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Molecular Evolution and Phylogeny

Molecular Evolution and Phylogeny. Early Evolutionary Studies. Anatomical features were the dominant criteria used to derive evolutionary relationships between species since Darwin till early 1960s

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Molecular Evolution and Phylogeny

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  1. Molecular Evolution and Phylogeny

  2. Early Evolutionary Studies • Anatomical features were the dominant criteria used to derive evolutionary relationships between species since Darwin till early 1960s • The evolutionary relationships derived from these relatively subjective observations were often inconclusive. Some of them were later proved incorrect

  3. Evolutionary Trees from DNA analysis • For roughly 100 years scientists were unable to figure out which family the giant panda belongs to • Giant pandas look like bears but have features that are unusual for bears and typical for raccoons, e.g., they do not hibernate • In 1985, Steven O’Brien and colleagues solved the giant panda classification problem using DNA sequences and algorithms

  4. 1 2 3 4 5 http://www.becominghuman.org Out of Africa Hypothesis • DNA-based reconstruction of the human evolutionary tree led to the Out of Africa Hypothesis thatclaims our most ancient ancestor lived in Africa roughly 200,000 years ago

  5. Human Evolutionary Tree http://www.mun.ca/biology/scarr/Out_of_Africa2.htm

  6. The Origin of Humans: ”Out of Africa” vs Multiregional Hypothesis Out of Africa: • Humans evolved in Africa ~150,000 years ago • Humans migrated out of Africa, replacing other humanoids around the globe Multiregional: • Humans evolved in the last two million years as a single species. Independent appearance of modern traits in different areas • Humans migrated out of Africa mixing with other humanoids on the way

  7. Evolutionary Tree of Humans (mtDNA) • African population is the most diverse (sub-populations had more time to diverge) • Evolutionary tree separates one group of Africans from a group containing all five populations. • Tree rooted on branch between groups of greatest difference. Vigilant, Stoneking, Harpending, Hawkes, and Wilson (1991)

  8. Evolutionary Tree of Humans: (microsatellites) • Neighbor joining tree for 14 human populations genotyped with 30 microsatellite loci.

  9. 1 4 3 5 2 5 2 3 1 4 Phylogenetic Trees How are these trees built from DNA sequences? • Leaves represent existing species • Internal vertices represent ancestors • Root represents the oldest evolutionary ancestor

  10. Rooted and Unrooted Trees • In the unrooted tree the position of the root (“oldest ancestor”) is unknown. Otherwise, they are like rooted trees

  11. 1 4 3 5 2 5 2 3 1 4 Phylogenetic Trees How are these trees built from DNA sequences? Methods • Distance • Parsimony Minimum number of mutations • Likelihood Probabilistic model of mutations

  12. Distances in Trees • Edges may have weights reflecting: • Number of mutations on evolutionary path from one species to another • Time estimate for evolution of one species into another • In a tree T, we often compute dij(T) - the length of a path between leaves i and j dij(T) – treedistance between i and j

  13. j i Distance in Trees: an Example d1,4 = 12 + 13 + 14 + 17 + 13 = 69

  14. Distance Matrix • n x n distance matrixDij for n species • Example: Dij = edit distance between a gene in species i and species j. Dij – editdistance between i and j Rat: ACAGTGACGCCCCAAACGT Mouse: ACAGTGACGCTACAAACGT Gorilla: CCTGTGACGTAACAAACGA Chimpanzee: CCTGTGACGTAGCAAACGA Human: CCTGTGACGTAGCAAACGA

  15. Edit Distance vs. Tree Distance • n x n distance matrixDij for n species • Example: Dij = edit distance between a gene in species i and species j. Dij – editdistance between i and j • Note the difference with dij(T) – tree distance between i and j Rat: ACAGTGACGCCCCAAACGT Mouse: ACAGTGACGCTACAAACGT Gorilla: CCTGTGACGTAACAAACGA Chimpanzee: CCTGTGACGTAGCAAACGA Human: CCTGTGACGTAGCAAACGA

  16. Fitting Distance Matrix • Given n species, we can compute the n x n distance matrixDij • Evolution of these genes is described by a tree that we don’t know. • We need an algorithm to construct a tree that best fits the distance matrix Dij Unknown topology of tree makes evolutionary tree reconstruction hard! # unrooted binary trees n leaves = (2n-3)! / ((n-2)! 2n-2) n = 24: 5.74 £ 1026

