1. Lectures 3-6: Pair-wise Sequence Alignment
CS 4233& 5263 Bioinformatics
Lectures 3-6: Pair-wise Sequence Alignment

2. Outline Part I: Algorithms Part II: Biological issues Part III: BLAST
Biological problem
Intro to dynamic programming
Global sequence alignment
Local sequence alignment
More efficient algorithms
Part II: Biological issues
Model gaps more accurately
Alignment statistics
Part III: BLAST

3. Evolution at the DNA levelC
…ACGGTGCAGTCACCA…
…ACGTTGC-GTCCACCA…
DNA evolutionary events (sequence edits):
Mutation, deletion, insertion

4. Sequence conservation implies function
next generation OK OK OK X X
Still OK?

5. Why sequence alignment?
Conserved regions are more likely to be functional
Can be used for finding genes, regulatory elements, etc.
Similar sequences often have similar origin and function
Can be used to predict functions for new genes / proteins
Sequence alignment is one of the most widely used computational tools in biology

6. Global Sequence Alignment
AGGCTATCACCTGACCTCCAGGCCGATGCCC
TAGCTATCACGACCGCGGTCGATTTGCCCGAC T S’
-AGGCTATCACCTGACCTCCAGGCCGA--TGCCC---
TAG-CTATCAC--GACCGC--GGTCGATTTGCCCGAC T’
Definition
An alignment of two strings S, T is a pair of strings S’, T’ (with spaces) s.t.
|S’| = |T’|, and (|S| = “length of S”)
removing all spaces in S’, T’ leaves S, T

7. What is a good alignment?
The “best” way to match the letters of one sequence with those of the other
How do we define “best”?

8. The score of aligning (characters or spaces) x & y is σ (x,y).
S’: -AGGCTATCACCTGACCTCCAGGCCGA--TGCCC---
T’: TAG-CTATCAC--GACCGC--GGTCGATTTGCCCGAC
The score of aligning (characters or spaces) x & y is σ (x,y).
Score of an alignment:
An optimal alignment: one with max score

9. Scoring Function Sequence edits: AGGCCTC Scoring Function:
Mutations AGGACTC
Insertions AGGGCCTC
Deletions AGG-CTC
Scoring Function:
Match: +m ~~~AAC~~~
Mismatch: -s ~~~A-A~~~
Gap (indel): -d

10. Match = 2, mismatch = -1, gap = -1
Score = 3 x 2 – 2 x 1 – 1 x 1 = 3

11. More complex scoring function
Substitution matrix
Similarity score of matching two letters a, b should reflect the probability of a, b derived from the same ancestor
It is usually defined by log likelihood ratio
Active research area. Especially for proteins.
Commonly used: PAM, BLOSUM

12. An example substitution matrixC G T 3 -2 -1

13. How to find an optimal alignment?
A naïve algorithm:
for all subseqs A of S, B of T s.t. |A| = |B| do
align A[i] with B[i], 1 ≤i ≤|A|
align all other chars to spaces
compute its value
retain the max
end
output the retained alignment
S = abcd A = cd
T = wxyz B = xz
-abc-d a-bc-d
w--xyz -w-xyz

14. Analysis Assume |S| = |T| = n Cost of evaluating one alignment: ≥n
How many alignments are there:
pick n chars of S,T together
say k of them are in S
match these k to the k unpicked chars of T
Total time:
E.g., for n = 20, time is > 240 >1012 operations

15. Intro to Dynamic Programming

16. Dynamic programming What is dynamic programming? Two simple examples:
A method for solving problems exhibiting the properties of overlapping subproblems and optimal substructure
Key idea: tabulating sub-problem solutions rather than re-computing them repeatedly
Two simple examples:
Computing Fibonacci numbers
Find the special shortest path in a grid

17. Example 1: Fibonacci numbers
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
F(0) = 1;
F(1) = 1;
F(n) = F(n-1) + F(n-2)
How to compute F(n)?

18. A recursive algorithm function fib(n) if (n == 0 or n == 1) return 1;
else return fib(n-1) + fib(n-2);
F(9)
F(8)
F(7)
F(7)
F(6)
F(5)
F(6)
F(5)
F(4)
F(3)

19. Why recursive Fib algorithm is inefficient?
Time complexity:
Between 2n/2 and 2n
O(1.62n), i.e. exponential
Why recursive Fib algorithm is inefficient?
Overlapping subproblems

20. An iterative algorithm
function fib(n)
F[0] = 1; F[1] = 1;
for i = 2 to n
F[i] = F[i-1] + F[i-2];
Return F[n];
Time complexity:
Time: O(n), space: O(n)

21. Example 2: shortest path in a grid
Each edge has a length (cost). We need to get to G from S. Can only move right or down. Aim: find a path with the minimum total length

22. Optimal substructures
Naïve algorithm: enumerate all possible paths and compare costs
Exponential number of paths
Key observation:
If a path P(S, G) is the shortest from S to G, any of its sub-path P(S,x), where x is on P(S,G), is the shortest from S to x

23. Proof Proof by contradiction
If the path between P(S,x) is not the shortest, i.e., P’(S,x) < P(S,x)
Construct a new path P’(S,G) = P’(S,x) + P(x, G)
P’(S,G) < P(S,G) => P(S,G) is not the shortest
Contradiction
Therefore, P(S, x) is the shortest S x G

24. Recursive solution Index each intersection by two indices, (i, j)
(0,0)
Index each intersection by two indices, (i, j)
Let F(i, j) be the total length of the shortest path from (0, 0) to (i, j). Therefore, F(m, n) is the shortest path we wanted.
To compute F(m, n), we need to compute both F(m-1, n) and F(m, n-1) m
(m, n) n
F(m-1, n) + length((m-1, n), (m, n))
F(m, n) = min
F(m, n-1) + length((m, n-1), (m, n))

25. Recursive solution F(i-1, j) + length((i-1, j), (i, j)) F(i, j) = min
(0,0)
But: if we use recursive call, many subpaths will be recomputed for many times
Strategy: pre-compute F values starting from the upper-left corner. Fill in row by row (what other order will also do?)
(i-1, j)
(i, j-1)
(i, j) m
(m, n) n

26. Dynamic programming illustration3 9 1 2 3 12 13 15 5 3 3 3 3 3 2 5 2 5 6 8 13 15 2 3 3 9 3 2 4 2 3 7 9 11 13 16 6 2 3 7 4 3 6 3 3 13 11 14 17 20 4 6 3 1 3 1 2 3 2 17 17 17 18 20 G
F(i-1, j) + length(i-1, j, i, j)
F(i, j) = min
F(i, j-1) + length(i, j-1, i, j)

