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. Perfect Phylogeny MLE for Phylogeny Lecture 14 Based on: Setubal&Meidanis 6.2, Durbin et. Al. 8.1
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2 Final Exam Details The Final Exam will take Place on Thursday, 3.2.04, 0900, at Taub 4. Allowed Material: Course&Tutorial slides+ the textbooks of the course (Durbin et el, Setubal&Meidanis, Gusfield).
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3 2. The perfect phylogeny problem u A character is assumed to be a property which distinguishes between species (e.g. dental structure). u A characters state is a value of the character (human dental structure). u Problem: Given set of species, specified by their characters, reconstruct their evolutionary tree.
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4 Characters as Colorings A coloring of a tree T=(V,E) is a mapping C:V [set of colors] A partial coloring of T is a mapping defined on a subset of the vertices U V: C:U [set of colors] U=
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5 Each character defines a (partial) coloring of the correspondeing phylogenetic tree: Characters as Colorings (2) Species ≡ Vertices States ≡ Colors
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6 Convex Colorings (and Characters) C Definition: A (partial/total) coloring of a tree is convex iff its d-carriers are mutually disjoint Let T=(V,E) be a partially colored tree, and d be a color. The d-carrier is the minimal subtree of T containing all vertices colored d
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7 A character is Homoplasy free (avoids reversal and convergence transitions) ↕ The corresponding (partial) coloring is convex Convexity Homoplasy Freedom
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8 The Perfect Phylogeny Problem u Input: a set of species, and many characters. u Question: is there a tree T containing the species as vertices, in which all the characters (colorings) are convex? (always possible for one chracter)
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9 Input: Partial colorings (C 1,…,C k ) of a set of vertices U (in the example: 3 total colorings: left, center, right, each by two colors). Problem: Is there a tree T=(V,E), s.t. U V and for i=1,…,k,, C i is a convex (partial) coloring of T? RBRRBRRRR BBRRRB The Perfect Phylogeny Problem (pure graph theoretic setting) NP-Hard In general, in P for some special cases
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10 Perfect Phylogeny for a 0-1 Matrix Rows correspond to objects, columns to characters. Each character has two states: 0 (non exists) or 1 (exists). A tree T is a perfect phylogeny for the matrix iff it has the following properties: A.Each of the n objects corresponds to a leaf of T. B.Each of the m characters labels exactly one edge of T. C.Object p has character i i labels an edge on the path from p to the root. Note: [B and C] [each character is convex on T] C1C1C2C3C4C5 A11000 B00100 C11001 D00110 E01000 A E D C B C4 C3 C2 C1 C5
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11 Perfect Phylogeny for a 0-1 Matrix By the definition, for each character C there is one edge in which it is converted from 0 to 1. In the below tree, the edge on which character C2 is converted to 1 is marked. The resulted tree is convex for this character. C1C2C3C4C5 A1 B0 C1 D0 E1 A E D C B C2
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12 The (Binary) Perfect Phylogeny Problem Problem: Given a 0-1 matrix M, determine if it has a perfect phylogeny in which the root has 0 for all characters, and construct one if it does. (Note: edges are labeled by characters: edge labeled by i represent changing character i’s state from 0 to 1). As we show below, the answer is yes for our matrix: C1C2C3C4C5 A11000 B00100 C11001 D00110 E01000 A E D C B C4 C3 C2 C1 C5
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13 Efficient algorithm for the Binary Perfect Phylogeny Problem Definition: Given a 0-1 matrix M, O k ={j:M jk =1}, ie: O k is the set of objects that have character Ck. Theorem: M has a perfect phylogenetic tree iff the sets {O i } are laminar, ie: for all i, j, either O i and O j are disjoint, or one includes the other. C1C2C3C4C5 A11000 B00100 C11001 D00110 E01000 C1C2 C3 C4C5 A11000 B00101 C11001 D00110 E01001 LaminarNot Laminar
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14 Proof : Assume M has a perfect phylogeny, and let Ci, Cj be given. Consider the edges labeled Ci and Cj. Case 1: There is a root to leaf path containing both edges. Then one is included in the other (C2 and C1 below). Case 2: not case 1. Then they are disjoint (C2 and C3). A E D C B C4 C3 C2 C1 C5
