1. . Sequence Alignment I Lecture #2 This class has been edited from Nir Friedman’s lecture. Changes made by Dan Geiger, then Shlomo Moran. Background Readings: The second chapter (pages 12-45) in the text book, Biological Sequence Analysis, Durbin et al., 2001.

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

3. 3 Sequence Comparison (cont) u Finding similarity between sequences is important for many biological questions For example: u Find similar proteins  Allows to predict function & structure u Locate similar subsequences in DNA  Allows to identify (e.g) regulatory elements u Locate DNA sequences that might overlap  Helps in sequence assembly g1g1 g2g2

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

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

6. 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. 7 Scoring Alignments Intuition: u Similar sequences evolved from a common ancestor u 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 u Scoring of sequence similarity should examine how many and which operations took place

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

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

10. 10 More General Scores u The choice of +1,-1, and -2 scores is quite arbitrary u 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 u Probabilistic interpretation: (e.g.) How likely is one alignment versus another ?

11. 11 Additive Scoring Rules u 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 u The score of an alignment is the sum of position scores

12. 12 The Optimal Score  The optimal (maximal) score between two sequences is the maximal score of all alignments of these sequences, namely, u Computing the maximal score or actually finding an alignment that yields the maximal score are closely related tasks with similar algorithms. u One more comment before we start...

13. 13 Alternative view: Edit Distance Instead of speaking about the score of an alignment, one often talks about an edit distance between two sequences, defined to be the “cost” of the “cheapest” set of edit operations needed to transform one sequence into the other.  Cheapest operation is “no change”  Next cheapest operation is “replace”  The most expensive operations are “add space/delete”. Our goal is now to minimize the cost of operations.

14. 14 Computing Optimal Score u How can we compute the optimal score ?  If | s | = m and | t | = n, the number A( m,n ) of possible “legal” alignments is large! Exercise: Show that u The additive form of the score allows us to perform dynamic programming to compute optimal score efficiently.

15. 15 Recursive Argument (Needleman Wunsch, 1970)  Suppose we have two sequences: s[1..m+1] and t[1..n+1] The last column in a best alignment must be one of : 1. Last column is ( s[m+1],t[n +1] ) 2. Last column is ( s[m +1], - ) 3. Last column is ( -, t[n +1] )

16. 16 Recursive Argument  Suppose we have two sequences: s[1..n+1] and t[1..m+1] The best alignment must be one of three cases: 1. Last match is ( s[m+1],t[n +1] ) 2. Last match is ( s[m +1], - ) 3. Last match is ( -, t[n +1] )

17. 17 Recursive Argument  Suppose we have two sequences: s[1..n+1] and t[1..m+1] The best alignment must be one of three cases: 1. Last match is ( s[m+1],t[n +1] ) 2. Last match is ( s[m +1], - ) 3. Last match is ( -, t[n +1] )

18. 18 Recursive Argument Define the notation:  Using our recursive argument, we get the following recurrence for V : V[i,j]V[i+1,j] V[i,j+1]V[i+1,j+1]

19. 19 Recursive Argument u Of course, we also need to handle the base cases in the recursion: AA - We fill the matrix using the recurrence rule: S T versus

20. 20 Dynamic Programming Algorithm We continue to fill the matrix using the recurrence rule S T

21. 21 Dynamic Programming Algorithm V[0,0]V[0,1] V[1,0]V[1,1] +1 -2 -A A- -2 (A- versus -A) versus S T

22. 22 Dynamic Programming Algorithm S T

23. 23 Dynamic Programming Algorithm Conclusion: d( AAAC, AGC ) = -1 S T

24. 24 Reconstructing the Best Alignment u To reconstruct the best alignment, we record which case(s) in the recursive rule maximized the score S T

25. 25 Reconstructing the Best Alignment u We now trace back a path that corresponds to the best alignment AAAC AG-C S T

26. 26 Reconstructing the Best Alignment u Sometimes, more than one alignment has the best score S T AAAC A-GC AAAC -AGC AAAC AG-C

27. 27 Time Complexity Space: O(mn) Time: O(mn)  Filling the matrix O(mn)  Backtrace O(m+n) S T

28. 28 Space Complexity  In real-life applications, n and m can be very large u 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 u We can afford to perform extra computation to save space  Looping over million operations takes less than seconds on modern workstations u Can we trade space with time?

