Comp. Genomics

1
Comp. Genomics

• Recitation 2
• 12/3/09
• Slides by Igor Ulitsky

2
Outline

• Alignment re-cap
• End-space free alignment
• Affine gap alignment algorithm and proof
• Bounded gap/spaces alignments

3
Dynamic programming

• Useful in many string-related settings
• Will be repeatedly used in the course
• General idea
• Confine the exponential number of possibilities
  into some hierarchy, such that the number of
  cases becomes polynomial

4
Dynamic programming for shortest paths

• Finding the shortest path from X to Y using the
  Floyd Warshall
• Idea if we know what is the shortest path using
  intermediate vertices 1,, k-1, computing
  shortest paths using 1,, k is easy
• wij if k0
• dij(k) mindij(k-1), dik(k-1)dkj(k-1) otherwise

5
Alignment reminder
Something1G
Something1G
Something1C
Something2C
Something1G
Something1C
Somethin g1G
Something2C-
Something1G
Something1G-
Something1C
Somethin g2C
6
Global alignment

• Input S1,S2
• Output Minimum cost alignment
• V(k,l) score of aligning S11..k with S21..l
• Base conditions
• V(i,0) ?k0..i?(sk,-)
• V(0,j) ?k0..j?(-,tk)
• Recurrence relation V(i-1,j-1) ?(si,tj)
• ?1?i?n, 1?j?m V(i,j) max V(i-1,j) ?(si,-)
• V(i,j-1) ?(-,tj)

7
Alignment reminder

• Global alignment
• All of S1 has to be aligned with all of S2
• Every gap is payed for
• Solution equals V(n,m)

Traceback all the way
Alignment score here
8
Local alignment

• Local alignment
• Subset of S1 aligned with a subset of S2
• Gaps outside subsets costless
• Solution equals the maximum score cell in the DP
  matrix
• Base conditions
• V(i,0) 0
• V(0,j) 0
• Recurrence relation V(i-1,j-1) ?(si,tj)
• ?1?i?n, 1?j?m V(i,j) max V(i-1,j) ?(si,-)
• V(i,j-1) ?(-,tj)
• 0

9
Ends-free alignment

• Something between global and local
• Consider aligning a gene to a (bacterial) genome
• Gaps in the beginning and end of S and T are
  costless
• But all of S,T should be aligned
• Base conditions
• V(i,0) 0
• V(0,j) 0
• Recurrence relation V(i-1,j-1) ?(si,tj)
• ?1?i?n, 1?j?m V(i,j) max V(i-1,j) ?(si,-)
• V(i,j-1) ?(-,tj)
• The optimal solution is found at the last
  row/column
• (not necessarily at bottom right corner)

10
Handling weird gaps

• Affine gap different cost for a new and old
  gaps

Something1G
Something1G
Something1C
Something2C
Something1G
Something1C
Somethin g1G
Something2C-
Now we care if there were gaps here
Two new things to keep track ? Two additional
matrices
Something1G
Something1G-
Something1C
Somethin g2C
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S.....i T.....j
G(i,j)
Alignment with Affine Gap Penalty

• Base Conditions
• V(i, 0) F(i, 0) Wg iWs
• V(0, j) E(0, j) Wg jWs
• Recursive Computation
• V(i, j) max E(i, j), F(i, j), G(i, j)
• where
• G(i, j) V(i-1, j-1) ?(si, tj)
• E(i, j) max E(i, j-1) Ws , G(i, j-1) Wg
  Ws , F(i, j-1) Wg Ws
• F(i, j) max F(i-1, j) Ws , G(i-1, j) Wg
  Ws , E(i-1, j) Wg Ws

S.....i------ T..............j
E(i,j)
S...............i T.....j-------

• Time complexity O(nm) - compute 4 matrices
  instead of one.
• Space complexity O(nm) - saving 3 (Why?)
  matrices. O(nm) w/ Hir.

