Molecular Evolution and Phylogenetic Tree Reconstruction

1. 1 4 3 5 2 5 2 3 1 4 Molecular Evolution and Phylogenetic Tree Reconstruction

2. Phylogenetic Trees • Nodes: species • Edges: time of independent evolution • Edge length represents evolution time • AKA genetic distance • Not necessarily chronological time

3. Inferring Phylogenetic Trees Trees can be inferred by several criteria: • Morphology of the organisms • Can lead to mistakes! • Sequence comparison Example: Mouse: ACAGTGACGCCCCAAACGT Rat: ACAGTGACGCTACAAACGT Baboon: CCTGTGACGTAACAAACGA Chimp: CCTGTGACGTAGCAAACGA Human: CCTGTGACGTAGCAAACGA

4. Inferring Phylogenetic Trees Sequence-based methods Deterministic (Parsimony) Probabilistic (SEMPHY) Distance-based methods UPGMA Neighbor-Joining Can compute distances from sequences

5. Distance Between Two Sequences Basic principles: Degree of sequence difference is proportional to length of independent sequence evolution Only use positions where alignment is certain – avoid areas with (too many) gaps

6. Distance Between Two Sequences Given sequences xi, xj, Define dij = distance between the two sequences One possible definition: dij = fraction f of sites u where xi[u]  xj[u] Better scores are derived by modeling evolution as a continuous change process

7. Outline • Molecular Evolution • Distance Methods • UPGMA / Average Linkage • Neighbor-Joining • Sequence Methods • Deterministic (Parsimony) • Probabilistic (SEMPHY)

8. Molecular Evolution Q: How can we model evolution on nucleotide level? (ignore gaps, focus on substitutions) A: Consider what happens at a specific position for small time interval Δt P(t) = vector of probabilities of {A,C,G,T} at time t μAC = rate of transition from A to C per unit time μA = μAC + μAG + μAT rate of transition out of A pA(t+Δt) = pA(t) – pA(t) μA Δt + pC(t) μCA Δt + …

9. Molecular Evolution In matrix/vector notation, we get P(t+Δt) = P(t) + Q P(t) Δt where Q is the substitution rate matrix

10. Molecular Evolution This is a differential equation: P’(t) = Q P(t) A substitution rate matrix Q implies a probability distribution over {A,C,G,T} at each position, including stationary (equilibrium) frequencies πA, πC, πG, πT Each Q is an evolutionary model (some work better than others)

11. Evolutionary Models Jukes-Cantor Kimura Felsenstein HKY

12. Estimating Distances Solve the differential equation and compute expected evolutionary time given sequences Jukes-Cantor Kimura

13. Outline Molecular Evolution Distance Methods UPGMA / Average Linkage Neighbor-Joining Sequence Methods Deterministic (Parsimony) Probabilistic (SEMPHY)

14. A simple clustering method for building tree UPGMA (unweighted pair group method using arithmetic averages) Or the Average Linkage Method Given two disjoint clusters Ci, Cj of sequences, 1 dij = ––––––––– {p Ci, q Cj}dpq |Ci|  |Cj| Claim that if Ck = Ci  Cj, then distance to another cluster Cl is: dil |Ci| + djl |Cj| dkl = –––––––––––––– |Ci| + |Cj|

15. Algorithm: Average Linkage 1 4 Initialization: Assign each xi into its own cluster Ci Define one leaf per sequence, height 0 Iteration: Find two clusters Ci, Cj s.t. dij is min Let Ck = Ci  Cj Define node connecting Ci, Cj, and place it at height dij/2 Delete Ci, Cj Termination: When two clusters i, j remain, place root at height dij/2 3 5 2 5 2 3 1 4

16. Average Linkage Example 4 3 2 1 v w x y z

17. Ultrametric Distances and Molecular Clock Definition: A distance function d(.,.) is ultrametric if for any three distances dij dik  dij, it is true that dij dik = dij The Molecular Clock: The evolutionary distance between species x and y is 2 the Earth time to reach the nearest common ancestor That is, the molecular clock has constant rate in all species The molecular clock results in ultrametric distances years 1 4 2 3 5

18. Ultrametric Distances & Average Linkage Average Linkage is guaranteed to reconstruct correctly a binary tree with ultrametric distances Proof: Exercise 5 1 4 2 3

19. Weakness of Average Linkage Molecular clock: all species evolve at the same rate (Earth time) However, certain species (e.g., mouse, rat) evolve much faster Example where UPGMA messes up: AL tree Correct tree 3 2 1 3 4 4 2 1

20. d1,4 Additive Distances 1 Given a tree, a distance measure is additive if the distance between any pair of leaves is the sum of lengths of edges connecting them Given a tree T & additive distances dij, can uniquely reconstruct edge lengths: • Find two neighboring leaves i, j, with common parent k • Place parent node k at distance dkm = ½ (dim + djm – dij) from any node m  i, j 4 12 8 3 13 7 9 5 11 10 6 2

21. Reconstructing Additive Distances Given T x T D y 5 4 3 z 3 4 w 7 6 v If we know T and D, but do not know the length of each leaf, we can reconstruct those lengths

22. Reconstructing Additive Distances Given T x T D y z w v

23. Reconstructing Additive Distances Given T D x T y z a w D1 v dax = ½ (dvx + dwx – dvw) day = ½ (dvy + dwy – dvw) daz = ½ (dvz + dwz – dvw)

