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. Intro to Phylogenetic Trees Computational Genomics Lecture 4b Sections 7.1, 7.2, in Durbin et al. Chapter 17 in Gusfield Slides by Shlomo Moran and Ido.

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Presentation on theme: ". Intro to Phylogenetic Trees Computational Genomics Lecture 4b Sections 7.1, 7.2, in Durbin et al. Chapter 17 in Gusfield Slides by Shlomo Moran and Ido."— Presentation transcript:

1 . Intro to Phylogenetic Trees Computational Genomics Lecture 4b Sections 7.1, 7.2, in Durbin et al. Chapter 17 in Gusfield Slides by Shlomo Moran and Ido Wexler. Slight modifications by Benny Chor

2 2 Evolution Evolution of new organisms is driven by u Diversity l Different individuals carry different variants of the same basic blue print u Mutations l The DNA sequence can be changed due to single base changes, deletion/insertion of DNA segments, etc. u Selection bias

3 3 The Tree of Life Source: Alberts et al

4 4 Daprès Ernst Haeckel, 1891 Tree of life- a better picture

5 5 Primate evolution A phylogeny is a tree that describes the sequence of speciation events that lead to the forming of a set of current day species; also called a phylogenetic tree.

6 6 Historical Note u Until mid 1950s phylogenies were constructed by experts based on their opinion (subjective criteria) u Since then, focus on objective criteria for constructing phylogenetic trees l Thousands of articles in the last decades u Important for many aspects of biology l Classification l Understanding biological mechanisms

7 7 Morphological vs. Molecular u Classical phylogenetic analysis: morphological features: number of legs, lengths of legs, etc. u Modern biological methods allow to use molecular features l Gene sequences l Protein sequences u Analysis based on homologous sequences (e.g., globins) in different species

8 8 Topology based on Morphology Archonta Glires Ungulata Carnivora Insectivora Xenarthra (Based on Mc Kenna and Bell, 1997)

9 9 RatQEPGGLVVPPTDA RabbitQEPGGMVVPPTDA GorillaQEPGGLVVPPTDA CatREPGGLVVPPTEG From Sequences to a Phylogenetic Tree Different genes/proteins may lead to different phylogenetic trees

10 10 RatQEPGGLVVPPTDA RabbitQEPGGMVVPPTDA GorillaQEPGGLVVPPTDA CatREPGGLVVPPTEG From sequences to a phylogenetic tree There are many possible types of sequences to use (e.g. mitochondrial vs. nuclear proteins).

11 11 Perissodactyla Carnivora Cetartiodactyla Rodentia 1 Hedgehogs Rodentia 2 Primates Chiroptera Moles+Shrews Afrotheria Xenarthra Lagomorpha + Scandentia Topology 1, based on Mitochondrial DNA (Based on Pupko et al.,)

12 12 Cetartiodactyla Afrotheria Chiroptera Eulipotyphla Glires Xenarthra Carnivora Perissodactyla Scandentia+ Dermoptera Pholidota Primate (tree by Madsenl) (Based on Pupko et al. slide) Topology 2,based on Nuclear DNA

13 13 Theory of Evolution u Basic idea l speciation events lead to creation of different species. l Speciation caused by physical separation into groups where different genetic variants become dominant u Any two species share a (possibly distant) common ancestor

14 14 Phylogenenetic trees u Leafs - current day species u Nodes - hypothetical most recent common ancestors u Edges length - time from one speciation to the next AardvarkBisonChimpDogElephant

15 15 Types of Trees A natural model to consider is that of rooted trees Common Ancestor

16 16 Types of trees Unrooted tree represents the same phylogeny without the root node Depending on the model, data from current day species often does not distinguish between different placements of the root.

17 17 Rooted versus unrooted trees Tree a a b Tree b c Tree c Represents all three rooted trees

18 18 Positioning Roots in Unrooted Trees u We can estimate the position of the root by introducing an outgroup: l a set of species that are definitely distant from all the species of interest AardvarkBisonChimpDogElephant Falcon Proposed root

19 19 Dangers of Gene Duplication Speciation events Gene Duplication 1A 2A 3A3B 2B1B If we happen to consider genes 1A, 2B, and 3A of species 1,2,3, we get a wrong tree that does not represent the phylogeny of the host species of the given sequences because duplication does not create new species. In the sequel we assume all given sequences are orthologs. S S S

20 20 Types of Data u Distance-based l Input is a matrix of distances between species. l Can be fraction of residue they disagree on, or alignment score between them, etc. u Character-based l Input is a multiple sequence alignment. Sequences consist of characters (e.g., residues) that are examined separately. u Genome/Proteome –based l Input is whole genome or proteome sequences. l No MSA or obvious distance definition.

21 21 Tree Construction: Two Popular Methods u Distance Based- A weighted tree that realizes the distances between the objects (or gets close to it). u Character Based – A tree that optimizes an objective function based on all characters in input sequences (major methods are parsimony and likelihood). We start with distance based methods, considering the following question: Given a set of species (leaves in a supposed tree), and distances between them – construct a phylogeny which best fits the distances.

