequence

• Σ = {A,C,D,E, F,G,H, I,K, L,

M,N, P,Q,R, S, T, V,W, Y } for proteins

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s(a, b): the score to substitute a by b

s(a,−): delete a

s(−, b): insert b

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Nomenclature

BIOLOGY COMPUTER SCIENCE- sequence - string, word- subsequence - substring (contiguous)- N/A - subsequence- N/A - exact matching- alignment - inexact matching

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Algorithm for PairwiseAlignment

To find the best alignment (with the highestscore) through

• Brute-force• Dynamic programming

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Brute-force Algorithm

Try all possible alignments of x and y.

F (m, n) = F (m − 1, n) + F (m, n − 1) + F (m − 1, n − 1)

k

l

=

k − 1

l − 1

+

k − 1

l

m + n

m

=

m + n − 1

m − 1

+

m + n − 1

m

C(m, n) = C(m − 1, n) + C(m, n − 1)

∴ F (m, n) ≥ C(m, n) =

m + n

m

,

2n

n

'22n

√πn

.

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Dynamic Programming Approach

F (i, j): the score for the best alignment betweenx1 . . . xi and y1 . . . yj.

F (i, j) = max

F (i − 1, j − 1) + 1, xi = yi (match)

F (i − 1, j − 1) − 1, xi 6= yi (substitution)

F (i − 1, j) − 2, align xi with a gap

F (i, j − 1) − 2, align yj with a gap

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{

x1x2 . . . xi−1 xi

y1y2 . . . yj−1 yj

⇒ F (i − 1, j − 1) + s(xi, yi)

{

x1x2 . . . xi−1 xi

y1y2 . . . yj −⇒ F (i − 1, j) − d

{

x1x2 . . . xi −

y1y2 . . . yj−1 yj

⇒ F (i, j − 1) − d

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Alignment Graph

F (i − 1, j − 1) F (i − 1, j)

F (i, j − 1) F (i, j)

+s(xi , y

j )

−d

−d

Initial value:

F (0, 0) = 0, F (0, j) = −jd, F (i, 0) = −id.

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Example- a c c t g a

- 0 -2 -4 -6 -8 -10 -12

a -2 1 -1 -3 -5 -7 -9

g -4 -1 0 -2 -4 -4 -6

c -6 -3 0 1 -1 -3 -5

t -8 -5 -2 -1 2 0 -2

a -10 -7 -4 -3 0 1 1

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Example- a c c t g a

- 0 -2 -4 -6 -8 -10 -12

a -2 1 -1 -3 -5 -7 -9

g -4 -1 0 -2 -4 -4 -6

c -6 -3 0 1 -1 -3 -5

t -8 -5 -2 -1 2 0 -2

a -10 -7 -4 -3 0 1 1backtrace

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a c c t g aa g c t - a

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Complexity

1. time = O(mn)

2. space= O(mn) if we need to find out theoptimal alignment

The problem for space is more serious when m

and n are very large.

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Linear-space AlignmentAlgorithm

B(i, j): the best alignment score of the suffixesxm−i+1 . . . xm and yn−j+1 . . . yn

F (i, j): forward matrix, B(i, j): backward matrixThen

F (m,n) = max0≤k≤n

{F (m

2, k) + B(

m

2, n − k)}.

m2

m2

k n − k

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Algorithm

1. Compute F while saving the m2

-th row.

2. Compute B while saving the m2

-th row.

3. Find the column k∗ such that

F (m

2, k∗) + B(

m

2, n − k∗) = F (m,n).

4. Recursively partition the problem to two sub-problems:

(a) Find the path from (0, 0) to (m2, k∗).

(b) Find the path from (m2, k∗) to (m,n).

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Example- a c c t g a

- 0 -2 -4 -6 -8 -10 -12

a -2 1 -1 -3 -5 -7 -9

g -4 -1 0 -2 -4 -4 -6

c -6 -3 0 1 -1 -3 -5

t -8 -5 -2 -1 2 0 -2

a -10 -7 -4 -3 0 1 1(F (i, j) matrix)

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- a g t c c a

- 0 -2 -4 -6 -8 -10 -12

a -2 1 -1 -3 -5 -7 -9

t -4 -1 0 0 -2 -4 -6

c -6 -3 -2 -1 1 -1 -3

g -8 -5 -2 -3 -2 0 -2

a -10 -7 -4 -3 -4 -2 1(B(i, j) matrix)

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-4 -1 0 -2 -4 -4 -6

-6 -3 -2 -1 1 -1 -3

F (m

2, k∗) + B(

m

2, n − k∗) = F (m,n).

In this case, F (m,n) = 1 and k∗ = 2.

Hence, the best alignment of (acctga,agcta) is the

concatenation of (ac,ag) and (ctga,cta).

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Analysis of Complexity

Clearly, the required space is O(min(m,n)). Fortime complexity, let T (m,n) be the time bound ofthe algorithm.Hence, we have

T (m,n) = T (bm

2c, k) + T (d

m

2e, n − k) + O(mn)

for some k.

