Lecture: Bioinformatics

Transcription

1 Lecture: Bioinformatics ENS Sacley, 2018 Some slides graciously provided by Daniel Huson & Celine Scornavacca

2 Phylogenetic Trees - Motivation 2 / 31

3 2 / 31 Phylogenetic Trees - Motivation Motivation - study relation between species - evolution of characteristics - co-evolution (host-parasite) - geological migration - genetic development of viruses/diseases

4 2 / 31 Phylogenetic Trees - Motivation Motivation - study relation between species - evolution of characteristics - co-evolution (host-parasite) - geological migration - genetic development of viruses/diseases Evolution genetic material changes over time new species branch off tree of life

5 Phylogenetic Trees - Motivation 2 / 31

6 Phylogenetic Trees - Motivation 2 / 31

7 3 / 31 Rooted Phylogenetic Trees evolution of species over time, leaves extant, hypothetical ancestors possibly branch lengths (time) Notation taxon, cluster, triplet

8 3 / 31 Rooted Phylogenetic Trees evolution of species over time, leaves extant, hypothetical ancestors possibly branch lengths (time) Notation taxon, cluster, triplet

9 3 / 31 Rooted Phylogenetic Trees evolution of species over time, leaves extant, hypothetical ancestors possibly branch lengths (time) Notation taxon, cluster, triplet

10 3 / 31 Rooted Phylogenetic Trees evolution of species over time, leaves extant, hypothetical ancestors possibly branch lengths (time) Notation taxon, cluster, triplet

11 Rooted Phylogenetic Trees evolution of species over time, leaves extant, hypothetical ancestors possibly branch lengths (time) Exercise Time Notation Polytomies taxon, cluster, triplet history not clear soft known fan out hard Exercise: use xy z LCA(xz)=LCA(yz)>LCA(xy) to prove ab c + bc d ac d 3 / 31

12 4 / 31 Unrooted Phylogenetic Trees similarity between genomes, leaves extant, internal vertices have no meaning possibly branch lengths (amount of change) Notation taxon, split, quartet

13 4 / 31 Unrooted Phylogenetic Trees similarity between genomes, leaves extant, internal vertices have no meaning possibly branch lengths (amount of change) Notation taxon, split, quartet

14 4 / 31 Unrooted Phylogenetic Trees similarity between genomes, leaves extant, internal vertices have no meaning possibly branch lengths (amount of change) Notation taxon, split, quartet

15 4 / 31 Unrooted Phylogenetic Trees similarity between genomes, leaves extant, internal vertices have no meaning possibly branch lengths (amount of change) Notation taxon, split, quartet

16 4 / 31 Unrooted Phylogenetic Trees similarity between genomes, leaves extant, internal vertices have no meaning possibly branch lengths (amount of change) Notation taxon, split, quartet

17 5 / 31 Reconstructing Phylogenetic Trees Group Species By... - morphology - behavior - geography Diptera = two wings

18 5 / 31 Reconstructing Phylogenetic Trees Group Species By... - morphology - behavior - geography - distance of sequences - genetic distance - etc. Diptera = two wings

19 6 / 31 Reconstructing Phylogenetic Trees Vertebrata has backbone

20 6 / 31 Reconstructing Phylogenetic Trees Vertebrata has backbone Tetrapoda has 4 legs

21 6 / 31 Reconstructing Phylogenetic Trees Vertebrata has backbone Tetrapoda has 4 legs Mammalia breast feeding

22 6 / 31 Reconstructing Phylogenetic Trees Vertebrata has backbone Tetrapoda has 4 legs Salientia can leap Mammalia breast feeding

23 6 / 31 Reconstructing Phylogenetic Trees Vertebrata has backbone Tetrapoda has 4 legs Salientia can leap Mammalia breast feeding Elephantidae elephants

24 6 / 31 Reconstructing Phylogenetic Trees Vertebrata has backbone Tetrapoda has 4 legs Salientia can leap Mammalia breast feeding Elephantidae elephants Loxodonta africana African Elephant Elephas maximus Asian Elephant

25 Reconstructing Phylogenetic Trees Tetrapoda has 4 legs Vertebrata has backbone Keep in Mind Automatic reconstruction should be - fast (deal with tons of species & genes) - consistent (optimal data evolution correctly reflected) - non-arbitrary Salientia can leap Mammalia breast feeding Elephantidae elephants Loxodonta africana African Elephant Elephas maximus Asian Elephant 6 / 31

26 Distance-Based Reconstruction Idea: cluster hierarchically 7 / 31

27 7 / 31 Distance-Based Reconstruction Idea: cluster hierarchically

28 7 / 31 Distance-Based Reconstruction Idea: cluster hierarchically Idea: merge closest clusters

29 7 / 31 Distance-Based Reconstruction Idea: cluster hierarchically Idea: merge closest clusters

30 7 / 31 Distance-Based Reconstruction Idea: cluster hierarchically Idea: merge closest clusters

31 Distance-Based Reconstruction Idea: cluster hierarchically Idea: merge closest clusters branch lengths?? assume molecular clock ultrametric 7 / 31

32 Distance-Based Reconstruction Idea: cluster hierarchically Idea: merge closest clusters update matrix d X Y,Z = X d X,Z + Y d Y,Z X + Y branch lengths?? assume molecular clock ultrametric / 31

33 Distance-Based Reconstruction Idea: cluster hierarchically Idea: merge closest clusters update matrix d X Y,Z = X d X,Z + Y d Y,Z X + Y branch lengths?? assume molecular clock ultrametric / 31

34 Distance-Based Reconstruction Idea: cluster hierarchically Idea: merge closest clusters update matrix d X Y,Z = X d X,Z + Y d Y,Z X + Y branch lengths?? assume molecular clock ultrametric / 31

35 Distance-Based Reconstruction Idea: cluster hierarchically 6 28/3 28/3 Idea: merge closest clusters update matrix d X Y,Z = X d X,Z + Y d Y,Z X + Y branch lengths?? assume molecular clock ultrametric / 31