22 22 Exact solution: Additive sets Given a set M of L objects with an L×L distance matrix: u d(i,i)=0, and for ij, d(i,j)>0 u d(i,j)=d(j,i). u For all i,j,k it holds that d(i,k) d(i,j)+d(j,k). Can we construct a weighted tree which realizes these distances?

23 23 Additive sets (cont) We say that the set M with L objects is additive if there is a tree T, L of its nodes correspond to the L objects, with positive weights on the edges, such that for all i,j, d(i,j) = d T (i,j), the length of the path from i to j in T. Note: Sometimes the tree is required to be binary, and then the edge weights are required to be non-negative.

24 24 Three objects sets always additive: For L=3: There is always a (unique) tree with one internal node. a b c i j k m Thus

25 25 How about four objects? L=4: Not all sets with 4 objects are additive: e.g., there is no tree which realizes the distances below. ijkl i 0222 j 022 k 03 l 0

26 26 The Four Points Condition Theorem: A set M of L objects is additive iff any subset of four objects can be labeled i,j,k,l so that: d(i,k) + d(j,l) = d(i,l) +d(k,j) d(i,j) + d(k,l) We call {{i,j},{k,l}} the split of {i,j,k,l}. i k l j Proof: Additivity 4 Points Condition: By the figure...

27 27 4P Condition Additivity: Induction on the number of objects, L. For L 3 the condition is empty and tree exists. Consider L=4. B = d(i,k) +d(j,l) = d(i,l) +d(j,k) d(i,j) + d(k,l) = A Let y = (B – A)/2 0. Then the tree should look as follows: We have to find the distances a,b, c and f. a b ij k m c y l n f

28 28 Tree construction for L=4 a b i j k m c y l n f Construct the tree by the given distances as follows: 1. Construct a tree for {i, j,k}, with internal vertex m 2. Add vertex n,d(m,n) = y 3. Add edge (n,l), c+f=d(k,l) n f n f n f Remains to prove: d(i,l) = d T (i,l) d(j,l) = d T (j,l)

29 29 Proof for L=4 a b i j k m c y l n f By the 4 points condition and the definition of y: d(i,l) = d(i,j) + d(k,l) +2y - d(k,j) = a + y + f = d T (i,l) (the middle equality holds since d(i,j), d(k,l) and d(k,j) are realized by the tree) d(j,l) = d T (j,l) is proved similarly.

30 30 Induction step for L>4: u Remove Object L from the set u By induction, there is a tree, T, for {1,2,…,L-1}. u For each pair of labeled nodes (i,j) in T, let a ij, b ij, c ij be defined by the following figure: a ij b ij c ij i j L m ij

31 31 Induction step: u Pick i and j that minimize c ij. u T is constructed by adding L (and possibly m ij ) to T, as in the figure. Then d(i,L) = d T (i,L) and d(j,L) = d T (j,L) Remains to prove: For each k i,j: d(k,L) = d T (k,L). a ij b ij c ij i j L m ij T

32 32 Induction step (cont.) Let k i,j be an arbitrary node in T, and let n be the branching point of k in the path from i to j. By the minimality of c ij, {{i,j},{k,L}} is not a split of {i,j,k,L}. So assume WLOG that {{i,L},{j,k}} is a split of {i,j, k,L}. a ij b ij c ij i j L m ij T k n

33 33 Induction step (end) Since {{i,L},{j,k}} is a split, by the 4 points condition d(L,k) = d(i,k) + d(L,j) - d(i,j) d(i,k) = d T (i,k) and d(i,j) = d T (i,j) by induction, and d(L,j) = d T (L,j) by the construction. Hence d(L,k) = d T (L,k). QED a ij b ij c ij i j L m ij T k n

34 34 From Additive Distance to a Tree By following the proof, the four point condition can be used to construct a tree from a distance matrix, or to decide that there is no such tree (namely that the distance is not additive). But this algorithm will go over all quartets, resulting in O(L 4 ) many steps for L species (too sllllllllllllow). The most popular method for constructing trees for additive sets uses the neighbor joining approach.

35 35 Constructing additive trees: The neighbor joining problem Let i, j be sisters (neighboring leaves) in a tree, let k be their father, and let m be any other vertex. Using eq. we can compute the distances from k to all other leaves. This suggest the following method to construct tree from an additive distance matrix: 1.Find sisters i,j in the tree, 2.Replace i,j by their father, k, and recursively construct a tree T for the smaller set. 3.Add i,j as children of k in T.

36 36 Neighbor Finding How can we find from distances alone a pair of sisters (neighboring leaves)? Closest nodes are not necessarily neighboring leaves. A B C D Next, we show a way to find neighbors from distances.

