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Chapter 7 Building Phylogenetic Trees

Jan 15, 2016

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Chapter 7 Building Phylogenetic Trees. Contents. Phylogeny Phylogenetic trees How to make a phylogenetic tree from pairwise distances UPGMA method (+ an example) Neighbor-Joining method (+ an example) Comparison of methods Conclusion. Phylogeny. - PowerPoint PPT Presentation
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Page 1: Chapter 7 Building Phylogenetic Trees

1

Chapter 7

Building Phylogenetic Trees

Page 2: Chapter 7 Building Phylogenetic Trees

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Contents

• Phylogeny • Phylogenetic trees• How to make a phylogenetic tree from pairwise

distances– UPGMA method (+ an example)– Neighbor-Joining method (+ an example)

• Comparison of methods• Conclusion

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Phylogeny• Phylogeny is the evolution of related

species/genes• Phylogenetic tree: diagram showing evol

utionary lineages of species/genes• The history of genes or species may be

very different• Genes can be homologous or

analogous, but still remind each other

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Phylogeny

• The similarity of molecular mechanisms of the organisms that have been studied strongly suggests that all organisms on Earth had a common ancestor

• Any set of species is related, and this relationship is called a phylogeny

• The relationship can be represented by a phylogenetic tree

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Phylogeny

• Traditionally, morphological characters (both from living and fossilized organisms) have been used for inferring phylogenies

• Zuckerkandel & Pauling (1962) showed that molecular sequences provide sets of characters that can carry a large amount information

• If we have a set of sequences from different species , we may be able to use them to infer a likely phylogeny of the species in question

• This assumes that the sequences have descended from some common ancestral gene in a common ancestral species

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Phylogeny

• The widespread occurrence of gene duplication means that the foregoing assumption needs to be checked carefully

• The phylogentic tree of a group of seqences does not necessarily reflect the phylogenetic tree of their host species, because gene duplication is another mechanism, in addition to speciation, by which two sequences can be separated and diverge from a common ancestor

• Genes which diverged because of speciation

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Phylogeny

• Genes which diverged because of speciation are called orthologues ( 直系同源 )

• Genes which diverged by gene duplication are called paralogues ( 平行進化同源 )

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Phylogeny• Homologous sequences can be divided into two

parts– Orthologous sequences diverged by

specification from a common ancestor– Paralogous sequences evolved by gene

dublication within species• Analogous sequences may appear and function

very similarly, but they do not have a common ancestor

• WHEN WE WANT TO EXPLORE EVOLUTIONARY RELATIONSHIPS, WE NEED TO HANDLE ORTHOLOGOUS SEQUENCES

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Genes

Homologous Analogous

Orthologous Paralogous

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Orthologues / Paralogues

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Orthology/paralogy

Orthologous genes are homologous (corresponding) genes in different species (genomes)

Paralogous genes are homologous genes within the same species (genome)

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Phylogenetic Trees

• WHY construct a phylogenetic tree?– to understand lineage of various species– to understand how various functions evolved– to inform multiple alignments

• Trees can be rooted (a common ancestor in known) or unrooted

• Leaves are the terminal nodes that correspond to the observed sequences of genes or species (A, B, C, D)

• Internal nodes are hypothetical ancestral nodes• All trees will be assumed to be binary, meaning that an

edge that branches splits into two daughter edges• Each edge has a certain amount of evolutionary

divergence associated to it, defined by some measure of distance between sequences, or from a model of substitution of residues over the course of evolution

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Phylogenetic Trees

• We adopt the general term “length” or “edge length” here, and represent this by the lengths of edge in the figures we draw

• A true biological phylogeny has a “root”, or ultimate ancestor of all the sequences

• The leaves of trees have names or numbers• A tree with a given labelling will be called a labell

ed branching pattern• We refer to this as the tree topology and denote i

t by the symbol T• The lengths of its edges are denoted by ti with a

suitable numbering scheme for the is

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Rooted / Unrooted Tree

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Types of treesUnrooted tree represents the same phylogeny w

ithout the root node

Depending on the model, data from current day species often does not distinguish between different placements of the root.