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T (m,n) = T (m

2, k) + T (

m

2, n − k) + cmn)

for some k.Suppose T (m,n) = αmn, then the right handside becomes

αm

2· k + α

m

2· (n − k) + cmn =

αmn

2+ cmn.

Let α = 2c, then it equals to the left-hand side.

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For more information on linear-space algorithmsin pairwise alignment, seeChao, K. M., Hardison, R. C., and Miller, W.1994. Recent developments in linear-spacealignment methods: a survey. Journal ofComputational Biology, 1:271–291.

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Revisiting Dynamic Programming

• Principle of optimality• Recurrence• Bottom up

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Substitution matrices

• Suppose we have two models:1. random model2. match model

• Given any two aligned sequencesx = x1 x2 . . . xn

y = y1 y2 . . . yn

where xi is aligned with yi.

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• In random model R, we suppose each letter a occursindependently with some frequency qa. Hence,

Pr(x, y|R) =∏

i

qxi

∏

j

qyj.

• In match model M, letters a and b are aligned with jointprobability pab. Suppose residues a and b have beenderived indep. from some unknown residue c. Hence,

Pr(x, y|M) =∏

i

pxiyi.

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• Define the odds ratio as

Pr(x, y|M)

Pr(x, y|R)=

∏

i pxiyi∏

i qxi

∏

j qyj

=∏

i

pxiyi

qxiqyi

.

• The log-odds ratio:

S =∑

i

s(xi, yi) where s(a, b) = log(pab

qaqb

).

• S > 0 means that x, y are more likely to be an instanceof the match model. (Maximum Likelihood)

• BLOSUM & PAM matrices for proteins

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PAM matrices

1. Dayhoff, Schwartz, Orcutt (1978)

2. The most widely used matrix is PAM250.

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BLOSUM Matrices

1. Henikoff & Henikoff (1992)

2. Derived from a set of aligned, ungappedregions from protein families called theBLOCKS database.

3. BLOSUM62 is the standard for ungappedmatching.

4. BLOSUM50 is better for alignment with gaps.

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BLOSUM50

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Pairwise Alignment Problems

1. Global alignment (Needleman & Wunsch,1970)

2. Local alignment (Smith-Waterman, 1981)

3. End-space free alignment

4. Gap penality

The version we currently used was due to Gotoh

(1982).

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Global Alignment

Given two sequences x and y, what is the maxi-

mum similarity between them? Find a best align-

ment.

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Local Alignment

Given two sequences x and y, what is the maxi-

mum similarity between a subsequence of x and

a subsequence of y? Find most similar subse-

quences.

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End-space Free Alignment

or

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Global Alignment

F (i, j) = max

F (i − 1, j − 1) + s(xi, yj),

F (i − 1, j) − d,

F (i, j − 1) − d.

with initial value

F (0, 0) = 0, F (0, j) = −jd, F (i, 0) = −id.

And F (m,n) is the score.

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Example

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Local Alignment

Motivation:• Ignore stretches of non-coding DNA.• Protein domains

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Local Alignment

F (i, j) = max

0,

F (i − 1, j − 1) + s(xi, yj),

F (i − 1, j) − d,

F (i, j − 1) − d.

with initial value F (0, 0) = F (0, j) = F (i, 0) = 0. And the

highest value of F (i, j) over the whole matrix is the score.

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Example

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Ends-free Alignment

Motivation:• shotgun sequence assembly

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Ends-free Alignment

F (i, j) = max

F (i − 1, j − 1) + s(xi, yj),

F (i − 1, j) − d,

F (i, j − 1) − d.

with initial value

F (0, 0) = F (0, j) = F (i, 0) = 0.

And the highest value of F (i, j) in the last column F (i∗, n)

or the last row F (m, j∗) is the score.

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Example

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Complexity

All of the above algorithms can be implemented

in time O(mn) and in space O(m + n).

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Gap Penality

• A gap is any maximal consecutive run ofspaces in an alignment.

• The length of a gap is the number of indeloperations in it.

a t t c - - g a - t g g a c ca - - c g t g a t t - - - c c

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Motivation:• Insertion or deletion of an entire sequence

often occurs as a single mutation event.• Two protein sequences might be relatively

similar over several intervals.• cDNA: the complement of mRNA

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Gap Penality Models

1. constant gap penalty model: Wg × #gaps

2. affine gap penalty model: (y = ax + b)Wg × #gaps + Ws × #spaces

3. convex gap penalty model: Wg + log(q) whereq is the length of the gap.

4. arbitrary gap penalty model

Wg: gap-open penalty, Ws: gap-extension penalty

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Complexity

1. constant gap penalty model:Time= O(mn)

2. affine gap penalty model:Time= O(mn)

3. convex gap penalty model:Time= O(mn lg(m + n))

4. arbitrary gap penalty model:Time = O(mn(m + n))

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Conclusion

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