Alignment of Long Sequences. BMI/CS Spring 2012 Colin Dewey

Transcription

1 Alignment of Long Sequences BMI/CS Spring 2012 Colin Dewey

2 Gols for Lecture the key concepts to understnd re the following how lrge-scle lignment differs from the simple cse the cnonicl three step pproch of lrge-scle ligners using suffix trees to find MUMs (lignment seeds) using tries nd threded tries to find lignment seeds constrined dynmic progrmming to lign between/ round nchors using sprse DP to find chin of locl lignments

3 Pirwise Lrge-Scle Alignment: Tsk Definition Given pir of lrge-scle sequences (e.g. chromosomes) method for scoring the lignment (e.g. substitution mtrices, insertion/deletion prmeters) Do construct globl lignment: identify ll mtching positions between the two sequences

4 Lrge Scle Alignment Exmple: Mouse Chr6 vs. Humn Chr12

5 Why the Problem is Chllenging sequences too big to mke O(n 2 ) dynmicprogrmming methods prcticl long sequences re less likely to be coliner becuse of rerrngements initilly we ll ssume colinerity we ll consider rerrngements in next lecture

6 Generl Strtegy Figure from: Brudno et l. Genome Reserch, perform pttern mtching to find seeds for globl lignment 2. find good chin of nchors 3. fill in reminder with stndrd but constrined lignment method

7 Comprison of Lrge-Scle Alignment Methods Method Pttern mtching Chining MUMmer suffix tree - MUMs LIS vrint AVID suffix tree - exct & wobble mtches Smith-Wtermn vrint LAGAN k-mer trie, inexct mtches sprse DP

8 The MUMmer System Delcher et l., Nucleic Acids Reserch, 1999 Given: genomes A nd B 1. find ll mximl, unique, mtching subsequences (MUMs) 2. extrct the longest possible set of mtches tht occur in the sme order in both genomes 3. close the gps

9 Step 1: Finding Seeds in MUMmer mximl unique mtch (MUM): occurs exctly once in both genomes A nd B not contined in ny longer MUM mismtches key insight: significntly long MUM is certin to be prt of the globl lignment

10 Suffix Trees substring problem: given text S of length m preprocess S in O(m) time such tht, given query string Q of length n, find occurrence (if ny) of Q in S in O(n) time suffix trees solve this problem, nd others

11 Suffix Tree Definition key property suffix tree T for string S of length m is tree with the following properties: rooted nd directed m leves, lbeled 1 to m ech edge lbeled by substring of S conctention of edge lbels on pth from root to lef i is suffix i of S (we will denote this by Si...m) ech internl non-root node hs t lest two children edges out of node must begin with different chrcters

12 Suffixes S = bnn$ suffixes of S $ $ n$ n$ nn$ nn$ bnn$

13 Suffix Tree Exmple S = bnn$ dd $ to end so tht suffix tree exists (no suffix is prefix of nother suffix) n $ $ n $ b n n $ n n $ $ $

14 Solving the Substring Problem ssume we hve suffix tree T FindMtch(Q, T): follow (unique) pth down from root of T ccording to chrcters in Q if ll of Q is found to be prefix of such pth return lbel of some lef below this pth else, return no mtch found

15 Solving the Substring Problem Q = nn Q = nb n $ $ n $ b n n $ n n $ $ $ 7 STOP n $ $ n $ b n n $ n n $ $ $ return return no mtch found

16 MUMs nd Generlized Suffix Trees build one suffix tree for both genomes A nd B lbel ech lef node with genome it represents Genome A: cccg# Genome B: cct$ cg# c g# t$ ech internl node represents repeted sequence A, 3 cg# c g# t$ A, 5 B, 3 A, 2 A, 4 B, 2 cg# t$ A, 1 B, 1 ech lef represents suffix nd its position in sequence

17 MUMs nd Suffix Trees unique mtch: internl node with 2 children, lef nodes from different genomes but these mtches re not necessrily mximl Genome A: cccg# Genome B: cct$ cg# c g# t$ A, 3 cg# c g# t$ A, 5 B, 3 A, 2 A, 4 B, 2 cg# t$ A, 1 B, 1 represents unique mtch

18 MUMs nd Suffix Trees to identify mximl mtches, cn compre suffixes following unique mtch nodes Genome A: ct# Genome B: c$ c t# A, 4 $ $ c t# t# $ B, 4 B, 3 A, 3 A, 2 B, 2 t# A, 1 $ B, 1 the suffixes following these two mtch nodes re the sme; the left one represents longer mtch (c)

19 Using Suffix Trees to Find MUMs O(n) time to construct suffix tree for both sequences (of lengths n) O(n) time to find MUMs - one scn of the tree (which is O(n) in size) O(n) possible MUMs in contrst to O(n 2 ) possible exct mtches min prmeter of pproch: length of shortest MUM tht should be identified (20 50 bses)

20 Step 2: Chining in MUMmer sort MUMs ccording to position in genome A solve vrition of Longest Incresing Subsequence (LIS) problem to find sequences in scending order in both genomes Figure from: Delcher et l., Nucleic Acids Reserch 27, 1999

21 Finding Longest Subsequence unlike ordinry LIS problems, MUMmer tkes into ccount lengths of sequences represented by MUMs overlps requires O( k log k) time where k is number of MUMs

22 Types of Gps in MUMmer Alignment Figure from: Delcher et l., Nucleic Acids Reserch 27, 1999

23 Step 3: Close the Gps SNPs: between MUMs: trivil to detect otherwise: hndle like repets inserts trnspositions (subsequences tht were deleted from one loction nd inserted elsewhere): look for out-of-sequence MUMs simple insertions: trivil to detect

