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Phylogenetic Trees Lecture 2

Phylogenetic Trees Lecture 2. Based on: Durbin et al Section 7.3, 7.4, 7.8. The Four Points Condition. Theorem: A set M of L objects is additive 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 )

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Phylogenetic Trees Lecture 2

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  1. Phylogenetic TreesLecture 2 Based on: Durbin et al Section 7.3, 7.4, 7.8 .

  2. The Four Points Condition Theorem: A set M of L objectsis additive 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) We call {{i,j},{k,l}} the “split” of {i,j,k,l}. • The four point condition doesn’t provides an algorithm to construct a tree from distance matrix, or to decide whether there is such a tree. • The first methods for constructing trees for additive sets used neighbor joining methods:

  3. Constructing additive trees:The neighbor joining problem • Let i, jbe 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: • Find neighboring leaves i, j in the tree, • Replace i, j by their parent kand recursively construct a tree T for the smaller set. • Add i, j as children of kinT.

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

  5. T1 T2 m l k i j Neighbor Finding: Saitou & 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. The proof is rather involved!

  6. m k i j Neighbor Joining Algorithm • Set L to contain all leaves Iteration: • Choose i, j such that D(i, j) is minimal • Create new node k, and set • Remove i, j from L, and add k Terminate:when |L| =2 , connect two remaining nodes

  7. 1 • • a d c e b • g f 2 • 4 3 5 • Saitou & Nei’s Idea: Let (i, j) = d(i, j) – (ri + rj) “ L-2 ” is crucial! D12 = (a+c+d) – (1/3)(a+b + a+c+d + a+c+e+f + a+c+e+g + d+c+a + d+c+b + d+e+f + d+e+g) D13 = (a+b) – (1/3)(a+b + a+c+d + a+c+e+f + a+c+e+f + b+a + b+c+d + b+c+e+f + b+c+e+g) Hence D12 - D13 = (4/3) c

  8. B A e2 F E e3 e1 C D Saitou & Nei’s proof Notations used in the proof : p(i, j) = the path from vertex i to vertex j; P(D,C) = (e1, e2, e3) = (D, E, F, C) For a vertex i, and an edge e=(i , j): Ni(e) = |{k : e is on p(i, k), k is a leave}|. e.g. ND(e1) = 3, ND(e2) = 2, ND(e3) = 1 NC(e1) = 1

  9. Saitou & Nei’s proof: Crucial Observation Rest of T l k i j

  10. T1 T2 l k i j Saitou & Nei’s proof Proof of Theorem:Assume for contradiction that D(i, j) is minimized for i, j which are not neighboring leaves. Let (i, l, ..., k, j) be the path from i to j. let T1 and T2 be the subtrees rooted at k and l which do not contain edges from P(i,j) (see figure). Notation: |T| = #(leaves in T).

  11. T2 m l k i j Saitou & Nei’s proof Case 1:i or j has a neighboring leaf. WLOG j has a neighbor leaf m. A. D(i,j) - D(m,j)=(L-2)(d(i,j) - d(j,m)) – (ri+rj) +(rm+ rj) =(L-2)(d(i,k)-d(k,m))+rm-ri B.rm-ri ≥ (L-2)(d(k,m)-d(i,l)) + (4-L)d(k,l) (since for each edge eP(k,l), Nm(e) ≥ 2 and Ni(e) L-2) Substituting B in A: D(i,j) - D(m,j) ≥ (L-2)(d(i,k)-d(i,l)) + (4-L)d(k,l) = 2d(k,l)> 0, contradicting the minimality assumption.

  12. T1 m n p T2 k l i j Saitou & Nei’s proof Case 2: Not case 1. Then both T1andT2contain 2 neighboring leaves. WLOG |T2|≥|T1|. Let n,m be neighboring leaves in T1. We shall prove that D(m,n) < D(i,j), which will again contradict the minimality assumption.