27. Trackback 3 9 1 2 3 12 13 15 5 3 3 3 3 3 2 5 2 5 6 8 13 15 2 3 3 9 3 2 4 2 3 7 9 11 13 16 6 2 3 7 4 3 6 3 3 13 11 14 17 20 4 6 3 1 3 1 2 3 2 17 17 17 18 20

28. Elements of dynamic programming
Optimal sub-structures
Optimal solutions to the original problem contains optimal solutions to sub-problems
Overlapping sub-problems
Some sub-problems appear in many solutions
Memorization and reuse
Carefully choose the order that sub-problems are solved

29. Dynamic Programming for sequence alignment
Suppose we wish to align
x1……xM
y1……yN
Let F(i,j) = optimal score of aligning
x1……xi
y1……yj
Scoring Function:
Match: +m
Mismatch: -s
Gap (indel): -d

30. Elements of dynamic programming
Optimal sub-structures
Optimal solutions to the original problem contains optimal solutions to sub-problems
Overlapping sub-problems
Some sub-problems appear in many solutions
Memorization and reuse
Carefully choose the order that sub-problems are solved

31. Optimal substructure ... x: y:1 2 i M j N x: y:
If x[i] is aligned to y[j] in the optimal alignment between x[1..M] and y[1..N], then
The alignment between x[1..i] and y[1..j] is also optimal
Easy to prove by contradiction

