. Intro to Phylogenetic Trees Lecture 5 Sections 7.1, 7.2, in Durbin et al. Chapter 17 in Gusfield Slides by Shlomo Moran. Slight modifications by Benny Chor

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 The Tree of Life Source: Alberts et al

4 D’après Ernst Haeckel, 1891 Tree of life- a better picture

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 Historical Note u Until mid 1950’s 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 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 Morphological topology Archonta Glires Ungulata Carnivora Insectivora Xenarthra (Based on Mc Kenna and Bell, 1997)

9 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).

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

11 Nuclear topology Cetartiodactyla Afrotheria Chiroptera Eulipotyphla Glires Xenarthra Carnivora Perissodactyla Scandentia+ Dermoptera Pholidota Primate (tree by Madsenl) (Based on Pupko et al. slide)

12 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

13 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

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

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

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

17 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

18 Type 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, or … u Character-based l Examine each character (e.g., residue) separately

19 Two Methods of Tree Construction u Distance- A weighted tree that realizes the distances between the objects. 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.

20 Exact solution: Additive sets Given a set M of L objects with an L×L distance matrix: u d(i,i)=0, and for i≠j, 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?

21 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.

22 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

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

24 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...

25 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

26 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)

27 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.

28 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

29 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’

30 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

31 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

32 Dangers of Paralogs 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