maximum parsimony.pdf

Lecture 11
Phylogenetic trees
Principles of Computational
Biology
Teresa Przytycka, PhD

Phylogenetic (evolutionary) Tree
• showing the evolutionary relationships among
various biological species or other entities that
are believed to have a common ancestor.
• Each node is called a taxonomic unit.
• Internal nodes are generally called hypothetical
taxonomic units
• In a phylogenetic tree, each node with
descendants represents the most recent
common ancestor of the descendants, and the
• edge lengths (if present) correspond to time
estimates.

Methods to construct phylogentic
trees
• Parsimony
• Distance matrix based
• Maximum likelihood

Parsimony methods
The preferred evolutionary tree is
the one that requires
“the minimum net amount of evolution”
[Edwards and Cavalli-Sforza, 1963]

Assumption of character based
parsimony
• Each taxa is described by a set of characters
• Each character can be in one of finite number
of states
• In one step certain changes are allowed in
character states
• Goal: find evolutionary tree that explains the
states of the taxa with minimal number of
changes

Example
Taxon1 Yes Yes No
Taxon 2 YES Yes Yes
Taxon 3 Yes No No
Taxon 4 Yes No No
Taxon 5 Yes No Yes
Taxon 6 No No Yes

Version parsimony models:
• Character states
– Binary: states are 0 and 1 usually interpreted as presence or
absence of an attribute (eg. character is a gene and can be
present or absent in a genome)
– Multistate: Any number of states (Eg. Characters are
position in a multiple sequence alignment and states are
A,C,T,G.
• Type of changes:
– Characters are ordered (the changes have to happen in
particular order or not.
– The changes are reversible or not.

Variants of parsimony
• Fitch Parsimony unordered, multistate characters with
reversibility
• Wagner Parsimony ordered, multistate characters
with reversibility
• Dollo Parsimony ordered, binary characters with
reversibility but only one insertion allowed per
character characters that are relatively chard to gain but easy
to lose (like introns)
• Camin-Sokal Parsimony- no reversals, derived states
arise once only
• (binary) prefect phylogeny – binary and non-
reversible; each character changes at most once.

Prefect – No
(triangle gained
and the lost)
Dollo – Yes
Camin-Sokal –
No (for the same
reason as perfect)

3 Changes
Camin-Sokal Parsimony
Triangle inserted twice!
But this is
not prefect
and not
Dollo

Homoplasy
Having some states arise more than once is
called homoplasy.
Example – triangle in the tree on the
previous slide

Finding most parsimonious tree
• There are exponentially many trees with n
nodes
• Finding most parsimonious tree is NP-
complete (for most variants of parsimony
models)
• Exception: Perfect phylogeny if exists can
be found quickly. Problem – perfect
phylogeny is to restrictive in practice.

Perfect phylogeny
• Each change can happen only once and is
not reversible.
• Can be directed or not
Example: Consider binary characters. Each character
corresponds to a gene.
0-gene absent
1-gene present
It make sense to assume directed changes only form 0 to 1.
The root has to be all zeros

Perfect phylogeny
Example: characters = genes; 0 = absent ; 1 = present
Taxa: genomes (A,B,C,D,E)
A 0 0 0 1 1 0
B 1 1 0 0 0 0
C 0 0 0 1 1 1
D 1 0 1 0 0 0
E 0 0 0 1 0 0
genes
B D E A C
Perfect phylogeny tree
Goal: For a given character state matrix construct a tree topology
that provides perfect phylogeny.
1
1
1
1 1
1

Does there exist prefect
parsimony tree for our example
with geometrical shapes?
There is a simple test

Character Compatibility
• Two characters A, B are
compatible if there do
not exits four taxa
containing all four
combinations as in the
table
• Fact: there exits perfect
phylogeny if and only if
and only if all pairs of
characters are
compatible
T1 1 1
T2 1 0
T3 0 1
T4 0 0
A B

Are not compatible
Taxon1 Yes Yes No
Taxon 2 YES Yes Yes
Taxon 3 Yes No No
Taxon 4 Yes No No
Taxon 5 Yes No Yes
Taxon 6 No No Yes

