Lecture 10: Phylogeny


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1 Computational Genomics Prof. Ron Shamir & Prof. Roded Sharan School of Computer Science, Tel Aviv University גנומיקה חישובית פרופ' רון שמיר ופרופ' רודד שרן ביה"ס למדעי המחשב,אוניברסיטת תל אביב Lecture 10: Phylogeny 25,27/12/12
2 Phylogeny Slides: Adi Akavia Nir Friedman s slides at HUJI (based on ALGMB 98) Anders Gorm Pedersen,Technical University of Denmark Sources: Joe Felsenstein Inferring Phylogenies (2004) 1
3 Phylogeny Phylogeny: the ancestral relationship of a set of species. Represented by a phylogenetic tree leaf branch?? Internal node??? Leaves  contemporary Internal nodes  ancestral Branch length  distance between sequences 2
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8 Phylogeny schools classical Classical vs. Modern: Classical  morphological characters Modern  molecular sequences. 7
9 Trees and Models rooted / unrooted topology / distance binary / general molecular clock 8
10 To root or not to root? Unrooted tree: phylogeny without direction. 9
11 Rooting an Unrooted Tree We can estimate the position of the root by introducing an outgroup: a species that is definitely most distant from all the species of interest Proposed root Falcon Aardvark Bison Chimp Dog Elephant 10
12 A Scally et al. Nature 483, (2012) doi: /nature10842
13 HOW DO WE FIGURE OUT THESE TREES? TIMES? 12
14 Dangers of Paralogs Right species distance: (1,(2,3)) Gene Duplication Sequence Homology Caused By: Orthologs  speciation, Paralogs  duplication Xenologs  horizontal (e.g., by virus) Speciation events 1A 2A 3A 3B 2B 1B 13
15 Dangers of Paralogs Right species distance: (1,(2,3)) If we only consider 1A, 2B, and 3A: ((1,3),2) Gene Duplication Speciation events 1A 2A 3A 3B 2B 1B 14
16 Type of Data Distancebased Input: matrix of distances between species Distance can be fraction of residue they disagree on, alignment score between them, Characterbased Examine each character (e.g., residue) separately 15
17 Distance Based Methods 16
18 Tree based distances d(i,j)=sum of arc lengths on the path i j Given d, can we find an exactly matching tree? An approximately matching tree? 17
19 The Problem Input: matrix d of distances between species Goal: Find a tree with leaves=chars and edge distances, matching d best. Quality measure: sum of squares: SSQ( T ) i j i t ij : distance in the tree = w ( d ij ij t ij 2 ) w ij : pair weighting. Options: (1) 1 (2) 1/d ij (3) 1/d ij 2 NPhard (Day 86). We ll describe common heuristics 18
20 UPGMA Clustering (Sokal & Michener 1958) (Unweighted pairgroup method with arithmetic mean) Approach: Form a tree; closer species according to input distances should be closer in the tree Build the tree bottom up, each time merging two smaller trees All leaves are at same distance from the root 19
21 Efficiency lemma Approach: gradually form clusters: sets of species Repeatedly identify two clusters and merge them. For clusters C i C j, define the distance between them to be the average dist betw their members: 1 d ( C = i, C j ) d ( p, q) C C i p C i q C j Lemma: If C k is formed by merging C i and C j then for every other cluster C l d(c k,c l ) = ( C i *d(c i,c l ) + C j *d(c j,c l )) / ( C i + C j ) Can update distances between clusters in time prop. to the number of clusters. j 20
22 UPGMA algorithm Initialize: each node is a cluster Ci={i}. d(ci,cj)=d(i,j) set height(i) = 0 i Iterate: Find Ci,Cj with smallest d(ci,cj) Introduce a new cluster node Ck that replaces Ci and Cj //Ck represents all the leaves in clusters Ci and Cj Introduce a new tree node Aij with height(aij) = d(ci,cj)/2 // d(ci,cj) is the average dist among leaves of Ci and Cj Connect the corresponding tree nodes Ci,Cj to Aij with length(ci, Aij) = height(aij)  height(ci) length(cj,aij) = height(aij)  height(cj) For all other Cl: d(ck,cl) = ( Ci *d(ci,cl) + Cj *d(cj,cl)) / ( Ci + Cj ) //dist to any old cluster is the ave dist between its leaves and leaves in Ci, Cj Time: Naïve: O(n 3 ); Can show O(n 2 logn) (ex.) and O(n 2 ) (harder ex.) 21
