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Vineet Bafna. How can we compute the local alignment itself?

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Presentation on theme: "Vineet Bafna. How can we compute the local alignment itself?"— Presentation transcript:

1 Vineet Bafna

2 How can we compute the local alignment itself?

3 T G C A A T C A T 1 1 0 0 1 0 1 2 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0

4 Gaps appear together

5 It is more likely for gaps to be contiguous The penalty for a gap of length l should be

6 What is the time taken for this? What are the values that l can take? Can we get rid of the extra Dimension?

7 Define D[i,j] : Score of the best alignment, given that the final column is a ‘deletion’ (s i is aligned to a gap) Define I[i,j]: Score of the best alignment, given that the final column is an insertion (t j is aligned to a gap) s[i] Optimum alignment of s[1..i-1], and t[1..j] - Optimum alignment of s[1..i], and t[1..j-1] t[j]

8

9 How much space do we need? Is the space requirement too much?

10 Copyright ©2004 by the National Academy of Sciences Istrail, Sorin et al. (2004) Proc. Natl. Acad. Sci. USA 101, 1916-1921 Fig. 1. Dot-plot representation of sample assembly comparison results

11 Score computation For i = 1 to n For j = 1 to m

12 In Linear Space, we can do each row of the D.P. We need to compute the optimum path from the origin (0,0) to (m,n) (0,0) (m,n)

13 Score computation For i = 1 to n For j = 1 to m

14 At i=n/2, we know scores of all the optimal paths ending at that row. Define F[j] = S[n/2,j] One of these j is on the true path. Which one?

15 Let S b [i,j] be the optimal score of aligning s[i+1..n] with t[j+1..m] Boundary cases? S b [n,j]? S b [m,j]?

16 Let S b [i,j] be the optimal score of aligning s[i+1..n] with t[j+1..m] Define B[j] = S b [n/2,j] One of these j is on the true path. Which one?

17 At the optimal coordinate, j – F[j]+B[j]=S[n,m] In O(nm) time, and O(m) space, we can compute one of the coordinates on the optimum path.

18 Align(1..n,1..m) – For all 1<=j <= m Compute F[j]=S(n/2,j) – For all 1<=j <= m Compute B[j]=S b (n/2,j) – j* = max j {F[j]+B[j] } – X = Align(1..n/2,1..j*) – Y = Align(n/2+1..n,j*+1..m) – Return X,j*,Y

19 T(nm) = c.nm + T(nm/2) = O(nm) Space = O(m)

20 Name an early Bioinformatics Researcher (preferably dead) Name an early Bioinformatics Researcher who is a woman

21 We have seen that affine gap penalties help concentrate the gaps in small regions. What about substitution errors. Are all substitutions alike?

22 Scoring protein sequence alignments is a much more complex task than scoring DNA – Not all substitutions are equal Problem was first worked on by Pauling and collaborators In the 1970s, Margaret Dayhoff created the first similarity matrices. – “One size does not fit all” – Homologous proteins which are evolutionarily close should be scored differently than proteins that are evolutionarily distant – Different proteins might evolve at different rates and we need to normalize for that

23 We have seen that affine gap penalties help concentrate the gaps in small regions. What about substitution errors. Are all substitutions alike?

24 DNA has structure.

25 So far, we considered a simple match/mismatch criterion. The nucleotides can be grouped into Purines (A,G) and Pyrimidines. Nucleotide substitutions within a group (transitions) are more likely than those across a group (transversions)

26 Transversions are more heavily penalized than transitions.

27 Suppose we are searching with a mouse protein. Blast returns proteins ranked by score – Top hit is to human – Somewhere below is Drosophila – Which one will you trust? hum mus dros


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