Building Phylogenetic Trees

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• 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 evolutionary
  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 (????)
• Genes which diverged by gene duplication are
  called paralogues (??????)

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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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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
  labelled branching pattern
• We refer to this as the tree topology and denote
  it 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 trees
Unrooted tree represents the same phylogeny
without 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 trees
Tree c
b
a
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
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
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How many trees?

• Number of unrooted trees (2n-5)! / 2n-3 (n-3)!
• 
  3x5xx(2n-5)
• Number of rooted trees (2n-3)! / 2n-32(n-2)!
• 
  3x5xx(2n-3)

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Combinatoric explosion

• sequences unrooted rooted
• trees trees
• 2 1 1
• 3 1 3
• 4 3 15
• 5 15 105
• 6 105 945
• 7 945 10,395
• 8 10,395 135,135
• 9 135,135 2,027,025
• 10 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
A
C
B

• 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 are equal
• dij dik djk or two of them are equal and one
  is smaller djk lt 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)

D
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Algorithm UPGMA
Initialisation 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
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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
  root
• at height

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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)
step 1 step 2
(AB) C d(AB)C
Distance matrix Tree
d(AB)C (dAC dAB) / 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 D
• Actually 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 lt 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 nC2
• 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

A B C D
A 0 8 7 12
B 0 9 14
C 0 11
D 0
3,5
A
C
3,5
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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)

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An example UPGMA (4)

• 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

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)

• 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
• ?

AC B D
AC 0 8.5 11.5
B 0 14
D 0
3.5
A
C
3.5
B
4.25
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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

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 0
Original distance matrix and final phylogenetic
tree(including the branch lengths)
3.5
A
0.75
A B C D
A 0 8 7 12
B 0 9 14
C 0 11
D 0
C
1.92
3.5
B
4.25
D
6.17
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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 lt 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) aa
? ab ab
i
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
a
b
k
a
j
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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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• dimdikdkm
• djmdjkdkm
• dimdjmdikdjk2dkmdij2dkm
• dkm(dimdjm-dij)/2

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Neighbour Joining why not pick the smallest
(i,j) pair?
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Neighbour Joining(3)
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Neighbour Joining Algorithm
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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, two
  of the distances dijdkl, dikdjl and dildjk
  must be equal and larger than the third
• dijdkl dikdjl gt dildjk

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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) (ab)(cd) ?
(amc)(bmd) (amd)(bmc)
k
i
a
c
m
b
d
j
l
Result all pairwise distances obtained by
traversing the tree
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An example N-J (1)
A B C D Step 1 - ri
A 0 8 7 12 (8712)/(4-2) 13.5
B 8 0 9 14 (8914)/(4-2)15.5
C 7 9 0 11 (7911)/(4-2)13.5
D 12 14 11 0 (121411)/(4-2)18.5
Step 1. Compute for each row in distance
matrix Step 2. Compute (the lower-diagonal
matrix) and choose the smallest (most
negative)
A B C D
A 0 8 7 12
B 8-(13.515.5)-21 0 9 14
C 7-(13.513.5)-20 9-(15.513.5) -20 0 11
D 12-(13.518.5)-20 14-(15.518.5)-20 11-(13.518.5)-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)
B
5
v1
3
A
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An example N-J (3)
New reduced distance matrix
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
in distance matrix Step 2 Compute and choose
the smallest (the lower-diagonal matrix)
AB C D Step 1 ri
AB 0 4 9 (49)/113
C 4 0 11 (411)/115
D 9 11 0 (911)/120
AB C D
AB 0 4 9
C 4-(1315)-24 0 11
D 9-(1320)-24 11-(1520)-24 0
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An example N-J (4)
AB C D Step 1 ui
AB 0 4 9 (49)/113
C 4 0 11 (411)/115
D 9 11 0 (911)/120
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)
B
5
v1
v2
1
3
3
A
C
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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
Original distance matrix and final phylogenetic
tree (including the edge lengths)
D
A B C D
A 0 8 7 12
B 0 9 14
C 0 11
D 0
8
B
5
1
3
3
A
C
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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

D
8
3.5
A
B
5
C
3.5
1
B
3
3
A
C
4.25
D
6.17
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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
a
a

• Solved independently for each position
• Linear time solution

a,g
a
68
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
A CAGGTA B CAGACA C CGGGTA D TGCACT E TGCGTA
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Parsimony Distance
parsimony
Sequences 1 2 3 4 5 6
7 Drosophila 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
Drosophila
mouse
1
6
4
5
2
3
7
human
fugu
distance
human x mouse 2 x fugu 4 4
x Drosophila 5 5 3 x
Drosophila
mouse
2
1
2
1
1
human
fugu
human
mouse
fugu
Drosophila
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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 S M A V R -
I F S M C L R L L F T 3 4 3 8 6 6 8 6 V K
V S I I S I V R V S I I S I L R L T L L T L
5
1 2 3
Original
4
2x
3x
1
1 2 3
Non-supportive
Scrambled
5