1. Dynamic Programming: Sequence alignment
CS 466
Saurabh Sinha

2. DNA Sequence Comparison: First Success Story
Finding sequence similarities with genes of known function is a common approach to infer a newly sequenced gene’s function
In 1984 Russell Doolittle and colleagues found similarities between cancer-causing gene and normal growth factor (PDGF) gene
A normal growth gene switched on at the wrong time causes cancer !

3. Cystic Fibrosis
Cystic fibrosis (CF) is a chronic and frequently fatal genetic disease of the body's mucus glands. CF primarily affects the respiratory systems in children.
Search for the CF gene was narrowed to ~1 Mbp, and the region was sequenced.
Scanned a database for matches to known genes. A segment in this region matched the gene for some ATP binding protein(s). These proteins are part of the ion transport channel, and CF involves sweat secretions with abnormal sodium content!

4. Role for Bioinformatics
Gene similarities between two genes with known and unknown function alert biologists to some possibilities
Computing a similarity score between two genes tells how likely it is that they have similar functions
Dynamic programming is a technique for revealing similarities between genes

5. Motivating Dynamic Programming

6. Dynamic programming example: Manhattan Tourist Problem
Imagine seeking a path (from source to sink) to travel (only eastward and southward) with the most number of attractions (*) in the Manhattan grid
Source * * * * * * * * * * * *
Sink

7. Dynamic programming example: Manhattan Tourist Problem
Imagine seeking a path (from source to sink) to travel (only eastward and southward) with the most number of attractions (*) in the Manhattan grid
Source * * * * * * * * * * * *
Sink

8. Manhattan Tourist Problem: Formulation
Goal: Find a “most weighted” path in a weighted grid.
Input: A weighted grid G with two distinct vertices, one labeled “source” and the other labeled “sink”
Output: A “most weighted” path in G from “source” to “sink”

9. MTP: An Example source sink 3 5 9 13 15 19 20 23 j coordinate1 2 3 4
j coordinate
source 3 2 4 3 5 9 1 4 3 2 3 2 4 2 13 1 1 4 6 5 2 7 3 4 15 19
i coordinate 2 4 4 5 2 1 3 3 2 3 20 3 5 6 8 5 1 3 2 2
sink 23 4

10. MTP: Greedy Algorithm Is Not Optimal1 2 5
source 22 5 3 10 5 2 1 5 3 5 3 1 2 3 4
promising start, but leads to bad choices! 5 2
sink 18

11. MTP: Simple Recursive Program
MT(n,m)
if n=0 or m=0
return MT(n,m)
x  MT(n-1,m)+
length of the edge from (n- 1,m) to (n,m)
y  MT(n,m-1)+
length of the edge from (n,m-1) to (n,m)
return max{x,y}
What’s wrong with this approach?

12. Here’s what’s wrong M(n,m) needs M(n, m-1) and M(n-1, m)
Both of these need M(n-1, m-1)
So M(n-1, m-1) will be computed at least twice
Dynamic programming: the same idea as this recursive algorithm, but keep all intermediate results in a table and reuse

13. MTP: Dynamic Programmingj 1
source 1 1 i
S0,1 = 1 5 1 5
S1,0 = 5
Calculate optimal path score for each vertex in the graph
Each vertex’s score is the maximum of the prior vertices score plus the weight of the respective edge in between

14. MTP: Dynamic Programming (cont’d)j 1 2
source 1 2 1 3 i
S0,2 = 3 5 3 -5 1 5 4
S1,1 = 4 3 2 8
S2,0 = 8

15. MTP: Dynamic Programming (cont’d)j 1 2 3
source 1 2 5 1 3 8 i
S3,0 = 8 5 3 10 -5 1 1 5 4 13
S1,2 = 13 3 5 -5 2 8 9
S2,1 = 9 3 8
S3,0 = 8

16. MTP: Dynamic Programming (cont’d)j 1 2 3
source 1 2 5 1 3 8 i 5 3 10 -5 -5 1 -5 1 5 4 13 8
S1,3 = 8 3 5 -3 -5 3 2 8 9 12
S2,2 = 12 3 8 9
S3,1 = 9
greedy alg. fails!