36 Distance-Based Reconstruction Idea: cluster hierarchically 6 28/3 28/3 Idea: merge closest clusters update matrix d X Y,Z = X d X,Z + Y d Y,Z X + Y branch lengths?? assume molecular clock ultrametric 7 / 31

37 Distance-Based Reconstruction Idea: cluster hierarchically 28/3 Idea: merge closest clusters update matrix d X Y,Z = X d X,Z + Y d Y,Z X + Y branch lengths?? assume molecular clock ultrametric 7 / 31

38 Distance-Based Reconstruction Idea: cluster hierarchically 28/3 5/3 Idea: merge closest clusters 5/3 update matrix d X Y,Z = X d X,Z + Y d Y,Z X + Y branch lengths?? assume molecular clock ultrametric 7 / 31

39 Distance-Based Reconstruction Idea: cluster hierarchically Unweighted Pair Group Method w/ Avg. - find closest pair - join them - update distances & recurse 28/3 5/3 Idea: merge closest clusters 5/3 update matrix d X Y,Z = X d X,Z + Y d Y,Z X + Y branch lengths?? assume molecular clock ultrametric 7 / 31

40 Distance-Based Reconstruction Idea: cluster hierarchically Unweighted Pair Group Method w/ Avg. - find closest pair - join them - update distances & recurse /3 Idea: merge closest clusters 5/3 update matrix d X Y,Z = X d X,Z + Y d Y,Z X + Y branch lengths?? assume molecular clock ultrametric 7 / 31

41 Distance-Based Reconstruction Idea: cluster hierarchically Unweighted Pair Group Method w/ Avg. - find closest pair - join them - update distances & recurse Idea: merge closest clusters update matrix d X Y,Z = X d X,Z + Y d Y,Z X + Y Exercise Time branch lengths?? assume molecular clock ultrametric 7 / 31

42 Distance-Based Reconstruction Idea: cluster hierarchically Unweighted Pair Group Method w/ Avg. - find closest pair - join them - update distances & recurse Idea: merge closest clusters update matrix d X Y,Z = X d X,Z + Y d Y,Z X + Y branch lengths?? assume molecular clock ultrametric 7 / 31

43 8 / 31 Distance-Based Reconstruction What about unrooted trees?

44 8 / 31 Distance-Based Reconstruction B 9 C 11 8 D A B C What about unrooted trees?

45 8 / 31 Distance-Based Reconstruction Problem: correct pairs may not be closest B 9 C 11 8 D A B C

46 9 / 31 Distance-Based Reconstruction Problem: correct pairs may not be closest B 9 C 11 8 D A B C B A D C

47 9 / 31 Distance-Based Reconstruction Problem: correct pairs may not be closest Neighbor Joining (unrooted) - build new matrix: Q X,Y = Z (d X,Z + d Y,Z d X,Y ) + 2d X,Y - find max in Q - join them - update distances & recurse

48 9 / 31 Distance-Based Reconstruction Problem: correct pairs may not be closest Neighbor Joining (unrooted) - build new matrix: Q X,Y = Z (d X,Z + d Y,Z d X,Y ) + 2d X,Y - find max in Q - join them - update distances & recurse

49 9 / 31 Distance-Based Reconstruction Problem: correct pairs may not be closest Neighbor Joining (unrooted) - build new matrix: Q X,Y = Z (d X,Z + d Y,Z d X,Y ) + 2d X,Y - find max in Q - join them - update distances & recurse Theorem Q X,Y max any tree T yielding Q has cherry (X, Y )

50 9 / 31 Distance-Based Reconstruction Problem: correct pairs may not be closest Neighbor Joining (unrooted) - build new matrix: Q X,Y = Z (d X,Z + d Y,Z d X,Y ) + 2d X,Y - find max in Q - join them - update distances & recurse update distances d X Y,Z = 1 /2 (d X,Z + d Y,Z d X,Y ) branch lengths 2b(X ) = Z / {X,Y } (d X,Z d Y,Z +d X,Y ) n 2

51 9 / 31 Distance-Based Reconstruction Problem: correct pairs may not be closest Neighbor Joining (unrooted) - build new matrix: Q X,Y = Z (d X,Z + d Y,Z d X,Y ) + 2d X,Y - find max in Q - join them - update distances & recurse update distances d X Y,Z = 1 /2 (d X,Z + d Y,Z d X,Y ) branch lengths 2b(X ) = Z / {X,Y } (d X,Z d Y,Z +d X,Y ) n 2 2 3

52 9 / 31 Distance-Based Reconstruction Problem: correct pairs may not be closest Neighbor Joining (unrooted) - build new matrix: Q X,Y = Z (d X,Z + d Y,Z d X,Y ) + 2d X,Y - find max in Q - join them - update distances & recurse update distances d X Y,Z = 1 /2 (d X,Z + d Y,Z d X,Y ) branch lengths 2b(X ) = Z / {X,Y } (d X,Z d Y,Z +d X,Y ) n 2 2 3

53 9 / 31 Distance-Based Reconstruction Problem: correct pairs may not be closest Neighbor Joining (unrooted) - build new matrix: Q X,Y = Z (d X,Z + d Y,Z d X,Y ) + 2d X,Y - find max in Q - join them - update distances & recurse update distances d X Y,Z = 1 /2 (d X,Z + d Y,Z d X,Y ) branch lengths 2b(X ) = Z / {X,Y } (d X,Z d Y,Z +d X,Y ) n 2 2 3

54 9 / 31 Distance-Based Reconstruction Problem: correct pairs may not be closest Neighbor Joining (unrooted) - build new matrix: Q X,Y = Z (d X,Z + d Y,Z d X,Y ) + 2d X,Y - find max in Q - join them - update distances & recurse update distances d X Y,Z = 1 /2 (d X,Z + d Y,Z d X,Y ) branch lengths 2b(X ) = Z / {X,Y } (d X,Z d Y,Z +d X,Y ) n