24. Reconstructing Additive Distances Given T D1 x T y 5 4 b 3 z 3 a 4 c w 7 D2 6 d(a, c) = 3 d(b, c) = d(a, b) – d(a, c) = 3 d(c, z) = d(a, z) – d(a, c) = 7 d(b, x) = d(a, x) – d(a, b) = 5 d(b, y) = d(a, y) – d(a, b) = 4 d(a, w) = d(z, w) – d(a, z) = 4 d(a, v) = d(z, v) – d(a, z) = 6 Correct!!! v D3

25. Neighbor-Joining • Guaranteed to produce the correct tree if distance is additive • May produce a good tree even when distance is not additive Step 1: Finding neighboring leaves Define Dij = (N – 2) dij – kidik – kjdjk Claim: The above “magic trick” ensures that Dij is minimal iffi, j are neighbors 1 3 0.1 0.1 0.1 0.4 0.4 4 2

26. Algorithm: Neighbor-Joining Initialization: Define T to be the set of leaf nodes, one per sequence Let L = T Iteration: Pick i, j s.t. Dij is minimal Define a new node k, and set dkm = ½ (dim + djm – dij) for all m  L Add k to T, with edges of lengths dik = ½ (dij + ri – rj), djk = dij – dik where ri = (N – 2)-1kidik Remove i, j from L; Add k to L Termination: When L consists of two nodes, i, j, and the edge between them of length dij

27. Outline Molecular Evolution Distance Methods UPGMA / Average Linkage Neighbor-Joining Sequence Methods Deterministic (Parsimony) Probabilistic (SEMPHY)

28. Parsimony • One of the most popular methods: • GIVEN multiple alignment • FIND tree & history of substitutions explaining alignment Idea: Find the tree that explains the observed sequences with a minimal number of substitutions Two computational subproblems: • Find the parsimony cost of a given tree (easy) • Search through all tree topologies (hard)

29. Example: Parsimony Cost of One Column {A} C = 1 {A} {A, B} C++ A B A A A A B A {B} {A} {A} {A}

30. Parsimony Scoring Given a tree, and an alignment column u Label internal nodes to minimize the number of required substitutions Initialization: Set cost C = 0; node k = 2N – 1 (last leaf) Iteration: If k is a leaf, set Rk = { xk[u] } // Rk is simply the character of kth species If k is not a leaf, Let i, j be the daughter nodes; Set Rk = Ri Rj if intersection is nonempty Set Rk = Ri  Rj, and increment C if intersection is empty Termination: Minimal cost of tree for column u, = C

31. Example {B} {A,B} {A} {B} {A} {A,B} {A} A A A A B B A B {A} {A} {A} {A} {B} {B} {A} {B}

32. Parsimony Traceback Traceback: • Choose an arbitrary nucleotide from R2N – 1 for the root • Having chosen nucleotide r for parent k, If r  Ri choose r for daughter i Else, choose arbitrary nucleotide from Ri Easy to see that this traceback produces some assignment of cost C

33. Another Parsimony Algorithm Let C(v) be cost for subtree rooted at node v Let C(v,x) be cost for subtree rooted at v if we force v to have value x Initialization: For each leaf v C(v) = 0 C(v,x) = 0 if x is input character that labels v; C(v,x) = ∞ otherwise Iteration: Let u, w be children of v C(v,x) = min(C(u) + 1, C(u,x)) + min(C(v) + 1, C(v,x)) C(v) = min C(v,x) Termination: Minimal cost is C(root)

34. Probabilistic Methods A more refined measure of evolution along a tree than parsimony P(x1, x2, xroot | t1, t2) = P(xroot) P(x1 | t1, xroot) P(x2 | t2, xroot) If we use Jukes-Cantor, for example, and x1 = xroot = A, x2 = C, t1 = t2 = 1, = pA¼(1 + 3e-4α) ¼(1 – e-4α) = (¼)3(1 + 3e-4α)(1 – e-4α) xroot t1 t2 x1 x2

35. Probabilistic Methods xroot • If we know all internal labels xu, P(x1, x2, …, xN, xN+1, …, x2N-1 | T, t) = P(xroot)jrootP(xj | xparent(j), tj, parent(j)) • Usually we don’t know the internal labels, therefore P(x1, x2, …, xN | T, t) = xN+1 xN+2 … x2N-1P(x1, x2, …, x2N-1 | T, t) xu x2 xN x1

36. Felsenstein’s Likelihood Algorithm Define: and recursively compute:

37. Felsenstein’s Likelihood Algorithm Now using u and U we can compute: and

38. Probabilistic Methods Given M (ungapped) alignment columns of N sequences, • Define likelihood of a tree: L(T, t) = P(Data | T, t) = m=1…M P(x1m, …, xnm | T, t) Maximum Likelihood Reconstruction: • Given data X = (xij), find a topology T and length vector t that maximize likelihood L(T, t)

39. Current popular methods HUNDREDS of programs available! http://evolution.genetics.washington.edu/phylip/software.html#methods Some recommended programs: • Discrete—Parsimony-based • Rec-1-DCM3 http://www.cs.utexas.edu/users/tandy/mp.html Tandy Warnow and colleagues • Probabilistic • SEMPHY http://www.cs.huji.ac.il/labs/compbio/semphy/ Nir Friedman and colleagues