23 UPGMA alg (2) Orange nodes represent the groups of nodes that they replaced, and maintain the average dist of the set from other leaf nodes/clusters A h 123 A 123 h 12 A 12 h 12 A C 12 C 45 C 123 C 45 1,
24 UPGMA Algorithm
25 UPGMA Algorithm
26 UPGMA Algorithm
27 UPGMA Algorithm
28 UPGMA Algorithm
29 Molecular Clock UPGMA implicitly assumes the tree is ultrametric: equal leafroot distances => common uniform clock Works reasonably well for nearby species 28
30 29 Additivity A weaker additivity assumption: distances between species are the sum of distances between intermediate nodes (even if the tree is not ultrametric) a b c i j k c b k j d c a k i d b a j i d + = + = + = ), ( ), ( ), (
31 30 Consequences of Additivity Suppose input distances are additive For any three leaves Thus a b c i j k c b k j d c a k i d b a j i d + = + = + = ), ( ), ( ), ( m )), ( ), ( ), ( ( ), ( j i d k j d k i d 2 1 m k d + =
32 31 Suppose input distances are additive For any four leaves Induced tree wt: W Largest sum must appear twice Consequences of Additivity II b W l j d k i d b W k j d l i d b W l k d j i d = + + = + + = + ), ( ), ( ), ( ), ( ), ( ), ( a b c i j k l d e
33 Consequences of Additivity III If we can identify neighbor leaves, then can use pairwise distances to reconstruct the tree: Remove neighbors i, j from the leaf set Add k i Set d km = (d im + d jm d ij )/2 m But can we find neighbor leaves? k j b 32
34 Closest pairs may not be neighbors! Closest pair: kj even though they are not neighbors k j 4 4 i l 33
35 Neighbor Joining (SaitouNei 87) Let where D i, j ) = d ( i, j ) ( r i + r ) ( j 1 r = i d( i, k) L 2 k Corrected average distance of i from all other nodes Theorem: if D(i,j) is minimum among all pairs of leaves, then i and j are neighbors in the tree 34
36 35 Neighbor Joining Set L to contain all leaves Iteration: Choose i,j such that D(i,j) is minimum Create new node k, and set remove i,j from L, and add k Update r, D Termination: when L =2 connect the two nodes 2 )) /, ( ), ( ), ( ( ), ( 2 ) / ), ( ( ), ( 2 ) / ), ( ( ), ( j i d m j d m i d m k d r r j i d k j d r r j i d k i d i j j i + = + = + = Thm: Opt tree guaranteed if distances match a tree Does not assume a clock Time: O(n 3 ) = k i k i d L r ), ( 2 1 ) ( ), ( ), ( j r i r j i d j i D + = j m i k
37 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B C D A B C  14 D 
38 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B C D A B C  14 D  i u i A ( )/2=32. 5 B ( )/2=23. 5 C ( )/2=23. 5 D ( )/2=29. 5
39 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B C D A B C  14 D  A B C D A B C D  i u i A ( )/2=32. 5 B ( )/2=23. 5 C ( )/2=23. 5 D ( )/2=29. 5 D ij  u i  u j
40 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B C D A B C  14 D  A B C D A B C D  i u i A ( )/2=32. 5 B ( )/2=23. 5 C ( )/2=23. 5 D ( )/2=29. 5 D ij  u i  u j
41 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B C D A B C  14 D  A B C D A B C D  i u i A ( )/2=32. 5 B ( )/2=23. 5 C ( )/2=23. 5 D ( )/2=29. 5 C D X D ij  u i  u j
42 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B C D A B C  14 D  A B C D A B C D  D ij  u i  u j i u i A ( )/2=32. 5 B ( )/2=23. 5 C ( )/2=23. 5 D ( )/2=29. 5 C 4 10 X D v C = 0. 5 x x ( ) = 4 v D = 0. 5 x x ( ) = 10
43 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B C D X A B C  14 D  X  C 4 10 X D
44 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B C D X A B C  14 D  X  D XA = ( D CA + D DA  D CD )/2 = ( )/2 = 17 D XB = ( D CB + D DB  D CD )/2 = ( )/2 = 8 C 4 10 X D
45 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B C D X A B C  14 D  X  D XA = ( D CA + D DA  D CD )/2 = ( )/2 = 17 D XB = ( D CB + D DB  D CD )/2 = ( )/2 = 8 C 4 10 X D
