comp. genomics recitation 2 12/3/09 slides by igor ulitsky

Comp. Genomics

Recitation 212/3/09

Slides by Igor Ulitsky

Outline

• Alignment re-cap• End-space free alignment• Affine gap alignment algorithm and proof

• Bounded gap/spaces alignments

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

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 k=0

• dij(k)= min{dij

(k-1), dik(k-1)+dkj

(k-1)} otherwise

Global alignment

• Input: S1,S2

• Output: Minimum cost alignment• V(k,l) – score of aligning S1[1..k] with S2[1..l]

• Base conditions: • V(i,0) = k=0..i(sk,-)• V(0,j) = k=0..j(-,tk)

• Recurrence relation: V(i-1,j-1) + (si,tj)

1in, 1jm: V(i,j) = max V(i-1,j) + (si,-)

V(i,j-1) + (-,tj)

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)

Alignment score here

Traceback all the way

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



V(i,j-1) + (-,tj)0

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



V(i,j-1) + (-,tj)

• The optimal solution is found at the last row/column

(not necessarily at bottom right corner)

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.....iT.....j

S.....i------T..............j

S...............iT.....j-------

G(i,j)

E(i,j)

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

When do constant and affine gap costs differ?

• Consider:AGAGACTGACGCTTA

ATATTA

AGAGACTGACGCTTA

----A-T-A---TTA

Constant penalty: Mismatch: -5Gap: -1

AGAGACTGACGCTTA

ATA---------TTA

Affine penalty: Mismatch: -5Gap open: -3Gap extend: -0.5

-9 -14

-12-14.5

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)

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?

Bounding the number of spaces

• Let’s 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

• What’s the complexity?

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?


( 1, 1) ( , )

( , ) min ( ( , ) ( , -) ( )}

min ( ( , ) (-, ) ( )k i

k j

V i j i j

V i j V i k j s f k

V i j k t f k

− − += − + +

− + +min

Special cases

• How about a logarithmic penalty? Wg+Ws*log(k)

• This is a special case of a convex penalty, which is solvable in O(mn*log(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(mn*log(K))

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

Reminder: LCS

• Longest common non-contigous subsequence:• Adjust global alignment with similarity scores

• 1 for match• 0 for gaps • -∞ for mismatches

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

Supersequence

• For S=“Pride” T=“Parade”: • P-R-IDE• PARA-DE• PARAIDE – Shortest common supersequence

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

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 j>i).

• Set diagonal values to 0

• Complexity: O(n2) time and O(n) space

Variant #2

• Specific requirement: the two sequences may not overlap

• Solution: Absence of overlap means that k exists such that one string is in S[1..k] and another in S[k+1..n] • Check local alignments between S[1..k] and S[k+1..n] for any 1<=k<n

• Pick the highest-scoring alignment


Variant #2

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 S[1..k] and S[k+1..n], • No penalties for gaps in the beginning of S[1..k]

• No penalties for gaps in the end of S[k+1..n]


Variant #3

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)

Variant #4

• Solution: Compute the local alignment between S and SRC

• Look for results on the diagonal i+j=n

• AGCTAACGCGTTCGAA (n=16)

• Complexity: O(n2) time, O(n) space

Index 8 Index 8

comp. genomics recitation 2 12/3/09 slides by igor ulitsky

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