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15 Proof (cont.) : Assume for all i, j, either O i and O j are disjoint, or one includes the other. We prove by induction on the number of characters that M has a perfect phylogenetic tree for the matrix. Basis: one character. Then there are at most two objects, one with and one without this character. C1 A1 B0 AB
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16 Proof (cont.) : Induction step: Assume correctness for n-1 characters, and consider a matrix with n characters (non-zero columns). WLOG assume that O 1 is not contained in O j for j > 1. Let S 1 be the set of objects j for which M j1 = 1, and S 2 be the remaining objects. Then each character belongs to objects in S 1 or S 2, but not both (prove!). By induction there are trees T 1 and T 2 for S 1 and S 2. Combining them as below gives the desired tree. C1C2C3C4C5 A11000 B00100 C11001 D00110 E10000 T1T1 T2T2 1
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17 Efficient Implementation 1 Sort the columns (characters) by decreasing value when considered as binary numbers. (Time complexity: O(mn), using radix sort). Claim: If the binary value of column i is larger than that of column j, then O i is not a proper subset of O j. Proof: O i – O j > 0 means the 1’s in O i are not covered by the 1’s in O j. C1C2C3C4C5 A11000 B00100 C11001 D00110 E01000 C2C1C3C5C4 A11000 B00100 C11010 D00101 E10000
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18 Efficient Implementation(2) 2. Make a backwards linked list of the 1’s in each row (leftmost 1 in each row points at itself). Time complexity: O(mn). C2C1C3C5C5C4 A11000 B00100 C11010 D00101 E10000 Claim: If the columns are sorted, then the set of columns is laminar iff for each column i, all the links leaving column i point at the same column. Can be checked in O(mn) time.
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19 Examples A11000 B00100 C11010 D00101 E10000 A11000 B00100 C11010 D00101 E10110 laminarNot laminar
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20 Efficient Implementation(3) 3. When the matrix is laminar, the tree edges corresponding to characters are defined by the backwards links in the matrix. C2C1C3C5C4 A11000 B00100 C11010 D00101 E10000 A E D C B C3 C2 C1 C5 remaining edges and leaves are determined by the characters of each object. Needs O(mn) time.
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21 A scenario where Maximum Parsimony (and Perfect Phylogeny) are misleading A AA 1 4 32 Consider a model with 4 letters (DNA), where the probability for a substitution is proportional to time. In the following topology, 2 and 3 are likely to be like the origin, but 4 and 5 can be different. In this case, Maximum Parsimony is misleading.
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22 Parsimony may be useless/misleading A A C G A G G G I Uninformative II Uninformative III Uninformative A AA 1 4 32 IV Misinformative For leaves 1,4 there are 4 combinations of substitution. In the first three, all three topologies will obtain the same parsimony score. In the fourth, a wrong topology will score best
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23 Parsimony may be Useless Case I A AA 1 4 32 AA 1 2 3 4 A A A A 1 3 2 4 A A A A 1 4 2 3 A A A A Score=0
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24 Parsimony Imay be useless Case II A AA 1 4 32 GA 1 2 3 4 A A A G 1 3 2 4 A A A G 1 4 2 3 A G A A Score=1
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25 Parsimony may be misleading Case III A AA 1 4 32 GC 1 2 3 4 A A C G 1 3 2 4 A A C G 1 4 2 3 A G C A Score=2
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26 Parsimony may be misleading Case III A AA 1 4 32 CC 1 2 3 4 A A C C 1 3 2 4 A A C C 1 4 2 3 A C C A Score=2 Score=1
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27 Parsimony may be misleading A CA 1 4 32 AC CA A CA 1 4 32 AC AA Will infer correctly only in the rare case of a change on the central edge, or In an even more rare case of a parallel change from A to C on the pendant edges to 1 and 2.
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28 3. Maximum Likelihood Approach Consider the phylogenetic tree to be a stochastic process. AGA GGA AAA AAG AAA AGA AAA The likelihood of transition from character a to charcter b is given by parameters b|a. The liklihood of a letter a in the root is q a. Given the complete tree, its probability is defined by the values of the b|a ‘s and the q a ’s.
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29 Maximum Likelihood Approach(2) When the data consists only of the leaves sequences (but the topology is fixed): AGA GGA AAA AAG Write down the likelihood of the data (leaves sequences) given the tree. Use EM to estimate the b|a parameters. When the tree is not given: Search for the tree that maximizes Prob(data|Tree, EM )
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