2. Ch.6 Phylogenetic Trees

3. 2 Contents Phylogenetic Trees Character State Matrix Perfect Phylogeny Binary Character States Two Characters Distance Matrix Additive Trees Ultrametric Trees Agreement (Isomorphic) between Phylogenies

4. 3 Phylogenetic Trees (Phylogenies) Explain the evolutionary history of today’s species (Figure 6.1) A hypothesis; do not have enough data about distant ancestors of present-day species Characteristic Leaf; an object or a set of objects, Interior node; hypothetical ancestor objects Unrooted tree Classify input data for phylogeny reconstruction into main categories Character state matrix Distance matrix

5. 4 Character State Matrix Character have following features Independent inheritance Homologous Character state matrix A matrix M with n rows (objects) and m columns (characters) M ij denotes the state the object i has for character j Each row is the state vector for an object

6. 5 Difficulties to create a phylogeny from a character state matrix Convergence or parallel evolution Objects that share the same state are genetically closer than objects that do not Reversal Gains and losses of the character ☞ assume convergence or reversal should not happen, or their number should be minimized Ordered or unordered, directed

7. 6 Perfect Phylogeny Problem For each state s of each character c, the set of all nodes u (leaves and interior nodes) for which the state is s with respect to c must form a subtree of T Characters are compatible If a set of objects defined by a character state matrix admits a perfect phylogeny

8. 7 Example

9. 8 Perfect Phylogeny Problem How many different trees can we build for n objects? Consider only unrooted binary trees

10. 9 Binary Character States Two phases algorithm (runs in time O(nm)) Decide whether the input matrix M admits a perfect phylogeny Construct one possible phylogeny Assume that state 0 is ancestral and state 1 is derived

11. 10 Deciding perfect phylogeny A rooted tree T is a perfect phylogeny for input matrix M, if Every character in input matrix M there corresponds an edge in T, and this edge marks the transition from state 0 to state 1 for that character Edges are labeled by their respective characters and root has character state vector (0, 0, …, 0)

12. 11 Deciding perfect phylogeny Definition 6.1 For each column j of M, let O j be the set of objects whose state is 1 for j. Let O j be the set of objects whose state is 0 for j Lemma 6.1 A binary matrix M admits a perfect phylogeny if and only if for each pair of character i and j the sets O i and O j are disjoint or one of them contains the other

13. 12 Deciding perfect phylogeny Example; Table 6.2 O 1 = {B, D}, O 2 = {B}, O 3 = {D} O 4 = {A, C, E}, O 5 = {A, C}, O 6 = {C} Lemma 6.1 for decision phase takes O(nm 2 ) Figure 6.5 Algorithm Perfect Binary Phylogeny Decision -> O(nm)

14. 13 Deciding perfect phylogeny if L ij ≠ L lj for some i, l and both L ij and L lj are nonzero then return FALSE Mc4c1c5c2c3c6 A101000 B010100 C101001 D010010 E100000 Lc4c1c5c2c3c6 A01000 B0 0200 C 01003 D0 0010 E 00000

15. 14 Construction perfect phylogeny Figure 6.6 Algorithm Perfect Binary Phylogeny Construction Running time O(nm)

16. 15 Unordered binary character The majority state becomes 0 and the other 1 If equal frequency, choose either one to be 0 and the other to be 1

17. 16 Two characters Allow characters can be unordered and have an arbitrary number of states, but restrict on the maximum number of characters two Definition 6.2 A triangulated graph is an undirected graph in which any cycle with four or more vertices has a chord, that is, an edge joining two nonconsecutive vertices of the cycle Theorem 6.1 To every collection of subtrees {T 1, T 2, …, T l } of a tree T there corresponds a triangulated graph and vice versa

