Phylogeny Tree Reconstruction

1 4 3 5 2 5 2 3 1 4 Phylogeny Tree Reconstruction

Inferring Phylogenies Trees can be inferred by several criteria: • Morphology of the organisms • Sequence comparison Example: Orc: ACAGTGACGCCCCAAACGT Elf: ACAGTGACGCTACAAACGT Dwarf: CCTGTGACGTAACAAACGA Hobbit: CCTGTGACGTAGCAAACGA Human: CCTGTGACGTAGCAAACGA

Modeling Evolution During infinitesimal time t, there is not enough time for two substitutions to happen on the same nucleotide So we can estimate P(x | y, t), for x, y  {A, C, G, T} Then let P(A|A, t) …… P(A|T, t) S(t) = … … … … P(T|A, t) …… P(T|T, t) x x t y

Modeling Evolution A C Reasonable assumption: multiplicative (implying a stationary Markov process) S(t+t’) = S(t)S(t’) That is, P(x | y, t+t’) = z P(x | z, t) P(z | y, t’) Jukes-Cantor: constant rate of evolution 1 - 3    For short time , S() = I+R =  1 - 3     1 - 3     1 - 3 T G

Modeling Evolution Jukes-Cantor: For longer times, r(t) s(t) s(t) s(t) S(t) = s(t) r(t) s(t) s(t) s(t) s(t) r(t) s(t) s(t) s(t) s(t) r(t) Where we can derive: r(t) = ¼ (1 + 3 e-4t) s(t) = ¼ (1 – e-4t) S(t+) = S(t)S() = S(t)(I + R) Therefore, (S(t+) – S(t))/ = S(t) R At the limit of   0, S’(t) = S(t) R Equivalently, r’ = -3r + 3s s’ = -s + r Those diff. equations lead to: r(t) = ¼ (1 + 3 e-4t) s(t) = ¼ (1 – e-4t)

Modeling Evolution Kimura: Transitions: A/G, C/T Transversions: A/T, A/C, G/T, C/G Transitions (rate ) are much more likely than transversions (rate ) r(t)s(t)u(t)s(t) S(t) = s(t)r(t)s(t)u(t) u(t)s(t)r(t)s(t) s(t)u(t)s(t)r(t) Where s(t) = ¼ (1 – e-4t) u(t) = ¼ (1 + e-4t – e-2(+)t) r(t) = 1 – 2s(t) – u(t)

Phylogeny and sequence comparison Basic principles: • Degree of sequence difference is proportional to length of independent sequence evolution • Only use positions where alignment is pretty certain – avoid areas with (too many) gaps

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 model (Jukes-Cantor): f = ¾ (1 – e-4t)  ¾ e-4t = ¾ - f  log (e-4t) = log (1 – 4/3 f) dij = t = - ¼ -1 log(1 – 4/3 f)

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| Proof Ci,Cl dpq + Cj,Cl dpq dkl = –––––––––––––––– (|Ci| + |Cj|) |Cl| |Ci|/(|Ci||Cl|) Ci,Cl dpq + |Cj|/(|Cj||Cl|) Cj,Cl dpq = –––––––––––––––––––––––––––––––––––– (|Ci| + |Cj|) |Ci| dil + |Cj| djl = ––––––––––––– (|Ci| + |Cj|)

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, & 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

Example 4 3 2 1 v w x y z

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

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

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

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 4 12 8 3 13 7 9 5 11 10 6 2

Additive Distances z For any four leaves x, y, z, w, consider the three sums d(x, y) + d(z, w) d(x, z) + d(y, w) d(x, w) + d(y, z) One of them is smaller than the other two, which are equal d(x, y) + d(z, w) < d(x, z) + d(y, w) = d(x, w) + d(y, z) x w y

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

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

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)

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

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 = dij – (ri + rj) Where 1 ri = –––––k dik |L| - 2 Claim: The above “magic trick” ensures that Dij is minimal iffi, j are neighbors Proof: Very technical, please read Durbin et al.! 1 3 0.1 0.1 0.1 0.4 0.4 4 2

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) 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

Parsimony • One of the most popular methods 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)

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; k = 2N – 1 Iteration: If k is a leaf, set Rk = { xk[u] } 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 C += 1, if intersection is empty Termination: Minimal cost of tree for column u, = C

Example {A, B} C+=1 {A} {A, B} C+=1 B A B A {B} {B} {A} {A}

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}

Traceback to find ancestral nucleotides 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

Example Admissible with Traceback x B Still optimal, but inadmissible with Traceback A {A, B} B A x {A} B A A B B {A, B} x B x A A B B A A A B B {B} {A} {A} {B} A x A x A A B B

Number of labeled unrooted tree topologies • How many possibilities are there for leaf 4? 2 1 4 4 4 3

Number of labeled unrooted tree topologies • How many possibilities are there for leaf 4? For the 4th leaf, there are 3 possibilities 2 1 4 3

Number of labeled unrooted tree topologies • How many possibilities are there for leaf 5? For the 5th leaf, there are 5 possibilities 2 1 4 5 3

Number of labeled unrooted tree topologies • How many possibilities are there for leaf 6? For the 6th leaf, there are 7 possibilities 2 1 4 5 3

Number of labeled unrooted tree topologies • How many possibilities are there for leaf n? For the nth leaf, there are 2n – 5 possibilities 2 1 4 5 3

Number of labeled unrooted tree topologies • #unrooted trees for n taxa: (2n-5)*(2n-7)*...*3*1 = (2n-5)! / [2n-3*(n-3)!] • #rooted trees for n taxa: (2n-3)*(2n-5)*(2n-7)*...*3 = (2n-3)! / [2n-2*(n-2)!] 2 1 N = 10 #unrooted: 2,027,025 #rooted: 34,459,425 N = 30 #unrooted: 8.7x1036 #rooted: 4.95x1038 4 5 3

Search through tree topologies: Branch and Bound Observation: adding an edge to an existing tree can only increase the parsimony cost Enumerate all unrooted trees with at most n leaves: [i3][i5][i7]……[i2N–5]] where each ik can take values from 0 (no edge) to k At each point keep C = smallest cost so far for a complete tree Start B&B with tree [1][0][0]……[0] Whenever cost of current tree T is > C, then: • T is not optimal • Any tree extending T with more edges is not optimal: Increment by 1 the rightmost nonzero counter

Bootstrapping to get the best trees Main outline of algorithm • Select random columns from a multiple alignment – one column can then appear several times • Build a phylogenetic tree based on the random sample from (1) • Repeat (1), (2) many (say, 1000) times • Output the tree that is constructed most frequently

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

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

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)

Phylogeny Tree Reconstruction