  17. Fitting Distance Matrix • Fitting means Dij = dij(T) Lengths of path in an (unknown) tree T Edit distance between species (known)

  18. Tree reconstruction for any 3x3 matrix is straightforward We have 3 leaves i, j, k and a center vertex c Reconstructing a 3 Leaved Tree Observe: dic + djc = Dij dic + dkc = Dik djc + dkc = Djk

  19. Reconstructing a 3 Leaved Tree(cont’d) dic + djc = Dij + dic + dkc = Dik 2dic + djc + dkc = Dij + Dik 2dic + Djk = Dij + Dik dic = (Dij + Dik – Djk)/2 Similarly, djc = (Dij + Djk – Dik)/2 dkc = (Dki + Dkj – Dij)/2

  20. Trees with > 3 Leaves • An tree with n leaves has 2n-3 edges • This means fitting a given tree to a distance matrix D requires solving a system of “n choose 2” equations with 2n-3 variables • This is not always possible to solve for n > 3

  21. Additive Distance Matrices Matrix D is ADDITIVE if there exists a tree T with dij(T) = Dij NON-ADDITIVE otherwise

  22. 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 edge, we can reconstruct those lengths

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

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

  25. Reconstructing Additive Distances Given T x T y 5 4 D1 b 3 z 3 a 4 c w 7 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 D2

  26. Distance Based Phylogeny Problem • Goal: Reconstruct an evolutionary tree from a distance matrix • Input: n x n distance matrix Dij • Output: weighted tree T with n leaves fitting D • If D is additive, this problem has a solution and there is a simple algorithm to solve it

  27. Find neighboring leavesi and j with parent k Remove the rows and columns of i and j Add a new row and column corresponding to k, where the distance from k to any other leaf m can be computed as: Using Neighboring Leaves to Construct the Tree Dkm = (Dim + Djm – Dij)/2 Compress i and j into k, iterate algorithm for rest of tree

  28. Finding Neighboring Leaves • To find neighboring leaves we simply select a pair of closest leaves.

  29. Finding Neighboring Leaves • To find neighboring leaves we simply select a pair of closest leaves. WRONG

  30. Finding Neighboring Leaves • Closest leaves aren’t necessarily neighbors • i and j are neighbors, but (dij= 13) > (djk = 12) • Finding a pair of neighboring leaves is • a nontrivial problem!

  31. Degenerate Triples • A degenerate triple is a set of three distinct elements 1≤i,j,k≤n where Dij + Djk = Dik • Element j in a degenerate triple i,j,k lies on the evolutionary path from i to k (or is attached to this path by an edge of length 0).

  32. Looking for Degenerate Triples • If distance matrix Dhas a degenerate triple i,j,k then j can be “removed” from D thus reducing the size of the problem. • If distance matrix Ddoes not have a degenerate triple i,j,k, one can “create” a degenerative triple in D by shortening all hanging edges (in the tree).

  33. Shortening Hanging Edges to Produce Degenerate Triples • Shorten all “hanging” edges (edges that connect leaves) until a degenerate triple is found

  34. Finding Degenerate Triples • If there is no degenerate triple, all hanging edges are reduced by the same amount δ, so that all pair-wise distances in the matrix are reduced by 2δ. • Eventually this process collapses one of the leaves (when δ = length of shortest hanging edge), forming a degenerate triple i,j,k and reducing the size of the distance matrix D. • The attachment point for j can be recovered in the reverse transformations by saving Dijfor each collapsed leaf.

  35. Reconstructing Trees for Additive Distance Matrices Trim(D, δ) for all 1 ≤ i ≠ j ≤ n Dij = Dij - 2δ

  36. AdditivePhylogeny Algorithm • AdditivePhylogeny(D) • ifD is a 2 x 2 matrix • T = tree of a single edge of length D1,2 • return T • ifD is non-degenerate • Compute trimming parameter δ • Trim(D, δ) • Find a triple i, j, k in D such that Dij + Djk = Dik • x = Dij • Remove jth row and jth column from D • T = AdditivePhylogeny(D) • Traceback