55 9 / 31 Distance-Based Reconstruction Problem: correct pairs may not be closest Neighbor Joining (unrooted) - build new matrix: Q X,Y = Z (d X,Z + d Y,Z d X,Y ) + 2d X,Y - find max in Q - join them - update distances & recurse update distances d X Y,Z = 1 /2 (d X,Z + d Y,Z d X,Y ) branch lengths 2b(X ) = Z / {X,Y } (d X,Z d Y,Z +d X,Y ) n

56 9 / 31 Distance-Based Reconstruction Problem: correct pairs may not be closest Neighbor Joining (unrooted) - build new matrix: Q X,Y = Z (d X,Z + d Y,Z d X,Y ) + 2d X,Y - find max in Q - join them - update distances & recurse update distances d X Y,Z = 1 /2 (d X,Z + d Y,Z d X,Y ) branch lengths 2b(X ) = Z / {X,Y } (d X,Z d Y,Z +d X,Y ) n

57 Parsimony Reconstructing 10 / 31

58 10 / 31 Parsimony Reconstructing fur AUS pouch land X X X X X X X X X

59 10 / 31 Parsimony Reconstructing fur AUS pouch land X X X X X X X X X

60 10 / 31 Parsimony Reconstructing fur AUS pouch land X X X X X X X 0001 X X

61 Parsimony Reconstructing fur AUS pouch land X X X X X X X 0001 X X sum distance of endpoints of each edge 10 / 31

62 Parsimony Reconstructing fur AUS pouch land X X X X X X X 0001 X X sum distance of endpoints of each edge cost 6 (Hamming) 10 / 31

63 10 / 31 Parsimony Reconstructing ACTACGGACCGCTGCG A_TACAAAC GCG ATTACGGACCGCTGCG Small Parsimony Input: character state matrix M, rooted tree T Task: assign characters to internal nodes minimizing total cost ATTACAAAC GCC TTTACAAAC CC

64 10 / 31 Parsimony Reconstructing ACTACGGACCGCTGCG A_TACAAAC GCG ATTACGGACCGCTGCG Small Parsimony Input: character state matrix M, rooted tree T Task: assign characters to internal nodes minimizing total cost ATTACAAAC GCC TTTACAAAC CC O(nm) time [Fitch 71]

65 10 / 31 Parsimony Reconstructing ACTACGGACCGCTGCG A_TACAAAC GCG ATTACGGACCGCTGCG Small Parsimony Input: character state matrix M, rooted tree T Task: assign characters to internal nodes minimizing total cost ATTACAAAC GCC TTTACAAAC CC Exercise Time O(nm) time [Fitch 71]

66 10 / 31 Parsimony Reconstructing ACTACGGACCGCTGCG A_TACAAAC GCG ATTACGGACCGCTGCG ATTACAAAC GCC TTTACAAAC CC Small Parsimony Input: character state matrix M, rooted tree T Task: assign characters to internal nodes minimizing total cost O(nm) time Large Parsimony Input: character state matrix M [Fitch 71] Task: find tree T & assign characters to internal nodes minimizing total cost

67 10 / 31 Parsimony Reconstructing ACTACGGACCGCTGCG A_TACAAAC GCG ATTACGGACCGCTGCG ATTACAAAC GCC TTTACAAAC CC Small Parsimony Input: character state matrix M, rooted tree T Task: assign characters to internal nodes minimizing total cost O(nm) time Large Parsimony Input: character state matrix M [Fitch 71] Task: find tree T & assign characters to internal nodes minimizing total cost NP-hard

68 10 / 31 Parsimony Reconstructing ACTACGGACCGCTGCG A_TACAAAC GCG ATTACGGACCGCTGCG ATTACAAAC GCC TTTACAAAC CC Small Parsimony Input: character state matrix M, rooted tree T Task: assign characters to internal nodes minimizing total cost O(nm) time Large Parsimony Input: character state matrix M [Fitch 71] Task: find tree T & assign characters to internal nodes minimizing total cost NP-hard Note: alignment is crucial!

69 11 / 31 Maximum Likelihood Reconstruction Idea: find a tree (with branch lengths) under which evolution is most likely to have produced the observed characters/genomes

70 11 / 31 Maximum Likelihood Reconstruction Idea: find a tree (with branch lengths) under which evolution is most likely to have produced the observed characters/genomes need model of evolution

71 11 / 31 Maximum Likelihood Reconstruction Idea: find a tree (with branch lengths) under which evolution is most likely to have produced the observed characters/genomes need model of evolution Jukes & Cantor Model - each base evolves individually - each base occurs with equal frequency in the genome - constant rate µ of mutation - each base is equally likely to be result of mutation

72 11 / 31 Maximum Likelihood Reconstruction Idea: find a tree (with branch lengths) under which evolution is most likely to have produced the observed characters/genomes need model of evolution Jukes & Cantor Model - each base evolves individually - each base occurs with equal frequency in the genome - constant rate µ of mutation - each base is equally likely to be result of mutation Generalized Time Reversible Model - each base evolves individually - each base X has a frequency π X to occur in the genome - each base-substitution has its own rate of occurance

73 11 / 31 Maximum Likelihood Reconstruction Idea: find a tree (with branch lengths) under which evolution is most likely to have produced the observed characters/genomes need model of evolution Jukes & Cantor Model - each base evolves individually - each base occurs with equal frequency in the genome - constant rate µ of mutation - each base is equally likely to be result of mutation Generalized Time Reversible Model - each base evolves individually - each base X has a frequency π X to occur in the genome - each base-substitution has its own rate of occurance compute likelihood, given tree & parameters O(mn) time

74 11 / 31 Maximum Likelihood Reconstruction Idea: find a tree (with branch lengths) under which evolution is most likely to have produced the observed characters/genomes need model of evolution Jukes & Cantor Model - each base evolves individually - each base occurs with equal frequency in the genome - constant rate µ of mutation - each base is equally likely to be result of mutation Generalized Time Reversible Model - each base evolves individually - each base X has a frequency π X to occur in the genome - each base-substitution has its own rate of occurance compute likelihood, given tree & parameters O(mn) time find best tree & parameters NP-hard