46 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B X A B  8 X  D XA = ( D CA + D DA  D CD )/2 = ( )/2 = 17 D XB = ( D CB + D DB  D CD )/2 = ( )/2 = 8 C 4 10 X D
47 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B X A B  8 X  i C 4 10 X D u i A ( 17+17)/1 = 34 B ( 17+8) /1 = 25 X ( 17+8) /1 = 25
48 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B X A B  8 X  A B X A B X  i C 4 10 X D u i A ( 17+17)/1 = 34 B ( 17+8) /1 = 25 X ( 17+8) /1 = 25 D ij  u i  u j
49 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B X A B  8 X  A B X A B X  i C 4 10 X D u i A ( 17+17)/1 = 34 B ( 17+8) /1 = 25 X ( 17+8) /1 = 25 D ij  u i  u j
50 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B X A B  8 X  A B X A B X  D ij  u i  u j i C 4 10 X D u i A ( 17+17)/1 = 34 B ( 17+8) /1 = 25 X ( 17+8) /1 = 25 v A = 0. 5 x x ( 3425) = 13 v D = 0. 5 x x ( 2534) = 4 A 13 Y 4 B
51 Neighbor Joining Algorithm A B X Y CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B  8 X  Y C 4 10 X D A 13 Y 4 B
52 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B X Y A B  8 X  4 Y D YX = ( D AX + D BX  D AB )/2 = ( )/2 = 4 C 4 10 X D A 13 Y 4 B
53 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS X Y X  4 Y  D YX = ( D AX + D BX  D AB )/2 = ( )/2 = 4 C 4 10 X D A 13 Y 4 B
54 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS X Y X  4 Y  D YX = ( D AX + D BX  D AB )/2 = ( )/2 = 4 C 4 10 D A 13 4 B 4
55 Neighbor Joining Algorithm CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS A B C D A B C  14 D  D 10 C B 13 A
56 Division of Population Genetics, National Institute of Genetics & Department of Genetics, Graduate University for Advanced Studies (Sokendai) Mishima, , Japan Naruya Saitou
57 Probabilistic approaches 56
58 Likelihood of a Tree Given: n aligned sequences M= X 1,,X n A tree T, leaves labeled with X 1,,X n reconstruction t: labeling of internal nodes branch lengths Goal: Find optimal reconstruction t* : One maximizing the likelihood P(M T,t*) 57
59 Likelihood (2) We need a model for computing P(M T,t*) Assumptions: Each character is independent The branching is a Markov process: The probability that a node x has a specific label is only a function of the parent node y and the branch length t between them. The probabilities P(x y,t) are known 58
60 59 Example x 1 x 2 x 3 x 4 x 5 t 1 t 2 t 3 t 5 ) ( ), ( ), ( ), ( ), ( *),,..., ( x P t x x P t x x P t x x P t x x P t T x x P =
61 = Calculating the Likelihood Assume that the branch lengths t uv are known. P(M T, character t*) = character j reconstrcuction P( root) P( M j j reconstrcuction R edge u v R, R T, t*) p u v ( t uv ) Independence of sites Markov property independence of each branch 60
62 Additional Assumed Properties Additivity: P ( a b, t) P( b c, s) = P( a c, s + t) b Reversibility (symmetry): q P( a b, t) q P( b a, t) b = Allows one to freely move the root a Provable under broad and reasonable assumptions 61
63 JukesCantor Model (JK 69) Assumptions: Each base in sequence has equal chance of changing Changes to other 3 bases with equal probability Characteristics Each base appears with equal frequency in DNA The quantity r is the rate of change NOTE: r is not the rate of mutation, it is the rate of substitution event of a nucleotide throughout a population. During each infinitesimal time t a substitution occurs with probability 3r t 62
64 a=r 63
65 JukesCantor Model P A(t) : prob of A in the character at time t Discrete case: Prob. change A B in one time unit is r P A(t+1) = (13r) P A(t) + r (1 P A(t) ) P A(t+1)  P A(t) = 3r P A(t) + r (1 P A(t) ) Continuous case: P A(t) = 3r P A(t) + r (1 P A(t) )= 4r P A(t) + r Solution to the differential equation: P A(t) = 1 4 rt (1 + 3e 4 ) 64
66 prob. that the nucleotide remains unchanged 1 rt over t time units: P 3 4 same = + e 4 4 Probability of specific change: Probability of change: Note: For t P change 3 4 P A B = 1 1 4rt 4 e rt P change = e