24 Step 3: Close the Gps polymorphic regions short ones: lign them with dynmic progrmming method long ones: cll MUMmer recursively w/ reduced min MUM length repets detected by overlpping MUMs Figure from: Delcher et l. Nucleic Acids Reserch 27, 1999

25 The LAGAN Method Brudno et l., Genome Reserch, 2003 Given: genomes A nd B nchors = find_nchors(a, B) step 3: finish globl lignment with DP constrined by nchors find_nchors(a, B) step 1: find locl lignments by mtching, chining k-mer seeds step 2: nchors = highest-weight sequence of locl lignments for ech pir of djcent nchors 1, 2 in nchors if 1, 2 re more thn d bses prt A, B = sequences between 1, 2 sub-nchors = find_nchors( A, B ) insert sub-nchors between 1, 2 in nchors return nchors

26 Step 1: Finding Seeds in LAGAN degenerte k-mers: mtching k-long sequences with smll number of mismtches llowed by defult, LAGAN uses 10-mers nd llows 1 mismtch ccg cgcgctct cct ct cgcggtct cgt

27 Finding Seeds in LAGAN exmple: trie to represent ll 3-mers of the sequence gccgcct c g c c g c c g t 2 3, c one sequence is used to build the trie the other sequence (the query) is wlked through to find mtching k-mers

28 Allowing Degenerte Mtches suppose we re llowing 1 bse to mismtch in looking for mtches to the 3-mer cc; need to explore green nodes c g c c g c c g t 2 3, c

29 LAGAN Uses Threded Tries in threded trie, ech lef for word w 1...w p hs bck pointer to the node for w 2...w p c g c c g c c g t 2 3, c

30 Trversing Threded Trie consider trversing the trie to find 3-mer mtches for the query sequence: ccgt c g c c g c c g t 2 3, c usully requires following only two pointers to mtch ginst the next k-mer, insted of trversing tree from root for ech

31 Step 1b: Chining Seeds in LAGAN cn chin seeds s 1 nd s 2 if the indices of s 1 > indices of s 2 (for both sequences) s 1 nd s 2 re ner ech other keep trck of seeds in the serch box s the query sequence is processed Figure from: Brudno et l. BMC Bioinformtics, 2003

32 Step 2: Chining in LAGAN use sprse dynmic progrmming to chin locl lignments

33 The Problem: Find Chin of Locl Alignments (x,y) (x,y ) requires x < x y < y Ech locl lignment hs weight FIND the chin with highest totl weight Slide from Serfim Btzoglou, Stnford University

34 Sprse DP for rectngle chining 1,, N: rectngles h (h j, l j ): y-coordintes of rectngle j w(j): weight of rectngle j l V(j): optiml score of chin ending in j L: list of triplets (l j, V(j), j) y L is sorted by l j : smllest (North) to lrgest (South) vlue L is implemented s blnced binry tree Slide from Serfim Btzoglou, Stnford University

35 Sprse DP for rectngle chining Min ide: Sweep through x- coordintes To the right of b, nything chinble to is chinble to b Therefore, if V(b) > V(), rectngle is useless for subsequent chining In L, keep rectngles j sorted with incresing l j - coordintes sorted with incresing V(j) score V() V(b) Slide from Serfim Btzoglou, Stnford University

36 Sprse DP for rectngle chining Go through rectngle x-coordintes, from lowest to highest: 1. When on the leftmost end of rectngle i: j. j: rectngle in L, with lrgest l j < h i b. V(i) = w(i) + V(j) k i 2. When on the rightmost end of i:. k: rectngle in L, with lrgest l k l i b. If V(i) > V(k): i. INSERT (l i, V(i), i) in L ii. REMOVE ll (l j, V(j), j) with V(j) V(i) & l j l i Slide from Serfim Btzoglou, Stnford University

37 Exmple x : 5 V b c d e b: c: 3 d: 4 e: 2 L l i V(i) i c b d e y 1. When on the leftmost end of rectngle i:. j: rectngle in L, with lrgest l j < h i b. V(i) = w(i) + V(j) Slide from Serfim Btzoglou, Stnford University 2. When on the rightmost end of i:. k: rectngle in L, with lrgest l k l i b. If V(i) > V(k): i. INSERT (l i, V(i), i) in L ii. REMOVE ll (l j, V(j), j) with V(j) V(i) & l j l i

38 Time Anlysis 1. Sorting the x-coords tkes O(N log N) 2. Going through x-coords: N steps 3. Ech of N steps requires O(log N) time: Serching L tkes log N Inserting to L tkes log N All deletions re consecutive, so log N per deletion Ech element is deleted t most once: N log N for ll deletions Recll tht INSERT, DELETE, SUCCESSOR, tke O(log N) time in blnced binry serch tree Slide from Serfim Btzoglou, Stnford University

39 Constrined Dynmic Progrmming if we know tht the i th element in one sequence must lign with the j th element in the other, we cn ignore two rectngles in the DP mtrix j i

40 Step 3: Computing the Globl Alignment in LAGAN given n nchor tht strts t (i, j) nd ends t (i, j ), LAGAN limits the DP to the unshded regions thus nchors re somewht flexible Figure from: Brudno et l. Genome Reserch, 2003

41 Step 3: Computing the Globl Alignment in LAGAN Figures from: Brudno et l. Genome Reserch, 2003

42 Exmple Alignment: E. Coli O157:H7 vs. E. coli K-12 Figure from: Pern et l. Nture, 2001