CS 5263 Bioinformatics

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CS 5263 Bioinformatics

• Lecture 4 Local Sequence Alignment, More
  Efficient Sequence Alignment Algorithms

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Roadmap

• Review of last lecture
• Local sequence alignment
• More efficient sequence alignment algorithms

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• Given a scoring scheme,
• Match m
• Mismatch -s
• Gap -d
• We can easily compute an optimal alignment by
  dynamic programming

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• Look at any column of an alignment between two
  sequences X x1x2xM, Y y1y1yN
• Only three cases
• xi is aligned to yj
• xi is aligned to a gap
• yj is aligned to a gap

F(i-1, j-1) ? (xi, yj) F(i, j) max
F(i-1, j) - d F(i, j-1) - d
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Example
F(i,j) j 0 1 2 3 4
i 0
A A
G -
T T
A A
A A
G -
T T
A A
1
2
3
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Equivalent graph problem
S1
G
A
T
A
(0,0)
? a gap in the 2nd sequence ? a gap in the 1st
sequence match / mismatch
1
1
A
S2
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.

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Variants of Needleman-Wunsch alg

• LCS longest common subsequence
• No penalty for gaps or mutations
• Change score function
• Overlapping variants
• No penalty for starting/ending gaps
• Change initial / termination step
• Other variants
• cDNA-genome alignment

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Local alignment
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The local alignment problem

• Given two strings X x1xM,
• Y y1yN
• 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
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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

C
D
B
A
D
A
B
C
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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.

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Reminder

• The overlap detection algorithm
• We do not give penalty to gaps in the ends

Free gap
Free gap
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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

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The Smith-Waterman algorithm

• Initialization F(0, j) F(i, 0) 0
• 0
• F(i 1, j) d
• F(i, j 1) d
• F(i 1, j 1) ?(xi, yj)

Iteration F(i, j) max
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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 gt t
• For all i, j find F(i, j) gt t, and trace back

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Match 2 Mismatch -1 Gap -1
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Match 2 Mismatch -1 Gap -1
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Match 2 Mismatch -1 Gap -1
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Match 2 Mismatch -1 Gap -1
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Match 2 Mismatch -1 Gap -1
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Match 2 Mismatch -1 Gap -1
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Match 2 Mismatch -1 Gap -1
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Trace back
Match 2 Mismatch -1 Gap -1
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Trace back
Match 2 Mismatch -1 Gap -1
cxde c-de
x-de xcde
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• 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

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Analysis

• Time
• O(MN) for finding the best alignment
• Time to report all alignments depends on the
  number of sub-opt alignments
• Memory
• O(MN)
• O(MN) possible

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• More efficient alignment algorithms

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• Given two sequences of length M, N
• Time O(MN)
• ok
• Space O(MN)
• bad
• 1Mb seq x 1Mb seq 1000G memory
• Can we do better?

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Bounded alignment

• Good alignment should appear near the diagonal

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Bounded Dynamic Programming

• If we know that x and y are very similar
• Assumption gaps(x, y) lt k
• xi
• Then, implies i j lt k
• yj

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Bounded Dynamic Programming

• Initialization
• F(i,0), F(0,j) undefined for i, j gt k
• Iteration
• For i 1M
• For j max(1, i k)min(N, ik)
• F(i 1, j 1) ?(xi, yj)
• F(i, j) max F(i, j 1) d, if j gt i k
• F(i 1, j) d, if j lt i k
• Termination same

x1 xM
yN y1
k
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Analysis

• Time O(kM) ltlt O(MN)
• Space O(kM) with some tricks

gt
M
M
2k
2k
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• Given two sequences of length M, N
• Time O(MN)
• ok
• Space O(MN)
• bad
• 1mb seq x 1mb seq 1000G memory
• Can we do better?

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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?
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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((MN)(MN))
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(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.
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(0,0)
(3,2)
(3,4)
(3,6)
(3,0)
(6,6)

• Longest path from (0, 0) to (6, 6) is max_k
  (LP(0,0,3,k) LP(6,6,3,k))

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Hirschbergs idea

• Divide and conquer!

Y
Forward algorithm Align x1x2xM/2 with Y
X
M/2
F(M/2, k) represents the best alignment between
x1x2xM/2 and y1y2yk
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Backward Algorithm
Y
Backward algorithm Align reverse(xM/21xM) with
reverse(Y)
X
M/2
B(M/2, k) represents the best alignment between
reverse(xM/21xM) and reverse(ykyk1yN )
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Linear-space alignment

• Using 2 (4) rows of space, we can compute
• for k 1N, F(M/2, k), B(M/2, k)

M/2
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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
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Linear-space alignment

• Iterate this procedure to the two sub-problems!

M/2
k
M/2
N-k
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Analysis

• Memory O(N) for computation, O(NM) 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
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MN
MN/2
MN/4
MN MN/2 MN/4 MN/8 MN (1 ½ ¼
1/8 1/16 ) 2MN O(MN)
MN/8