Chapter 7 Building Phylogenetic Trees

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• 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

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

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

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

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 )

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

Homologous Analogous

Orthologous Paralogous

Orthologues / Paralogues

Orthology/paralogy

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

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

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

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

Rooted / Unrooted Tree

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.

Rooted versus unrooted treesTree a

Tree b

Tree c

Represents all three rooted trees

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

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

Counting Trees

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

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)

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

Phylogenetic trees

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

HRV100

HRV1bGenba

HRV89Genba

HRV2Genban

HRV16Genba

HRVHanks

HRV65HRV46

HRV8HRV95

HRV36HRV89GenbaHRV7

HRV23HRV30

HRV2Genban

HRV16Genba

HRV57HRV55

HRVHanks

HRV21HRV11

HRV33HRV76

HRV90HRV18HRV34HRV50

HRV96 HRV15HRV74

HRV67HRV32HRV9HRV19

HRV82HRV22

HRV94HRV1A

HRV1bGenbaHRV39

HRV59HRV63

HRV25 HRV62

HRV100HRV10

HRV100

HRV1bGenba

HRV89Genba

HRV2Genban

HRV16Genba

HRVHanks

HRV100

HRV25HRV62

HRV1AHRV1bGenba

HRV53HRV71

HRV89Genba

HRV2Genban

HRV16Genba

HRVHanks

HRV33HRV76

HRV13HRV41

HRV38HRV60

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

Distance matrix methods

• UPGMA– Algorithm introduced by Sokal and Michener

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

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

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)

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

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

ikij d

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

dAB / 2

d(AB)C / 2

step 1 step 2

UPGMA -- Ilustrations

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

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

2ACd A

A B C D

A 0 8 7 12

B 0 9 14

C 0 11

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

5.89*2

ABAC d

5.1111*2

ADAC d

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

New distance matrix

An example UPGMA (5)AC B D

AC 0 8.5 11.5

B 0 14

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

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*

ACDACB d

ACB 0 12.33

New distance matrix

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

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

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

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

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

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).

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

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

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

• dim=dik+dkm

• djm=djk+dkm

• dim+djm=dik+djk+2dkm=dij+2dkm

• dkm=(dim+djm-dij)/2

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

Neighbour Joining(3)i j

i j ik jk

ij ik jk

r r d d

Neighbour Joining: Algorithmik jk i j

ik jk ij

d d r r

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).

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

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)

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)

Result: all pairwise distances obtained by traversing the tree

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

)( 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

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)

5.155.13

5.135.15

ABBDADDAB

ABBCACCAB

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

)( 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

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)

ABCABCC

CABABC

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

ABCCDABD

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 0 8

8A B C D

A 0 8 7 12

B 0 9 14

C 0 11

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

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

6.17 D

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

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

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:

Parsimony

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

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)

Traditional Parsimony

nodesinternalmin);,...,(

vu xxEvu

n TssPar

•Solved independently for each position

•Linear time solution

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

Traditional Parsimony is not “complete”

Weighted Parsimony

Example

Aardvark Bison Chimp Dog Elephant

A: CAGGTAB: CAGACAC: CGGGTAD: TGCACTE: TGCGTA

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

fuguDrosophila

Drosophila

parsimony

distance

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

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

Original

Scrambled

Non-supportive

Chapter 7 Building Phylogenetic Trees

molecular sequences

set of sequences

observed sequences of

speciesanalogous sequences

partsorthologous sequences

homologous genes

common ancestral gene

orthologues genes

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