12
When do constant and affine gap costs differ?
AGAGACTGACGCTTA ATATTA

• Consider

AGAGACTGACGCTTA ATA---------TTA
AGAGACTGACGCTTA ----A-T-A---TTA
Constant penalty Mismatch -5 Gap -1
-9
-14
Affine penalty Mismatch -5 Gap open -3 Gap
extend -0.5
-12
-14.5
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Bounding the number of gaps

• Lets say we are allowed to have at most K gaps
• (Gaps ? Spaces ? Gap can contain many spaces)
• Now we keep track of the number of gaps we opened
  so far
• Also still need to keep track of whether a gap is
  currently open in S or T (E/F matrices)

14
Bounding the number of gaps

• A multi-layer DP matrix
• Actually separate functions V,E,F, on every
  layer, keeping track of layer no.
• Every time we open or close a gap we jump to
  the next layer
• Where to look for the solution? (not only
• at last layer!)
• What is the complexity?

15
Bounding the number of spaces

• Lets say that no gap can exceed k spaces
• Of course now cannot also bound number
• of gaps as well (why?)
• How many matrices do we need now?
• Here, no monotone notion of layer like before
• Whats the complexity?

16
What about arbitrary gap functions?

• If the gap cost is an arbitrary function of its
  length f(k)
• Thus, when computing Dij, we need to look at j
  places back and i places up
• Complexity?

Something1G
Something1C
min
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Special cases

• How about a logarithmic penalty? WgWslog(k)
• This is a special case of a convex penalty, which
  is solvable in O(mnlog(m))
• The logarithmic case can be done in O(mn)
• For a piece-wise linear gap function made of K
  lines, DP can be done in O(mnlog(K))

18
Supersequence

• Exercise A is called a non-contiguous
  supersequence of B if B is a non-contiguous
  subsequence of A.
• e.g., YABADABADU is a non-contigous supersequence
  of BABU (YABADABADU)
• Given S and T, find their shortest common
  supersequence

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Reminder LCS

• Longest common non-contigous subsequence
• Adjust global alignment with similarity scores
• 1 for match
• 0 for gaps
• -8 for mismatches

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Supersequence

• Find the longest common sub-sequence of S,T
• Generate the string as follows
• for every column in the alignment
• Match add the matching character (once!)
• Gap add the character aligned against the gap

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Supersequence

• For SPride TParade
• P-R-IDE
• PARA-DE
• PARAIDE Shortest common supersequence

22
Exercise Finding repeats

• Basic objective find a pair of subsequences
  within S with maximum similarity
• Simple (albeit wrong) idea Find an optimal
  alignment of S with itself! (Why wrong?)
• But using local alignment is still a good idea

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Variant 1

• Specific requirement the two sequences may
  overlap
• Solution Change the local alignment algorithm
• Compute only the upper triangular submatrix
  (V(i,j), where jgti).
• Set diagonal values to 0
• Complexity O(n2) time and O(n) space

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Variant 2

• Specific requirement the two sequences may not
  overlap
• Solution Absence of overlap means that k exists
  such that one string is in S1..k and another in
  Sk1..n
• Check local alignments between S1..k and
  Sk1..n for any 1ltkltn
• Pick the highest-scoring alignment
• Complexity O(n3) time and O(n) space

25
Variant 2
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Variant 3

• Specific requirement the two sequences must be
  consequtive (tandem repeat)
• Solution Similar to variant 2, but somewhat
  ends-free seek a global alignment between
  S1..k and Sk1..n,
• No penalties for gaps in the beginning of S1..k
• No penalties for gaps in the end of Sk1..n
• Complexity O(n3) time and O(n) space

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Variant 3
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Variant 4

• Specific requirement the two sequences must be
  consequtive and the similarity is measured
  between the first sequence and the reverse
  complement of the second - SRC (inverted repeat)
• Tempting (albeit wrong) to use something in the
  spirit of variant 3 will give complexity O(n3)

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Variant 4

• Solution Compute the local alignment between S
  and SRC
• Look for results on the diagonal ijn
• AGCTAACGCGTTCGAA (n16)
• Complexity O(n2) time, O(n) space

?Index 8
Index 8 ?