Perfect Phylogeny MLE for Phylogeny Lecture 14

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Perfect Phylogeny MLE for Phylogeny Lecture 14
Based on SetubalMeidanis 6.2, Durbin et. Al. 8.1
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Some Announcements

• The Final Exam will take Place on Friday,
  17.2.04, 0900, at Taub 8.
• Allowed Material CourseTutorial slides the
  textbooks of the course (Durbin et el,
  SetubalMeidanis, Gusfield).

• Lab offered next semester
• algorithms for constructing phylogenetic trees
• http//www.cs.technion.ac.il/moran/lab06.htm

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2. The perfect phylogeny problem

• A character is assumed to be a property which
  distinguishes between species (e.g. dental
  structure).
• A characters state is a value of the character
  (human dental structure).
• Problem Given set of species, specified by their
  characters, reconstruct their evolutionary tree.

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The Perfect Phylogeny Problem(pure graph
theoretic setting)
Input Partial colorings (C1,,Ck) 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 i1,,k,, Ci is a convex (partial)
coloring of T?
NP-Hard In general, in P for some special cases
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Perfect Phylogeny for directed binary characters

• Input a matrix where rows correspond to objects
  (species), columns to characters.
• Each character has two states 0 (non exists) or
  1 (exists).
• Question Is there a directed perfect phylogeny
  tree for the given species, in which all the
  characters have value 0 at the root?

(00000)
C1 C2 C3 C4 C5
A 1 1 0 0 0
B 0 0 1 0 0
C 1 1 0 0 1
D 0 0 1 1 0
E 0 1 0 0 0
E
B
(01000)
D
(00100)
(00110)
A
C
(11000)
(11001)
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Perfect Phylogeny for directed binary characters

• 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.

the edge on which character C2 is converted to 1
C1 C2 C3 C4 C5
A 1
B 0
C 1
D 0
E 1
0
C2
E
B
1
D
0
0
A
C
1
1
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Perfect Phylogeny for directed binary characters

• A tree is a directed perfect phylogeny for a
  given 0-1 matrix M iff we can map each character
  to an edge s.t. edge labeled by Ci represent
  changing character Cis state from 0 to 1. Below
  we show such a tree for the given matrix

C1 C2 C3 C4 C5
A 1 1 0 0 0
B 0 0 1 0 0
C 1 1 0 0 1
D 0 0 1 1 0
E 0 1 0 0 0
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Efficient algorithm for the Binary Perfect
Phylogeny Problem

• Definition Given a 0-1 matrix M, OkjMjk1,
  ie Ok is the set of objects that have character
  Ck.
• Theorem M has a perfect phylogenetic tree iff
  the sets Oi are laminar, ie for all i, j,
  either Oi and Oj are disjoint, or one includes
  the other.

Laminar
Not Laminar
C1 C2 C3 C4 C5
A 1 1 0 0 0
B 0 0 1 0 0
C 1 1 0 0 1
D 0 0 1 1 0
E 0 1 0 0 0
C1 C2 C3 C4 C5
A 1 1 0 0 0
B 0 0 1 0 1
C 1 1 0 0 1
D 0 0 1 1 0
E 0 1 0 0 1
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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).

C2
C3
C1
C4
E
D
B
C5
A
C
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Proof (cont.)

• ? Assume for all i, j, either Oi and Oj 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
A 1
B 0
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Proof (cont.)

• ? Induction step Assume correctness for n-1
  characters, and consider a matrix with n
  characters (non-zero columns). WLOG assume that
  O1 is not contained in Oj for j gt 1.
• Let S1 be the set of objects j for which Mj1 1,
  and S2 be the remaining objects. Then each
  character belongs to objects in S1 or S2, but not
  both (prove!). By induction there are trees T1
  and T2 for S1 and S2. Combining them as below
  gives the desired tree.

C1 C2 C3 C4 C5
A 1 1 0 0 0
B 0 0 1 0 0
C 1 1 0 0 1
D 0 0 1 1 0
E 1 0 0 0 0
S1A,C,E S2B,D
1
T1
T2
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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 Oi is not a proper
  subset of Oj.
• Proof Oi Oj gt 0 means the 1s in Oi are not
  covered by the 1s in Oj.

C1 C2 C3 C4 C5
A 1 1 0 0 0
B 0 0 1 0 0
C 1 1 0 0 1
D 0 0 1 1 0
E 0 1 0 0 0
C2 C1 C3 C5 C4
A 1 1 0 0 0
B 0 0 1 0 0
C 1 1 0 1 0
D 0 0 1 0 1
E 1 0 0 0 0
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Efficient Implementation(2)

• 2. Make a backwards linked list of the 1s in
  each row (leftmost 1 in each row points at
  itself). Time complexity O(mn).

C2 C1 C3 C5 C4
A 1 1 0 0 0
B 0 0 1 0 0
C 1 1 0 1 0
D 0 0 1 0 1
E 1 0 0 0 0
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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Examples
laminar
Not laminar

A 1 1 0 0 0
B 0 0 1 0 0
C 1 1 0 1 0
D 0 0 1 0 1
E 1 0 0 0 0

A 1 1 0 0 0
B 0 0 1 0 0
C 1 1 0 1 0
D 0 0 1 0 1
E 1 0 1 1 0
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Efficient Implementation(3)

• 3. When the matrix is laminar, the tree edges
  corresponding to characters are defined by the
  backwards links in the matrix.

remaining edges and leaves are determined by the
characters of each object. Needs O(mn) time.
C2 C1 C3 C5 C4
A 1 1 0 0 0
B 0 0 1 0 0
C 1 1 0 1 0
D 0 0 1 0 1
E 1 0 0 0 0
C2
C3
C1
C4
E
D
B
C5
A
C
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A scenario where Maximum Parsimony (and Perfect
Phylogeny) are misleading
Consider a model with 4 letters (DNA), where the
probability for a substitution is proportional to
time.
1
4
In the following topology, 2 and 3 are likely to
be as the origin, but 4 and 5 are likely to be
different. In this case, Maximum Parsimony
principle may be useless or misleading.
A
A
3
2
A
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Parsimony may be useless/misleading
A
I Uninformative
II Uninformative
A
III Uninformative
C
Assume the (likely) scenario where leaves 2 and 3
are the same. There are 4 combinations of
substitution for leaves 1,4. In the first three,
all three topologies will obtain the same
parsimony score.
G
In the fourth, a wrong topology will score best
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Case I Parsimony is Useless
A
A
1
4
A
A
3
2
A
Score0
Score0
Score0
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Case II Parsimony is Useless
G
A
1
4
A
A
3
2
A
Score1
Score1
Score1
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Case III Parsimony is useless
G
C
1
4
A
A
3
2
A
Score2
Score2
Score2
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Case III Parsimony is misleading
C
C
1
4
A
A
3
2
A
Score1
Score2
Score2
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Parsimony is correct only in rare cases
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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3. Maximum Likelihood Approach
Consider the phylogenetic tree to be a stochastic
process.
A simple model assumes that in each edge,
likelihood of transition from character a to
charcter b is given by parameters ?ba . The
liklihood of a letter a in the root is qa. Given
the complete tree, its probability is defined by
the values of the ?ba s and the qas.
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Maximum Likelihood Approach(2)
When the data consists only of the leaves
sequences (but the topology is fixed)
Write down the likelihood of the data (leaves
sequences) given the tree. Use EM to estimate the
?ba parameters. When the tree is not given
Search for the tree that maximizes
Prob(dataTree, ?EM)