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Rooted versus unrooted treesTree a

ab

Tree b

c

Tree c

Represents all three rooted trees

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Rrooting the tree:

To root a tree mentally, imagine that the tree is made of string. Grab the string at the root and tug on it until the ends of the string (the taxa) fall opposite the root: A

BC

Root D

A B C D

RootNote that in this rooted tree, taxon A is no more closely related to taxon B than it is to C or D.

Rooted tree

Unrooted tree

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Counting Trees

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Counting Trees

(2N - 5)!! = # unrooted trees for N taxa(2N- 3)!! = # rooted trees for N taxa

CA

B D

A B

C

A D

B E

C

A D

B E

C

F

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How many trees?

• Number of unrooted trees = (2n-5)! / 2n-3 (n-3)!

=3x5x…x(2n-5)

• Number of rooted trees = (2n-3)! / 2n-32(n-2)!

=3x5x…x(2n-3)

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Combinatoric explosion

# sequences # unrooted # rooted trees trees

2 1 13 1 34 3 155 15 1056 105 9457 945 10,3958 10,395 135,1359 135,135 2,027,02510 2,027,025 34,459,425

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Phylogenetic trees

• Different ways to represent a phylogenetic tree (illustrated by Treeview)

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Making a tree from pairwise distances

• Distances dij between each pair of sequences i and j are calculated in the given dataset

• Different ways defining distances– For nucleotide sequences:

Jukes-Cantor, Kimura-2-parameter K2P, HKY (Hasegawa-Kishino-Yano), F84, Tamura-Nei, General time-reversible model, General 12-parameter model

– For amino acid sequences:PAM-matrices, BLOSUM-matrices

A B C D

A 0 32 44 46

B 32 0 29 43

C 44 29 0 30

D 46 43 30 0

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Distance matrix methods

• UPGMA– Algorithm introduced by Sokal and Michener

1958

• Neighbor-Joining– Algorithm introduced by Saitou and Nei 1987– Modified by Studier and Keppler 1988

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Clustering method: UPGMA

• UPGMA = Unweighted pair group method using arithmetic averages

• Simple method • It works by clustering the sequences, at each

stage connecting two clusters and finally creating a new node on a tree

• Method assumes equal rate of evolutionary change along branches Molecular clock assumption

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UPGMA• UPGMA produces a rooted tree• Branch lengths satisfy a molecular clock The divergence of sequences is assumed to occur at the same

constant rate at all points in the tree• Trees that are clocklike are rooted and the total branch length

from the root up to any leaf is equal• Trees are often referred to be ultrametric• A distance measures are ultrametric if either all three distances a

re equal dij = dik = djk or two of them are equal and one is smaller: djk < dij = dik

UPGMA is guaranteed to build the correct tree if distances are ultrametric

• Method can be used for reconstructing phylogenies if evolutionary rates are assumed to be same in all lineages criticism in the phylogeny literature– Suitable for the species closely related

• Running time O(n2)

A

C

B

D

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Algorithm: UPGMAInitialisation:

Assign each sequence i in dataset to its own cluster

Define one leaf of T for each sequence, and place at height zero

Iteration:

Find the two clusters i and j for which dij is the smallest (pick randomly if several equal distances)

Define a new cluster ij by Cij = Ci U Cj. Cluster ij has nij = ni + nj

members ( initially ni = 1 )

Connect i and j on the tree to a new node v

The branch lengths from new node to i and j are

placed at height

2ijd

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Algorithm: UPGMA (cont.)

Iteration (cont.)Compute the distances between the new cluster and the remaining clusters by using

Add ij to the current clusters and remove i and j Termination:

When only two clusters i and j remain, place the rootat height

2ijd

jkji

jik

ji

ikij d

nn

nd

nn

nd

),(

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UPGMA -- Unweighted Pair Group Method with Arithmetic mean

simplest method - uses sequential clustering algorithm(assumption of rate constancy among lineages - often violated)

A BB dABC dAC dBC

(AB)C d(AB)C d(AB)C = (dAC + dAB) / 2Distance matrix

Tree

dAB / 2

A

B

A

d(AB)C / 2

B

C

step 1 step 2

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UPGMA -- Ilustrations

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An example UPGMA (1)

• Distance matrix (arbitrary) for four items (sequences) A, B, C and DActually distances are not ultrametric, because three distances are not equal

dij ≠ dik ≠ djk or two of them are not equal and one is smaller:

djk < dij ≠ dik

A B C D

A 0 8 7 12

B 8 0 9 14

C 7 9 0 11

D 12 14 11 0

Step 1. Find the smallest distance, dij, between two clusters A and C, where dij is 7

