1. Post-processing long pairwise alignments 陳啟煌 93/4/28 Zheng Zhang et al., Bioinformatics Vol.15 no. 12 1999

2. Outline Motivation Theoretical basis of the proposed algorithms How to build up Useful Tree An application

3. Motivation Avoid local alignment problems Smith-Waterman lead to inclusion of an arbitrarily poor internal segment. Others approaches may generate an alignment score less than some internal segment

4. Smith-Waterman approach 8 10 11 13 2 3 5 0 103580000 18101320350 7 352580 82580000 115800020 135800350 28002580 00000000 C G G A T C A T CTTAACTCTTAACT The best score

5. Inclusion of a poor segment Inclusion of an arbitrarily poor region in an alignment Smith-Waterman approach potential flaws.

6. X-Alignments An X-Drop within an alignment, where X>0 is fixed in advance. A region of consecutive columns scoring less than <-X Alignments contain no X-Drop, we call X-alignments

7. BLAST In Blast Step 3: Extend hits. Terminate if the score of the extension fades away. (That is, when we reach a segment pair whose score falls a certain distance below the best score found for shorter extensions.) hit

8. 2X-Drop

9. Non-normal alignment The HSP has been extended to the right side in such a way that the entire alignment score less than the section from a to b

10. The Proposed Approach Provide techniques for decomposing a long alignment into sub-alignments that avoid the both problems. Show how to scan an alignment to collect information from which a decomposition corresponding any X can be found almost instantaneously. Provide a method for detecting variations in the rate of genome evolution

11. Useful Tree

12. X-full alignment An alignment are normal if each of its prefixes or suffixes has non-negative score. An alignment is not contained in any longer normal alignment is called full X-alignment + maximal X-normal is called X-full

13. X-full alignment 0-full alignment is maximal runs of columns of A with non-negative scores. For every X, X-full alignments are pairwise disjoint. If X<Y  X-full alignment contained in Y- full alignment.  -full alignments are just full alignments

14. Useful Tree Encode X-full alignments for all X≥ 0 in tree data structure. Leaves: 0-full alignments & maximal runs of negative score columns alternately Terminal Leaves: add two special leaves with score -  Each internal node is a disjoint union of its three children. Keep alignment ’ s score and the minimum sub-alignment ’ s score

15. Time complexity Construct time: O(N) Search Time: If k such alignments,need inspect at most 3k+1 nodes (2k+1) leaves+((2k+1-1)/2) internal nodes =3k+1 nodes

16. Decompose rules Alignment A  A 1,A 2, …., A 2n-1 # of sub-alignment is odd  i :score of A i Negative & Non-negative score alternately  0=  2n= -∞

17. Theoretical basis

19. Lemma1: X is consistent Lemma2:A normal drop is consistent with X

20. Lemma 3

21. Useful tree definition Each node of T is a segment consistent with X. Each leaf of T is of the form [i,i+1) Each internal node [a,d) has exactly three children. [a,b),[b,c) and [c,d) and the signs of their scores alternate.

22. Lemma 4

23. Possible negative merge LEMMA 5. Assume that three consecutive roots in our sequence, [a,b),[b,c),and [c,d), satisfy 0 ≤  (b,c)< min(-  (a,b),-  (c,d)) Then merging these trees into a single tree with root [a,d) creates a useful tree and the resulting sequence still satisfies P1 and P2. If a,b,c and d satisfy this lemma,[a,d) is a possible negative merger.

24. Possible positive merge LEMMA 6. Assume that five consecutive roots in our sequence, [a,b),[b,c),[c,d),[d,e) and [e,f) satisfy 0 >  (c,d) ≥ max(  (a,b),  (e,f)) neither [a,d) nor [c,f) is a possible negative Then merging these trees into a single tree with roots[b,c),[c,d),[d,e)into a single root[b,e) creates a useful tree and the resulting sequence still satisfies P1 and P2. If a,b,c,d,e and f satisfy this lemma,[a,d) is a possible positive merger.