18. 17 Two characters Definition 6.3 An intersection graph for a collection C of sets is the graph G that we get by mapping each set in C to a vertex of G, and linking two vertices in G by an edge if the corresponding sets have a nonempty intersection Definition 6.4 Given a graph G = (V, E) with a coloring c on V, we say that G can be c-triangulated if there exists a triangulated graph H = (V, E’), such that E ⊆ E’ and c is a valid coloring for H. In other words, any edge present in E’ but not in E must link two vertices with different colors

19. 18 Two characters Theorem 6.2 A character state matrix M, with a character set defining a coloring c, admits a perfect phylogeny if and only if its corresponding SIG can be c-triangulated Theorem 6.3 A character state matrix M with only two characters admits a perfect phylogeny if and only if its corresponding SIG is acyclic

20. 19 Example x1 y1 x2 z2 x3 y3 z3y2 {B}{A, B}{A} {B, C} {C}{C, D}{D} {A, D}

21. 20 Reconstruction algorithm for two characters Running time O(n) Test for acyclicity -> O(n) Reconstruction of the perfect phylogeny -> O(n)

22. 21 Parsimony and Compatibility Real character state matrices are unlikely to admit perfect phylogenies Experimental data always carries errors The assumptions (no reversals and no convergence) sometimes are violated Two approach Parsimony criterion Allow reversal and convergence events, but to try to minimize their occurrence Compatibility criterion Find a maximum set of characters that are compatible -> exclude characters that cause such “problem”

23. 22 Algorithms for Distance Matrices Problem of reconstructing trees based on comparative numerical data between n objects, distance matrix M Consider two problems Reconstructing Additive Trees Reconstructing Ultrametric Trees

24. 23 Reconstructing Additive Trees Metric space A set of objects O such that to every pair i, j ∈ O and associated a nonnegative real number d ij with the following properties: d ij > 0 for i ≠ j, d ij = 0 for i= j, d ij = d ji for all i and j, d ij ≤ d ik + d kj for all i, j, and k (the triangle inequality) M and T are additive Tree must have n leaves Leaves are nodes with degree one; the others with degree three All edges in the tree have nonnegative weight The weight of the path between any two leaves i and j must be equal to Mij

25. 24 Reconstructing Additive Trees Lemma 6.2 A metric space O is additive if and only if given any four objects of O labeled i, j, k, and l such that dij + dkl = dik + djl ≥ dil + djk If M is additive, T is unique (algorithm runs in time O(n 2 )) Real-life distance matrices are rarely additive due to errors in the distance measurement Obtain a tree that is as close as possible to an additive tree Approaching the problem that is tractable

26. 25 Reconstructing Ultrametric Trees Given two distance matrices, M l and M h, reconstruct an evolutionary tree such that the distances measured on the tree fit “between” these two input matrices (sandwich constraints, ) A tree is ultrametric when it is additive and can be rooted in such a way that the lengths of all leaf-root paths are equal -> the objects being studied have evolved at equal rate from a common ancestor

27. 26 Reconstructing Ultrametric Trees link of a and b in MST T; (a, b) max The largest-weight edge in the unique path from a to b in T Definition 6.5 The cut-weight of an edge e of the minimum spanning tree of G h is given by

28. 27 Reconstructing Ultrametric Trees Reconstruction algorithm -> runs in time O(n 2 ) Compute a MST T of G h ; Construction of R; Compute CW(e); Build ultrametric tree U

29. 28 Agreement between Phylogenies In practice it occurs quite often that two different methods applied on the same data yield different trees (in the topological sense) Definition 6.6 We say that a tree T r refines another tree T s whenever T r can be transformed into T s by contracting selected edges from T r. Two trees T 1 and T 2 agree when there exists a tree T 3 that refines both

30. 29 Isomorphic Two trees T1 and T2 are isomorphic when there is an one-to-one correspondence between their nodes such that for every pair u, v of corresponding nodes, u ∈ T1 and v ∈ T2, the objects contained in leaves below u are the same as the objects contained in leaves below v Binary Tree Isomorphism Figure 6.21 runs in time O(n) General case (leaves contain several objects) Figure 6.22 runs in time O(n)