1. Phylogenetic trees Sushmita Roy BMI/CS 576 www.biostat.wisc.edu/bmi576/ sroy@biostat.wisc.edu Sep 23 rd, 2014

2. Key concepts in this section What are phylogenies or phylogenetic trees? – Terminology such as extant, ancestral, branch point, branch length, orthologs, paralogs Why build phylogenetic trees? Algorithms to build phylogenetic trees – Distance-based methods – Parsimony methods Minimize the number of changes – Probabilistic methods Find the tree that best explains the data using probabilistic models

3. Readings Chapter 7 – 7.1-7.5

4. What are phylogenetic trees? A tree that describes evolutionary relationships among entities – Species, genes, strains This relationship is called “phylogeny” Leaves represent extant (current day) species Internal nodes represent ancestral species Phylogenetics: – The task for inferring the phylogenetic tree from observations in existing organisms

5. Why phylogenetic trees? Inform multiple sequence alignments Identify signatures of conservation of sequence Understand how organisms are related – Do humans and chimpanzees share a common ancestor or do humans and gorillas? Ask how closely organisms are related – Humans and chimpanzees shard a common ancestor 5mya How specific functions/traits have evolved – What made us human? Conjecture the fate of specific regions of the genome – Will the human Y disappear?

6. From http://tellapallet.com/tree_of_life.htm Tree of life aims to represents the phylogeny of all species on earth

7. Tracing the evolution of the Ebola virus Ebola virus: a lethal human pathogen, fatality rate 78% Ebola is spreading now in Africa – Until recently the largest known case happened in 1976 (318 cases) – This year’s outbreak reported in Feb 2014 – As of 19 Aug 2014, 1229 deaths have been reported Largest known in history Key questions – Where did the pathogen come from? – How is it evolving? In a 2014 Science paper, researchers reported whole genome sequence alignment of 78 Ebola virus samples

8. Phylogenetic tree of the Ebola virus Gire et al, Science 2014 Three recent outbreaks from the same ancestor

9. Insights gained from sequence comparison “Genetic similarity across the sequenced 2014 samples suggests a single transmission from the natural reservoir, followed by human-to-human transmission during the outbreak” “..data suggest that the Sierra Leone outbreak stemmed from the introduction of two genetically distinct viruses from Guinea around the same time…” “..the catalog of 395 mutations, including 50 fixed nonsynonymous changes with 8 at positions with high levels of conservation across ebola viruses, provides a starting point for such studies” Gire et al., Science 2014

10. Phylogenetic tree basics Leaves represent entities(genes, species, individuals/strains) being compared – the term taxon (taxa plural) is used to refer to these when they represent species and broader classifications of organisms – For example if taxa are species, the tree is a species tree Internal nodes are ancestral units Phylogenetic trees can be rooted or unrooted – the root represents the common ancestor In a rooted tree, path from root to a node represents an evolutionary path – Gives directionality to evolutionary time An unrooted tree specifies relationships among taxa, but not from an ancestor

11. Tree basics 6 6 2 2 1 1 8 8 5 5 7 7 3 3 4 4 Branch Leaf node: Extant Internal node: Ancestral For a species tree, internal nodes represent speciation events 1 1 2 2 6 6 Unrooted treeRooted tree Each tree topology represents a different evolutionary history 3 3 4 4 5 5 7 7 8 8 9 9 Branch length Branch length describes the evolutionary divergence between two nodes

12. Orthologs and paralogs Orthologs: – Two sequences in two species that have a a common ancestor – Diverged due to a speciation event – Used to create a “species tree” Paralogs: – Two sequences in the same species that arose from a gene duplication event – Captured in a “gene tree”.

13. Tree counting A rooted tree with n leaf nodes has – n-1 internal nodes – 2n-2 edges/branches An unrooted tree with n leaf nodes has – n-2 internal nodes – 2n-3 edges/branches – A root can be added to any of these branches to give 2n-3 rooted trees for any unrooted tree E.g. for n=3 there is one unrooted tree and three rooted trees

14. Tree counting 1 2 3 1 2 3 1 2 3 1 2 3 3 1 2 1 2 3 2 1 3 An unrooted tree Possible positions for rootRooted trees

15. Tree counting Instead of adding a root we could add a branch for the n+1 th taxon 1 2 3 1 2 3 1 2 3 1 2 3 3 2 1 4 1 2 3 4 1 2 3 4

16. Tree counting A tree with 3 nodes can be grown in (2*3)-3=3 ways to make a tree of 4 nodes Each tree with 4 nodes can be grown in (2*4)-3=5 ways to make a tree of 5 nodes – So we have 3*5 trees Each tree of 5 nodes can be grown in (2*5)-3=7 – So we have 3*5*7 In general for n nodes we can have – (1)*(3)*(5)*..(2n-5) unrooted trees

17. Tree counting This grows very fast – For n=10, we have 2 million unrooted trees – For n=20, we have 2.2*10 20

18. Constructing phylogenetic trees Three types of methods – Distance based methods – Parsimony methods – Probabilistic approaches Most methods start with pairwise distance methods

19. Methods for phylogenetic tree reconstruction Distance-based methods – UPGMA – Neighbor joining – Assume additivity and sometimes a “molecular clock” – Additivity means we can add up the branch lengths of the tree connecting two nodes and get their distances. Alignment-based methods – Parsimony – Probabilistic