32. Recursive solution ~~~~~~~ yN ~~~~~~~  ~~~~~~~ yN max
Notice three possible cases:
xM aligns to yN
~~~~~~~ xM
~~~~~~~ yN
2. xM aligns to a gap
~~~~~~~ xM
~~~~~~~ 
yN aligns to a gap
~~~~~~~ yN
m, if xM = yN
F(M,N) = F(M-1, N-1) +
-s, if not
max
F(M,N) = F(M-1, N) - d
F(M,N) = F(M, N-1) - d

33. Recursive solution Generalize: F(i,j) = max F(i-1, j) – d
F(i-1, j-1) + (Xi,Yj)
F(i,j) = max F(i-1, j) – d
F(i, j-1) – d
(Xi,Yj) = m if Xi = Yj, and –s otherwise
Boundary conditions:
F(0, 0) = 0.
F(0, j) = ?
F(i, 0) = ?
-jd: y[1..j] aligned to gaps.
-id: x[1..i] aligned to gaps.

34. What order to fill? F(0,0) F(M,N) F(i, j) F(i, j-1) F(i-1, j)2 3 i j

35. What order to fill?
F(0,0)
F(M,N)

36. Example A G T F(i,j) i = 0 1 2 3 4 j = 0 1 2 3 x = AGTA m = 1
y = ATA s = 1
d = 1
F(i-1, j-1) + (Xi,Yj)
F(i,j) = max F(i-1, j) – d
F(i, j-1) – d
F(i,j) i = A G T
j = 0 1 2 3

37. Example A G T -1 -2 -3 -4 F(i,j) i = 0 1 2 3 4 j = 0 1 2 3
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i-1, j-1) + (Xi,Yj)
F(i,j) = max F(i-1, j) – d
F(i, j-1) – d
F(i,j) i = A G T -1 -2 -3 -4
j = 0 1 2 3

38. Example A G T -1 -2 -3 -4 1 F(i,j) i = 0 1 2 3 4 j = 0 1 2 3
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i-1, j-1) + (Xi,Yj)
F(i,j) = max F(i-1, j) – d
F(i, j-1) – d
F(i,j) i = A G T -1 -2 -3 -4 1
j = 0 1 2 3

39. Example A G T -1 -2 -3 -4 1 F(i,j) i = 0 1 2 3 4 j = 0 1 2 3
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i-1, j-1) + (Xi,Yj)
F(i,j) = max F(i-1, j) – d
F(i, j-1) – d
F(i,j) i = A G T -1 -2 -3 -4 1
j = 0 1 2 3

40. Example A G T -1 -2 -3 -4 1 2 F(i,j) i = 0 1 2 3 4 Optimal Alignment:
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i-1, j-1) + (Xi,Yj)
F(i,j) = max F(i-1, j) – d
F(i, j-1) – d
F(i,j) i = A G T -1 -2 -3 -4 1 2
j = 0
Optimal Alignment:
F(4,3) = 2 1 2 3

41. Example A G T -1 -2 -3 -4 1 2 F(i,j) i = 0 1 2 3 4 j = 0
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i-1, j-1) + (Xi,Yj)
F(i,j) = max F(i-1, j) – d
F(i, j-1) – d
F(i,j) i = A G T -1 -2 -3 -4 1 2
j = 0
Optimal Alignment:
F(4,3) = 2
This only tells us the best score 1 2 3

42. Trace-back A G T -1 -2 -3 -4 1 2 A F(i,j) i = 0 1 2 3 4 j = 0 1 2 3
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i-1, j-1) + (Xi,Yj)
F(i,j) = max F(i-1, j) – d
F(i, j-1) – d
F(i,j) i = A G T -1 -2 -3 -4 1 2
j = 0 1 A 2 3

43. Trace-back A G T -1 -2 -3 -4 1 2 T A F(i,j) i = 0 1 2 3 4 j = 0 1 2 3
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i-1, j-1) + (Xi,Yj)
F(i,j) = max F(i-1, j) – d
F(i, j-1) – d
F(i,j) i = A G T -1 -2 -3 -4 1 2
j = 0 1 T A 2 3

44. Trace-back A G T -1 -2 -3 -4 1 2 G T A - F(i,j) i = 0 1 2 3 4 j = 0 1
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i-1, j-1) + (Xi,Yj)
F(i,j) = max F(i-1, j) – d
F(i, j-1) – d
F(i,j) i = A G T -1 -2 -3 -4 1 2
j = 0 1 G T A - 2 3

45. Trace-back A G T -1 -2 -3 -4 1 2 A G T - F(i,j) i = 0 1 2 3 4 j = 0 1
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i-1, j-1) + (Xi,Yj)
F(i,j) = max F(i-1, j) – d
F(i, j-1) – d
F(i,j) i = A G T -1 -2 -3 -4 1 2
j = 0 1 A G T - 2 3

46. Trace-back A G T -1 -2 -3 -4 1 2 F(i,j) i = 0 1 2 3 4
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i-1, j-1) + (Xi,Yj)
F(i,j) = max F(i-1, j) – d
F(i, j-1) – d
F(i,j) i = A G T -1 -2 -3 -4 1 2
j = 0
Optimal Alignment:
F(4,3) = 2
AGTA
ATA 1 2 3

47. Using trace-back pointers
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i,j) i = A G T -1 -2 -3 -4
j = 0 1 2 3

48. Using trace-back pointers
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i,j) i = A G T -1 -2 -3 -4 1
j = 0 1 2 3

49. Using trace-back pointers
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i,j) i = A G T -1 -2 -3 -4 1
j = 0 1 2 3

50. Using trace-back pointers
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i,j) i = A G T -1 -2 -3 -4 1 2
j = 0 1 2 3

51. Using trace-back pointers
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i,j) i = A G T -1 -2 -3 -4 1 2
j = 0 1 2 3

52. Using trace-back pointers
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i,j) i = A G T -1 -2 -3 -4 1 2
j = 0 1 2 3

53. Using trace-back pointers
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i,j) i = A G T -1 -2 -3 -4 1 2
j = 0 1 2 3

54. Using trace-back pointers
x = AGTA m = 1
y = ATA s = 1
d = 1
F(i,j) i = A G T -1 -2 -3 -4 1 2
j = 0
Optimal Alignment:
F(4,3) = 2
AGTA
ATA 1 2 3

55. The Needleman-Wunsch Algorithm
Initialization.
F(0, 0) = 0
F(0, j) = - j  d
F(i, 0) = - i  d
Main Iteration. Filling in scores
For each i = 1……M
For each j = 1……N
F(i-1,j) – d [case 1]
F(i, j) = max F(i, j-1) – d [case 2]
F(i-1, j-1) + σ(xi, yj) [case 3]
UP, if [case 1]
Ptr(i,j) = LEFT if [case 2]
DIAG if [case 3]
Termination. F(M, N) is the optimal score, and from Ptr(M, N) can trace back optimal alignment

56. Complexity Time: Space:
O(NM)
Space:
Linear-space algorithms do exist (with the same time complexity)

57. Equivalent graph problem
S1 = A G T A
(0,0)
: a gap in the 2nd sequence
: a gap in the 1st sequence
: match / mismatch 1 1
S2 = A 1 T
Value on vertical/horizontal line: -d
Value on diagonal: m or -s 1 1 A
(3,4)
Number of steps: length of the alignment
Path length: alignment score
Optimal alignment: find the longest path from (0, 0) to (3, 4)
General longest path problem cannot be found with DP. Longest path on this graph can be found by DP since no cycle is possible.

58. Question
If we change the scoring scheme, will the optimal alignment be changed?
Old: Match = 1, mismatch = gap = -1
New: match = 2, mismatch = gap = 0
New: Match = 2, mismatch = gap = -2?

59. Question What kind of alignment is represented by these paths? A- BC
Alternating gaps are impossible if –s > -2d

60. A variant of the basic algorithm
Scoring scheme: m = s = d: 1
Seq1: CAGCA-CTTGGATTCTCGG
|| |:|||
Seq2: ---CAGCGTGG
Seq1: CAGCACTTGGATTCTCGG
|||| | | ||
Seq2: CAGC-----G-T----GG
The first alignment may be biologically more realistic in some cases (e.g. if we know s2 is a subsequence of s1)
Score = -7
Score = -2

61. A variant of the basic algorithm
Maybe it is OK to have an unlimited # of gaps in the beginning and end:
CTATCACCTGACCTCCAGGCCGATGCCCCTTCCGGC
