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Sequence comparison: Dynamic programming Genome 559: Introduction to Statistical and Computational Genomics Prof. James H. Thomas http://faculty.washington.edu/jht/GS559_2013/ Sequence comparison overview Problem: Find the best

  1. Sequence comparison: Dynamic programming Genome 559: Introduction to Statistical and Computational Genomics Prof. James H. Thomas http://faculty.washington.edu/jht/GS559_2013/

  2. Sequence comparison overview • Problem: Find the “best” alignment between a query sequence and a target sequence. • To solve this problem, we need – a method for scoring alignments, and – an algorithm for finding the alignment with the best score. • The alignment score is calculated using – a substitution matrix – gap penalties. • The algorithm for finding the best alignment is dynamic programming (DP).

  3. A simple alignment problem • Problem: find the best pairwise alignment of GAATC and CATAC . • Use a linear gap penalty of -4. • Use the following substitution matrix: A C G T A 10 -5 0 -5 C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10

  4. How many possibilities? GAATC GAAT-C -GAAT-C CATAC C-ATAC C-A-TAC GAATC- GAAT-C GA-ATC CA-TAC CA-TAC CATA-C • How many different possible alignments of two sequences of length n exist?

  5. How many possibilities? GAATC GAAT-C -GAAT-C CATAC C-ATAC C-A-TAC GAATC- GAAT-C GA-ATC CA-TAC CA-TAC CATA-C • How many different alignments of two sequences of length n exist? 5 2.5x10 2 2 n 2 n ! 10 1.8x10 5 FYI for two sequences of length m and n, possible 20 1.4x10 11 2 alignments number: n n ! 30 1.2x10 17 40 1.1x10 23 mn ( mn )! 2n choose n - the 2 min( , ) m n (min( , )!) m n binomial coefficient

  6. A C G T GA DP matrix A 10 -5 0 -5 CA C -5 10 -5 0 G 0 -5 10 -5 j 0 1 2 3 etc. T -5 0 -5 10 i G A A T C 0 C 1 A 5 2 T 3 A 4 Value at ( i,j ) will be the score of the best alignment of the first i characters 5 C of one sequence to the first j characters of the other sequence. init. row and column

  7. A C G T DP matrix GAA A 10 -5 0 -5 C -5 10 -5 0 CA- G 0 -5 10 -5 T -5 0 -5 10 G A A T C C A 5 1 T Moving horizontally in the A matrix introduces a gap in the sequence along the left edge. C

  8. A C G T GA- DP matrix A 10 -5 0 -5 CAT C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C Moving vertically in the matrix introduces a gap in C the sequence along the top edge. A 5 T 1 A C

  9. A C G T GAA DP matrix A 10 -5 0 -5 CAT C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C Moving diagonally in the C matrix aligns two residues A 5 T 0 A C

  10. A C G T Initialization A 10 -5 0 -5 Start at top C -5 10 -5 0 left and move G 0 -5 10 -5 progressively T -5 0 -5 10 G A A T C 0 C A T A C

  11. A C G T G Introducing a gap A 10 -5 0 -5 - C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 C A T A C

  12. A C G T - Introducing a gap A 10 -5 0 -5 C C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 C -4 A T A C

  13. Complete first row A C G T ----- A 10 -5 0 -5 and column CATAC C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 C -4 A -8 T -12 A -16 C -20

  14. A C G T Three ways to get G- A 10 -5 0 -5 to i=1 , j=1 C -5 10 -5 0 -C G 0 -5 10 -5 j 0 1 2 3 etc. T -5 0 -5 10 i G A A T C 0 -4 0 C -8 1 A 2 T 3 A 4 5 C

  15. Three ways to get A C G T -G A 10 -5 0 -5 to i=1 , j=1 C- C -5 10 -5 0 G 0 -5 10 -5 j 0 1 2 3 etc. T -5 0 -5 10 i G A A T C 0 0 C -4 -8 1 A 2 T 3 A 4 5 C

  16. Three ways to get A C G T G A 10 -5 0 -5 to i=1 , j=1 C C -5 10 -5 0 G 0 -5 10 -5 j 0 1 2 3 etc. T -5 0 -5 10 i G A A T C 0 0 C -5 1 A 2 T 3 A 4 5 C

  17. A C G T Accept the highest scoring A 10 -5 0 -5 C -5 10 -5 0 of the three G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 C -4 -5 A -8 T -12 A -16 C -20 Then simply repeat the same rule progressively across the matrix

  18. A C G T DP matrix A 10 -5 0 -5 C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 C -4 -5 A -8 ? T -12 A -16 C -20