75 11 / 31 Maximum Likelihood Reconstruction Idea: find a tree (with branch lengths) under which evolution is most likely to have produced the observed characters/genomes need model of evolution Jukes & Cantor Model - each base evolves individually - each base occurs with equal frequency in the genome - constant rate µ of mutation - each base is equally likely to be result of mutation Generalized Time Reversible Model - each base evolves individually - each base X has a frequency π X to occur in the genome - each base-substitution has its own rate of occurance compute likelihood, given tree & parameters O(mn) time find best tree & parameters NP-hard local search in the tree space

76 12 / 31 ML Reconstruction - Tree Spaces Observe: a tree and a rearrangement operation span a space Nearest Neighbor Interchange change any configuration of 4 (3) neighboring subtrees into another

77 12 / 31 ML Reconstruction - Tree Spaces Observe: a tree and a rearrangement operation span a space Nearest Neighbor Interchange change any configuration of 4 (3) neighboring subtrees into another

78 12 / 31 ML Reconstruction - Tree Spaces Observe: a tree and a rearrangement operation span a space Nearest Neighbor Interchange change any configuration of 4 (3) neighboring subtrees into another

79 12 / 31 ML Reconstruction - Tree Spaces Observe: a tree and a rearrangement operation span a space Nearest Neighbor Interchange change any configuration of 4 (3) neighboring subtrees into another Subtree Prune & Regraft break any edge uv & connect v to any edge of the component of u

80 12 / 31 ML Reconstruction - Tree Spaces Observe: a tree and a rearrangement operation span a space Nearest Neighbor Interchange change any configuration of 4 (3) neighboring subtrees into another Subtree Prune & Regraft break any edge uv & connect v to any edge of the component of u

81 12 / 31 ML Reconstruction - Tree Spaces Observe: a tree and a rearrangement operation span a space Nearest Neighbor Interchange change any configuration of 4 (3) neighboring subtrees into another Subtree Prune & Regraft break any edge uv & connect v to any edge of the component of u

82 12 / 31 ML Reconstruction - Tree Spaces Observe: a tree and a rearrangement operation span a space Nearest Neighbor Interchange change any configuration of 4 (3) neighboring subtrees into another Subtree Prune & Regraft break any edge uv & connect v to any edge of the component of u

83 12 / 31 ML Reconstruction - Tree Spaces Observe: a tree and a rearrangement operation span a space Nearest Neighbor Interchange change any configuration of 4 (3) neighboring subtrees into another Subtree Prune & Regraft break any edge uv & connect v to any edge of the component of u

84 ML Reconstruction - Tree Spaces Observe: a tree and a rearrangement operation span a space Nearest Neighbor Interchange change any configuration of 4 (3) neighboring subtrees into another Subtree Prune & Regraft break any edge uv & connect v to any edge of the component of u Tree Bisection & Reconnection break any edge & insert a new reconnecting edge between any 2 edges 12 / 31

85 ML Reconstruction - Tree Spaces Observe: a tree and a rearrangement operation span a space Nearest Neighbor Interchange change any configuration of 4 (3) neighboring subtrees into another Subtree Prune & Regraft break any edge uv & connect v to any edge of the component of u Tree Bisection & Reconnection break any edge & insert a new reconnecting edge between any 2 edges 12 / 31

86 ML Reconstruction - Tree Spaces Observe: a tree and a rearrangement operation span a space Nearest Neighbor Interchange change any configuration of 4 (3) neighboring subtrees into another Subtree Prune & Regraft break any edge uv & connect v to any edge of the component of u Tree Bisection & Reconnection break any edge & insert a new reconnecting edge between any 2 edges 12 / 31

87 ML Reconstruction - Tree Spaces Observe: a tree and a rearrangement operation span a space Nearest Neighbor Interchange change any configuration of 4 (3) neighboring subtrees into another Subtree Prune & Regraft break any edge uv & connect v to any edge of the component of u Tree Bisection & Reconnection break any edge & insert a new reconnecting edge between any 2 edges 12 / 31

88 ML Reconstruction - Tree Spaces Observe: a tree and a rearrangement operation span a space Nearest Neighbor Interchange change any configuration of 4 (3) neighboring subtrees into another Subtree Prune & Regraft break any edge uv & connect v to any edge of the component of u Tree Bisection & Reconnection break any edge & insert a new reconnecting edge between any 2 edges 12 / 31

89 13 / 31 ML Reconstruction - Tree Spaces Exercise: turn into (any) caterpillar: Exercise Time Exercise: how are the distances related?

90 14 / 31 Checking Robustness Bootstrap Method suppose: method X yields tree T from n m character-state matrix M repeat k times the following experiment: 1. draw m columns from M (with repetition) 2. use X to compute T i Finally, for each branch of T, check how often it occurs in the T i bootstrap value measures robustness ( support ) of each branch

91 Reconstruction by Gene Trees 15 / 31

92 15 / 31 Reconstruction by Gene Trees A Common Method For Reconstructing Trees 1. get genomes of multiple species 2. extract genes using START & STOP codons 3. cluster genes in families of similar genes 4. within each family, infer a gene tree using dissimilarities 5. build a consensus among the gene trees species tree (Note: species tree may differ significantly from individual gene trees) 6. reconcile all gene trees with the species tree to learn the evolution of those genes

93 15 / 31 Reconstruction by Gene Trees A Common Method For Reconstructing Trees 1. get genomes of multiple species 2. extract genes using START & STOP codons 3. cluster genes in families of similar genes 4. within each family, infer a gene tree using dissimilarities 5. build a consensus among the gene trees species tree (Note: species tree may differ significantly from individual gene trees) 6. reconcile all gene trees with the species tree to learn the evolution of those genes

