Ch.6 Phylogenetic Trees

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Ch.6 Phylogenetic Trees
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Contents

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

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Phylogenetic Trees (Phylogenies)

• Explain the evolutionary history of todays
  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

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Character State Matrix

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

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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

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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

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Example
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Perfect Phylogeny Problem

• How many different trees can we build for n
  objects?
• Consider only unrooted binary trees

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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

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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)

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Deciding perfect phylogeny

• Definition 6.1 For each column j of M, let Oj be
  the set of objects whose state is 1 for j. Let Oj
  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 Oi and Oj are disjoint
  or one of them contains the other

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Deciding perfect phylogeny

• Example Table 6.2
• O1 B, D, O2 B, O3 D
• O4 A, C, E, O5 A, C, O6 C
• Lemma 6.1 for decision phase takes O(nm2)
• Figure 6.5 Algorithm Perfect Binary Phylogeny
  Decision -gt O(nm)

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Deciding perfect phylogeny

• if Lij ? Llj for some i, l and both Lij and Llj
  are nonzero then
• return FALSE

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Construction perfect phylogeny

• Figure 6.6 Algorithm Perfect Binary Phylogeny
  Construction
• Running time O(nm)

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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

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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 T1,
  T2, , Tl of a tree T there corresponds a
  triangulated graph and vice versa

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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

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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

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Example
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Reconstruction algorithm for two characters

• Running time O(n)
• Test for acyclicity -gt O(n)
• Reconstruction of the perfect phylogeny -gt O(n)

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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 -gt exclude characters that cause such
  problem

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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

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Reconstructing Additive Trees

• Metric space
• A set of objects O such that to every pair i, j ?
  O and associated a nonnegative real number dij
  with the following properties
• dij gt 0 for i ? j,
• dij 0 for i j,
• dij dji for all i and j,
• dij dik dkj 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

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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(n2))
• 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

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Reconstructing Ultrametric Trees

• Given two distance matrices, Ml and Mh,
  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 -gt the objects being
  studied have evolved at equal rate from a common
  ancestor

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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 Gh is given by

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Reconstructing Ultrametric Trees

• Reconstruction algorithm -gt runs in time O(n2)
• Compute a MST T of Gh
• Construction of R
• Compute CW(e)
• Build ultrametric tree U

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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 Tr refines
  another tree Ts whenever Tr can be transformed
  into Ts by contracting selected edges from Tr.
  Two trees T1 and T2 agree when there exists a
  tree T3 that refines both

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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)