  19. A C G T -G G- --G DP matrix A 10 -5 0 -5 CA CA CA- C -5 10 -5 0 -4 G 0 -5 10 -5 -9 -12 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 C -4 -5 0 -4 -4 A -8 ? T -12 A -16 C -20

  20. A C G T -G G- --G DP matrix A 10 -5 0 -5 CA CA CA- C -5 10 -5 0 -4 G 0 -5 10 -5 -9 -12 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 C -4 -5 0 -4 -4 A -8 -4 T -12 A -16 C -20

  21. A C G T DP matrix A 10 -5 0 -5 C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 C -4 -5 A -8 -4 T -12 ? A -16 ? C -20 ?

  22. A C G T DP matrix A 10 -5 0 -5 C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 C -4 -5 A -8 -4 T -12 -8 A -16 -12 C -20 -16

  23. A C G T DP matrix A 10 -5 0 -5 C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 ? C -4 -5 A -8 -4 ? T -12 -8 ? A -16 -12 ? C -20 -16 ?

  24. A C G T Traceback A 10 -5 0 -5 C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 -9 C -4 -5 A -8 -4 5 T -12 -8 1 What is the alignment associated with this entry? A -16 -12 2 C -20 -16 -2 (just follow the arrows back - this is called the traceback)

  25. A C G T DP matrix A 10 -5 0 -5 C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 -9 C -4 -5 A -8 -4 5 -G-A T -12 -8 1 CATA A -16 -12 2 C -20 -16 -2

  26. A C G T DP matrix A 10 -5 0 -5 C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 C -4 -5 -9 A -8 -4 5 Continue and find the optimal global T -12 -8 1 alignment, and its score. A -16 -12 2 C -20 -16 -2 ?

  27. A C G T Best alignment starts at bottom right A 10 -5 0 -5 and follows traceback arrows to top left C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 C -4 -5 -9 -13 -12 -6 A -8 -4 5 1 -3 -7 T -12 -8 1 0 11 7 A -16 -12 2 11 7 6 C -20 -16 -2 7 11 17

  28. A C G T GA-ATC One best traceback A 10 -5 0 -5 C -5 10 -5 0 CATA-C G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 C -4 -5 -9 -13 -12 -6 A -8 -4 5 1 -3 -7 T -12 -8 1 0 11 7 A -16 -12 2 11 7 6 C -20 -16 -2 7 11 17

  29. A C G T GAAT-C A 10 -5 0 -5 -CATAC Another best traceback C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 C -4 -5 -9 -13 -12 -6 A -8 -4 5 1 -3 -7 T -12 -8 1 0 11 7 A -16 -12 2 11 7 6 C -20 -16 -2 7 11 17

  30. A C G T GAAT-C GA-ATC A 10 -5 0 -5 -CATAC CATA-C C -5 10 -5 0 G 0 -5 10 -5 T -5 0 -5 10 G A A T C 0 -4 -8 -12 -16 -20 C -4 -5 -9 -13 -12 -6 A -8 -4 5 1 -3 -7 T -12 -8 1 0 11 7 A -16 -12 2 11 7 6 C -20 -16 -2 7 11 17

  31. Multiple solutions GA-ATC • When a program returns a single CATA-C sequence alignment, it may not be the only best alignment but it GAAT-C is guaranteed to be one of them. CA-TAC GAAT-C • In our example, all of the C-ATAC alignments at the left have equal scores. GAAT-C -CATAC

  32. DP in equation form • Align sequence x and y . • F is the DP matrix; s is the substitution matrix; d is the linear gap penalty. F 0 , 0 0 F i 1 , j 1 s x , y i j F i , j max F i 1 , j d F i , j 1 d

  33. DP equation graphically F i , j 1 F i 1 j , 1 d s x i y , j F i 1 , j d F , i j take the max of these three

  34. Dynamic programming • Yes, it’s a weird name. • DP is closely related to recursion and to mathematical induction. • We can prove that the resulting score is optimal.

  35. What you should know • Scoring a pairwise alignment requires a substitution matrix and gap penalties. • Dynamic programming (DP) is an efficient algorithm for finding an optimal alignment. • Entry ( i,j ) in the DP matrix stores the score of the best-scoring alignment up to that position. • DP iteratively fills in the matrix using a simple mathematical rule. • How to use DP to find an alignment.

  36. Practice problem: find a best pairwise alignment of GAATC and AATTC A C G T G A A T C A 10 -5 0 -5 C -5 10 -5 0 0 G 0 -5 10 -5 T -5 0 -5 10 A d = -4 A T T C

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