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Class 2: Basic Sequence Alignment

Class 2: Basic Sequence Alignment. Sequence Comparison. Much of bioinformatics involves sequences DNA sequences RNA sequences Protein sequences We can think of these sequences as strings of letters DNA & RNA: alphabet of 4 letters Protein: alphabet of 20 letters.

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Class 2: Basic Sequence Alignment

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  1. Class 2: Basic Sequence Alignment .

  2. Sequence Comparison Much of bioinformatics involves sequences • DNA sequences • RNA sequences • Protein sequences We can think of these sequences as strings of letters • DNA & RNA: alphabet of 4 letters • Protein: alphabet of 20 letters

  3. Sequence Comparison (cont) • Finding similarity between sequences is important for many biological questions For example: • Find genes/proteins with common origin • Allows to predict function & structure • Locate common subsequences in genes/proteins • Identify common “motifs” • Locate sequences that might overlap • Help in sequence assembly

  4. Sequence Alignment Input: two sequences over the same alphabet Output: an alignment of the two sequences Example: • GCGCATGGATTGAGCGA • TGCGCCATTGATGACCA A possible alignment: -GCGC-ATGGATTGAGCGA TGCGCCATTGAT-GACC-A

  5. Alignments -GCGC-ATGGATTGAGCGA TGCGCCATTGAT-GACC-A Three elements: • Perfect matches • Mismatches • Insertions & deletions (indel)

  6. Choosing Alignments There are many possible alignments For example, compare: -GCGC-ATGGATTGAGCGA TGCGCCATTGAT-GACC-A to ------GCGCATGGATTGAGCGA TGCGCC----ATTGATGACCA-- Which one is better?

  7. Scoring Alignments Rough intuition: • Similar sequences evolved from a common ancestor • Evolution changed the sequences from this ancestral sequence by mutations: • Replacements: one letter replaced by another • Deletion: deletion of a letter • Insertion: insertion of a letter • Scoring of sequence similarity should examine how many operations took place

  8. Simple Scoring Rule Score each position independently: • Match: +1 • Mismatch : -1 • Indel -2 Score of an alignment is sum of positional scores

  9. Example Example: -GCGC-ATGGATTGAGCGA TGCGCCATTGAT-GACC-A Score: (+1x13) + (-1x2) + (-2x4) = 3 ------GCGCATGGATTGAGCGA TGCGCC----ATTGATGACCA-- Score: (+1x5) + (-1x6) + (-2x11) = -23

  10. More General Scores • The choice of +1,-1, and -2 scores was quite arbitrary • Depending on the context, some changes are more plausible than others • Exchange of an amino-acid by one with similar properties (size, charge, etc.) vs. • Exchange of an amino-acid by one with opposite properties

  11. Additive Scoring Rules • We define a scoring function by specifying a function • (x,y) is the score of replacing x by y • (x,-) is the score of deleting x • (-,x) is the score of inserting x • The score of an alignment is the sum of position scores

  12. Edit Distance • The edit distance between two sequences is the “cost” of the “cheapest” set of edit operations needed to transform one sequence into the other • Computing edit distance between two sequences almost equivalent to finding the alignment that minimizes the distance

  13. Computing Edit Distance • How can we compute the edit distance?? • If |s| = n and |t| = m, there are more than alignments • The additive form of the score allows to perform dynamic programming to compute edit distance efficiently

  14. Recursive Argument • Suppose we have two sequences:s[1..i+1] and t[1..j+1] The best alignment must be in one of three cases: 1. Last position is (s[i+1],t[j+1] ) 2. Last position is (s[i+1],-) 3. Last position is (-, t[j +1] )

  15. Recursive Argument • Suppose we have two sequences:s[1..i+1] and t[1..j+1] The best alignment must be in one of three cases: 1. Last position is (s[i+1],t[j +1] ) 2. Last position is (s[i +1],-) 3. Last position is (-, t[j +1] )