GCGAGTTCATCTATCAC--GACCGC--GGTCG
Then, we don’t want to penalize gaps in the ends

62. The Overlap Detection variant
Changes:
Initialization
For all i, j,
F(i, 0) = 0
F(0, j) = 0
Termination
maxi F(i, N)
FOPT = max maxj F(M, j)
x1 ……………………………… xM
yN ……………………………… y1

63. Different types of overlapsx x y y

64. The local alignment problem
Given two strings X = x1……xM,
Y = y1……yN
Find substrings x’, y’ whose similarity (optimal global alignment value) is maximum
e.g. X = abcxdex X’ = cxde
Y = xxxcde Y’ = c-de x y

65. Why local alignment Conserved regions may be a small part of the whole
Global alignment might miss them if flanking “junk” outweighs similar regions
Genes are shuffled between genomes A B C D B D A C

66. Naïve algorithm for all substrings X’ of X and Y’ of Y
Align X’ & Y’ via dynamic programming
Retain pair with max value
end ;
Output the retained pair
Time: O(n2) choices for A, O(m2) for B, O(nm) for DP, so O(n3m3 ) total.

67. Reminder The overlap detection algorithm
We do not give penalty to gaps at either end
Free gap
Free gap

68. The local alignment idea
Do not penalize the unaligned regions (gaps or mismatches)
The alignment can start anywhere and ends anywhere
Strategy: whenever we get to some low similarity region (negative score), we restart a new alignment
By resetting alignment score to zero

69. The Smith-Waterman algorithm
Initialization: F(0, j) = F(i, 0) = 0
F(i – 1, j) – d
F(i, j – 1) – d
F(i – 1, j – 1) + (xi, yj)
Iteration: F(i, j) = max

70. The Smith-Waterman algorithm
Termination:
If we want the best local alignment…
FOPT = maxi,j F(i, j)
If we want all local alignments scoring > t
For all i, j find F(i, j) > t, and trace back

71. x c d e a b
Match: 2
Mismatch: -1
Gap: -1

72. x c d e a b
Match: 2
Mismatch: -1
Gap: -1

73. x c d e a b 2 1
Match: 2
Mismatch: -1
Gap: -1

74. x c d e a b 2 1
Match: 2
Mismatch: -1
Gap: -1

75. x c d e a b 2 1 3
Match: 2
Mismatch: -1
Gap: -1

76. x c d e a b 2 1 3 5
Match: 2
Mismatch: -1
Gap: -1

77. x c d e a b 2 1 3 5 4
Match: 2
Mismatch: -1
Gap: -1

78. Trace back x c d e a b 2 1 3 5 4
Match: 2
Mismatch: -1
Gap: -1

79. Trace back x c d e a b 2 1 3 5 4 cxde | || c-de x-de | || xcdea b 2 1 3 5 4
Match: 2
Mismatch: -1
Gap: -1
cxde
| ||
c-de
x-de
| ||
xcde

80. No negative values in local alignment DP array
Optimal local alignment will never have a gap on either end
Local alignment: “Smith-Waterman”
Global alignment: “Needleman-Wunsch”

81. Analysis Time: Memory: O(MN) for finding the best alignment
Time to report all alignments depends on the number of sub-opt alignments
Memory:
O(MN)
O(M+N) possible

82. More efficient alignment algorithms

83. Given two sequences of length M, N Time: O(MN) Space: O(MN)
Ok, but still slow for long sequences
Space: O(MN)
bad
1Mb seq x 1Mb seq = 1TB memory
Can we do better?

84. Bounded alignment
Good alignment should appear near the diagonal

85. Bounded Dynamic Programming
If we know that x and y are very similar
Assumption: # gaps(x, y) < k xi
Then, | implies | i – j | < k yj

86. Bounded Dynamic Programming
Initialization:
F(i,0), F(0,j) undefined for i, j > k
Iteration:
For i = 1…M
For j = max(1, i – k)…min(N, i+k)
F(i – 1, j – 1)+ (xi, yj)
F(i, j) = max F(i, j – 1) – d, if j > i – k
F(i – 1, j) – d, if j < i + k
Termination: same
x1 ………………………… xM
yN ………………………… y1 k

87. Analysis Time: O(kM) << O(MN) Space: O(kM) with some tricks
=> M M 2k 2k

89. Given two sequences of length M, N Time: O(MN) Space: O(MN)ok
Space: O(MN)
bad
1mb seq x 1mb seq = 1TB memory
Can we do better?

90. Linear space algorithm
If all we need is the alignment score but not the alignment, easy!
We only need to keep two rows
(You only need one row, with a little trick)
But how do we get the alignment?

91. Linear space algorithm
When we finish, we know how we have aligned the ends of the sequences XM YN
Naïve idea: Repeat on the smaller subproblem F(M-1, N-1)
Time complexity: O((M+N)(MN))

92. (0, 0)
M/2
(M, N)
Key observation: optimal alignment (longest path) must use an intermediate point on the M/2-th row. Call it (M/2, k), where k is unknown.

93. (0,0)
(3,0)
(3,2)
(3,4)
(3,6)
(6,6)
Longest path from (0, 0) to (6, 6) is max_k (LP(0,0,3,k) + LP(6,6,3,k))

94. Hirschberg’s idea Divide and conquer! Forward algorithmY X
Forward algorithm
Align x1x2…xM/2 with Y
M/2
F(M/2, k) represents the best alignment between x1x2…xM/2 and y1y2…yk

95. Backward Algorithm Backward algorithmY X
Backward algorithm
Align reverse(xM/2+1…xM) with reverse(Y)
M/2
B(M/2, k) represents the best alignment between reverse(xM/2+1…xM) and reverse(ykyk+1…yN )

96. Linear-space alignment
Using 2 (4) rows of space, we can compute
for k = 1…N, F(M/2, k), B(M/2, k)
M/2

97. Linear-space alignment
Now, we can find k* maximizing F(M/2, k) + B(M/2, k)
Also, we can trace the path exiting column M/2 from k*
Conclusion: In O(NM) time, O(N) space, we found optimal alignment path at row M/2

98. Linear-space alignment
Iterate this procedure to the two sub-problems!
M/2 k*
M/2
N-k*

99. Analysis
Memory: O(N) for computation, O(N+M) to store the optimal alignment
Time:
MN for first iteration
k M/2 + (N-k) M/2 = MN/2 for second … k
M/2
M/2
N-k

100. MN + MN/2 + MN/4 + MN/8 + … = MN (1 + ½ + ¼ + 1/8 + 1/16 + …)
= 2MN = O(MN)
MN/8

101. Outline Part I: Algorithms Part II: Biological issues Part III: BLAST
Biological problem
Intro to dynamic programming
Global sequence alignment
Local sequence alignment
More efficient algorithms
Part II: Biological issues
Model gaps more accurately
Alignment statistics
Part III: BLAST

102. How to model gaps more accurately

103. What’s a better alignment?
GACGCCGAACG
||||| |||
GACGC---ACG
GACGCCGAACG
|||| | | ||
GACG-C-A-CG
Score = 8 x m – 3 x d
Score = 8 x m – 3 x d
However, gaps usually occur in bunches.
During evolution, chunks of DNA may be lost entirely
Aligning genomic sequences vs. cDNAs (reverse complimentary to mRNAs)

104. Model gaps more accurately
Current model:
Gap of length n incurs penalty nd
General:
Convex function
E.g. (n) = c * sqrt (n)  n  n

105. General gap dynamic programming
Initialization: same
Iteration:
F(i-1, j-1) + s(xi, yj)
