Sequence Alignment

1. Sequence Alignment Arthur W. Chou Tunghai University Fall 2005

2. 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

3. Why align sequences? Lots of sequences don’t have known ancestry, structure, or function. A few of them do. If they align, they are similar. If they are similar, they might have the same ancestry, similar structure or function. If one of them has known ancestry, structure, or function, then alignment to the others yields insight about them.

4. Alignments -GCGC-ATGGATTGAGCGA TGCGCCATTGAT-GACC-A Three kinds of match: Exact matches Mismatches Indels (gaps)

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

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

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

8. Example -GCGC-ATGGATTGAGCGA TGCGCCATTGAT-GACC-A Score: (+113) + (-1  2) + (-2  4) = 3 ------GCGCATGGATTGAGCGA TGCGCC----ATTGATGACCA-- Score: (+1  5) + (-1  6) + (-2  11) = -23

9. More General Scores • The choice of +1,-1, and -2 scores is 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 • Probabilistic interpretation: How likely is one alignment versus another ?

10. Dot Matrix Method • A dot is placed at each position where two residues match. • It's a visual aid. The human eye can rapidly identify similar regions in sequences. • It's a good way to explore sequence organization: e.g. sequence repeats. • It does not provide an alignment. • This method produces dot-plots with too much noise • to be useful • The noise can be reduced by calculating a score using a window of residues. • The score is compared to a threshold or stringency. THEFA-TCAT ||||| |||| THEFASTCAT

11. Tissue-Type plasminogen Activator Urokinase-Type plasminogen Activator Dot Matrix Representation • Produces a graphical representation of similarity regions • The horizontal and vertical dimensions correspond to the compared sequences • A region of similarity stands out as a diagonal

12. HEF THE THE ||| THE THE HEF CAT THE Score: -5 Score: 23 Score: -5 Score: -4 Dot Matrix or Dot-plot • Each window of the first sequence is aligned (without gaps) to each window of the 2nd sequence • A colour is set into a rectangular array according to the score of the aligned windows

13. Dot Matrix Display • Diagonal rows ( ) of dots reveal sequence similarity or repeats. • Anti-diagonal rows ( ) of dots represent inverted repeats. • Isolated dots represent random similarity. H C G E T F G R W F T P E W K C • G • P • T • • F • • G • R • I A C • G • • E • • M

14. Dot matrix web server http://www.isrec.isb-sib.ch/java/dotlet/Dotlet.html We can filter it by using a sliding window looking for longer strings of matches and eliminates random matches

15. Longest Common Subsequence Sequence A:nematode_knowledge Sequence B:empty_bottle n e m a t o d e _ k n o w l e d g e | | | | | | | e m p t y _ b o t t l e • LCS Alignment with match score 1, mismatch score 0, and gap penalty 0

16. What is an algorithm? • A step-by-step description of the procedures to accomplish a task. • Properties: • Determination of output for each input • Generality • Termination • Criteria: • Correctness (proof, test, etc.) • Time efficiency (no. of steps is small) • Space efficiency (spaced used is small)

17. Naïve algorithm: exhaustive search sequences of length “n” i G C G A A T G G A T T G A G C G T T G A G C C A T T G A T G A C C A j i j j i j i j j i i . . . . . . . . . . . . . . 2n Worst case time complexity is ~ 2

18. Dynamic programming algorithms for pairwise sequence alignment • Similar to Longest Common Subsequence • Introduced for biological sequences by • S. B. Needleman & C. D. Wunsch.A general method applicable to the search for similarities in the amino acid sequence of two proteins.J. Mol. Biol. 48:443-453 (1970)

19. - G C G C – A T G G A T T G A G C G A T G C G C C A T T G A T – G A C C - A Dynamic Programming • Optimality substructure • Reduction to a “small” number of sub-problems • Memorization of solutions to sub-problems in a table • Table look-up and tracing Optimality Sub-structure

20. - G C G C – A T G G A T T G A G C G A T G C G C C A T T G A T – G A C C - A 二陽指TM sequences of length “n” i G C G A A T G G A T T G A G C G T T G A G C C A T T G A T G A C C A j

21. Recursive LCS lcs_len( i , j): length of LCS from i-th position onward in String A and from j-th position onward in String B int lcs_len ( i , j ) { if (A[ i ] == ‘\0’ || B[ j ] == ‘\0’ ) return 0 ; else if (A[ i ] == B[ j ] ) return ( 1 + lcs_len ( i+1, j+1 ) ) ; else return max ( lcs_len ( i+1, j ) , lcs_len ( i, j+1 ) ); }

