Sequence Alignment I Lecture #2

.

Sequence Alignment ILecture #2

This class has been edited from Nir Friedman’s lecture which is available at www.cs.huji.ac.il/~nir. Changes made by Dan Geiger. Thanks to Shlomo Moran for several improvements.

Background Readings: The second chapter (pages 12-45) in the text book, Biological Sequence Analysis, Durbin et al., 2001.

2

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

3

Sequence Comparison (cont)

Finding similarity between sequences is important for many biological questions

For example: Find similar proteins

Allows to predict function & structure Locate similar subsequences in DNA

Allows to identify (e.g) regulatory elements Locate DNA sequences that might overlap

Helps in sequence assembly

g1

g2

4

Sequence Alignment

Input: two sequences over the same alphabet

Output: an alignment of the two sequences

Example: GCGCATGGATTGAGCGA TGCGCCATTGATGACCA

A possible alignment:

-GCGC-ATGGATTGAGCGA

TGCGCCATTGAT-GACC-A

5

Alignments

-GCGC-ATGGATTGAGCGA

TGCGCCATTGAT-GACC-A

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

6

Choosing Alignments

There are many possible alignments

For example, compare:

-GCGC-ATGGATTGAGCGA

TGCGCCATTGAT-GACC-A

to

------GCGCATGGATTGAGCGA

TGCGCC----ATTGATGACCA--

Which one is better?

7

Scoring Alignments

Intuition: Similar sequences evolved from a common

ancestor Evolution changed the sequences from this

ancestral sequence by mutations: Replacements: one letter replaced by another Deletion: deletion of a letter Insertion: insertion of a letter

Scoring of sequence similarity should examine how many and which operations took place

8

Simple Scoring Rule

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

Score of an alignment is sum of position scores

9

Example

Example:

-GCGC-ATGGATTGAGCGA

TGCGCCATTGAT-GACC-A

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

------GCGCATGGATTGAGCGA

TGCGCC----ATTGATGACCA--

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

10

More General Scores

The choice of +1,-1, and -2 scores is quite arbitrary Depending on the context, some changes are more

plausible than others Exchange of an amino-acid by one with similar

properties (size, charge, etc.)

vs. Exchange of an amino-acid by one with opposite

properties Probabilistic interpretation: (e.g.) How likely is one

alignment versus another ?

11

Additive Scoring Rules

We define a scoring function by specifying a function

(x,y) is the score of replacing x by y (x,-) is the score of deleting x (-,x) is the score of inserting x

The score of an alignment is the sum of position scores

}){(}){(:

12

The Optimal Score

The optimal (maximal) score between two sequences is the maximal score of all alignments of these sequences, namely,

Computing the maximal score or actually finding an alignment that yields the maximal score are closely related tasks with similar algorithms.

We now address these problems.

nment)score(aligmax),d( & of alignment 21 ss21 ss

13

Computing Optimal Score

How can we compute the optimal score ? If |s| = n and |t| = m, there are many alignments !

Exercise : Show that the number of legal alignments

denoted by A(m,n), where n > m, satisfies (legal means not to a “-” versus another “-”) :

The additive form of the score allows us to perform dynamic programming to compute optimal score efficiently.

),( nmAm

nm

14

Recursive Argument

Suppose we have two sequences:s[1..n+1] and t[1..m+1]

The best alignment must be one of three cases:

1. Last match is (s[n+1],t[m +1] )2. Last match is (s[n +1],-)3. Last match is (-, t[m +1] )

])1[],1[(

])..1[],..,1[(])1..1[],1..1[(

mtns

mtnsdmtnsd

15

Recursive Argument

Suppose we have two sequences:s[1..n+1] and t[1..m+1]

The best alignment must be one of three cases:

1. Last match is (s[n+1],t[m +1] )2. Last match is (s[n +1],-)3. Last match is (-, t[m +1] )

)],1[(

])1..1[],..,1[(])1..1[],1..1[(

ns

mtnsdmtnsd

16

Recursive Argument

Suppose we have two sequences:s[1..n+1] and t[1..m+1]

The best alignment must be one of three cases:

1. Last match is (s[n+1],t[m +1] )2. Last match is (s[n +1],-)3. Last match is (-, t[m +1] )

])1[,(

])..1[],1..,1[(])1..1[],1..1[(

mt

mtnsdmtnsd

17

Recursive Argument

Define the notation:

Using our recursive argument, we get the following recurrence for V:

])[,(],[

)],[(],[

])[],[(],[

max],[

1jtj1iV

1is1jiV

1jt1isjiV

1j1iV

])..[],..[(],[ j1ti1sdjiV

V[i,j] V[i+1,j]

V[i,j+1] V[i+1,j+1]