67 Other Models Kimura s 2parameter model: A,G  purines; C,T  pyrmidines Two different rates purinepurine or pyrmidinepyrimidine (transitions) purinepyrmidine or pyrmidinepurine (transversions) Felsenstein 84 and Yano, Hasegawa & Kishino 85 extend the Kimura model to asymmetric base frequencies. C T A G 66
68 Charles Cantor Boston University Professor, Biomedical Engineering Professor of Pharmacology, School of Medicine Center for Advanced Biotechnology Ph.D., Biophysical Chemistry, University of California, Berkeley Dr. Cantor is currently Chief Scientific Officer at Sequenom, Inc in San Diego, California. 67
69 Efficient Likelihood Calculation (Felsenstein 73) Use dynamic programming DefineS j (a,v) = Pr(subtree rooted in v v j = a) Initialization: leafv set S j (a,v) = 1 if v is labeled by a, else S j (a,v) = 0 Recursion: Traverse the tree in postorder: for each node v with children u and w, for each state x S j ( x, v) Final Soln: = S j ( y, u) px y ( tvu ) S y y j ( y, w) p L = S j ( x, root) P( x) j x x y ( t vw ) Complexity: O(nmk 2 ) n species, m chars, k states 68
70 69
71 Finding Optimal Branch lengths Optimizing the length of a single rootchild branch rv can be done using standard optimization techniques log L m = log j= 1 x, y p( x) S" j ( x, r) p x y ( t rv ) S j ( y, v) Under the symmetry assumption, each node can be made (temporarily) the root To heuristically optimize all the branch lengths: repeatedly optimize one branch at a time No guaranteed convergence, but often works 70
72 Character Based Methods 71
73 Inferring a Phylogenetic Tree Generic problem: Optimal Phylogenetic Tree: Input: n species, set of characters, for each species, the state of each of the characters. (parameters) Goal: find a fullylabeled phylogenetic tree that best explains the data. (maximizes a target function). Assumptions: characters  mutually independent species evolve independently A: CAGGTA B: CAGACA C: CGGGTA D: TGCACT E: TGCGTA A B C D E 72
74 A Simple Example Five species, three have C and two T at a specified position. A minimal tree has one evolutionary change: C T C C T C C T T C 73
75 Inferring a Phylogenetic Tree Naive Solution  Enumeration: No. of nonisomorphic, labeled, binary, unrooted trees, containing n leaves: (2n 5)!! = Π i=3 n (2i5) Rooted: x(2n3) for n=20 this is 10 21! n leaves n2 internal nodes 2n3 edges Pf: induction on n a b Each new species adds 2 new edges a b c a c b b c a Adding 3rd species 74
76 Parsimony Goal: explain data with min. no. of evolutionary changes ( mutations, or mismatches) Parsimony: S(T) Σ j Σ {v, u} E(T) {j: v j u j } Small parsimony problem : Input: leaf sequences + a leaflabeled tree T Goal: Find ancestral sequences implying minimum no. of changes (most parsimonious) Large parsimony problem : Input: leaf sequences Gaol: Find a most parsimonious tree (topology, leaf labeling and internal seqs.) 75
77 Algorithm for the Small Parsimony Problem (Fitch `71) Initialization: scan T in postorder, assign: leaf vertex m: S m ={state at node m} internal node m with children l, r : S m = S l S r if S l S r =φ (i) S l S r o/w (ii) Solution Construction: scan tree in preorder, choose: for the root choose x S root at node m with child k: pick same state as k if possible; o/w  pick arbitrarily cost = no. of case (i) events. time: O(n k) (k = #states s) 76
78 Fitch s Alg T T,C T >C C A,T T >Α T C C A T T T 77
79 Walter Fitch May 21, March 10, 2011 One of the most influential evolutionary biologists in the world, who established a new scientific field: molecular phylogenetics. He was a member in the National Academy of Sciences, the American Academy of Arts and Sciences and the American Philosophical Society. He cofounded and was the first president of the Society for Molecular Biology, which established the annually awarded Fitch Prize. Additionally, he was a founding editor of Molecular Biology and Evolution. Fitch was at the University of California, Irvine, until his death, preceded by three years at the University of Southern California and 24 years University of Wisconsin Madison. 78