22. Reduction to Subproblems intlcs_len ( String A , String B ) { returnsubproblem ( 0, 0 ); } intsubproblem ( int i, intj) { if (A[i] == ‘\0’ ||B[j] == ‘\0’) return 0; else if ( A[i] == B[j] ) return (1 +subproblem(i+1, j+1)); else return max (subproblem(i+1, j) , subproblem ( i,j+1) ); }

23. Memorizing the solutions : Matrix L[ i , j ] = -1 ; // initializing the memory device int subproblem ( int i, int j ) { if ( L[i, j] < 0 ) { if (A[ i ] == ‘\0’ || B[ j ] == ‘\0’) L[i , j] = 0; else if ( A[ i ] == B[ j ] ) L[i, j] = 1 + subproblem(i+1, j+1); else L[i, j] = max( subproblem(i+1, j), subproblem(i, j+1)); } return L[ i, j ] ; }

24. Iterative LCS: Table Look-up To get the length of LCS of A and B { first allocate storage for the matrix L; for each row i from m downto 0 for each column j from n downto 0 if (A[ i ] == ‘\0’ or B[ j ] == ‘\0’) L[ i, j ] = 0; else if (A[ i ] == B[ j ]) L[ i, j ] = 1 + L[i+1, j+1]; else L[ i, j ] = max(L[i+1, j], L[i, j+1]); } return L[0, 0]; }

25. Iterative LCS: Table Look-up int lcs_len ( String A , String B ) // find the length { // First allocate storage for the matrix L; for ( i = m ; i >= 0 ; i-- ) // A has length m+1 for ( j = n ; j >= 0 ; j-- ) { // B has length n+1 if (A[ i ] == ‘\0’ || B[ j ] == ‘\0’) L[ i, j ] = 0; else if (A[ i ] == B[ j ]) L[ i, j ] = 1 + L[i+1, j+1]; else L[ i, j ] = max(L[i+1, j], L[i, j+1]); } return L[0, 0]; }

26. Dynamic Programming Algorithm B j j+1 A Matrix L L[i, j] = 1 + L[i+1, j+1] , if A[ i ] == B[ j ] ; L[i, j] = max ( L[i+1, j], L[i, j+1] ) otherwise L[ i, j ] L[ i, j+1 ] i L[ i+1, j ] L[i+1, j+1] i+1

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28. Obtain the subsequence Sequence S = empty; // the LCS i = 0; j = 0; while ( i < m && j < n) { if ( A[ i ] == B[ j ] ) { add A[i] to end of S; i++; j++; } else if ( L[i+1, j] >= L[i, j+1]) i++; else j++; }

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30. i x Dynamic Programming with scores and penalties y j

31. s : score w : penalty function best score from i, j onward Dynamic Programming with scores and penalties • from ‘i-th’ pos. in A and ‘j-th’ pos. in B onward s ( A[i] , B[j] ) + S[i+1, j+1] S[i , j] = max max { S[i+x, j] – w( x ); gap x in sequence B } max { S[i, j+y] – w( y ); gap y in sequence A }

32. Algorithm for simple gap penalty If for each gap, the penalty is a fixed constant “c”, then • s(A[ i ] , B[ j ]) + S[i+1, j+1]; S[i , j] = max• S[ i+1, j ] – c ; // one gap • S[ i, j+1 ] – c ; // one gap

33. Table Tracing To do table tracing based on similarity matrix of amino acids, we re-define S[i , j] to be the optimal score of choosing the match of A[i] with B[j]. S[ i , j ] = s (A[ i ] , B[ j ]) + // s : score S[i+1, j+1] // w : gap penalty max { S[i+1+x, j+1] – w( x ); + max gap x in sequence B } max { S[i+1, j+1+y] – w( y ); gap y in sequence A }

34. Diagram Matrix S: j j+1 i i+1

35. Summation operation 1. Start at lower right corner. 2. Move diagonally up one position. 3. Find largest value in either row segment starting diagonally below current position and extending to the right or column segment starting diagonally below current position and extending down. 4. Add this value to the value in the current cell. 5. Repeat steps 3 and 4 for all cells to the left in current row and all cells above in current column. 6. If we are not in the top left corner, go to step 2.