94 15 / 31 Reconstruction by Gene Trees A Common Method For Reconstructing Trees 1. get genomes of multiple species 2. extract genes using START & STOP codons 3. cluster genes in families of similar genes 4. within each family, infer a gene tree using dissimilarities 5. build a consensus among the gene trees species tree (Note: species tree may differ significantly from individual gene trees) 6. reconcile all gene trees with the species tree to learn the evolution of those genes

95 15 / 31 Reconstruction by Gene Trees A Common Method For Reconstructing Trees 1. get genomes of multiple species 2. extract genes using START & STOP codons 3. cluster genes in families of similar genes 4. within each family, infer a gene tree using dissimilarities 5. build a consensus among the gene trees species tree (Note: species tree may differ significantly from individual gene trees) 6. reconcile all gene trees with the species tree to learn the evolution of those genes

96 16 / 31 Supertrees - Build Algorithm Idea: find root partition & recurse (as long as there are 3 taxa) a c b f e a d b c

97 16 / 31 Supertrees - Build Algorithm Idea: find root partition & recurse (as long as there are 3 taxa) b c a c b f e a d b c d a e f

98 16 / 31 Supertrees - Build Algorithm Idea: find root partition & recurse (as long as there are 3 taxa) b c a c b f e a d b c d a e f

99 16 / 31 Supertrees - Build Algorithm Idea: find root partition & recurse (as long as there are 3 taxa) b c a c b f e a d b c d a e f a, b, c, d e f

100 16 / 31 Supertrees - Build Algorithm Idea: find root partition & recurse (as long as there are 3 taxa) b c a c b f e a d b c d a e f a, b, c, d e f

101 16 / 31 Supertrees - Build Algorithm Idea: find root partition & recurse (as long as there are 3 taxa) b c a c b f e a d b c d a e f a, b, c, d e f

102 16 / 31 Supertrees - Build Algorithm Idea: find root partition & recurse (as long as there are 3 taxa) b c a c b f e a d b c d a e f a, c, d b e f

103 16 / 31 Supertrees - Build Algorithm Idea: find root partition & recurse (as long as there are 3 taxa) b c a c b f e a d b c d a e f a, c, d b e f

104 Supertrees - Build Algorithm Idea: find root partition & recurse (as long as there are 3 taxa) b c a c b f e a d b c d a e f b c e f a d 16 / 31

105 Supertrees - Build Algorithm Idea: find root partition & recurse (as long as there are 3 taxa) a c b f e a d b c Note: always works if trees are compatible/consistent b c e f a d 16 / 31

106 Supertrees - Build Algorithm Idea: find root partition & recurse (as long as there are 3 taxa) a c b f e a d b c Note: always works if trees are compatible/consistent incompatible? c b e f - largest compatible subset NP-hard (even for triplets) - voting schemes (each tree votes for their clades) - reinterpret clades as characters, combine into matrix & reconstruct a d 16 / 31

107 17 / 31 Consensi of Non-Agreeing Trees strict consensus consensus subtree majority consensus

108 18 / 31 Reconstruction by Gene Trees A Common Method For Reconstructing Trees 1. get genomes of multiple species 2. extract genes using START & STOP codons 3. cluster genes in families of similar genes 4. within each family, infer a gene tree using dissimilarities 5. build a consensus among the gene trees species tree (Note: species tree may differ significantly from individual gene trees) 6. reconcile all gene trees with the species tree to learn the evolution of those genes

109 18 / 31 Reconstruction by Gene Trees A Common Method For Reconstructing Trees 1. get genomes of multiple species 2. extract genes using START & STOP codons 3. cluster genes in families of similar genes 4. within each family, infer a gene tree using dissimilarities 5. build a consensus among the gene trees species tree (Note: species tree may differ significantly from individual gene trees) 6. reconcile all gene trees with the species tree to learn the evolution of those genes

110 The History of a Gene Family Recall gene = functional element of DNA, clustered into gene-families each family yields a tree depicting its history gene tree consensus of the gene trees yields species tree But: what really happened??? Mouse Bat Dog Rat Mouse Rat Bat Dog 19 / 31

111 The History of a Gene Family Recall gene = functional element of DNA, clustered into gene-families each family yields a tree depicting its history gene tree consensus of the gene trees yields species tree But: what really happened??? Mouse Bat Dog Rat Mouse Rat Bat Dog 19 / 31

112 20 / 31 Reconciliation A B C a a b b c Embedding Rules gene tree G, species tree S - mapping ρ : V (G) V (S) - l is leaf in G ρ(l) corresponds to l (a A, a A, etc.) - u V (G) is called duplication if ρ(u) = ρ(c) for any child c of u in G - all non-leaves of G that not duplications are called speciations - each edge uv of G incurs a loss-cost equal to the number of edges in the ρ(u)-ρ(v)-path in S minus 1 if v is a speciation or 0 if v is a duplication

113 Reconciliation A B C a a b b c DL-model = speciation = duplication = loss a a b b c 20 / 31

114 Reconciliation Goal: embed gene tree into species tree (extant genes must map to their species) Max. Likelihood find most probable embedding (computationally expensive) Parsimony find embedding minimizing #events (possibly weighted) DL-model = speciation = duplication = loss a a b b c 20 / 31

115 Reconciliation Parsimonious Reconciliation Input: species tree S, gene tree G, δ, λ N Task: embed G in S, minimizing the weighted sum of events Result: LCA-assignment solves this optimally in O( S + G ) DL-model = speciation (0) = duplication (δ) = loss (λ) a a b b c 20 / 31

116 20 / 31 Reconciliation Parsimonious Reconciliation Input: species tree S, gene tree G, δ, λ, τ N Task: embed G in S, minimizing the weighted sum of events Result: LCA-assignment solves this optimally in O( S + G ) Exercise Time A B C D a b c d a b

117 Reconciliation Parsimonious Reconciliation Input: species tree S, gene tree G, δ, λ, τ N Task: embed G in S, minimizing the weighted sum of events Result: LCA-assignment solves this optimally in O( S + G ) DTL-model = speciation (0) = duplication (δ) = loss (λ) = transfer (τ) a a b b c c 20 / 31