F(i, j) = max maxk=0…i-1F(k,j) – (i-k)
maxk=0…j-1F(i,k) – (j-k)
Termination: same
Running Time: O((M+N)MN) (cubic)
Space: O(NM) (linear-space algorithm not applicable)

106. Compromise: affine gaps
(n) = d + (n – 1)e
| |
gap gap
open extension e d
Match: 2
Gap open: -5
Gap extension: -1
GACGCCGAACG
||||| |||
GACGC---ACG
GACGCCGAACG
|||| | | ||
GACG-C-A-CG
8x2-5-2 = 9
8x2-3x5 = 1
We want to find the optimal alignment with affine gap penalty in
O(MN) time
O(MN) or better O(M+N) memory

107. Allowing affine gap penalties
Still three cases
xi aligned with yj
xi aligned to a gap
Are we continuing a gap in x? (if no, start is more expensive)
yj aligned to a gap
Are we continuing a gap in y? (if no, start is more expensive)
We can use a finite state machine to represent the three cases as three states
The machine has two heads, reading the chars on the two strings separately
At every step, each head reads 0 or 1 char from each sequence
Depending on what it reads, goes to a different state, and produces different scores

108. Finite State Machine ? / ? ? / ? ? / ? ? / ? ? / ? ? / ? ? / ?
Input
Output
? / ?
? / ? Ix
? / ?
? / ? F
? / ? Iy
? / ?
State
? / ?
F: have just read 1 char from each seq (xi aligned to yj )
Ix: have read 0 char from x. (yj aligned to a gap)
Iy: have read 0 char from y (xi aligned to a gap)

109. (-, yj) / e (xi,yj) /  (xi,yj) /  (-, yj) / d (xi,-) / d (xi,-) / e
Input
Output
(-, yj) / e
(xi,yj) /  Ix
(xi,yj) / 
(-, yj) / d F
(xi,-) / d Iy
(xi,-) / e
Start state
(xi,yj) / 
Current state
Input
Output
Next state F
(xi,yj) 
(-,yj) d Ix
(xi,-) Iy e …

110. (-, yj) / e (xi,yj) /  (-, yj) / d (xi,-) / d (xi,-) / eF Ix Iy
(xi,yj) / 
(xi,-) / d
(xi,-) / e
(-, yj) / d
(-, yj) / e
start
state
F-F-F-F
F-Iy-F-F-Ix
F-F-Iy-F-Ix
AAC
ACT
AAC
|||
ACT
AAC- ||
-ACT
AAC-
| |
A-CT
Given a pair of sequences, an alignment (not necessarily optimal) corresponds to a state path in the FSM.
Optimal alignment: find a state path to read the two sequences such that the total output score is the highest

111. Dynamic programming
We encode this information in three different matrices
For each element (i,j) we use three variables
F(i,j): best alignment (score) of x1..xi & y1..yj if xi aligns to yj
Ix(i,j): best alignment of x1..xi & y1..yj if yj aligns to gap
Iy(i,j): best alignment of x1..xi & y1..yj if xi aligns to gap xi xi xi yj yj yj
F(i, j)
Ix(i, j)
Iy(i, j)

112. (-, yj)/e (xi,yj) / (xi,yj) / (-, yj) /d (xi,-) /d (xi,-)/eIx
(xi,yj) /
(-, yj) /d F
(xi,-) /d Iy
(xi,-)/e
(xi,yj) /
F(i-1, j-1) + (xi, yj)
F(i, j) = max Ix(i-1, j-1) + (xi, yj)
Iy(i-1, j-1) + (xi, yj) xi yj

113. F(i, j-1) + d Ix(i, j) = max Ix(i, j-1) + e (-, yj)/e (xi,yj) /
(-, yj) /d F
(xi,-) /d Iy
(xi,-)/e
(xi,yj) /
F(i, j-1) + d
Ix(i, j) = max
Ix(i, j-1) + e xi yj
Ix(i, j)

114. F(i-1, j) + d Iy(i, j) = max Iy(i-1, j) + e (-, yj)/e (xi,yj) /Ix
(xi,yj) /
(-, yj) /d F
(xi,-) /d Iy
(xi,-)/e
(xi,yj) /
F(i-1, j) + d
Iy(i, j) = max
Iy(i-1, j) + e xi yj
Iy(i, j)

115. F(i, j) = (xi, yj) + max Ix(i – 1, j – 1) Iy(i – 1, j – 1)
F(i, j – 1) + d
Ix(i, j) = max
Ix(i, j – 1) + e
F(i – 1, j) + d
Iy(i, j) = max
Iy(i – 1, j) + e
Continuing alignment
Closing gaps in x
Closing gaps in y
Opening a gap in x
Gap extension in x
Opening a gap in y
Gap extension in y

116. Data dependency F i Ix Iy j
i-1
i-1
j-1
j-1

117. Data dependency F i Ix Iy j i i j j

118. Data dependency If we stack all three matrices No cyclic dependency
Therefore, we can fill in all three matrices in order

119. Algorithm for i = 1:m F(M, N) = max (F(M, N), Ix(M, N), Iy(M, N))
for j = 1:n
Fill in F(i, j), Ix(i, j), Iy(i, j)
end
F(M, N) = max (F(M, N), Ix(M, N), Iy(M, N))
Time: O(MN)
Space: O(MN) or O(N) when combined with the linear-space algorithm

120. Exercise
x = GCAC
y = GCC
m = 2
s = -2
d = -5
e = -1

121. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A-
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A
F: aligned on both
Iy: Insertion on y
y =
G C C
F(i-1, j-1)
Iy(i-1, j-1)
x = - -5 -6 -7
Iy(i-1,j)
(xi, yj)
F(i-1,j) G C A e
Ix(i-1, j-1) d
F(i, j)
F(i,j-1)
Iy(i,j) d
Ix(i,j)
Ix(i,j-1) e
Ix: Insertion on x

122. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = - -5 -6 -7
F(i-1, j-1)
Iy(i-1, j-1) G C A
(xi, yj) = 2
Ix(i-1, j-1)
F(i, j) Ix

123. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = - -5 -6 -7
F(i-1, j-1)
Iy(i-1, j-1) G C A
(xi, yj) = -2
Ix(i-1, j-1)
F(i, j) Ix

124. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = - -5 -6 -7
F(i-1, j-1)
Iy(i-1, j-1) G C A
(xi, yj) = -2
Ix(i-1, j-1)
F(i, j) Ix

125. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 G C A
F(i,j-1)
d = -5
Ix(i,j)
Ix(i,j-1)
e = -1 Ix

126. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 -4 G C A
F(i,j-1)
d = -5
Ix(i,j)
Ix(i,j-1)
e = -1 Ix

127. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 -4
Iy(i-1,j) G C A
F(i-1,j)
e=-1
d=-5
Iy(i,j) Ix

128. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 -4
F(i-1, j-1)
Iy(i-1, j-1) G C A
(xi, yj) = -2
Ix(i-1, j-1)
F(i, j) Ix

129. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8 4
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 -4
F(i-1, j-1)
Iy(i-1, j-1) G C A
(xi, yj) = 2
Ix(i-1, j-1)
F(i, j) Ix

130. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8 4 -1
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 -4
F(i-1, j-1)
Iy(i-1, j-1) G C A
(xi, yj) = 2
Ix(i-1, j-1)
F(i, j) Ix

131. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8 4 -1
x = - -5 -6 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 -4
-12 -1 G C A
F(i,j-1)
d = -5
Ix(i,j)
Ix(i,j-1)
e = -1 Ix

132. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8 4 -1
x = - -5 -6 -3 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 -4
-12 -1
Iy(i-1,j) G C A
F(i-1,j)
e=-1
d=-5
Iy(i,j) Ix

133. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8 4 -1
x = - -5 -6 -3
-12
-13 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
F(i-1, j-1)
Iy(i-1, j-1)
x =
Iy(i-1,j) -5 -6 -7 - -3 -4
-12 -1
(xi, yj)
F(i-1,j) G C A e
Ix(i-1, j-1) d
F(i, j)
F(i,j-1)
Iy(i,j) d
Ix(i,j)
Ix(i,j-1) e Ix

134. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8 4 -5
x = - -5 -6 -3
-12