18

Recursive Argument

Of course, we also need to handle the base cases in the recursion:

])[,(],[],[

)],[(],[],[

],[

1jtj0V1j0V

1is0iV01iV

000V

0 A 1

G 2

C 3

0 0 -2 -4 -6

A 1 -2

A 2 -4

A 3 -6

C 4 -8

AA- -

We fill the matrix using the recurrence rule:

ST

versus

19

Dynamic Programming Algorithm

We continue to fill the matrix using the recurrence rule

0

A 1

G 2

C 3

0 0 -2 -4 -6

A 1 -2

A 2 -4

A 3 -6

C 4 -8

ST

20

Dynamic Programming Algorithm

0

A 1

G 2

C 3

0 0 -2 -4 -6

A 1 -2 1

A 2 -4

A 3 -6

C 4 -8

V[0,0] V[0,1]

V[1,0] V[1,1]

+1-2 -A A-

-2 (A- versus -A)

versus

ST

21

Dynamic Programming Algorithm

0

A 1

G 2

C 3

0 0 -2 -4 -6

A 1 -2 1 -1 -3

A 2 -4 -1 0

A 3 -6 -3

C 4 -8 -5

ST

22

Dynamic Programming Algorithm

0

A 1

G 2

C 3

0 0 -2 -4 -6

A 1 -2 1 -1 -3

A 2 -4 -1 0 -2

A 3 -6 -3 -2 -1

C 4 -8 -5 -4 -1

Conclusion: d(AAAC,AGC) = -1

ST

23

Reconstructing the Best Alignment

To reconstruct the best alignment, we record which case(s) in the recursive rule maximized the score

0A1

G2

C3

0 0 -2 -4 -6

A 1 -2 1 -1 -3

A 2 -4 -1 0 -2

A 3 -6 -3 -2 -1

C 4 -8 -5 -4 -1

ST

24

Reconstructing the Best Alignment

We now trace back a path that corresponds to the best alignment

0A1

G2

C3

0 0 -2 -4 -6

A 1 -2 1 -1 -3

A 2 -4 -1 0 -2

A 3 -6 -3 -2 -1

C 4 -8 -5 -4 -1

AAACAG-C

ST

25

Reconstructing the Best Alignment

Sometimes, more than one alignment has the best score

0A1

G2

C3

0 0 -2 -4 -6

A 1 -2 1 -1 -3

A 2 -4 -1 0 -2

A 3 -6 -3 -2 -1

C 4 -8 -5 -4 -1

ST

AAACA-GC

AAAC-AGC

AAACAG-C

26

Time Complexity

Space: O(mn)Time: O(mn) Filling the matrix O(mn) Backtrace O(m+n)

0A1

G2

C3

0 0 -2 -4 -6

A 1 -2 1 -1 -3

A 2 -4 -1 0 -2

A 3 -6 -3 -2 -1

C 4 -8 -5 -4 -1

ST

27

Space Complexity

In real-life applications, n and m can be very large The space requirements of O(mn) can be too

demanding If m = n = 1000, we need 1MB space If m = n = 10000, we need 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?

28

Why Do We Need So Much Space?

Compute V(i,j), column by column, storing only two columns in memory (or line by line ). 0

-2

-4

-6

-8

-2

1

-1

-3

-5

-4

-1

0

-2

-4

-6

-3

-2

-1

-1

0A1

G2

C3

0

A 1

A 2

A 3

C 4

Note however that This “trick” fails when we

need to reconstruct the optimizing sequence.

Trace back information requires O(mn) memory bytes.

To compute V[n,m]=d(s[1..n],t[1..m]), we need only O(min(n,m)) space:

29

Space Efficient Version: Outline

Find position (n/2, j) at which the best alignment crosses a midpoint

Input: Sequences s[1,n] and t[1,m] to be aligned.

Idea: perform divide and conquer

Construct two alignments A= s[1,n/2] vs t[1,j] B= s[n/2+1,n] vs

t[j+1,m] Return the concatenated alignment AB

t

s

30

Finding the Midpoint The score of the best alignment that goes through j equals:

d(s[1,n/2],t[1,j]) + d(s[n/2+1,n],t[j+1,m])

t

s

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

We need to find j that maximizes this score. Such j determines a point (n/2,j) through which the best alignment passes.