118 Reconciliation Parsimonious Reconciliation Input: species tree S, gene tree G, δ, λ, τ N Task: embed G in S, minimizing the weighted sum of events Result: LCA-assignment solves this optimally in O( S + G ) T events only between co-existing species time constraints NP-hard DTL-model = speciation (0) = duplication (δ) = loss (λ) = transfer (τ) a a b b c c 20 / 31

119 Reconciliation Parsimonious Reconciliation Input: species tree S, gene tree G, δ, λ, τ N Task: embed G in S, minimizing the weighted sum of events Result: LCA-assignment solves this optimally in O( S + G ) T events only between co-existing species time constraints NP-hard Idea: dated species tree O( S 2 G ) [Doyon et al. 10] DTL-model = speciation (0) = duplication (δ) = loss (λ) = transfer (τ) a a b b c c 20 / 31

120 21 / 31 Comparing Phylogenetic Trees Distance Measures - Nearest Neighbor Interchange - Subtree Prune & Regraft - Tree Bisection & Reconnection

121 21 / 31 Comparing Phylogenetic Trees Distance Measures - Nearest Neighbor Interchange - Subtree Prune & Regraft - Tree Bisection & Reconnection - Robinson-Foulds distance - quartet/triplet distance

122 22 / 31 Agreement Forests Definition A forest F is called agreement forest of trees T 1 and T 2 if F can be obtained from T 1 and T 2 by removing edges. D C B D A E C E F F B A

123 22 / 31 Agreement Forests Definition A forest F is called agreement forest of trees T 1 and T 2 if F can be obtained from T 1 and T 2 by removing edges. D C B D A E C E F F B A

124 22 / 31 Agreement Forests Definition A forest F is called agreement forest of trees T 1 and T 2 if F can be obtained from T 1 and T 2 by removing edges. F C E A F E D C B A D B

125 22 / 31 Agreement Forests Definition A forest F is called agreement forest of trees T 1 and T 2 if F can be obtained from T 1 and T 2 by removing edges. F C E A F E D C B A D B

126 22 / 31 Agreement Forests Definition A forest F is called agreement forest of trees T 1 and T 2 if F can be obtained from T 1 and T 2 by removing edges. Theorem [Allen & Steel, 01] TBR-distance(T 1, T 2 ) = #trees in smallest agreement forest - 1 NP-hard to compute Theorem [Bordewich & Semple, 04] rspr-distance(t 1, T 2) = #trees in smallest rooted agreement forest - 1 NP-hard to compute

127 Robinson-Foulds Distance Definition RF(T 1,T 2 ) = #splits/clusters occuring in exactly one of T 1 and T 2 = edge-contraction distance a common tree Note: observe relation to NNI: RF(T 1,T 2 ) 2 NNI(T 1,T 2 ) E F F C D B A trivial splits, {A,B} {C,D,E,F}, {E,F} {A,B,E,F}, {C,D} {A,B,E,F} A B trivial splits, {A,B} {C,D,E,F} {E,F} {A,B,C,D} {A,B,D} {C,E,F} 23 / 31 D C E

128 Robinson-Foulds Distance Definition RF(T 1,T 2 ) = #splits/clusters occuring in exactly one of T 1 and T 2 = edge-contraction distance a common tree Note: observe relation to NNI: RF(T 1,T 2 ) 2 NNI(T 1,T 2 ) E F F C D B A trivial splits, {A,B} {C,D,E,F}, {E,F} {A,B,E,F}, {C,D} {A,B,E,F} A B trivial splits, {A,B} {C,D,E,F} {E,F} {A,B,C,D} {A,B,D} {C,E,F} 23 / 31 D C E

129 Robinson-Foulds Distance Definition RF(T 1,T 2 ) = #splits/clusters occuring in exactly one of T 1 and T 2 = edge-contraction distance a common tree Note: observe relation to NNI: RF(T 1,T 2 ) 2 NNI(T 1,T 2 ) (C) Leonardo de Oliveira Martins 23 / 31

130 23 / 31 Robinson-Foulds Distance Definition RF(T 1,T 2 ) = #splits/clusters occuring in exactly one of T 1 and T 2 = edge-contraction distance a common tree Note: observe relation to NNI: RF(T 1,T 2 ) 2 NNI(T 1,T 2 ) Note: splits correspond to clusters when rooted at last leaf Day s Algorithm (common clusters in O(n)) [Day 85] 1. relabel all leaves s.t. leaves continuous in T 1 2. each vertex in T 1 knows: L: smallest leaf in cluster R: largest leaf in cluster T 1 s clusters in table [L, R] (O(n) sparse set/perfect hashing) 3. each vertex in T 2 knows L & R & size N of its cluster 4. each vertex in T 2 only checks for [L, R] if R L = N 1 (lookup in T 1 s table in O(1) (average) time)

131 23 / 31 Robinson-Foulds Distance C D B A C D E A B F Day s Algorithm (common clusters in O(n)) [Day 85] F E 1. relabel all leaves s.t. leaves continuous in T 1 2. each vertex in T 1 knows: L: smallest leaf in cluster R: largest leaf in cluster T 1 s clusters in table [L, R] (O(n) sparse set/perfect hashing) 3. each vertex in T 2 knows L & R & size N of its cluster 4. each vertex in T 2 only checks for [L, R] if R L = N 1 (lookup in T 1 s table in O(1) (average) time)

132 23 / 31 Robinson-Foulds Distance C D B A C D E A B F Day s Algorithm (common clusters in O(n)) [Day 85] F E 1. relabel all leaves s.t. leaves continuous in T 1 2. each vertex in T 1 knows: L: smallest leaf in cluster R: largest leaf in cluster T 1 s clusters in table [L, R] (O(n) sparse set/perfect hashing) 3. each vertex in T 2 knows L & R & size N of its cluster 4. each vertex in T 2 only checks for [L, R] if R L = N 1 (lookup in T 1 s table in O(1) (average) time)