  13. Saitou & Nei’s proof A. 0 ≤ D(m,n) - D(i,j)= (L-2)(d(m,n) - d(i,j) ) + (ri+rj) – (rm+rn) B. rj-rm< (L-2)(d(j,k) – d(m,p)) + (|T1|-|T2|)d(k,p) C. ri-rn <(L-2)(d(i,k) – d(n,p)) + (|T1|-|T2|)d(l,p) Adding B and C, noting that d(l,p)>d(k,p): D. (ri+rj) – (rm+rn) < (L-2)(d(i,j)-d(n,m)) + 2(|T1|-|T2|)d(l,p) T1 m n p T2 k Substituting D in the right hand side of A: D(m,n ) - D(i,j)< 2(|T1|-|T2|)d(l,p) ≤ 0, as claimed. QED l i j

  14. A simpler neighbor finding method Select an arbitrary node r. • For each pair of labeled nodes (i, j)let C(i, j) be defined by the following figure: r C(i,j) j Claim: Let i, j be such that C(i, j)is maximized. Then i and j are neighboring leaves. i

  15. m k i j Neighbor Joining Algorithm • Set M to contain all leaves, and select a root r. |M|=L • If L =2, return tree of two vertices Iteration: • Choose i, j such that C(i, j) is maximal • Create new vertex k, and set • remove i, j, and add k to M • Recursively construct a tree on the smaller set, then add i, j as children on k, at distances d(i,k) and d(j,k).

  16. m k i j Complexity of Neighbor Joining Algorithm Naive Implementation: Initialization:Θ(L2) to compute the C(i, j)’s. Each Iteration: • O(L) to update {C(i, k): i L} for the new node k. • O(L2) to find the maximal C(i, j). Total of O(L3).

  17. Complexity of Neighbor Joining Algorithm Using Heap to store the C(i, j)’s: Initialization:Θ(L2) to compute and heapify the C(i,j)’s. Each Iteration: • O(1) to find the maximal C(i,j). • O(L logL) to delete {C(m,i), C(m,j)} and add C(m,k) for all vertices m. Total of O(L2log L). (implementation details are omitted)

  18. 8 3 5 5 3 2 5 3 3 3 3 3 0: A E D B C Ultrametric trees as special weighted trees Definition:An Ultrametric tree is a rooted weighted tree all of whose leaves are at the same depth. Edge weights can be represented by the distances of internal vertices from the leaves. E.g., the tree produced by UPGMA. Note: each internal vertex has at least two children

  19. 8 3 5 B 3 C D A E Ultrametric trees A more recent (and more efficient) way for constructing and identifying additive trees. Idea: Reduce the problem to constructing trees by the “heights” of the internal nodes. For leaves i, j, D(i, j) represent the “height” of the common ancestor of i and j.

  20. 8 3 5 B 3 C D A E Ultrametric Trees Definition: T is an ultrametric tree for a symmetric positive real matrix D (called ultrmetric matrix) if: • The leaves of T correspond to the rows and columns of D • Internals nodes have at least two children, and the Least Common Ancestor of i and j is labeled by D(i, j). • The labels decrease along paths from root to leaves

  21. Centrality of Ultrametric Trees We will study later the following question: Given a symmetric positive real matrix D, Is there an ultrametric tree T for D? But first we show ultrametric trees can be used to construct trees for additive sets and other related problems.

  22. 8 3 5 5 3 3 2 3 3 B 5 C 3 3 D A E Transforming Ultrametric Trees to Weighted Trees Use the labels to define weights for all internal edges in the natural way. For this, consider the labels of leaves to be 0. We get an additive ultrametric tree whose height is the label of the root. Note that in this tree all leaves are at the same height. This is why it is called ultrametric.