?
One cannot add triangle
to the tree so that no
character changes it state
twice:
If we add it to on of the
left branches it will be
inserted twice if to the
right most – circle would
have to be deleted
(insertion and the
deletion of the circle)

Ordered characters and perfect
phylogeny
• Assume that we in the last common
ancestor all characters had state 0.
• This assumption makes sense for many
characters, for example genes.
• Then compatibility criterion is even
simpler: characters are compatible if and
only if there do not exist three taxa
containing combinations (1,0),(0,1),(1,1)

Example
Under assumption that states are directed
form 0 to 1: if i and j are two different genes
then the set of species containing i is either
disjoint with set if species containing j or
one of this sets contains the other.
A 0 0 0 1 1 0
B 1 1 0 0 0 0
C 0 0 0 1 1 1
D 1 0 1 0 0 0
E 0 0 0 1 0 0
• The above property is necessary and sufficient for prefect
phylogeny under 0 to 1 ordering
• Why works: associated with each character is a subtree.
These subtrees have to be nested.

Simple test for prefect phylogeny
• Fact: there exits perfect phylogeny if and only if and
only if all pairs of characters are compatible
• Special case: if we assume directed parsimony (0!1
only) then characters are compatible if and only if
there do not exist tree taxa containing combinations
(1,0),(0,1),(1,1)
• Observe the last one is equivalent to non-overlapping
criterion
• Optimal algorithm: Gusfield O(nm):
n = # taxa; m= #characters

Two version optimization problem:
Small parsimony: Tree is given and we want to find the
labeling that minimizes #changes – there are good
algorithms to do it.
Large parsimony: Find the tree that minimize number of
evolutionary changes. For most models NP complete
One approach to large parsimony requires:
- generating all possible trees
- finding optimal labeling of internal nodes for each tree.
Fact 1: #tree topologies grows exponentially with #nodes
Fact 2: There may be many possible labels leading to the
same score.

Clique method for large parsimony
• Consider the following
graph:
– nodes – characters;
– edge if two characters
are compatible
1 2 3 4 5 6
α

1 0 0 1 1 0
β

0 0 1 0 0 0
γ

1 1 0 0 0 0
δ

1 1 0 1 1 1
ε

0 0 1 1 1 0
ω

0 0 0 0 0 0
characters
2
1
3
4
5
6
3,5 INCOMPATIBLE
Max. compatible set

Clique method (Meacham 1981) -
• Find maximal compatible clique (NP-
complete problem)
• Each characters defines a partition of the
set into two subsets
α,β,γ,

δ,ε,ω

α,γ,

δ

β

ε,ω

1 2
β

ε,ω

γ,

δ

α,

ε,ω

γ,

δ

α,

3
β

Small parsimony
• Assumptions: the tree is known
• Goal: find the optimal labeling of the tree
(optimal = minimizing cost under given
parsimony assumption)

“Small” parsimony
Infer nodes labels

Application of small parsimony
problem
• errors in data
• loss of function
• convergent evolution (a trait
developed independently by two
evolutionary pathways e.g. wings
in birds an bats)
• lateral gene transfer (transferring
genes across species not by
inheritance)
From paper: Are There Bugs in Our Genome:
Anderson, Doolittle, Nesbo, Science 292 (2001) 1848-51
Red – gene encoding
N-acetylneuraminate
lyase

Dynamic programming algorithm for small
parsimony problem
• Sankoff (1975) comes with the DP approach
(Fitch provided an earlier non DP algorithm)
• Assumptions
– one character with multiple states
- The cost of change from state v to w is δ(v,w)
(note that it is a generalization, so far we talk about
cost of any change equal to 1)

DP algorithm continued
st(v) = minimum parsimony cost for node v under
assumption that the character state is t.
st(v) = 0 if v is a leaf. Otherwise let u, w be children
of u
st(v) = min i {si(u)+ δ(i,t)}+ min j {sj(w)+ δ(j,t)}
t
δ(i,t) δ(j,t)
u w
Try all possible states in u and v
O(nk) cost where n=number of nodes
k = number of sates

Exercise
1 2
5
4
6
7
3
St(v) 1 2 3 4 5 6 7
t
Left and right characters are independent,
We will compute the left.