17. MTP: Dynamic Programming (cont’d)j 1 2 3
source 1 2 5 1 3 8 i 5 3 10 -5 -5 1 -5 1 5 4 13 8 3 5 -3 2 -5 3 3 2 8 9 12 15
S2,3 = 15 -5 3 8 9 9
S3,2 = 9

18. MTP: Dynamic Programming (cont’d)j 1 2 3
source 1 2 5 1 3 8
Done! i 5 3 10 -5 -5 1 -5 1 5 4 13 8
(showing all back-traces) 3 5 -3 2 -5 3 3 2 8 9 12 15 -5 1 3 8 9 9 16
S3,3 = 16

19. MTP: Recurrence
Computing the score for a point (i,j) by the recurrence relation:
si, j =
max
si-1, j + weight of the edge between (i-1, j) and (i, j)
si, j-1 + weight of the edge between (i, j-1) and (i, j)
The running time is n x m for a n by m grid
(n = # of rows, m = # of columns)

20. Traveling in the Grid
By the time the vertex x is analyzed, the values sy for all its predecessors y should be computed – otherwise we are in trouble.
We need to traverse the vertices in some order
For a grid, can traverse vertices row by row, column by column, or diagonal by diagonal

21. Traversing the Manhattan Gridb)
3 different strategies:
a) Column by column
b) Row by row
c) Along diagonals c)

22. Alignment

23. Aligning DNA Sequences
Alignment : 2 x k matrix ( k  m, n )
V = ATCTGATG
n = 8 4 1 2
matches
mismatches
deletions
insertions
m = 7
W = TGCATAC
match
mismatch V A T C G
deletion W
insertion
indels

24. Longest Common Subsequence (LCS) – Alignment without Mismatches
Given two sequences
v = v1 v2…vm and w = w1 w2…wn
The LCS of v and w is a sequence of positions in
v: 1 < i1 < i2 < … < it < m
and a sequence of positions in
w: 1 < j1 < j2 < … < jt < n
such that it -th letter of v equals to jt-letter of w and t is maximal

25. LCS: Example Every common subsequence is a path in 2-D grid 1 1 2 2 31 1 2 2 3 4 3 5 4 5 6 6 7 7 8
i coords:
elements of v A T -- C -- T G A T C
elements of w -- T G C A T -- A -- C
j coords:
(0,0)
(1,0)
(2,1)
(2,2)
(3,3)
(3,4)
(4,5)
(5,5)
(6,6)
(7,6)
(8,7)
positions in v:
2 < 3 < 4 < 6 < 8
Matches shown in red
positions in w:
1 < 3 < 5 < 6 < 7
Every common subsequence is a path in 2-D grid

26. Computing LCS Let vi = prefix of v of length i: v1 … vi
and wj = prefix of w of length j: w1 … wj
The length of LCS(vi,wj) is computed by:
si, j =
max
si-1, j
si, j-1
si-1, j if vi = wj

27. LCS Problem as Manhattan Tourist ProblemG A T C j 1 2 3 4 5 6 7 8 i T 1 G 2 C 3 A 4 T 5 A 6 C 7

28. Edit Graph for LCS Problemj 1 2 3 4 5 6 7 8 i T 1 G 2 C 3 A 4 T 5 A 6 C 7

29. Edit Graph for LCS Problemj 1 2 3 4 5 6 7 8
Every path is a common subsequence.
Every diagonal edge adds an extra element to common subsequence
LCS Problem: Find a path with maximum number of diagonal edges i T 1 G 2 C 3 A 4 T 5 A 6 C 7

30. Backtracking si,j allows us to compute the length of LCS for vi and wj
sm,n gives us the length of the LCS for v and w
How do we print the actual LCS ?
At each step, we chose an optimal decision si,j = max (…)
Record which of the choices was made in order to obtain this max

31. Computing LCS Let vi = prefix of v of length i: v1 … vi
and wj = prefix of w of length j: w1 … wj
The length of LCS(vi,wj) is computed by:
si, j =
max
si-1, j
si, j-1
si-1, j if vi = wj

