1. . Phylogenetic Trees Lecture 11 Sections 7.1, 7.2, in Durbin et al.

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 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.

5. 5 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

6. 6 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

7. 7 Morphological topology Archonta Glires Ungulata Carnivora Insectivora Xenarthra (Based on Mc Kenna and Bell, 1997)

8. 8 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).

9. 9 Perissodactyla Carnivora Cetartiodactyla Rodentia 1 Hedgehogs Rodentia 2 Primates Chiroptera Moles+Shrews Afrotheria Xenarthra Lagomorpha + Scandentia Mitochondrial topology (Based on Pupko et al.,)

10. 10 Nuclear topology Cetartiodactyla Afrotheria Chiroptera Eulipotyphla Glires Xenarthra Carnivora Perissodactyla Scandentia+ Dermoptera Pholidota Primate (tree by Madsenl) (Based on Pupko et al. slide)

11. 11 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

12. 12 Phylogenenetic trees u Leaves - current day species (or taxa – plural of taxon) u Internal vertices - hypothetical common ancestors u Edges length - “time” from one speciation to the next AardvarkBisonChimpDogElephant

13. 13 Dangers in Molecular Phylogenies u We have to emphasize that gene/protein sequence can be homologous for several different reasons: u Orthologs -- sequences diverged after a speciation event u Paralogs -- sequences diverged after a duplication event u Xenologs -- sequences diverged after a horizontal transfer (e.g., by virus)

14. 14 Dangers of Paralogs 1 2 3 Consider evolutionary tree of three taxa: …and assume that at some point in the past a gene duplication event occurred. Gene Duplication

15. 15 Dangers of Paralogs Speciation events Gene Duplication 1A 2A 3A3B 2B1B The gene evolution is described by this tree (A, B are the copies of the same gene).

16. 16 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 – created from a common ancestor by specification events. S S S

17. 17 Types of Trees A natural model to consider is that of rooted trees Common Ancestor

18. 18 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.

19. 19 Rooted versus unrooted trees Tree a a b Tree b c Tree c Represents the three rooted trees

20. 20 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

21. 21 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

22. 22 Two Methods of Tree Construction u Distance- A weighted tree that realizes the distances between the objects. u Parsimony – A tree with a total minimum number of character changes between nodes. 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.

23. 23 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?

24. 24 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.

25. 25 Three objects sets are additive: For L=3: There is always a (unique) tree with one internal node. a b c i j k m For instance ijk i 0 a+ba+c j 0 b+c k 0

26. 26 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

27. 27 The Four Points Condition A necessary condition for a set of four objects to be additive: its 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) Proof: By the figure... {{i,j},{k,l}} is a “split” of {i,j,k,l}. i k l j

28. 28 The Four Points Condition Definition: A set M of L objects satisfies the four points condition 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) i k l j

29. 29 The Four Points Condition Theorem: The following 3 conditions are equivalent for a distance matrix D on a set M of L objects 1. D is additive 2. D satisfies the four points condition for all quartets in M. 3. There is an object r in M, s.t. D satisfies the 4 points condition for all quartets that include r. i k l j

30. 30 The Four Points Condition Proof: we’ll show that 1  2  3  1. 1  2 Additivity  4P Condition satisfied by al quartets: By the figure... i k l j 2  3: trivial

31. 31 Proof that 3  1 Induction on the number of objects, L. For L ≤ 3 the condition is trivially true and a tree exists. For L=4: Consider 4 points which satisfy d(i,k) +d(j,l) = d(i,l) +d(j,k) ≥ d(i,j) + d(k,l) a b ij k m c y l n f We will construct a tree T with 4 leaves, s.t. d T (,x,y) = d(x,y) for each pair x,y in {i,j,k,l},

32. 32 Tree construction for L=4 i j k m l Assume split {{i,j},{k,l}}: d (i,j)+d (k,l)  d (j,k)+d (i,l) 1.Construct a tree for {i, j,k}, with internal vertex m 2.Construct a tree for {i,k,l}, by adding the vertex n and the edge (n,l). n The construction guarantees that d T (,x,y)=d(x,y) for all (x,y) except (j,l).

33. 33 Tree construction for L=4 i j k m l n d T (,x,y)=d(x,y) for all (x,y) except (j,l). Thus, since d T (i,j) + d T (k,l)  d T (j,k) + d T (i,l), {{i,j},{k,l}} is a split of the tree T. By the proof that 1  2, we have for the tree T: d(j,l) = d(i,l)+ d(j,k)- d(i,k)= d T (i,l)+ d T (j,k)- d T (i,k)= d T (j,l) And hence d T (x,y)=d(x,y) for all x,y.

34. 34 i j k l Corollary from the construction Corollary F: If d(i,k) +d(j,l) = d(i,l) +d(j,k) ≥ d(i,j) + d(k,l), then there is a unique tree which realizes all the distances except d(j,l), and this tree realizes also the distance d(j,l).* *(j,l) can be replaced by any pair in {i,j}  {k,l}.

35. 35 Induction step for L>4: u For each pair of labeled nodes (i,j) in T’, let c ij be defined by the following figure: c ij i j r m ij u Pick i and j that maximize c ij.

36. 36 Induction step: u Construct (by induction) T’ on M \ {i}. u Add i (and possibly m ij ) to T’, as in the figure. Then d(i,r) = d T (i,L) and d(j,r) = d T (j,r) Remains to prove: For each k  {r,j} it holds that : d(i,k) = d T (i,k). c ij i j r m ij T’

37. 37 Induction step (cont.) Let k ≠i,r be an arbitrary node in T’. The maximality of c ij means that {{r,k},{i,j}} is a split of {i,j,k,r}. Thus, by Corollary F, since d(x,y)=d T (x,y) for each x,y in {i,j,k,r}, except d(k,i), we have also that d(k,i)=d T (k,i) too. c ij i j r m ij T’ k

38. 38 Constructing additive trees: The neighbor joining problem Let i, j be neighboring leaves in a tree, let k be their parent, and let m be any other vertex. The formula shows that we can compute the distances of k to all other leaves. This suggest the following method to construct tree from a distance matrix: 1.Find neighboring leaves i,j in the tree, 2. Replace i,j by their parent k and recursively construct a tree T for the smaller set. 3.Add i,j as children of k in T.

39. 39 Neighbor Finding How can we find from distances alone a pair of nodes which are neighboring leaves? Closest nodes aren’t necessarily neighboring leaves. A B C D Next we show one way to find neighbors from distances.

40. 40 Neighbor Finding: Seitou&Nei method Theorem (Saitou&Nei) Assume all edge weights are positive. If D(i,j) is minimal (among all pairs of leaves), then i and j are neighboring leaves in the tree. Definitions