  23. Transforming Weighted Trees to Ultrametric Trees A weighted Tree T can be transformed to an ultrametric tree T’ as follows: Step 1: Pick a node k as a root, and “hang” the tree at k. a c a 2 k=a 2 4 1 3 3 b 2 1 b d 2 4 c d

  24. 9 a 2 7 1 3 b 4 2 4 c d Transforming Weighted Trees to Ultrametric Trees Step 2: Let M = maxid(i, k). M is taken to be the height of T’. Label the root by M, and label each internal node j by M - d(k, j). “ k ” is the root. c a k = a, M = 9 2 4 3 2 1 b d

  25. k 9 2 9 m 7 a 7 3 b 4 4 4 c d Transforming Weighted Trees to Ultrametric Trees Step 3: “Stretch” edges of leaves so that they are all at distance M from the root 9 (9) a 2 7 M = 9 i 1 3 (6) b 4 ( M-d(k,i) ) 2 4 c d (0) (2)

  26. k 2 0 m a 1 3 b i 4 2 c d Re-constructing Weighted Trees from Ultrametric Trees Weight of an internal edge is the difference between the labels (heights) its endpoints. Assume that the distance matrix D = [d(i, j)] of the original unrooted tree is given. Weights of an edge to leaf i is obtained by subtracting “M - d(k, i)” from its current weight. 9 M = 9 9 (-9) 7 a 7(-6) b 4 4 (M–d(k,m))–(M–d(k,i)) = d(i,m) 4(-2) c d

  27. 9 a k 2 7 m 1 3 b i 2 4 c d How D’ is constructed from D D’(i, j) should be the height of the Least Common Ancestor of i and j in T’, the ultrametric tree hanged at k. Let M = maxi d(i, k) and m is the LCA of i and j. Thus, D’(i, j) = M - d(k, m), where d(k, m) is computed by: Note that this can be computed without the additive tree! j

  28. The transformation of D to D’ a M=9 9 T’ T 2 7 1 3 a b 4 b 4 2 c d d c Distance matrix D Ultrametric matrix D’

  29. Identifying Ultrametric Trees Definition: A distance matrix D is ultrametric if for each 3 indices i, j, k D( i, j ) ≤ max {D( i, k ), D( j, k )}. (i.e., there is a tie for the maximum value) Theorem (U): D has an ultrametric tree iff it is ultrametric. (to be proved later)

  30. Theorem:D is an additive distance matrix if and only if D’ is an ultrametric matrix. Note that the construction of D’ is independent of the additive tree. Proof. ( ) Use the conversion from an additive tree to an ultrametric tree and Theorem (U) . ( ) Use Theorem (U) and the conversion from an ultrametric tree to an additive tree and check that the additive tree indeed realizes the distance matrix.

  31. Solving the Additive Tree Problem by the Ultrametric Problem: Outline • We solve the additive tree problem by reducing it to the ultrametric problem as follows: • Given an input matrix D = D(i, j) of distances, transform it to a matrix D’= D’(i, j) , where D’(i,j) is the height of the Least Common Ancestor of i and j in the corresponding ultrametric tree T’. (If not ultrametric, then the input matrix is not additive!) • Construct the ultrametric tree, T’, for D’. • Reconstruct the additive tree T from T’.

  32. 8 5 3 3 C B A D E LCA and distances in Ultrametric Tree Let LCA(i, j) denote the lowest common ancestor of leaves i and j. Let D(i, j) be the height of LCA(i, j), and dist(i,j) be the distance from i to j. Claim: For any pair of leaves i, j in an ultrametric tree: D(i, j)= 0.5 dist(i, j).

  33. Identifying Ultrametric Distances Definition: A distance matrix D of dimension L by L is ultrametric iff for each 3 indices i, j, k : D( i, j ) ≤ max { D( i, k ), D( j, k ) }. Theorem(U): The following conditions are equivalent for an LL symmetric matrix D: • D is ultrametric • There is an ultrametric tree of L leaves such that for each pair of leaves i, j : D(i, j) = height(LCA(i, j)) = ½ dist(i, j). Note: D(i, j) ≤ max {D(i, k), D(j, k)} is easier to check than the 4 points condition. Therefore the theorem implies that ultrametric sets are easier to characterize then an additive sets.