29. 29 Why Do We Need So Much Space?  Compute V(i,j), column by column, storing only two columns in memory (or row by row if rows are shorter). 0 -2 -4 -6 -8 -2 1 -3 -5 -4 0 -2 -4 -6 -3 -2 0 A 1 G 2 C 3 0 A 1 A 2 A 3 C 4 Note however that u This “trick” fails when we need to reconstruct the optimizing sequence.  Trace back information requires O(mn) memory bytes. To compute V[m,n]=d(s[1..m],t[1..n]), we need only O(min(m,n)) space:

30. 30 Space Efficient Version: Outline u If m=1 align s[1,1] and t[1,m]  Else, find position (i,n/2) at which some best alignment crosses a midpoint s t u Construct alignments  A=s[1,n/2] vs t[1,i]  B=s[n/2+1,m] vs t[i+1,n]  Return AB Input: Sequences s[1,m] and t[1,n] to be aligned. Idea: perform divide and conquer

31. 31 Finding a Midpoint The score of the best alignment that goes through j equals: d(s[1,m/2],t[1,i]) + d(s[m/2+1,n],t[i+1,m])  Thus, we need to compute these two quantities for all values of i s t

32. 32 Finding a Midpoint (Algorithm) Define u F[i,j] = d(s[1,i],t[1,j]) u 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] u Requires linear space complexity

33. 33 Time Complexity Analysis  Time to find a mid-point: cnm ( c - a constant) u Size of recursive sub-problems is (n/2,i) and (n/2,m-i), hence T(n,m) = cnm + T(n/2,i) + T(n/2,m-i) Lemma: T(n,m)  2cnm Proof (by induction): T(n,m)  cnm + 2c(n/2)i + 2c(n/2)(m-i) = 2cnm. Thus, time complexity is linear in size of the problem At worst, twice the cost of the regular solution.

34. 34 Local Alignment Consider now a different question:  Can we find similar substrings of s and t  Formally, given s[1..m] and t[1..n] find i,j,k, and l such that d(s[i..j],t[k..l]) is maximal

35. 35 Local Alignment u As before, we use dynamic programming  We now want to set V[i,j] to record the best alignment of a suffix of s[1..i] and a suffix of t[1..j] u How should we change the recurrence rule?  Same as before but with an option to start afresh u The result is called the Smith-Waterman algorithm

36. 36 Local Alignment New option: u We can start a new match instead of extending a previous alignment Alignment of empty suffixes

37. 37 Local Alignment Example s = TAATA t = TACTAA S T

38. 38 Local Alignment Example s = TAATA t = TACTAA S T

39. 39 Local Alignment Example s = TAATA t = TACTAA S T

40. 40 Local Alignment Example s = TAATA t = TACTAA S T

41. 41 Local Alignment Example s = TAATA t = TACTAA S T

42. 42 Alignment in Real Life u One of the major uses of alignments is to find sequences in a “database.” u Today’s protein database contains more than 10 8 residues! So searching a 10 3 long target sequence requires to evaluate about 10 11 matrix cells which will take about three hours in the rate of 10 millions evaluations per second. u Quite annoying when, say, one thousand target sequences need to be searched because it will take about four months to run.

43. 43 Heuristic Search u Instead, most searches rely on heuristic procedures u These are not guaranteed to find the best match u Sometimes, they will completely miss a high-scoring match Next week we’ll describe the main ideas used by the best known(?) of these heuristic procedures.