133 23 / 31 Robinson-Foulds Distance C D A,E,5 C,D B A,D,4 A,B A,D A,D,3 A C A,B,2 A,E D E A B F Day s Algorithm (common clusters in O(n)) [Day 85] F E 1. relabel all leaves s.t. leaves continuous in T 1 2. each vertex in T 1 knows: L: smallest leaf in cluster R: largest leaf in cluster T 1 s clusters in table [L, R] (O(n) sparse set/perfect hashing) 3. each vertex in T 2 knows L & R & size N of its cluster 4. each vertex in T 2 only checks for [L, R] if R L = N 1 (lookup in T 1 s table in O(1) (average) time)

134 23 / 31 Robinson-Foulds Distance C D A,E,5 C,D B A,D,4 A,B A,D A,D,3 A C A,B,2 A,E D E A B F Day s Algorithm (common clusters in O(n)) [Day 85] F E 1. relabel all leaves s.t. leaves continuous in T 1 2. each vertex in T 1 knows: L: smallest leaf in cluster R: largest leaf in cluster T 1 s clusters in table [L, R] (O(n) sparse set/perfect hashing) 3. each vertex in T 2 knows L & R & size N of its cluster 4. each vertex in T 2 only checks for [L, R] if R L = N 1 (lookup in T 1 s table in O(1) (average) time)

135 24 / 31 Quartet/Triplet Distance Definition Q/T(T 1,T 2 ) = #quartets/triplets occur. in exactly one of T 1 and T 2

136 24 / 31 Quartet/Triplet Distance Definition Q/T(T 1,T 2 ) = #quartets/triplets occur. in exactly one of T 1 and T 2 computing Q-distance (binary trees) [Bryant et al. 00] 1. each edge uv has 4 sets (2 clusters for each of u & v) 2. quartet AB CD belongs to edge e if e splits AB CD & e touches AB-path or CD-path each split is owned exactly once 3. uv T 1 & ab T 2: intersect 4 sets of uv with split of ab in T 2 4. sizes of intersections can be precomputed bottom-up in O(n 2 ) time

137 24 / 31 Quartet/Triplet Distance Definition Q/T(T 1,T 2 ) = #quartets/triplets occur. in exactly one of T 1 and T 2 computing Q-distance (binary trees) [Bryant et al. 00] 1. each edge uv has 4 sets (2 clusters for each of u & v) 2. quartet AB CD belongs to edge e if e splits AB CD & e touches AB-path or CD-path each split is owned exactly once 3. uv T 1 & ab T 2: intersect 4 sets of uv with split of ab in T 2 4. sizes of intersections can be precomputed bottom-up in O(n 2 ) time State of the Art count conflict quartets/triplets O(n log n) time [Brodal et al. 13] enumerate conflict quartets O(n 2 + d) [Bryant et al. 00] enumerate conflict triplets O(n + d) [Weller 17]

138 25 / 31 Phylogenetic Networks Observation Trees cannot capture hybridization

139 25 / 31 Phylogenetic Networks Observation Trees cannot capture hybridization phylogenetic network

140 25 / 31 Phylogenetic Networks Observation Trees cannot capture hybridization phylogenetic network Definition evolutionary network N = rooted DAG, leaves labeled (taxa) reticulations R = vertices of in-degree 2 binary = all inner vertices degree 3 block = component without cut-vertex display T = subdivision of T is a subgraph

141 25 / 31 Phylogenetic Networks Observation Trees cannot capture hybridization phylogenetic network Definition evolutionary network N = rooted DAG, leaves labeled (taxa) reticulations R = vertices of in-degree 2 binary = all inner vertices degree 3 block = component without cut-vertex display T = subdivision of T is a subgraph

142 Split Networks split = bipartition of set of taxa splits A B & X Y incompatible if both A & B intersect both X & Y Convex Hull Algorithm [Holland et al., 04] fur AUS pouch land X X X XX X XX c.f.: Neighbor Net [Bryant & Moulton, 03] (for circular splits) 26 / 31

143 Split Networks split = bipartition of set of taxa splits A B & X Y incompatible if both A & B intersect both X & Y Convex Hull Algorithm [Holland et al., 04] fur AUS pouch land X X X XX X XX vs c.f.: Neighbor Net [Bryant & Moulton, 03] (for circular splits) 26 / 31

144 Split Networks split = bipartition of set of taxa splits A B & X Y incompatible if both A & B intersect both X & Y Convex Hull Algorithm [Holland et al., 04] fur AUS pouch land X X X XX X XX vs c.f.: Neighbor Net [Bryant & Moulton, 03] (for circular splits) 26 / 31

145 Split Networks split = bipartition of set of taxa splits A B & X Y incompatible if both A & B intersect both X & Y Convex Hull Algorithm [Holland et al., 04] fur AUS pouch land X X X XX X XX vs vs c.f.: Neighbor Net [Bryant & Moulton, 03] (for circular splits) 26 / 31

146 Split Networks split = bipartition of set of taxa splits A B & X Y incompatible if both A & B intersect both X & Y Convex Hull Algorithm [Holland et al., 04] fur AUS pouch land X X X XX X XX vs vs c.f.: Neighbor Net [Bryant & Moulton, 03] (for circular splits) 26 / 31

147 Split Networks split = bipartition of set of taxa splits A B & X Y incompatible if both A & B intersect both X & Y Convex Hull Algorithm [Holland et al., 04] fur AUS pouch land X X X XX X XX vs vs c.f.: Neighbor Net [Bryant & Moulton, 03] (for circular splits) 26 / 31

148 Split Networks split = bipartition of set of taxa splits A B & X Y incompatible if both A & B intersect both X & Y Convex Hull Algorithm [Holland et al., 04] fur AUS pouch land X X X XX X XX vs vs vs c.f.: Neighbor Net [Bryant & Moulton, 03] (for circular splits) 26 / 31

