1. CS262 Lecture 9, Win07, Batzoglou Real-world protein aligners MUSCLE  High throughput  One of the best in accuracy ProbCons  High accuracy  Reasonable speed

2. CS262 Lecture 9, Win07, Batzoglou MUSCLE at a glance 1.Fast measurement of all pairwise distances between sequences D DRAFT (x, y) defined in terms of # common k-mers (k~3) – O(N 2 L logL) time 2.Build tree T DRAFT based on those distances, with UPGMA 3.Progressive alignment over T DRAFT, resulting in multiple alignment M DRAFT Only perform alignment steps for the parts of the tree that have changed 4.Measure new Kimura-based distances D(x, y) based on M DRAFT 5.Build tree T based on D 6.Progressive alignment over T, to build M 7.Iterative refinement; for many rounds, do: Tree Partitioning: Split M on one branch and realign the two resulting profiles If new alignment M’ has better sum-of-pairs score than previous one, accept

3. CS262 Lecture 9, Win07, Batzoglou PROBCONS at a glance 1.Computation of all posterior matrices M xy : M xy (i, j) = Prob(x i ~ y j ), using a HMM 2.Re-estimation of posterior matrices M’ xy with probabilistic consistency M’ xy (i, j) = 1/N  sequence z  k M xz (i, k)  M yz (j, k);M’ xy = Avg z (M xz M zy ) 3.Compute for every pair x, y, the maximum expected accuracy alignment A xy : alignment that maximizes  aligned (i, j) in A M’ xy (i, j) Define E(x, y) =  aligned (i, j) in Axy M’ xy (i, j) 4.Build tree T with hierarchical clustering using similarity measure E(x, y) 5.Progressive alignment on T to maximize E(.,.) 6.Iterative refinement; for many rounds, do: Randomized Partitioning: Split sequences in M in two subsets by flipping a coin for each sequence and realign the two resulting profiles

4. CS262 Lecture 9, Win07, Batzoglou Some Resources Genome Resources Annotation and alignment genome browser at UCSC http://genome.ucsc.edu/cgi-bin/hgGateway Specialized VISTA alignment browser at LBNL http://pipeline.lbl.gov/cgi-bin/gateway2 ABC—Nice Stanford tool for browsing alignments http://encode.stanford.edu/~asimenos/ABC/ Protein Multiple Aligners http://www.ebi.ac.uk/clustalw/ CLUSTALW – most widely used http://phylogenomics.berkeley.edu/cgi-bin/muscle/input_muscle.py MUSCLE – most scalable http://probcons.stanford.edu/ PROBCONS – most accurate

5. CS262 Lecture 9, Win07, Batzoglou Rapid Global Alignments How to align genomic sequences in (more or less) linear time

6. CS262 Lecture 9, Win07, Batzoglou

7. Motivation Genomic sequences are very long:  Human genome = 3 x 10 9 –long  Mouse genome = 2.7 x 10 9 –long Aligning genomic regions is useful for revealing common gene structure  It is useful to compare regions > 1,000,000-long

8. CS262 Lecture 9, Win07, Batzoglou The UCSC Browser http://genome.ucsc.edu/cgi-bin/hgGateway

9. CS262 Lecture 9, Win07, Batzoglou Main Idea Genomic regions of interest contain islands of similarity, such as genes 1.Find local alignments 2.Chain an optimal subset of them 3.Refine/complete the alignment Systems that use this idea to various degrees: MUMmer, GLASS, DIALIGN, CHAOS, AVID, LAGAN, TBA, & others

10. CS262 Lecture 9, Win07, Batzoglou Saving cells in DP 1.Find local alignments 2.Chain -O(NlogN) L.I.S. 3.Restricted DP

11. CS262 Lecture 9, Win07, Batzoglou Methods to CHAIN Local Alignments Sparse Dynamic Programming O(N log N)

12. CS262 Lecture 9, Win07, Batzoglou The Problem: Find a Chain of Local Alignments (x,y)  (x’,y’) requires x < x’ y < y’ Each local alignment has a weight FIND the chain with highest total weight

13. CS262 Lecture 9, Win07, Batzoglou Quadratic Time Solution Build Directed Acyclic Graph (DAG):  Nodes: local alignments [(x a,x b )  (y a,y b )] & score  Directed edges: local alignments that can be chained edge ( (x a, x b, y a, y b ), (x c, x d, y c, y d ) ) x a < x b < x c < x d y a < y b < y c < y d Each local alignment is a node v i with alignment score s i

14. CS262 Lecture 9, Win07, Batzoglou Quadratic Time Solution Initialization: Find each node v a s.t. there is no edge (u, v a ) Set score of V(a) to be s a Iteration: For each v i, optimal path ending in v i has total score: V(i) = ma x j s.t. there is edge (v j, v i ) ( weight(v j, v i ) + V(j) ) Termination: Optimal global chain: j = argmax ( V(j) ); trace chain from v j Worst case time: quadratic

15. CS262 Lecture 9, Win07, Batzoglou Sparse Dynamic Programming Back to the LCS problem: Given two sequences  x = x 1, …, x m  y = y 1, …, y n Find the longest common subsequence  Quadratic solution with DP How about when “hits” x i = y j are sparse?

