Post-processing long pairwise alignments

Post-processing long pairwise alignments

陳啟煌 93/4/28

Zheng Zhang et al., Bioinformatics Vol.15 no. 12 1999

Outline Motivation Theoretical basis of the proposed

algorithms How to build up Useful Tree An application

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

Smith-Waterman approach

810

11132350

103580000

1810132035071335258082580000

1158000201358003502800258000000000

C G G A T C A T

C

T

T

A

A

C

T

The best

score

Inclusion of a poor segment Inclusion of an arbitrarily poor

region in an alignment Smith-Waterman approach potential

flaws.

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

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

2X-Drop

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

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

Useful Tree

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

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<YX-full alignment contained in

Y-full alignment. -full alignments are just full

alignments

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

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

Decompose rules Alignment A A1,A2,…., A2n-1

# of sub-alignment is odd i :score of Ai Negative & Non-negative score

alternately 0=2n= -∞

Theoretical basis

Theoretical basis

Theoretical basis Lemma1:X is consistent Lemma2:A normal drop is

consistent with X

Lemma 3

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.

Lemma 4

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.

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.

Lemma 7

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

Decompose rules Alignment A A1,A2,…., A2n-1

# of sub-alignment is odd i :score of Ai Negative(odd i) & Non-

negative(even i) score alternately 0=2n= -∞

Useful Tree build up procedure

1.Push the first leaf on the stack2.While the stack size exceeds 1 or there is an

unvisited leaf do3. 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 then6. pop an item{e,f} perform line 4. and push {e,f} back7. else

8. push the next two leaves onto the stack

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

1 2 3 4 7 8 9 11S1 A C A A C A G A A A C T

| | | | | | |S2 A T A - - A G C A - C T

S 0 0 0 0 0 0 0 -1 0 0 1 -1 1 -1 1 -1 -13 -13 1 0 1 0 1 0 -13

22

5 6 10

-2-2-1

21

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

1 2 3 4 7 8 9 11S1 A C A A C A G A A A C T

| | | | | | |S2 A T A - - A G C A - C T

S 0 0 0 0 0 0 0 -1 0 0 1 -1 1 -1 1 -1 -13 -13 1 0 1 0 1 0 -13

22

5 6 10

-2-2-1

21

-2 -1 -131 1 2-1 -2 -1 -1 -131 1 2 2 3

-13 -13 -14 -14 -14 -14

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

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}

1 2 3 4 7 8 9 11S1 A C A A C A G A A A C T

| | | | | | |S2 A T A - - A G C A - C T

S 0 0 0 0 0 0 0 -1 0 0 1 -1 1 -1 1 -1 -13 -13 1 0 1 0 1 0 -13

22

5 6 10

-2-2-1

21

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

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

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.

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

Q&A Thank you!

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 (ai-3,ai-2) (ai-4,ai-1)

Proof 1

Lemma1:X is consistent

Proof of lemma2

Lemma 3

Proof of lemma3

Lemma 4

Proof of lemma4

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.

Proof of lemma5

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.

Proof of lemma6

Lemma 7

Proof of lemma7