-13 -7 -8
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
F(i-1, j-1)
Iy(i-1, j-1)
x =
Iy(i-1,j) -5 -6 -7 - -3 -4
-12 -1
-13
-10
(xi, yj)
F(i-1,j) G C A e
Ix(i-1, j-1) d
F(i, j)
F(i,j-1)
Iy(i,j) d
Ix(i,j)
Ix(i,j-1) e Ix

135. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8 4 -1 -5
x = - -5 -6 -3
-12
-13 -7 -8 -1
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 -4
-12 -1
-13
-10
Iy(i-1,j) G C A
F(i-1,j)
e=-1
d=-5
Iy(i,j) Ix

136. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8 4 -1 -5
x = - -5 -6 -3
-12
-13 -7 -8 -1
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
x = -5 -6 -7 - -3 -4
-12 -1
-13
-10
Iy(i-1,j) G C A
F(i-1,j)
e=-1
d=-5
Iy(i,j) Ix

137. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8 4 -1 -5 -9 -6 1
x = - -5 -6 -3
-12
-13 -7 -8 -1 -2
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
F(i-1, j-1)
Iy(i-1, j-1)
x =
Iy(i-1,j) -5 -6 -7 - -3 -4
-12 -1
-13
-10
-14
-11
(xi, yj)
F(i-1,j) G C A e
Ix(i-1, j-1) d
F(i, j)
F(i,j-1)
Iy(i,j) d
Ix(i,j)
Ix(i,j-1) e Ix

138. y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1 G C A G C A F- 2 -7 -8 4 -1 -5 -9 -6 1
x = - -5 -6 -3
-12
-13 -7 -8 -1 -2
m = 2
s = -2
d = -5
e = -1 G C A G C A F Iy
y =
G C C
F(i-1, j-1)
Iy(i-1, j-1)
x =
Iy(i-1,j) -5 -6 -7 - -3 -4
-12 -1
-13
-10
-14
-11
(xi, yj)
F(i-1,j) G C A e
Ix(i-1, j-1) d
F(i, j)
F(i,j-1)
Iy(i,j) d
Ix(i,j)
Ix(i,j-1) e Ix

139. GCAC || | GC-C y = G C C y = G C C x = x = m = 2 s = -2 d = -5 e = -1- 2 -7 -8 4 -1 -5 -9 -6 1
x = - -5 -6 -3
-12
-13 -7 -8 -1 -2
m = 2
s = -2
d = -5
e = -1 G C A G C A x
GCAC
|| |
GC-C y F Iy
y =
G C C
y =
G C C
x = -5 -6 -7 - -3 -4
-12 -1
-13
-10
-14
-11
x = G C A G C A Ix

140. Statistics of alignment
Where does (xi, yj) come from?
Are two aligned sequences actually related?

141. Probabilistic model of alignments
We’ll first focus on protein alignments without gaps
Given an alignment, we can consider two possible models
R: the sequences are related by evolution
U: the sequences are unrelated
How can we distinguish these two models?
How is this view related to amino-acid substitution matrix?

142. Model for unrelated sequences
Assume each position of the alignment is independently sampled from some distribution of amino acids
ps: probability of amino acid s in the sequences
Probability of seeing an amino acid s aligned to an amino acid t by chance is
Pr(s, t | U) = ps * pt
Probability of seeing an ungapped alignment between x = x1…xn and y = y1…yn randomly is i

143. Model for related sequences
Assume each pair of aligned amino acids evolved from a common ancestor
Let qst be the probability that amino acid s in one sequence is related to t in another sequence
The probability of an alignment of x and y is give by

144. Probabilistic model of Alignments
How can we decide which model (U or R) is more likely?
One principled way is to consider the relative likelihood of the two models (the odds ratio)
A higher ratio means that R is more likely than U

145. Log odds ratio Taking logarithm, we get
Recall that the score of an alignment is given by

146. Therefore, if we define
We are actually defining the alignment score as the log odds ratio between the two models R and U

147. How to get the probabilities?
ps can be counted from the available protein sequences
But how do we get qst? (the probability that s and t have a common ancestor)
Counted from trusted alignments of related sequences

148. Protein Substitution Matrices
Two popular sets of matrices for protein sequences
PAM matrices [Dayhoff et al, 1978]
Better for aligning closely related sequences
BLOSUM matrices [Henikoff & Henikoff, 1992]
For both closely or remotely related sequences

149. BLOSUM-N matrices Constructed from a database called BLOCKS
Contain many closely related sequences
Conserved amino acids may be over-counted
N = 62: the probabilities qst were computed using trusted alignments with no more than 62% identity
identity: % of matched columns
Using this matrix, the Smith-Waterman algorithm is most effective in detecting real alignments with a similar identity level (i.e. ~62%)

150. : Scaling factor to convert score to integer.
Important: when you are told that a
scoring matrix is in half-bits =>  = ½ ln2
Positive for chemically similar substitution
Common amino acids get low weights
Rare amino acids get high weights

151. BLOSUM-N matrices
If you want to detect homologous genes with high identity, you may want a BLOSUM matrix with higher N. say BLOSUM75
On the other hand, if you want to detect remote homology, you may want to use lower N, say BLOSUM50
BLOSUM-62: good for most purposes
Weak homology
Strong homology

152. For DNAs No database of trusted alignments to start with
Specify the percentage identity you would like to detect
You can then get the substitution matrix by some calculation

153. For example Suppose pA = pC = pT = pG = 0.25 We want 88% identity
qAA = qCC = qTT = qGG = 0.22, the rest = 0.12/12 = 0.01
(A, A) = (C, C) = (G, G) = (T, T)
= log (0.22 / (0.25*0.25)) = 1.26
(s, t) = log (0.01 / (0.25*0.25)) = for s ≠ t.

154. Substitution matrix A C G T
1.26
-1.83

155. A C G T 5 -7
Scale won’t change the alignment
Multiply by 4 and then round off to get integers

156. Arbitrary substitution matrix
Say you have a substitution matrix provided by someone
It’s important to know what you are actually looking for when you use the matrix

157. Which one should I use for my sequences?
NCBI-BLAST
WU-BLAST A C G T 1 -2 A C G T 5 -4
What’s the difference?
Which one should I use for my sequences?

158. We had
Scale it, so that
Reorganize:

159. Since all probabilities must sum to 1,
We have
Suppose again ps = 0.25 for any s
We know (s, t) from the substitution matrix
We can solve the equation for λ
Plug λ into to get qst

160. A C G T 1 -2 A C G T 5 -4 Translate: 95% identity
NCBI-BLAST
WU-BLAST A C G T 1 -2 A C G T 5 -4
 = 1.33
qst = 0.24 for s = t, and for s ≠ t
Translate: 95% identity
 = 0.19
qst = 0.16 for s = t, and 0.03 for s ≠ t
Translate: 65% identity

161. Details for solving  A C G T 1 -2
Known: (s,t) = 1 for s=t, and (s,t) = -2 for s t.
Since
and s,t qst = 1, we have