  34. Properties of ultrametric matrix used in the Proof of the Theorem (U) Definition: Let D be an L by L matrix, and let S  {1,...,L}. D[S] is the submatrix of D consisting of the rows and columns with indices from S. Claim 1: D is ultrametric iff for every S  {1,...,L}, D[S] is ultrametric. Claim 2: If D is ultrametric and maxi,jD(i, j)=m, , then m appears in every row of D above the row where the max occurs. One of the “?” Must be m

  35. Ultrametric tree Ultrametric matrix There is an ultrametric tree s.t. D(i, j) = ½ dist(i, j).  D is an ultrametric matrix: By properties of Least Common Ancestors in trees D(k, i) = D(j, i)≥ D(k, j) i j k

  36. 9 i j Ultrametric matrix  Ultrametric tree Proof of D is an ultrametric matrix D has an ultrametric tree : By induction on L, the size of D. Basis: L= 1: T is a leaf L= 2:T is a tree with two leaves i i  i i j i j

  37. Induction step Inductive Hyp.: Assume that it’s true for 1, 2, … , L-1. Induction step: L > 2. Let m = m1 be the maximum distance. Let Si ={l: D(1, l) = mi}, and { S1 , S2 , … Sk} form a partition of the leaves into k classes. (note: |S1| > 0) By Claim 1, D[Si],i = 1, 2, …, k are all ultrametric and hence we can construct tree T1 for S1, rooted at m and trees Ti for Siwith root labeled mi < m for i = 2, …, k. (if mi = 0 then Ti is a leaf).

  38. 6 T1 5,8 4 3 2,4,6 T2 3,7 T3 1 • Notice that on any ultrametric tree the path from the root to the leave “1” must have exactly k+1 nodes, where k is the number of classes. • Each node on this path must be labeled by one of the distinct entries in row 1, and those labels must appear in decreasing order on the path. 1 2 3 4 5 6 7 8 1 2 3 4 . . 0 4346436 0 4 2 6 1 4 6

  39. Correctness Proof By Inductive Hypothesis, Ti ’s are all ultrmetric trees, and we assemble them along the path from the root to leave “1” to form the tree T. To prove that T is an ultrametric tree for D, need to check that D(i, j) is the label of the LCA of i and j in T. If i and j are in the same subtree, this holds by induction; otherwise the label of the node that the higher tree attaches to the path, which is the LCA, is indeed D(i, j) . QED

  40. Complexity Analysis Let f (L) be the time complexity for L×L matrix. f (1)= f (2) = constant. For L > 2: • Constructing S1 and S2: O(L). Let |S1| = k, |S2| = L-k. • Constructing T1 and T2: f (k) + f (L-k). • Joining T1 and T2 to T: Constant. Thus we have: f (L) ≤ maxk[ f (k) + f (L-k)] +cL, 0 < k < L. f (L) = cL2satisfies the above. Need an appropriate data structure!

  41. Recall: identifying Additive Trees via Ultrametric trees • We solve the additive tree problem by reducing it to the ultrametric problem as follows: • Given an input matrix D = D(i, j) of distances, transform it to a matrix D’= D’(i, j), where D’(i, j) is the height of the LCA of i and j in the corresponding ultrametric tree T’. • Construct the ultrametric tree, T’, for D’. • Reconstruct the additive tree T from T’.

  42. How D’ is constructed from D D’(i, j) should be the height of the Least Common Ancestror of i and j in T’, the ultrametric tree hanged at k: Thus, D’(i,j) = M - d(k, m), where d(k, m) is computed by: 9 a 2 7 1 3 b 2 4 c d

  43. a 2 1 3 b 2 4 d c The transformation D  D’  T’T M=9 9 T’ T 7 a 4 b c d D’ D

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