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An example UPGMA (2)Step 2. Define new cluster ij, which has nij = ni + nj members (initially ni = 1)

New cluster A and C nAC = nA+ nC=2

Step 3. Connect A and C on the tree to a new node v1

Step 4. The branch lengths from new node v1 to A and C

5,32

7

2ACd A

C3,5

3,5

A B C D

A 0 8 7 12

B 0 9 14

C 0 11

D 0

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An example UPGMA (3)

Step 5. Compute the distances between the new cluster AC and the remaining clusters (B and D):

5.89*2

18*

2

1,

CBCA

CAB

CA

ABAC d

nn

nd

nn

nd

5.1111*2

112*

2

1,

CDCA

CAD

CA

ADAC d

nn

nd

nn

nd

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Step 6. Delete the columns and rows of the distance matrix that correspond to clusters A and C, and add a column and a row for cluster AC

An example UPGMA (4)

AC B D

AC 0 8.5 11.5

B 0 14

D 0

New distance matrix

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An example UPGMA (5)AC B D

AC 0 8.5 11.5

B 0 14

D 0

2nd iteration process

Step 1. Find the two sequences i and j for which dij is the smallest (randomly if several equal distances)AC-B

Step 2. Define new cluster (ij), which has nij = ni + nj members ( initially ni = 1 ) New cluster AC and B nACB = nAC+ nB = 2 + 1 = 3

Step 3. Connect AC and B on the tree to a new node v2

Step 4. The branch lengths from new node v2 to AC and B

25.42

5.8

2ACBd

A

C3.5

3.5

B4.25

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An example UPGMA (6)Step 5. Compute the distances between the new

cluster and the remaining cluster (D)

Step 6. Delete the columns and rows of the distance matrix that correspond to clusters AC and B, and add a column and a row for cluster ACB

33.1214*3

15.11*

3

2),(

BDBAC

BACD

BAC

ACDACB d

nn

nd

nn

nd

ACB D

ACB 0 12.33

D 0

New distance matrix

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An example UPGMA (7)Termination:

Only two clusters (ACB and D) remaining

Place the root height ACB D

ACB 0 12.33

D 017.62

33.12

2ijd

A

C3.5

3.5

B4.25

6.17D

Original distance matrix and final phylogenetic tree(including thebranch lengths)

1.92A B C D

A 0 8 7 12

B 0 9 14

C 0 11

D 0

0.75

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When UPGMA fails …

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When UPGMA fails …

• The closest leaves are not neighboring leaves; they do not have a common parent node

• A test of whether reconstruction is likely to be correct is the ultrametric condition

• A distance measures are ultrametric if either all three distances are equal dij = dik = djk or two of them are equal and one is smaller: djk < dij = dik

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Ultrametric Distances

Given three leaves, two distances are equal while a third is smaller:

d(i,j) d(i,k) = d(j,k)

a+a a+b = a+b

a

a

b

i

j

k

nodes i and j are at same evolutionary distance from k – the dendrogram will therefore have ‘aligned’ leaves; i.e. they are all at the same distance from root

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Evolutionary clock speeds

Uniform clock: Ultrametric distances lead to identical distances from root to leaves

Non-uniform evolutionary clock: leaves have different distances to the root -- an important property is that of additive trees. These are trees where the distance between any pair of leaves is the sum of the lengths of edges connecting them. Such trees obey the so-called 4-point condition (next slide).

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Additivity

• Given a tree, its edge lengths are said to be additive if the distance between any pair of leaves is the sum of the lengths of the edges on the path connecting them

• This property is built in automatically as the UMGMA tree is constructed

• It is possible for the molecular clock property to fail but for additivity to hold, and in that case there are algorithms that can be used to reconstruct the tree correctly

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Neighbor Joining

• Very popular method• Does not make molecular clock assumption :

modified distance matrix constructed to adjust for differences in evolution rate of each taxon

• Produces unrooted tree• Assumes additivity: distance between pairs of

leaves = sum of lengths of edges connecting them

• Like UPGMA, constructs tree by sequentially joining subtrees

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Neighbor Joining: Once we know the correct (i,j) pair

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• dim=dik+dkm

• djm=djk+dkm

• dim+djm=dik+djk+2dkm=dij+2dkm

• dkm=(dim+djm-dij)/2

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Neighbour Joining: why not pick the smallest (i,j) pair?