16. CS262 Lecture 9, Win07, Batzoglou Sparse Dynamic Programming 15324162042431118 4 20 24 3 11 15 11 4 18 20 Imagine a situation where the number of hits is much smaller than O(nm) – maybe O(n) instead

17. CS262 Lecture 9, Win07, Batzoglou Sparse Dynamic Programming – L.I.S. Longest Increasing Subsequence Given a sequence over an ordered alphabet  x = x 1, …, x m Find a subsequence  s = s 1, …, s k  s 1 < s 2 < … < s k

18. CS262 Lecture 9, Win07, Batzoglou Sparse Dynamic Programming – L.I.S. Let input be w: w 1,…, w n INITIALIZATION: L:last LIS elt. array L[0] = -inf L[1] = w 1 L[2…n] = +inf B:array holding LIS elts; B[0] = 0 P:array of backpointers // L[j]: smallest j th element w i of j-long LIS seen so far ALGORITHM for i = 2 to n { Find j such that L[j – 1] < w[i] ≤ L[j] L[j]  w[i] B[j]  i P[i]  B[j – 1] } That’s it!!! Running time?

19. CS262 Lecture 9, Win07, Batzoglou Sparse LCS expressed as LIS Create a sequence w Every matching point (i, j), is inserted into w as follows: For each column j = 1…m, insert in w the points (i, j), in decreasing row i order The 11 example points are inserted in the order given a = (y, x), b = (y’, x’) can be chained iff  a is before b in w, and  y < y’ 15324162042431118 6 4 27 18 10 9 5 11 3 4 20 24 3 11 15 11 4 18 20 x y

20. CS262 Lecture 9, Win07, Batzoglou Sparse LCS expressed as LIS Create a sequence w w = (4,2) (3,3) (10,5) (2,5) (8,6) (1,6) (3,7) (4,8) (7,9) (5,9) (9,10) Consider now w’s elements as ordered lexicographically, where (y, x) < (y’, x’) if y < y’ Claim: An increasing subsequence of w is a common subsequence of x and y 15324162042431118 6 4 27 18 10 9 5 11 3 4 20 24 3 11 15 11 4 18 20 x y

21. CS262 Lecture 9, Win07, Batzoglou Sparse Dynamic Programming for LIS Example: w = (4,2) (3,3) (10,5) (2,5) (8,6) (1,6) (3,7) (4,8) (7,9) (5,9) (9,10) L = [L1] [L2] [L3] [L4] [L5] … 1.(4,2) 2.(3,3) 3.(3,3) (10,5) 4.(2,5) (10,5) 5.(2,5) (8,6) 6.(1,6) (8,6) 7.(1,6) (3,7) 8.(1,6) (3,7) (4,8) 9.(1,6) (3,7) (4,8) (7,9) 10.(1,6) (3,7) (4,8) (5,9) 11.(1,6) (3,7) (4,8) (5,9) (9,10) Longest common subsequence: s = 4, 24, 3, 11, 18 15324162042431118 6 4 27 18 10 9 5 11 3 4 20 24 3 11 15 11 4 18 20 x y

22. CS262 Lecture 9, Win07, Batzoglou Sparse DP for rectangle chaining 1,…, N: rectangles (h j, l j ): y-coordinates of rectangle j w(j):weight of rectangle j V(j): optimal score of chain ending in j L: list of triplets (l j, V(j), j)  L is sorted by l j : smallest (North) to largest (South) value  L is implemented as a balanced binary tree y h l

23. CS262 Lecture 9, Win07, Batzoglou Sparse DP for rectangle chaining Main idea: Sweep through x- coordinates To the right of b, anything chainable to a is chainable to b Therefore, if V(b) > V(a), rectangle a is “useless” for subsequent chaining In L, keep rectangles j sorted with increasing l j - coordinates  sorted with increasing V(j) score V(b) V(a)

24. CS262 Lecture 9, Win07, Batzoglou Sparse DP for rectangle chaining Go through rectangle x-coordinates, from lowest to highest: 1.When on the leftmost end of rectangle i: a.j: rectangle in L, with largest l j < h i b.V(i) = w(i) + V(j) 2.When on the rightmost end of i: a.k: rectangle in L, with largest l k  l i b.If V(i) > V(k): i.INSERT (l i, V(i), i) in L ii.REMOVE all (l j, V(j), j) with V(j)  V(i) & l j  l i i j k Is k ever removed?

25. CS262 Lecture 9, Win07, Batzoglou Example x y a: 5 c: 3 b: 6 d: 4 e: 2 2 5 6 9 10 11 12 14 15 16 1.When on the leftmost end of rectangle i: a.j: rectangle in L, with largest l j < h i b.V(i) = w(i) + V(j) 2.When on the rightmost end of i: a.k: rectangle in L, with largest l k  l i b.If V(i) > V(k): i.INSERT (l i, V(i), i) in L ii.REMOVE all (l j, V(j), j) with V(j)  V(i) & l j  l i abcde V 5 L lili V(i) i 5 5 a 8 11 8 c 12 9 11 b 15 12 d 13 16 13 3

26. CS262 Lecture 9, Win07, Batzoglou Time Analysis 1.Sorting the x-coords takes O(N log N) 2.Going through x-coords: N steps 3.Each of N steps requires O(log N) time: Searching L takes log N Inserting to L takes log N All deletions are consecutive, so log N per deletion Each element is deleted at most once: N log N for all deletions Recall that INSERT, DELETE, SUCCESSOR, take O(log N) time in a balanced binary search tree

27. CS262 Lecture 9, Win07, Batzoglou Examples Human Genome Browser ABC