12 * ¼ * ¼ * e-2 + 4 * ¼ * ¼ * e = 1
Let e = x, we have
¾ x-2 + ¼ x = 1. Hence,
x3 – 4x2 + 3 = 0;
X has three solutions: 3.8, 1, -0.8
Only the first solution leads to a positive 
 = ln (3.8) = 1.33 A C G T 1 -2

162. Statistics of alignment
Where does (xi, yj) come from?
Are two aligned sequences actually related?

163. Statistics of Alignment Scores
Q: How do we assess whether an alignment provides good evidence for homology (i.e., the two sequences are evolutionarily related)?
Is a score 82 good? What about 180?
A: determine how likely it is that such an alignment score would result from chance

164. P-value of alignment p-value Model-based vs simulation-based
The probability that the alignment score can be obtained from aligning random sequences
Small p-value means the score is unlikely to happen by chance
The most common thresholds are 0.01 and 0.05,
also depend on purpose of comparison and cost of misclaim
Model-based vs simulation-based

165. Statistics of global seq alignment
Theory only applies to local alignment
For global alignment, your best bet is to do Monte-Carlo simulation
What’s the chance you can get a score as high as the real alignment by aligning two random sequences?
Procedure
Given sequence X, Y
Compute a global alignment (score = S)
Randomly shuffle sequence X (or Y) N times, obtain X1, X2, …, XN
Align each Xi with Y, (score = Ri)
P-value: the fraction of Ri >= S

166. Human HEXA
Fly HEXO1
Score = -74

167. -74
Distribution of the alignment scores between fly HEXO1 and 200 randomly shuffled human HEXA sequences
There are 88 random sequences with alignment score >= -74.
So: p-value = 88 / 200 = 0.44 => alignment is not significant

168. ……………………………………………………
Mouse HEXA
Human HEXA
Score = 732
……………………………………………………

169. No random sequences with alignment score >= 732
Distribution of the alignment scores between mouse HEXA and 200 randomly shuffled human HEXA sequences
732
No random sequences with alignment score >= 732
So: the P-value is less than 1 / 200 = 0.05
To get smaller p-value, have to align more random sequences
Very slow
Unless we can fit a distribution (e.g. normal distribution)
Such distribution may not be generalizable
No theory exists for global alignment score distribution

170. Statistics for local alignment
Elegant theory exists
Score for ungapped local alignment follows extreme value distribution (Gumbel distribution)
Normal distribution
Extreme value distribution
An example extreme value distribution:
Randomly sample 100 numbers from a normal distribution, and compute max
Repeat 100 times.
The max values will follow extreme value distribution

171. Statistics for local alignment
Given two unrelated sequences of lengths M, N
Expected number of ungapped local alignments with score at least S can be calculated by
E(S) = KMN exp[-S]
Known as E-value
: scaling factor as computed in last lecture
K: empirical parameter ~ 0.1
Depend on sequence composition and substitution matrix

172. P-value for local alignment score
P-value for a local alignment with score S
when P is small.

173. Example You are aligning two sequences, each has 1000 bases
m = 1, s = -1, d = -inf (ungapped alignment)
You obtain a score 20
Is this score significant?

174.  = ln3 = 1.1 (computed as discussed on slide #41)
E(S) = K MN exp{- S}
E(20) = 0.1 * 1000 * 1000 * 3-20 = 3 x 10-5
P-value = 3 x 10-5 << 0.05
The alignment is significant
Distribution of 1000 random sequence pairs 20

175. Multiple-testing problem
Searching a 1000-base sequence against a database of 106 sequences (each of length 1000)
How significant is a score 20 now?
You are essentially comparing 1000 bases with 1000x106 = 109 bases (ignore edge effect)
E(20) = 0.1 * 1000 * 109 * 3-20 = 30
By chance we would expect to see 30 matches
The P-value (probability of seeing at least one match with score >= 30) is 1 – e-30 =
The alignment is not significant
Caution: it does NOT mean that the two sequences are unrelated. Rather, it simply means that you have NO confidence to say whether the two sequences are related.

176. Score threshold to determine significance
You want a p-value that is very small (even after taking into consideration multiple-testing)
What S will guarantee you a significant p-value?
E(S)  P(S) << 1
=> KMN exp[-S] << 1
=> log(KMN) -S < 0
=> S > T + log(MN) /  (T = log(K) / , usually small)

177. Score threshold to determine significance
In the previous example
m = 1, s = -1, d = -inf =>  = 1.1
Aligning 1000bp vs 1000bp
S > log(106) / 1.1 = 13.
So 20 is significant.
Searching 1000bp against 106 x 1000bp
S > log(1012) / 1.1 = 25.
so 20 is not significant.

178. Statistics for gapped local alignment
Theory not well developed
Extreme value distribution works well empirically
Need to estimate K and  empirically
Given the database and substitution matrix, generate some random sequence pairs
Do local alignment
Fit an extreme value distribution to obtain K and 

179. Alignment statistics summary
How to obtain a substitution matrix?
Obtain qst and ps from established alignments (for DNA: from your knowledge)
Computing score:
How to understand arbitrary substitution matrix?
Solve function to obtain  and target qst
Which tells you what percent identity you are expecting
How to understand alignment score?
probability that a score can be expected from chance.
Global alignment: Monte-Carlo simulation
Local alignment: Extreme Value Distribution
Estimate p-value from a score
Determine a score threshold without computing a p-value

180. Part III: Heuristic Local Sequence Alignment: BLAST

181. State of biological databases
Sequenced Genomes:
Human 109 Yeast 1.2107
Mouse 2.7109 Rat 2.6109
Neurospora 4107 Fugu fish 3.3108
Tetraodon 3108 Mosquito 2.8108
Drosophila 1.2108 Worm 1.0108
Rice 1.0109 Arabidopsis 1.2108
sea squirts 108
Current rate of sequencing (before new-generation sequencing):
4 big labs  3 109 bp /year/lab
10s small labs
Private sectors
With new-generation sequencing:
Easily generating billions of reads daily

182. Some useful applications of alignments
Given a newly discovered gene,
- Does it occur in other species?
Assume we try Smith-Waterman:
Our new gene
104
The entire genomic database
May take several weeks!

183. Some useful applications of alignments
Given a newly sequenced organism,
- Which subregions align with other organisms?
- Potential genes
- Other functional units
Assume we try Smith-Waterman:
Our newly sequenced mammal
3109
The entire genomic database
> 1000 years ???

184. BLAST Basic Local Alignment Search Tool
Altschul, Gish, Miller, Myers, Lipman, J Mol Biol 1990
The most widely used bioinformatics tool
Which is better: long mediocre match or a few nearby, short, strong matches with the same total score?
Score-wise, exactly equivalent
Biologically, later may be more interesting, & is common
At least, if must miss some, rather miss the former
BLAST is a heuristic algorithm emphasizing the later
speed/sensitivity tradeoff: BLAST may miss former, but gains greatly in speed

185. BLAST
Available at NCBI (National Center for Biotechnology Information) for download and online use.
Along with many sequence databases
Main idea:
Construct a dictionary of all the words in the query
Initiate a local alignment for each word match between query and DB
Running Time: O(MN)
However, orders of magnitude faster than Smith-Waterman
query DB

186. BLAST  Original Version
Dictionary:
All words of length k (~11 for DNA, 3 for proteins)
Alignment initiated between words of alignment score  T (typically T = k)
Alignment:
Ungapped extensions until score
below statistical threshold
Output:
All local alignments with score
> statistical threshold ……
query ……
scan DB
query

187. BLAST  Original Version
A C G A A G T A A G G T C C A G T
Example:
k = 4, T = 4
The matching word GGTC initiates an alignment
Extension to the left and right with no gaps until alignment falls < 50%
Output:
GTAAGGTCC
GTTAGGTCC
C C C T T C C T G G A T T G C G A

188. Gapped BLAST Added features: Pairs of words can initiate alignment
A C G A A G T A A G G T C C A G T
Added features:
Pairs of words can initiate alignment
Extensions with gaps in a band around anchor
Output:
GTAAGGTCCAGT
GTTAGGTC-AGT
C T G A T C C T G G A T T G C G A

189. Example Query: gattacaccccgattacaccccgattaca (29 letters) [2 mins]
Database: All GenBank+EMBL+DDBJ+PDB sequences (but no EST, STS, GSS, or phase 0, 1 or 2 HTGS sequences) 1,726,556 sequences; 8,074,398,388 total letters
>gi| |gb|AC | Oryza sativa chromosome 3 BAC OSJNBa0087C10 genomic sequence, complete sequence Length = Score = 34.2 bits (17), Expect = 4.5 Identities = 20/21 (95%) Strand = Plus / Plus
Query: tacaccccgattacaccccga 24
||||||| |||||||||||||
Sbjct: tacacccagattacaccccga
Score = 34.2 bits (17),
Expect = 4.5 Identities = 20/21 (95%) Strand = Plus / Plus
Query: tacaccccgattacaccccga 24
Sbjct: tacacccagattacaccccga
>gi| |gb|AC | Oryza sativa chromosome 3 BAC OSJNBa0052F07 genomic sequence, complete sequence Length = Score = 34.2 bits (17), Expect = 4.5 Identities = 20/21 (95%) Strand = Plus / Plus
Sbjct: tacacccagattacaccccga 3911

190. Example Query: Human atoh enhancer, 179 letters [1.5 min]
Result: 57 blast hits
gi| |gb|AF |AF Homo sapiens ATOH1 enhanc e-95
gi| |gb|AC | Mus musculus Strain C57BL6/J ch e-68
gi| |gb|AF |AF Mus musculus Atoh1 enhanc e-66
gi| |gb|AF | Gallus gallus CATH1 (CATH1) gene e-12
gi| |emb|AL | Zebrafish DNA sequence from clo e-05
gi| |gb|AC | Oryza sativa chromosome 10 BAC O
gi| |ref|NM_ | Mus musculus suppressor of Ty
gi| |gb|BC | Mus musculus, Similar to suppres
gi| |gb|AF |AF Mus musculus Atoh1 enhancer sequence Length = 1517
Score = 256 bits (129), Expect = 9e-66 Identities = 167/177 (94%),
Gaps = 2/177 (1%) Strand = Plus / Plus
Query: 3 tgacaatagagggtctggcagaggctcctggccgcggtgcggagcgtctggagcggagca 62
||||||||||||| ||||||||||||||||||| ||||||||||||||||||||||||||
Sbjct: 1144 tgacaatagaggggctggcagaggctcctggccccggtgcggagcgtctggagcggagca 1203
Query: 63 cgcgctgtcagctggtgagcgcactctcctttcaggcagctccccggggagctgtgcggc 122
|||||||||||||||||||||||||| ||||||||| |||||||||||||||| |||||
Sbjct: 1204 cgcgctgtcagctggtgagcgcactc-gctttcaggccgctccccggggagctgagcggc 1262
Query: 123 cacatttaacaccatcatcacccctccccggcctcctcaacctcggcctcctcctcg 179
||||||||||||| || ||| |||||||||||||||||||| |||||||||||||||
Sbjct: 1263 cacatttaacaccgtcgtca-ccctccccggcctcctcaacatcggcctcctcctcg 1318

191. BLAST Score: bit score vs raw score
Bit score is converted from raw score by taking into account K and :
S’ = ( S – log K) / log 2
To compute E-value from bit score:
E = KM’N’ e-S = M’N’ 2-S’
Critical score is now:
S* = log2(M’N’)
If S’ >> S*: significant
If S’ << S*: not significant
(M’ ~ M, N’ ~ N)

192. Different types of BLAST
blastn: search nucleic acid databases
blastp: search protein databases
blastx: you give a nucleic acid sequence, search protein databases
tblastn: you give a protein sequence, search nucleic acid databases
tblastx: you give a nucleic sequence, search nucleic acid database, implicitly translate both into protein sequences

193. BLAST cons and pros Advantages Disadvantages New improvement Fast!!!!
A few minutes to search a database of 1011 bases
Disadvantages
Sensitivity may be low
Often misses weak homologies
New improvement
Make it even faster
Mainly for aligning very similar sequences or really long sequences
E.g. whole genome vs whole genome
Make it more sensitive
PSI-BLAST: iteratively add more homologous sequences
PatternHunter: discontinuous seeds

194. Variants of BLAST NCBI-BLAST: most widely used version
WU-BLAST: (Washington University BLAST): another popular version
Optimized, added features
MEGABLAST:
Optimized to align very similar sequences. Linear gap penalty
BLAT: Blast-Like Alignment Tool
BlastZ:
Optimized for aligning two genomes
PSI-BLAST:
BLAST produces many hits
Those are aligned, and a pattern is extracted
Pattern is used for next search; above steps iterated
Sensitive for weak homologies
Slower

195. Summary Part I: Algorithms Part II: Biological issues
Global sequence alignment: Needleman-Wunsch
Local sequence alignment: Smith-Waterman
Improvement on space and time
Part II: Biological issues
Model gaps more accurately: affine gap penalty
Alignment statistics
Part III: Heuristic algorithms – BLAST family