80 Weighted Parsimony k states S 1,, S k C ij = cost of changing from state i to j Algorithm (SankoffCedergren 88): Need: S k (x) best cost for the subtree rooted at x if state at x is k For leaf x, S k (x) = 0 if state of x is k o/w Scan T in postorder. At node a with children l, r S k (a) = min m (S m (l)+ C mk ) + min m (S m (r)+ C mk ) Opt=min m (S m (root)) time: O(n k 2 ) 79
81 C G C C 80
82 David Sankoff Over the past 30 years, Sankoff formulated and contributed to many of the fundamental problems in computational biology. In sequence comparison, he introduced the quadratic version of the NeedlemanWunsch algorithm, developed the first statistical test for alignments, initiated the study of the limit behavior of random sequences with Vaclav Chvatal and described the multiple alignment problem, based on minimum evolution over a phylogenetic tree. In the study of RNA secondary structure, he developed algorithms based on general energy functions for multiple loops and for simultaneous folding and alignment, and performed the earliest studies of parametric folding and automated phylogenetic filtering. Sankoff and Robert Cedergren collaborated on the first studies of the evolution of the genetic code based on trna sequences. His contributions to phylogenetics include early models for horizontal transfer, a general approach for optimizing the nodes of a given tree, a method for rapid bootstrap calculations, a generalization of the nearest neighbor interchange heuristic, various constraint, consensus and supertree problems, the computational complexity of several phylogeny problems with William Day, and a general technique for phylogenetic invariants with Vincent Ferretti. Over the last fifteen years he has focused on the evolution of genomes as the result of chromosomal rearrangement processes. Here he introduced the computational analysis of genomic edit distances, including parametric versions, the distribution of gene numbers in conserved segments in a random model with Joseph Nadeau, phylogeny based on gene order with Mathieu Blanchette and David Bryant, generalizations to include multigene families, including algorithms for analyzing genome duplication and hybridization with Nadia ElMabrouk, and the statistical analysis of gene clusters with Dannie Durand. Sankoff is also well known in linguistics for his methods of studying grammatical variation and 81 change in speech communities, the quantification of discourse analysis and production models of bilingual speech.
83 Large Parsimony Problem Input: n x m matrix M: M ij = state of j th character of species i. M i = label of i (all labels are distinct) Goal: Construct a phylogenetic tree T with n leaves and a label for each node, s.t. 11 correspondence of leaves and labels cost of tree is minimum. NPhard 82
84 Branch & Bound (HendyPenny 89) enumerate all unrooted trees with increasing no. of leaves Note: cost of tree with all leaves cost of subtree with some leaves pruned (and same labeling) => If cost of subtree best cost for full tree so far, then: can prune (ignore) all refinements of the subtree. enumeration & pruning can be done in O(1) time per visited subtree. 83
85 Branch swapping C D C D A B B A Each internal edge defines 4 subtrees: Can swap two such nonadjacent subtrees 84
86 Nearest Neighbor Interchanges handles nlabeled trees T and T are neighbors if one can get T by following operation on T: S R S R U V S U V U R V SV RU SU RV SR UV Use the neighborhood structure on the set of solutions (all trees) via hill climbing, annealing, other heuristics... 85
87 NN interchange graph 86
88 Wilson & Cann, Sci Amer 92 87
89 FIN 88
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