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Neighbour Joining(3)i j

i j ik jk

ij ik jk

r r d d

d d d

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Neighbour Joining: Algorithmik jk i j

ik jk ij

d d r r

d d d

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Neighbor-Joining: Complexity

• The method performs a search using time O(n2) and using time O(n2) to update distance matrix.

• Giving a total time complexity of O(n3),and a space complexity of O(n2).

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Neighbor-Joining

• We can use neighboring-joining even lengths are not additive, but reconstruction of the correct tree is no longer guaranteed

• We can test for additivity• For every set of four leaves, i, j, k, and l, t

wo of the distances dij+dkl, dik+djl and dil+djk must be equal and larger than the third

• dij+dkl= dik+djl > dil+djk

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Additivity

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Additivity

Theorem: A set M of L objects is additive iff any subset of four objects can be labeled i,j,k,l so that:

d(i,k) + d(j,l) = d(i,l) +d(k,j) ≥ d(i,j) + d(k,l)

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Additive trees

All distances satisfy 4-point condition:

For all leaves i,j,k,l:

d(i,j) + d(k,l) d(i,k) + d(j,l) = d(i,l) + d(j,k)

(a+b)+(c+d) (a+m+c)+(b+m+d) = (a+m+d)+(b+m+c)

i

j

k

l

a

b

mc

d

Result: all pairwise distances obtained by traversing the tree

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Step 1. Computefor each row indistance matrix

Step 2. Compute(the lower-diagonal matrix) and choose the smallest (most negative)

An example N-J (1)

A B C D Step 1 - ri

A 0 8 7 12 =(8+7+12)/(4-2) = 13.5

B 8 0 9 14 =(8+9+14)/(4-2)=15.5

C 7 9 0 11 =(7+9+11)/(4-2)=13.5

D 12 14 11 0 =(12+14+11)/(4-2)=18.5

n

ij

iji n

dr

2

)( jiij rrd

A B C D

A 0 8 7 12

B 8-(13.5+15.5)=-21 0 9 14

C 7-(13.5+13.5)=-20 9-(15.5+13.5)= -20 0 11

D 12-(13.5+18.5)=-20 14-(15.5+18.5)=-20 11-(13.5+18.5)=-21 0

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An example N-J (2)

Step 3. Join A and B together with a new node v1. Compute the edge lengths, from A to node v and from B to node v1

Step 4. Compute distances between the new node v1 and remaining items (C and D)

3

2

5.155.13

2

8

2

)(

2

BAAB

A

rrdv

5

2

5.135.15

2

8

2

)(

2

ABAB

B

rrdv

92

81412

2

)(

42

897

2

)(

),(

),(

ABBDADDAB

ABBCACCAB

dddd

dddd

v1

B

A

5

3

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57

An example N-J (3)

Step 5. Delete A and B from the distance matrix and replace them by new item AB

Step 6. Continue from step 1, because more than two items remain

Step 1. Compute for each row indistance matrix

Step 2 Computeand choose the smallest (the lower-diagonal matrix)

AB C D Step 1 = ri

AB 0 4 9 (4+9)/1=13

C 4 0 11 (4+11)/1=15

D 9 11 0 (9+11)/1=20

New reduced distance matrix

n

ij

iji n

dr

2

)( jiij rrd AB C D

AB 0 4 9

C 4-(13+15)=-24 0 11

D 9-(13+20)=-24 11-(15+20)=-24 0

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An example N-J (4)

Step 3 Join v1 and C together with a new node v2. Compute the edge lengths, from v1 to node v2 and from C to node v2

Step 4 Compute distances between the new node v2 and remaining items (D)

3

2

1315

2

4

22

12

1513

2

4

2

)(

21

ABCABCC

CABABC

rrdv

rrdv

AB C D Step 1 = ui

AB 0 4 9 (4+9)/1=13

C 4 0 11 (4+11)/1=15

D 9 11 0 (9+11)/1=20

82

4119

2

)(),(

ABCCDABD

DABC

dddd

v1B

A

5

3

v21

3C

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59

An example N-J (5)