Branch lengths
• Numbers that indicate the
number of changes in each
branch
• Problem – there may by many
most parsimonious trees
• Method 1: Average over all
most parsimonious trees.
• Still a problem – the branch
lengths are frequently
underestimated
1
4
6
1

Character patterns and parsimony
• Assume 2 state characters (0/1) and four
taxa A,B,C,D
• The possible topologies are:
A
A A
B
B
B
C
C C
D D D
A B C D
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
Changes in each topology
0, 0, 0
1, 1, 1
1, 1, 1
1, 2, 2
Informative character
(helps to decide the tree topology)
Informative characters: xxyy, xyxy,xyyx

Inconsistency
• Let p, q character change probability
• Consider the three informative patters xxyy, xyxy,
xyyx
• The tree selected by the parsimony depends
which pattern has the highest fraction;
• If q(1-q) < p2 then the most frequent pattern is
xyxy leading to incorrect tree.
p
p
q
q
q
A
B
C
D

Distance based methods
• When two sequences are similar they are
likely to originate from the same ancestor
• Sequence similarity can approximate
evolutionary distances
GAATC GAGTT
GA(A/G)T(C/T)

Distance Method
• Assume that for any pair of species we
have an estimation of evolutionary
distance between them
– eg. alignment score
• Goal: construct a tree which best
approximates these distance

Tree from distance matrix
A B
C D
E
1 1
3
2 2
1
3
5
A B C D E
0
0
0
0
0
2 7 7 12
2 7 7 12
7 7 4 11
7 7 4 11
11
11
12
12
A
B
C
D
E
length of the path from A to D = 1+3+1+2=7
Consider weighted trees: w(e) = weight of edge e
Recall: In a tree there is a unique path between any two nodes.
Let e1,e2,…ek be the edges of the path connecting u and v then the
distance between u and v in the tree is:
d(u,v) = w(e1) + w(e2) + … + w(ek)
M T

Can one always represent a
distance matrix as a weighted
tree?
0 10 5 10
10 0 9 5
5 9 0 8
10 5 8 0
a c
b
3
2
7
a
b
c
d
a b c d
d
?
There is no way to add d
to the tree and preserve
the distances

Quadrangle inequality
• Matrix that satisfies quadrangle inequality (called
also the four point condition) for every four taxa is
called additive.
• Theorem: Distance matrix can be represented
precisely as a weighted tree if and only if it is
additive.
a c
b d
d(a,c) + d(b,d) = d(a,d) + d(b,c) >= d(a,b) + d(d,c)

Constructing the tree representing an additive
matrix (one of several methods)
1. Start form 2-leaf tree a,b where a,b are
any two elements
2. For i = 3 to n (iteratively add vertices)
1. Take any vertex z not yet in the tree
and consider 2 vertices x,y that are
in the tree and compute
d(z,c) = (d(z,x) + d(z,y) - d(x,y) )/2
d(x,c) = (d(x,z) + d(x,y) – d(y,z))/2
2. From step 1 we know position of c
and the length of brunch (c,z).
If c did not hit exactly a brunching
point add c and z
else take as y any node from sub-tree
that brunches at c and repeat steps
1,2.
x
y
x y
c
z
x
y
z

Example
0 10 5 9
10 0 9 5
5 9 0 9
9 5 9 0
u
u
v
x
y
u v x y
v
10
Adding x:
d(x,c) = (d(u,x) + d(v,x) – d(u,v))/2 = (5+9-10)/2= 2
d(u,c) = (d(u,x) + d(u,v) - d(x,v))/2 = (5+10-9)/2 = 3
u v
x
3
2
7
c
Adding y:
d(y,c’) = (d(u,y) + d(v,y) – d(u,w))/2 = (5+9-10)/2= 2
d(u,c’) = (d(u,y) + d(u,v) - d(y,v))/2 = (10+9-5)/2 = 7
u v
x
3
2
3
c c’
y
2
4

Real matrices are almost never
additive
• Finding a tree that minimizes the error
Optimizing the error is hard
• Heuristics:
– Unweighted Pair Group Method with
Arithmetic Mean (UPGMA)
– Neighborhood Joining (NJ)