149 Split Networks split = bipartition of set of taxa splits A B & X Y incompatible if both A & B intersect both X & Y Convex Hull Algorithm [Holland et al., 04] fur AUS pouch land X X X XX X XX vs vs vs c.f.: Neighbor Net [Bryant & Moulton, 03] (for circular splits) 26 / 31

150 Split Networks split = bipartition of set of taxa splits A B & X Y incompatible if both A & B intersect both X & Y Convex Hull Algorithm [Holland et al., 04] fur AUS pouch land X X X XX X XX vs vs vs vs c.f.: Neighbor Net [Bryant & Moulton, 03] (for circular splits) 26 / 31

151 27 / 31 Consensus Split Networks Strategy 1. list all splits of all input trees 2. extend splits to full taxa using Z-closure 3. build consensus

152 27 / 31 Consensus Split Networks Strategy 1. list all splits of all input trees 2. extend splits to full taxa using Z-closure 3. build consensus Experimental Study 106 gene trees (yeast) [Rokas et al. 03, Holland et al. 04]

153 27 / 31 Consensus Split Networks Strategy 1. list all splits of all input trees 2. extend splits to full taxa using Z-closure 3. build consensus Experimental Study 106 gene trees (yeast) [Rokas et al. 03, Holland et al. 04]

154 27 / 31 Consensus Split Networks Strategy 1. list all splits of all input trees 2. extend splits to full taxa using Z-closure 3. build consensus Experimental Study 106 gene trees (yeast) [Rokas et al. 03, Holland et al. 04]

155 27 / 31 Consensus Split Networks Strategy 1. list all splits of all input trees 2. extend splits to full taxa using Z-closure 3. build consensus Experimental Study 106 gene trees (yeast) [Rokas et al. 03, Holland et al. 04]

156 28 / 31 Rooted Network Reconstruction Observation rooted network: cluster of u cluster of v u v rooted network is hasse diagram of its clusters

157 28 / 31 Rooted Network Reconstruction Observation rooted network: cluster of u cluster of v u v rooted network is hasse diagram of its clusters Example {a,b,c,d},{c,d,e,f,g,h}, {c,d,e,f,g}, {e,f,g,h}, {c,d,e}, {e,f,g}, {a,b}, {c,d}, {f,g} Exercise Time

158 28 / 31 Rooted Network Reconstruction Observation rooted network: cluster of u cluster of v u v rooted network is hasse diagram of its clusters Example {a,b,c,d},{c,d,e,f,g,h}, {c,d,e,f,g}, {e,f,g,h}, {c,d,e}, {e,f,g}, {a,b}, {c,d}, {f,g} a,b,c,d c,d,e,f,g c,d,e,f,g,h e,f,g,h c,d,e e,f,g a,b c,d f,g a b c d e f g h

159 28 / 31 Rooted Network Reconstruction Observation rooted network: cluster of u cluster of v u v rooted network is hasse diagram of its clusters Example {a,b,c,d},{c,d,e,f,g,h}, {c,d,e,f,g}, {e,f,g,h}, {c,d,e}, {e,f,g}, {a,b}, {c,d}, {f,g} a b c d e f g h

160 Rooted Network Reconstruction Observation rooted network: cluster of u cluster of v u v rooted network is hasse diagram of its clusters Example {a,b,c,d},{c,d,e,f,g,h}, {c,d,e,f,g}, {e,f,g,h}, {c,d,e}, {e,f,g}, {a,b}, {c,d}, {f,g} a b c d e c.f. cluster popping [Huson & Rupp, 08] f g h 28 / 31

161 28 / 31 Rooted Network Reconstruction Observation rooted network: cluster of u cluster of v u v rooted network is hasse diagram of its clusters Problem may produce more reticulations than necessary to explain the data

162 28 / 31 Rooted Network Reconstruction Observation rooted network: cluster of u cluster of v u v rooted network is hasse diagram of its clusters Problem may produce more reticulations than necessary to explain the data Hybridization Number Input: set of trees T, int k Question: Is there a network with k reticulations displaying all trees in T?

163 28 / 31 Rooted Network Reconstruction Observation rooted network: cluster of u cluster of v u v rooted network is hasse diagram of its clusters Problem may produce more reticulations than necessary to explain the data Hybridization Number Input: set of trees T, int k Question: Is there a network with k reticulations displaying all trees in T? NP-hard for 2 trees [Bordewich & Semple, 07]

164 28 / 31 Rooted Network Reconstruction Observation rooted network: cluster of u cluster of v u v rooted network is hasse diagram of its clusters Problem may produce more reticulations than necessary to explain the data Hybridization Number Input: set of trees T, int k Question: Is there a network with k reticulations displaying all trees in T? NP-hard for 2 trees [Bordewich & Semple, 07] Note: HN(T 1,T 2 ) = max. acyclic agreement forest - 1 [Baroni et al., 05]

165 29 / 31 Networks Display Trees Observation A network may display up to 2 R trees.

166 29 / 31 Networks Display Trees Observation A network may display up to 2 R trees.

167 29 / 31 Networks Display Trees Observation A network may display up to 2 R trees. But: how to decide if a given tree is displayed?

168 29 / 31 Networks Display Trees Tree Containment Input: a network N, a tree T Question: Does N display T?

169 29 / 31 Networks Display Trees Tree Containment Input: a network N, a tree T Question: Does N display T? NP-hard (from Disjoint Paths) [Kanj et al. 08]

170 29 / 31 Networks Display Trees Tree Containment Input: a network N, a tree T Question: Does N display T? NP-hard (from Disjoint Paths) [Kanj et al. 08] Note: linear time on reticulation visible N [Gunawan, 18][Weller, 18]

171 Networks Display Trees Tree Containment Input: a network N, a tree T Question: Does N display T? NP-hard (from Disjoint Paths) [Kanj et al. 08] Note: linear time on reticulation visible N [Gunawan, 18][Weller, 18] Can you see it? A F G B C E D D E C F A G B 29 / 31

172 Small Taxonomy of Network Classes c.f. Who is Who in Phylogenetic Networks ( 30 / 31

173 Thanks & Enjoy Part III 31 / 31