25. Lemma 7

26. Theoretical basis Normal rise and normal drop Useful Tree contains every segments Possible negative merger Possible positive merger Always exists possible negative merger or possible positive merger

27. Decompose rules Alignment A  A 1,A 2, …., A 2n-1 # of sub-alignment is odd  i :score of A i Negative(odd i) & Non-negative(even i) score alternately  0=  2n= -∞

28. Useful Tree build up procedure 1.Push the first leaf on the stack 2.While the stack size exceeds 1 or there is an unvisited leaf do 3. if the top three stack items indicate a negative merger then 4. pop three items,merge them and push the result onto the stack 5. else if the top five segments indicate a positive merge then 6. pop an item{e,f} perform line 4. and push {e,f} back 7. else 8. push the next two leaves onto the stack

29. Construct Useful Tree ACAACAGAAACT | | || ||| ATA--AG-CACT Gop:0 Gep:1 Match/mismatch: 1/-1

31. Push 1 Push 2,3 Push 4,5, Merge 2,3,4 as a Merge 1,a,5 as b Push 6,7 Push 8,9 Merge 6,7,8 as c Push 10,11 Merge 9,10,11 as d Merge b,c,d as e

32. Source code of this paper http://globin.cse.psu.edu/dist/decom/

33. Alignment file #:lav d { "simu elegans briggsae M = 10, I = -10, V = -10, O = 60, E = 2" } s { "s1" 1 12 "s2" 1 9 } h { ">SUPERLINK_RWXL 2782216-2889703" ">dna -c briggsae.dna " } a { s 562 b 1 1 e 3 3 l 1 1 3 3 99 l 6 4 9 7 99 l 11 8 12 9 99 }

34. An Application Different regions of a mammalian genome evolve at different rates. Provide a method for detecting variations in the rate of genome evolution To compare the rates of evolution in different genomic regions from humans and mice. Align each pair of homologous regions and determined

35. Pitfalls Tally statistics only at sequence not in exons Regions adjacent to an exon maybe be aligned Remove the exons before producing the alignment The alignment program is unable to differentiate the biologically meaning alignment

36. Proposed approach First align the sequences using the exons as guideposts Then re-score the alignment where positions within exons are masked,so that they cannot be aligned to another nucleotide.

38. References Zheng Zhang et al., “ Post-processing long pairwise alignments ”,Bioinformatics, Vol.15 no. 12 1999 http://globin.cse.psu.edu/dist/decom/ Kun-Mao Chao,Algorithms for Biological Sequence Analysis Lecture Notes, National Taiwan University, Spring 2004

39. Q&A Thank you!

40. Possible mistakes, but maybe not P.1015 left col., last 2 row ∑ k=1  ∑ k=i P.1015 Right col. [i,i) should be [i,j) P.1016 proof of lemma4 4 [i,i) should be [i,j) P.1017 proof of lemma5  (b,c)  (e,c) should be  (b,c)-  (e,c) P.1017 lemma7 (a i-3,a i-2 )  (a i-4,a i-1 )

43. Proof 1 Lemma1: X is consistent

44. Proof of lemma2

45. Lemma 3

46. Proof of lemma3

47. Lemma 4

48. Proof of lemma4

49. Possible negative merge LEMMA 5. Assume that three consecutive roots in our sequence, [a,b),[b,c),and [c,d), satisfy 0 ≤  (b,c)< min(-  (a,b),-  (c,d)) Then merging these trees into a single tree with root [a,d) creates a useful tree and the resulting sequence still satisfies P1 and P2. If a,b,c and d satisfy this lemma,[a,d) is a possible negative merger.

50. Proof of lemma5

51. Possible positive merge LEMMA 6. Assume that five consecutive roots in our sequence, [a,b),[b,c),[c,d),[d,e) and [e,f) satisfy 0 >  (c,d) ≥ max(  (a,b),  (e,f)) neither [a,d) nor [c,f) is a possible negative Then merging these trees into a single tree with roots[b,c),[c,d),[d,e)into a single root[b,e) creates a useful tree and the resulting sequence still satisfies P1 and P2. If a,b,c,d,e and f satisfy this lemma,[a,d) is a possible positive merger.

52. Proof of lemma6

53. Lemma 7

54. Proof of lemma7