Hierarchical Clustering
• Clustering problem: Group items (e.g. genes) with
similar properties (e.g. expression pattern, sequence
similarity) so that
– The clusters are homogenous (the items in each cluster are
highly similar, as measured by the given property –
sequence similarity, expression pattern)
– The clusters are well separated (the items in different
clusters are different)
• Hierarchical clustering Many clusters have natural
sub-clusters which are often easier to identify e.g. cuts
are sub-cluster of carnivore sub-cluster of mammals
Organize the elements into a tree rather than forming
explicit portioning

The basic algorithm
Input: distance array d; cluster to cluster distance function
Initialize:
1. Put every element in one-element cluster
2. Initialize a forest T of one-node trees (each tree
corresponds to one cluster)
while there is more than on cluster
1. Find two closest clusters C1 and C2 and merge them into
C
2. Compute distance from C to all other clusters
3. Add new vertex corresponding to C to the forest T and
make nodes corresponding to C1, C2 children of this
node.
4. Remove from d columns corresponding to C1,C2
5. Add to d column corresponding to C

A distance function
• dave(C1,C2) = 1/(|C1||C2|)Σ d(x,y)
x in C1 y in C2
Average over all distances
***

Example (on blackboard)
A B C D E
0
0
0
0
0
2 7 7 12
2 7 7 12
7 7 4 11
7 7 4 11
11
11
12
12
A
B
C
D
E
d

Unweighted Pair Group Method with
Arithmetic Mean (UPGMA)
• Idea:
– Combine hierarchical clustering with a method
to put weights on the edges
– Distance function used:
dave(C1,C2) = 1/(|C1||C2|)Σ d(x,y)
– We need to come up with a method of
computing brunch lengths
x in C1
y in C2

Ultrametric trees
• The distance from
any internal node C
to any of its leaves
is constant and
equal to h(C)
• For each node (v)
we keep variable h –
height of the node in
the tree. h(v) = 0 for
all leaves.

UPGMA algorithm
Initialization (as in hierarchical clustering); h(v) = 0
1. Find two closest clusters C1 and C2 and
merge them into C
2. Compute dave from C to all other clusters
3. Add new vertex corresponding to C to the
forest T and make nodes corresponding to C1,
C2 children of this node.
4. Remove from d columns corresponding to
C1,C2
6. h(C) = d(C1, C2) /2
7. Assign length h(C)-h(C1) to edge (C1,C)
8. Assign length h(C)-h(C2) to edge (C2,C)

Neighbor Joining
• Idea:
– Construct tree by
iteratively combing first
nodes that are neighbors
in the tree
• Trick: Figuring out a pair
of neighboring vertices
takes a trick – the closest
pair want always do:
• B and C are the closest
but are NOT neighbors.
B
A
C
D
0 5 7 10
5 0 4 7
7 4 0 5
10 7 5 0
A
B
C
D
2
4
4
1

Finding Neighbors
• Let u(C) = 1/(#clusters-2)Σ d(C,C’)
• Find a pair C1C2 that minimizes
f(C1,C2)= d(C1,C2)-(u(C1)+u(C2))
• Motivation: keep d(C1,C2) small while
(u(C1)+u(C2)) large
all clusters C’

Finding Neighbors
• Let u(C) = 1/(#clusters-2)Σ d(C,C’)
• Find a pair C1C2 that minimizes
f(C1,C2)= d(C1,C2)-(u(C1)+u(C2))
• For the data from example:
u(CA) = u(CD) = 1/2(5+7+10) = 11
u(CB) = u(CC) = 1/2(5+4+7) = 8
f(CA,CB) = 5-11 -8 = -14
f(CB,CC) = 4- 8 -8 = -12
all clusters C’

NJ algorithm
Initialization (as in hierarchical clustering); h(v) = 0
1. Find clusters C1 and C2 minimizing f(C1C2) and
merge then into C
2. Compute for all C*: d(C,C*) = (d(C1C)+ d(C2C))/2
3. Add new vertex corresponding to C to the forest T
and connect it to C1, C2
4. Remove from d columns corresponding to C1,C2
6. Assign length ½(d(C1C2)+u(C1)-u(C2) to edge C1C
7. Assign length ½(d(C1C2)+u(C2)-u(C1) to edge C2C