20. Defining distance between sequences Fractional alignment difference for two sequences i and j – p ij = m ij /L ij Gives an estimate of changes per site – m ij : Number of mismatches – Assumes that changes have happened only once Underestimates the distance between sequences – Assumes all sequences change at the same rate Jukes Cantor distance – The simplest evolutionary distance d ij between sequences i and j, p ij fractional mismatch

21. UPGMA algorithm for phylogenetic tree reconstruction UPGMA: Unweighted pair group method using arithmetic averages Represent all sequences as the leaf nodes of a tree Merge two closest nodes at a time to create a new node in the tree – Set new node at height determined by nodes being merged – Recompute distance between new node and all other nodes Leaf nodes have one sequence Intermediate nodes have multiple sequences We will call sequences associated with an intermediate node i cluster C i Need to compute – Distance between two clusters of sequences – Height

22. Computing distance between clusters Let i and j be two nodes Let C i be the cluster of sequences for node i Let C j be the cluster of sequences for node j |C j | : Number of sequences in C j Distance between nodes i and j

23. Computing distance from a new node Let k be a new node to be created from merging i and j Let C i be the cluster of sequences for node i Let C j be the cluster of sequences for node j Distance between nodes k and l, l!=i and l!=j

24. UPGMA algorithm Input – n sequences – Distance matrix for the n sequences Output – T Initialization – Assign each sequence i to its own cluster C i – Define one leaf of T for each sequence Iterate until only two clusters remain – Find two nodes Ci and Cj that have the smallest dij – Define new cluster Ck = Ci Ucj – Define daughters of k as I and j, place at dij/2 – Add k to cluster set. Remove I and j from the set of clusters Terminate – When only two clusters Ci and Cj remain, place root at d ij /2

25. Example of creating distance matrix Consider four sequences 1.AAAC 2.AGC 3.ACC 4.GAC Generate pairwise alignments for all pairs of sequences Define distance as 1-%identity

26. Pairwise alignment for all the pairs of sequences AAAC _AGC AAAC _ACC AAAC _GAC 1. and 2. 1. and 3. 1. and 4. 2. and 3. 2. and 4. AGC ACC AGC GAC 3. and 4. ACC GAC Sequence pairAlignment X0.5 X0.330.67 X X 1. 2.3.4. 2. 3. 4. % similarity 2/4 2/3 1/3 Distance 0.5 0.33 0.67

27. UPGMA in practice 1234 X0.5 X0.330.67 X X 1 1234 2 3 4 5 X0.5 X0.67 X 1 4 5 145 d 23 /2=1/6 Place new node at height d 23 /2

28. UPGMA in practice 12 3 4 5 X0.5 X0.67 X 1 4 5 145 1/6 d 14 /2=0.25 6 X0.59 X 5 6 12 3 4 5 1/6 d 14 /2=0.25 6 7 d 56 /2=0.29

29. UPGMA relies on the molecular clock assumption Sequences diverge at the same rate at different points in the phylogeny Distance from any leaf to root is the same. If this is true the distances are said to have an “ultrametric” property This assumption is rarely true in practice

30. The molecular clock assumption & ultrametric data Ultrametric data: for any triplet of sequences, i, j, k, the distances are either all equal, or two are equal and the remaining one is smaller ABCDE A08853 B0388 C088 D05 E0 AEDBC 1 2 3 4

31. Problems with the molecular clock assumption 1 2 3 4 Actual tree 2341 Constructed by UPGMA

32. Neighbor joining The ultra-metric property is too strong – Most sequences diverge at different rates A more relaxed requirement is that of additivity – Distance between a pair of species/nodes is equal to the sum of the branch lengths Uses a similar idea to construct trees as UPGMA – That is consider pairs of nodes and joins them Produces unrooted trees

33. How to select nodes for merging? Given all pairwise distances for n sequences d ij denote the distance between node i and j Should we select node pairs with the smallest dij? A B C D 0.4 0.1 Should we merge A and B?

34. Need to correct for long branches r i : Average distance from other nodes L : current set of leaves

35. Defining the distance to a new node i j m k d km ? New node Given d ij, d im, d jm, how to calculate distance to new node k?

36. Algorithm for NJ Initialization – T be set the of leaf nodes – L = T – Estimate r i for all i in L – Estimate D ij Iteration – Pick a pair i, j from L such that D ij is smallest – Define new node k – Estimate d ik, d jk, add edge between k and i, and between k to j – Add k to T, remove i and j from L – Estimate D mn for all nodes m, n in L Terminate – If L has two nodes, add the edge between these two.

37. An example with neighbor joining Consider 5 sequences: A, B, C, D, E Distance matrix What is the tree inferred by the Neighbor joining algorithm? 5498 5109 76 7 A B C D E BCDE

38. Can we check for additivity? Check for additivity: For four leaves, i, j, k, l and the distances d ij, d ik, d il, d jk, d jl, d kl i j k l The three sums of two distances i j k l i j k l i j k l Should be such that two of these are equal, and larger than the third.

39. Rooting a tree An unrooted tree can be converted to a rooted tree using an outgroup species Outgroup: a species known to be more distantly related all the species than each of the species themselves Find the branch where the outgroup is selected to be added That gives the root