  37. AdditivePhylogeny (cont’d) Traceback • Add a new vertex v to T at distance x from i to k • Add j back to T by creating an edge (v,j) of length 0 • for every leaf l in T • if distance from l to v in the tree ≠ Dl,j • output “matrix is not additive” • return • Extend all “hanging” edges by length δ • returnT

  38. The Four Point Condition • AdditivePhylogeny provides a way to check if distance matrix D is additive • An even more efficient additivity check is the “four-point condition” • Let 1 ≤ i,j,k,l ≤ n be four distinct leaves in a tree

  39. The Four Point Condition (cont’d) Compute: 1. Dij + Dkl, 2. Dik + Djl, 3. Dil + Djk 2 3 1 2 and3represent the same number:the length of all edges + the middle edge (it is counted twice) 1represents a smaller number:the length of all edges – the middle edge

  40. The Four Point Condition: Theorem • The four point condition for the quartet i,j,k,l is satisfied if two of these sums are the same, with the third sum smaller than these first two • Theorem : An n x n matrix D is additive if and only if the four point condition holds for every quartet 1 ≤ i,j,k,l ≤ n

  41. Least Squares Distance Phylogeny Problem • If the distance matrix D is NOT additive, then we look for a tree T that approximates D the best: Squared Error : ∑i,j (dij(T) – Dij)2 • Squared Error is a measure of the quality of the fit between distance matrix and the tree: we want to minimize it. • Least Squares Distance Phylogeny Problem: Find approximation tree T with minimum squared error for a non-additive matrix D. (NP-hard)

  42. 1 4 3 5 2 5 2 3 1 4 UPGMA Tree construction as clustering

  43. 1 4 3 5 2 5 2 3 1 4 UPGMAUnweighted Pair Group Method • Computes the distance between clusters using average pairwise distance • Assigns height to every vertex in the tree, effectively dating every vertex (Assumes molecular clock)

  44. Clustering in UPGMA Given two disjoint clusters Ci, Cj of sequences, 1 dij = ––––––––– {p Ci, q Cj}dpq |Ci|  |Cj| Note that if Ck = Ci  Cj, then distance to another cluster Cl is: dil |Ci| + djl |Cj| dkl = –––––––––––––– |Ci| + |Cj|

  45. 1 4 3 5 2 5 2 3 1 4 UPGMA Algorithm Initialization: Assign each xi to its own cluster Ci Define one leaf per sequence, each at height 0 Iteration: Find two clusters Ci andCj such that dij is min Let Ck = Ci  Cj Add a vertex connecting Ci, Cj and place it at height dij /2 Delete Ci and Cj Termination: When a single cluster remains

  46. UPGMA Weakness UPGMA produces an ultrametric tree; distance from the root to any leaf is the same 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

  47. Correct tree UPGMA 3 2 4 1 3 4 2 1 UPGMA’s Weakness: Example

  48. Neighbor Joining Algorithm • In 1987 Naruya Saitou and Masatoshi Nei developed a neighbor joining algorithm for phylogenetic tree reconstruction • Finds a pair of leaves that are close to each other but far from other leaves: implicitly finds a pair of neighboring leaves • Advantages: works well for additive and other non-additive matrices, it does not have the flawed molecular clock assumption

  49. Neighbor-Joining • Guaranteed to produce the correct tree if distance is additive • May produce a good tree even when distance is not additive Let C = current clusters. Step 1: Finding neighboring clusters Define: u(C) =1/(|C|-2) C’ 2C D(C, C0 ) u(C) measures separation of C from other clusters Want to minimize D(C1, C2) and maximize u(C1) + u(C2) Magic trick: Choose C1 and C2 that minimize D(C1, C2) - (u(C1) + u(C2) ) Claim: Above 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

  50. Algorithm: Neighbor-joining Initialization: For n clusters, one for each leaf node Define T to be the set of leaf nodes, one per sequence Iteration: Pick Ci, Cj s.t. D(Ci, Cj) – (u(C1) + u(C2)) is minimal Merge C1 and C2 into new cluster with |C1| + |C2| elements Add a new vertex C to T and connect to vertices C1 and C2 Assign length 1/2 (D(C1, C2) + (u(C1) - u(C2) ) to edge (C1, C) Assign length 1/2 (D(C1, C2) + (u(C2) - u(C1) ) to edge (C2, C) Remove rows and columns from D corresponding to C1 and C2; Add row and column to D for new cluster C Termination: When only one cluster

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