37 37 Neighbor Finding: Seitou & Nei method Theorem (Saitou&Nei) Assume d is additive, with all tree edge weights positive. If D(i,j) is minimal (among all pairs of leaves), then i and j are sister taxa in the tree. i j kl m T1T1 T2T2 The proof is rather involved, and will be skipped (no tears pls).skipped

38 38 Saitou & Nei proof (to be skipped) Notations used in the proof p(i,j) = the path from vertex i to vertex j; P(D,C) = (e 1,e 2,e 3 ) = (D,E,F,C) A B C D e1e1 e3e3 e2e2 For a vertex i, and an edge e=(i,j): N i (e) = |{k : e is on p(i,k)}|. N D (e 1 ) = 3, N D (e 2 ) = 2, N D (e 3 ) = 1 N C (e 1 ) = 1 EF

39 39 A simpler neighbor finding method: Select an arbitrary (fixed) node r. u For each pair of labeled nodes (i,j) let C(i,j) be defined by the following expression (also see figure): C(i,j) i j r Claim: Let i, j be such that C(i,j) is maximized. Then i and j are neighboring leaves.

40 40 Sisters Identification: Example A B C D Select arbitrarily r=A. C(B,C)=( )/2=5 C(B,D)=( )/2=8 C(C,D)=( )/2=5 Claim: Let i, j be such that C(i,j) is maximized. Then i and j are neighboring leaves.

41 41 Neighbor Joining Algorithm u Set M to contain all leaves, and select a root r. |M|=L u If L =2, return a tree of two vertices Iteration: u Choose i,j such that C(i,j) is maximal u Create a new vertex k, and update distances u remove i,j, and add k to M u Recursively construct a tree on the smaller set. u When done, add i,j as children on k, at distances d(i,k) and d(j,k). i j k m

42 42 Complexity of Neighbor Joining Algorithm Naive Implementation: Initialization: θ(L 2 ) to compute the C(i,j)s. Each Iteration: u O(L) to update {C(i,k):i L} for the new node k. u O(L 2 ) to find the maximal C(i,j). Total of O(L 3 ). i j k m

43 43 Complexity of Neighbor Joining Algorithm Using a Heap to store the C(i,j)s: Initialization: θ(L 2 ) to compute and heapify the C(i,j)s. Each Iteration: u O(1) to find the maximal C(i,j). u O(L log L) to delete {C(m,i), C(m,j)} and add C(m,k) for all vertices m. Total of O(L 2 log L). (implementation details are omitted)

44 44 Reconstructing Trees from Additive Matrices ABCDE A02747 B0747 C076 D07 E0 A C 1 B D E 3 3 Given a distance matrix constituting an additive metric, the topology of the corresponding additive tree is unique. Q: Do we have to test additivity before running NJ? A: This would be bad news, as this takes O(L 4 ) time!

45 45 Reconstructing Trees from Additive Matrices ABCDE A02747 B0747 C076 D07 E0 A C 1 B D E 3 3 Q: Do we have to test additivity before running NJ? A: By Seito-Nei, if matrix is additive, NJ will construct the correct tree. Algorithm does not care about awareness and need not know anything about the matrix!

46 46 NJ Algorithm: Example Identify i,j as neighbours if their divergence is minimal. Combine i,j into a new node u. update the distance matrix. If only 3 nodes are left – finish. Let r i be the sum of distances from i to every other node im jn kl

47 47 Distance Matrix ABCD A0236 B2035 C3306 D5660 U B A

48 48 Distance Matrix UCD U035.5 C306 D 60 U B A Y C

49 49 Distance Matrix YD Y05.6 D 0 U B A Y C D Z

50 50 Reconstructing Trees from non Additive Matrices Q: What if the distance matrix is not additive? A: We could still run NJ! Q: But can anything be said about the resulting tree? A: Not really. Resulting tree topology could even vary according to way ties are resolved on the way.

51 51 Almost Additive Matrix A distance matrix d is almost additive if there exists an additive matrix d such that Atteson: If d is almost additive with respect to a tree T, then the output of NJ is a tree T with the same topology as T

52 52 Distance Matrix

53 53 Unrooted Tree - NJ Root

54 54 Output - NJ Branch length is proportional to distance

55 55 N-J Method produces an Unrooted, Additive tree

56 56 What is required for the Neighbour joining method? Distance matrix 0. Distance Matrix Neighbor-Joining Method An Example

57 57 PAM distance 3.3 (Human - Monkey) is the minimum. So we'll join Human and Monkey to MonHum and we'll calculate the new distances. Mon-Hum MonkeyHumanSpinachMosquitoRice 1. First Step

58 58 After we have joined two species in a subtree we have to compute the distances from every other node to the new subtree. We do this with a simple average of distances: Dist[Spinach, MonHum] = (Dist[Spinach, Monkey] + Dist[Spinach, Human])/2 = ( )/2 = Mon-Hum MonkeyHumanSpinach 2. Calculation of New Distances

59 59 HumanMosquito Mon-Hum MonkeySpinachRice Mos-(Mon-Hum) 3. Next Cycle

60 60 HumanMosquito Mon-Hum MonkeySpinachRice Mos-(Mon-Hum) Spin-Rice 4. Penultimate Cycle

61 61 HumanMosquito Mon-Hum MonkeySpinachRice Mos-(Mon-Hum) Spin-Rice (Spin-Rice)-(Mos-(Mon-Hum)) 5. Last Joining

62 62 Human Monkey Mosquito Rice Spinach The result: Unrooted Neighbor-Joining Tree


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