  16. Recursive Argument • Suppose we have two sequences:s[1..i+1] and t[1..j+1] The best alignment must be in one of three cases: 1. Last position is (s[i+1],t[j +1] ) 2. Last position is (s[i +1],-) 3. Last position is (-, t[j +1] )

  17. Recursive Argument Define the notation: • Using the recursive argument, we get the following recurrence for V:

  18. Recursive Argument • Of course, we also need to handle the base cases in the recursion:

  19. Dynamic Programming Algorithm We fill the matrix using the recurrence rule

  20. Conclusion: d(AAAC,AGC) = -1 Dynamic Programming Algorithm

  21. Reconstructing the Best Alignment • To reconstruct the best alignment, we record which case in the recursive rule maximized the score

  22. Reconstructing the Best Alignment • We now trace back the path the corresponds to the best alignment AAAC AG-C

  23. Reconstructing the Best Alignment • Sometimes, more than one alignment has the best score AAAC A-GC

  24. Complexity Space: O(mn) Time: O(mn) • Filling the matrix O(mn) • Backtrace O(m+n)

  25. Space Complexity • In real-life applications, n and m can be very large • The space requirements of O(mn) can be too demanding • If m = n = 1000 we need 1MB space • If m = n = 10000, we need 100MB space • We can afford to perform extra computation to save space • Looping over million operations takes less than seconds on modern workstations • Can we trade off space with time?

  26. A G C 1 2 3 0 0 0 -2 -4 -6 1 -2 1 -1 -3 A 2 -4 -1 0 -2 A 3 -6 -3 -2 -1 A 4 -8 -5 -4 -1 C Why Do We Need So Much Space? To find d(s,t), need O(n) space • Need to compute V[i,m] • Can fill in V, column by column, storing only two columns in memory Note however • This “trick” fails when we need to reconstruct the sequence • Trace back information “eats up” all the memory

  27. Construct alignments • s[1,n/2] vs t[1,j] • s[n/2+1,n] vs t[j+1,m] Space Efficient Version: Outline Idea: perform divide and conquer • Find position (n/2, j) at which the best alignment crosses s midpoint s t

  28. Finding the Midpoint Suppose s[1,n] and t[1,m] are given • We can write the score of the best alignment that goes through j as: d(s[1,n/2],t[1,j]) + d(s[n/2+1,n],t[j+1,m]) • Thus, we need to compute these two quantities for all values of j

  29. Finding the Midpoint (cont) Define • F[i,j] = d(s[1,i],t[1,j]) • B[i,j] = d(s[i+1,n],t[j+1,m]) • F[i,j] + B[i,j] = score of best alignment through (i,j) • We compute F[i,j] as we did before • We compute B[i,j] in exactly the same manner, going “backward” from B[n,m]

  30. Time Complexity Analysis • Finding mid-point: cmn (c - a constant) • Recursive sub-problems of sizes (n/2,j) and (n/2,m-j-1) T(m,n) = cmn + T(j,n/2) + T(m-j-1, n/2) Lemma: T(m,n)  2cmn Time complexity is linear in size of the problem • At worse, twice the time of regular solution.

  31. Local Alignment Consider now a different question: • Can we find similar substring of s and t • Formally, given s[1..n] and t[1..m] find i,j,k, and l such that d(s[i..j],t[k..l]) is maximal

  32. Local Alignment • As before, we use dynamic programming • We now want to setV[i,j] to record the best alignment of a suffix of s[1..i] and a suffix of t[1..j] • How should we change the recurrence rule?

  33. Local Alignment New option: • We can start a new match instead of extend previous alignment Alignment of empty suffixes

  34. Local Alignment Example s =TAATA t =TACTAA

  35. Local Alignment Example s =TAATA t =TACTAA

  36. Local Alignment Example s =TAATA t =TACTAA

  37. Local Alignment Example s =TAATA t =TACTAA

  38. Sequence Alignment We saw two variants of sequence alignment: • Global alignment • Local alignment Other variants: • Finding best overlap (exercise) All are based on the same basic idea of dynamic programming

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