Step 5 Delete AB and C from the distance matrix and replace them by ABC

Step 6 Only two nodes remaining connect them

ABC D

ABC 0 8

D 0

B

A

5

3C

D

8A B C D

A 0 8 7 12

B 0 9 14

C 0 11

D 0

13

Original distance matrix and final phylogenetic tree (including the edge lengths)

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Comparison• UPGMA

– The total branch length from the root up to any leaf is equal

– Produces a rooted tree, where the root is hypothesized ancestor of the sequences in the tree

– Suitable for closely related sequences

– Can be used to infer phylogenies if one can assume that evolutionary rates are the same in all lineages

• Neighbor-joining– Unrooted tree, where the

direction of evolution is unknown

– Suitable for datasets with largely varying rates of evolution

– Suitable for large datasets

B

A

5

3C

D

8

13

A

C3.5

3.5

B4.25

6.17 D

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Comparison• UPGMA method constructs a rooted phylogenetic tree correctly

if there is a molecular clock with a constant rate of mutation• UPGMA method is rarely used, because molecular clock

assumption is not generally true: selection pressures vary across time periods, genes within organisms, organisms, regions within gene

• N-J method produces an unrooted tree without molecular clock hypothesis

• N-J method is one of the most popular and widely used by molecular evolutionist

• Distance methods are strongly dependent on the model of evolution used

• Sequence information is reduced when transforming sequence data into distances

• Distance methods are computationaly fast

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Parsimony

• Find the tree which can explain the observed sequences with a minimal number of substitutions

• It assigns a cost to a tree, and it is necessary to search through all topologies, or to pursue a more efficient search strategy that achieves this effect, in order to identify the ‘best’ tree

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Parsimony

• The computation of a cost for a given tree• A search through all trees, to find the overall

minimum of this cost• Suppose we have the following four aligned

nucleotide sequences:

AAG

AAA

GGA

AGA

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Parsimony

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Cost of Evaluating Parsimony

• Score is evaluated on each position independetly. Scores are then summed over all positions.

• If there are n nodes, m characters, and k possible values for each character, then complexity is O(nmk)

• By keeping traceback information, we can reconstruct most parsimonious values at each ancestor node

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Evaluating Parsimony Scores

• How do we compute the Parsimony score for a given tree?

• Traditional Parsimony– Each base change has a cost of 1

• Weighted Parsimony– Each change is weighted by the score c(a,b)

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Traditional Parsimony

}{},{

nodesinternalmin);,...,(

vu xxEvu

n TssPar

11

a g a

{a,g}

{a}

•Solved independently for each position

•Linear time solution

a

a

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Traditional Parsimony

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Traditional Parsimony

• There is a traceback procedure for finding ancestral assignments in traditional parsimony

• We choose a residue from R2n-1, then proceed down the tree

• Having chosen a residue from the set Rk, we pick the same residue from the daughter set Ri if possible, and otherwise pick a residue at random from Ri

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Traditional Parsimony is not “complete”

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Weighted Parsimony

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Example

Aardvark Bison Chimp Dog Elephant

A: CAGGTAB: CAGACAC: CGGGTAD: TGCACTE: TGCGTA

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Parsimony & DistanceSequences 1 2 3 4 5 6 7Drosophila t t a t t a a fugu a a t t t a a mouse a a a a a t a human a a a a a a t

human x

mouse 2 x

fugu 4 4 x

Drosophila 5 5 3 x

human

mouse

fuguDrosophila

Drosophila

fugu

mouse

human

12

3 7

64 5

Drosophila

fugu

mouse

human

2

11

12

parsimony

distance

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How to assess confidence in tree

• Distance method – bootstrap:– Select multiple alignment columns with

replacement– Recalculate tree– Compare branches with original (target) tree– Repeat 100-1000 times, so calculate 100-

1000 different trees– How often is branching (point between 3

nodes) preserved for each internal node?– Uses samples of the data

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The Bootstrap -- example

1 2 3 4 5 6 7 8 - C V K V I Y SM A V R - I F SM C L R L L F T

3 4 3 8 6 6 8 6 V K V S I I S IV R V S I I S IL R L T L L T L

1

2

3

1

2

3

Original

Scrambled

4

5

1

5

2x 3x

Non-supportive