31

Finding the Midpoint (Algorithm)Define V[i,j] = d(s[1..i],t[1..j]) (As before) B[i,j] = d(s[i+1..n],t[j+1..m]) (Symmetrically) F[i,j] + B[i,j] = score of best alignment through (i,j)

t

s

32

Computing V[i,j]

])[,(],[

)],[(],[

])[],[(],[

max],[

1jtj1iV

1is1jiV

1jt1isjiV

1j1iV

V[i,j] V[i+1,j]

V[i,j+1] V[i+1,j+1]

As before:

V[i,j] = d(s[1..i],t[1..j])

Requires linear space complexity

33

Computing B[i,j]

])[,(]1,[

)],[(],1[

])[],[(]1,1[

max],[

jtjiB

isjiB

jtisjiB

jiB

B[i,j] B[i+1,j]

B[i,j+1] B[i+1,j+1]

Symmetrically (replacing i with i+1, and j with j+1):

B[i,j] = d(s[i+1..n],t[j+1..m])

Requires linear space complexity

34

Time Complexity Analysis Time to find a mid-point: cnm (c - a constant) Size of recursive sub-problems is

(n/2,j) and (n/2,m-j-1), hence

T(n,m) = cnm + T(n/2,j) + T(n/2,m-j-1)

Lemma: T(n,m) 2cnmProof (by induction):

T(n,m) cnm + 2c(n/2)j + 2c(n/2)(m-j-1) 2cnm.

Thus, time complexity is linear in size of the problem.

At worse, twice the cost of the regular solution.

t

s

35

Local Alignment

Consider now a different question: Can we find similar substrings of s and t Formally, given s[1..n] and t[1..m] find i,j,k, and l

such that d(s[i..j],t[k..l]) is maximal

36

Local Alignment

As before, we use dynamic programming We now want to setV[i,j] to record the best

alignment of a suffix of s[1..i] and a suffix of t[1..j]

How should we change the recurrence rule? Same as before but with an option to start afresh

The result is called the Smith-Waterman algorithm

37

Local Alignment

New option: We can start a new match instead of extending a

previous alignment

0

1jtj1iV1is1jiV

1jt1isjiV

1j1iV])[,(],[)],[(],[

])[],[(],[

max],[

Alignment of empty suffixes

]))1[,(],0[,0max(]1,0[

))],1[(]0,[,0max(]0,1[

0]0,0[

jtjVjV

isiViV

V

38

Local Alignment Example

0

A 1

T 2

C 3

T 4

A 5

A 6

0 0 0 0 0 0 0 0

T 1 0

A 2 0

A 3 0

T 4 0

A 5 0

s = TAATAt = TACTAA

ST

39

Local Alignment Example

0

T 1

A 2

C 3

T 4

A 5

A 6

0 0 0 0 0 0 0 0

T 1 0 1 0 0 1 0 0

A 2 0 0 2 0 0 2 1

A 3 0

T 4 0

A 5 0

s = TAATAt = TACTAA

ST

40

Local Alignment Example

0T1

A2

C3

T4

A5

A6

0 0 0 0 0 0 0 0

T 1 0 1 0 0 1 0 0

A 2 0 0 2 0 0 2 1

A 3 0 0 1 1 0 1 3

T 4 0 0 0 0 2 0 1

A 5 0 0 1 0 0 3 1

s = TAATAt = TACTAA

ST

41

Local Alignment Example

0T1

A2

C3

T4

A5

A6

0 0 0 0 0 0 0 0

T 1 0 1 0 0 1 0 0

A 2 0 0 2 0 0 2 1

A 3 0 0 1 1 0 1 3

T 4 0 0 0 0 2 0 1

A 5 0 0 1 0 0 3 1

s = TAATAt = TACTAA

ST

A maximal alignment starts at a maximum entry and follows backwards till a zero entry.

42

Local Alignment Example

0T1

A2

C3

T4

A5

A6

0 0 0 0 0 0 0 0

T 1 0 1 0 0 1 0 0

A 2 0 0 2 0 0 2 1

A 3 0 0 1 1 0 1 3

T 4 0 0 0 0 2 0 1

A 5 0 0 1 0 0 3 1

s = TAATAt = TACTAA

ST

43

Variants of Sequence Alignment

We have seen two variants of sequence alignment: Global alignment Local alignment

Other variants in the book and in tutorial time:1. Finding best overlap

2. Using an affine cost d(g) = -d –(g-1)e for gaps of length g. The –d is for introducing a gap and –e for continuing the gap. We used d=e=2. We could use smaller e.

These variants are based on the same basic idea of dynamic programming.

44

Remark: Edit Distance

Instead of speaking about the score of an alignment, one often talks about an edit distance between two sequences, defined to be the “cost” of the “cheapest” set of edit operations needed to transform one sequence into the other.

Cheapest operation is “no change” Next cheapest operation is “replace” The most expensive operation is “add space”.

Our goal is now to minimize the cost of operations, which is equivalent to maximizing the corresponding score (e.g., change scores to edit costs as follows: 1 -1,-1 1,-22).