32. Printing LCS: Backtracking
PrintLCS(b,v,i,j)
if i = 0 or j = 0
return
if bi,j = “ “
PrintLCS(b,v,i-1,j-1)
print vi
else
PrintLCS(b,v,i-1,j)
PrintLCS(b,v,i,j-1)

33. From LCS to Alignment
The Longest Common Subsequence problem—the simplest form of sequence alignment – allows only insertions and deletions (no mismatches).
In the LCS Problem, we scored 1 for matches and 0 for indels
Consider penalizing indels and mismatches with negative scores
Simplest scoring scheme:
+1 : match premium
-μ : mismatch penalty
-σ : indel penalty

34. Simple Scoring
When mismatches are penalized by –μ, indels are penalized by –σ,
and matches are rewarded with +1,
the resulting score is:
#matches – μ(#mismatches) – σ (#indels)

35. The Global Alignment Problem
Find the best alignment between two strings under a given scoring schema
Input : Strings v and w and a scoring schema
Output : Alignment of maximum score
→ = -σ
= 1 if match
= -µ if mismatch
si-1,j if vi = wj
si,j = max s i-1,j-1 -µ if vi ≠ wj
s i-1,j - σ
s i,j-1 - σ
m : mismatch penalty
σ : indel penalty {

36. Scoring Matrices
To generalize scoring, consider a (4+1) x(4+1) scoring matrix δ.
In the case of an amino acid sequence alignment, the scoring matrix would be a (20+1)x(20+1) size. The addition of 1 is to include the score for comparison of a gap character “-”.
This will simplify the algorithm as follows:
si-1,j-1 + δ (vi, wj)
si,j = max s i-1,j + δ (vi, -)
s i,j-1 + δ (-, wj) {

37. Making a Scoring Matrix
Scoring matrices are created based on biological evidence.
Alignments can be thought of as two sequences that differ due to mutations.
Some of these mutations have little effect on the function of the sequence, therefore some penalties, δ(vi , wj), will be less harsh than others.

38. Scoring Matrix: Example for protein sequencesK 5 -2 -1 - 7 3 6
Notice that although R and K are different amino acids, they have a positive score.
Why? They are both positively charged amino acids will not greatly change function of protein.
AKRANR
KAAANK
-1 + (-1) + (-2) = 11

39. Conservation
Amino acid changes that tend to preserve the physico-chemical properties of the original residue
Polar to polar
aspartate  glutamate
Nonpolar to nonpolar
alanine  valine
Similarly behaving residues
leucine to isoleucine

40. Local vs. Global Alignment
The Global Alignment Problem tries to find the longest path between vertices (0,0) and (n,m) in the edit graph.
The Local Alignment Problem tries to find the longest path among paths between arbitrary vertices (i,j) and (i’, j’) in the edit graph.
In the edit graph with negatively-scored edges, Local Alignment may score higher than Global Alignment

41. Local Alignment: Example
Compute a “mini” Global Alignment to get Local Alignment
Local alignment
Global alignment

42. Local Alignments: Why?
Two genes in different species may be similar over short conserved regions and dissimilar over remaining regions.
Example:
Homeobox genes have a short region called the homeodomain that is highly conserved between species.
A global alignment would not find the homeodomain because it would try to align the ENTIRE sequence

43. The Local Alignment Problem
Goal: Find the best local alignment between two strings
Input : Strings v, w and scoring matrix δ
Output : Alignment of substrings of v and w whose alignment score is maximum among all possible alignment of all possible substrings

44. The Problem Is … Long run time O(n4):
- In the grid of size n x n there are ~n2 vertices (i,j) that may serve as a source.
- For each such vertex computing alignments from (i,j) to (i’,j’) takes O(n2) time.
This can be remedied by allowing every point to be the starting point

45. The Local Alignment Recurrence
The largest value of si,j over the whole edit graph is the score of the best local alignment.
The recurrence:
Notice there is only this change from the original recurrence of a Global Alignment
si,j = max si-1,j-1 + δ (vi, wj)
s i-1,j + δ (vi, -)
s i,j-1 + δ (-, wj) {