NJ tree is not rooted
The order of construction of internal nodes of NJ does not suggest an
ancestral relation:
1 2 3 4 5

Rooting a tree
• Choose one distant organism as an out-
group
Species of interests
out-group
root

Bootstraping
• Estimating confidence in the tree topology
•Are we sure if this is
correct?
•Is there enough
evidence that A is a
successor of B not the
other way around? A
B

Bootstrapping, continued
• Assume that the tree is build form multiple
sequence alignment
A
B
1 2 3 4 5 67 8 9 10 11
Columns
of the alignment
Select columns randomly
(with replacement)
13 2 1 10 6 5 4 5 0 11
Initial tree
A
B
New tree
Repeat, say 1000 times,
For each edge of initial tree calculate % times
it is present in the new tree
59%

Summary
• Assume you have multiple alignment of length N.
Let T be the NJ tree build from this alignment
• Repeat, say 1000 times the following process:
– Select randomly with replacement N columns of the
alignment to produce a randomized alignment
– Build the tree for this randomized alignment
• For each edge of T report % time it was present
in a tree build form randomized alignment . This
is called the bootstrap value.
• Trusted edges: 80% or better bootstrap.

Maximum Likelihood Method
• Given is a multiple sequence alignment
and probabilistic model of for substitutions
(like PAM model) find the tree which has
the highest probability of generating the
data.
• Simplifying assumptions:
– Positions involved independently
– After species diverged they evolve
independently.

Formally:
• Find the tree T such that assuming evolution
model M
Pr[Data| T,M] is maximized
• From the independence of symbols:
Pr[Data| T,M] = P i Pr[Di| T,M]
Where the product is taken over all characters i
and Di value of the character i is over all taxa

Computing Pr[Di| T,M]
Pr[Di| T,M]
= Σ x Σ y Σ z
p(x)p(x,y,t1)p(y,A,t2)p(y,B,t6)p(x,z,t3)p(z,C,t4)p(z,D,t5)
A B C D
Column Di
x
y z
consider all possible
assignments here
t1
t2
t3
t4
t5 time
p(x,y,t) = prob. of
mutation x to y in time t
(from the model)
t6

Discovering the tree of life
• “Tree of life” – evolutionary tree of all
organisms
• Construction: choose a gene universally present
in all organisms; good examples: small rRNA
subunit, mitochondrial sequences.
• Things to keep in mind while constructing tree of
life from sequence distances:
– Lateral (or horizontal) gene transfer
– Gene duplication: genome may contain similar genes
that may evolve using different pathways. Phylogeny
tree need to be derived based on orthologous genes.

Where we go with it…
• We now how to compute for given column
and given tree Pr[Di| T,M]
• Sum up over all columns to get
Pr[Data| T,M]
Now, explore the space of possible trees
Problem:
• Bad news: the space of all possible trees is HUGE
Various heuristic approaches are used.

Metropolis algorithm:Random Walk
in Energy Space
Goal design transition probabilities so that the
probability of arriving at state j is
P(j) = q(j) / Z
(typically, q(S) = e
–E(S)/kT
, where E – energy)
Z – partition function = sum over all states S of
terms q(s). Z cannot be computed analytically since the
space is to large
State i
(in our case tree)
State j
(in our case another tree)
Tree i can be obtained
from tree j by a “local” change
The network
of possible
trees, there is
an edge
between
“similar
trees”
temperature
const.

Monte Carlo, Metropolis
Algorithm
• At each state i choose uniformly at random
one of neighboring conformations j.
• Compute p(i,j) = min(1, q(i)/q(j) )
• With probability p(i,j) move to state j.
• Iterate

MrBayes
• Program developed by J.Huelsenbeck &
F.Ronquist.
• Assumption:
• q(i) = Pr[D| Ti,M]
Prior probabilities: All trees are equally
likely.
• Proportion of time a given tree is visited
approximates posterior probabilities.

Most Popular Phylogeny Software
• PAUP
• PHYLIP

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