Phylogenetic Trees

Lecture 3

.

Based on: Durbin et al 7.4; Gusfield 17

Character-based methods for constructing phylogenies

In this approach, trees are constructed by comparing the characters of the corresponding species. Characters may be morphological (teeth structures) or molecular (homologous DNA sequences). One common approach is Maximum Parsimony.

Assumptions:

 Independence of characters (no interactions)  Best tree is one where minimal changes take place 2

1. Maximum Parsimony

Input: four nucleotide sequences: AAG, AAA, GGA, AGA taken from four species.

Question: Which evolutionary tree best explains these sequences ?

One Answer (the parsimony principle): Pick a tree that has a minimum total number of substitutions of symbols between species and their originator in the phylogenetic tree.

AAA

1

AAG AAA

2

GGA AAA AAA

1

AGA

Total #substitutions = 4 3

Example Continued

There are many trees possible. For example:

1

AAA AAG AAA AAA

1 1

GGA AGA AGA

1

AAG AAA

1

AGA AAA AAA AAA

2

GGA

Total #substitutions = 3 Total #substitutions = 4 The left tree is preferred over the right tree.

The total number of changes is called the

parsimony score

.

4

Simple Example

 Suppose we have five species, such that three have ‘C’ and two ‘T’ at a specified position  Minimal tree has one evolutionary change: C C C C C T  C T T T 5

Extension to Many Letters

 What is the parsimony score of Aardvark Bison Chimp Dog A:

CAGGTA

B:

CAGACA

C: D:

CGGGTA TGCACT

E:

TGCGTA

Elephant We do it character after character; each score is computed independently of the others. 6

Fitch’s Algorithm of Evaluating Trees Traverse tree from leaves to root determining set of possible

states

(e.g. nucleotides) for each internal node Traverse tree from root to leaves picking ancestral states for internal nodes 7

Fitch’s Algorithm – Step 1

 # of changes = # union operations

C CT T T T G GT AGT T A T

8

Fitch’s Algorithm – Step 1

 D o a post-order (from leaves to root) traversal of tree  Determine possible states

R i

children

j

and

k

of internal node

i

with

R

i

  

R R

j j



R

k



R

k

if R

j



R

k

otherwise

    9

Fitch’s Algorithm – Step 2

10

Fitch’s Algorithm – Step 2

Do a pre-order (from root to leaves) traversal of tree Select state

r j

of internal node

j

with parent

i r j

  

r i if r i



arbitrary R j state



R j otherwise

  11

Weighted Version of Fitch’s Algorithm

I nstead of assuming all state changes are equally likely, use different costs

c

(

a, b

) for different changes

a



b

1 st step of algorithm is to propagate costs up through tree 12

Weighted Version of Fitch’s Algorithm

Want to determine minimal cost

S

(

i, a

) of assigning character

a

to node

i

For leaves:

S(i, a)

   0 

if a is a character at otherwise leaf

  13

Weighted Version of Fitch’s Algorithm

W ant to determine min. cost

S

(

i, a

) of assigning character

a

to node

i

For internal nodes:

S

(

i

,

a

)  min

b

(

S

(

j

,

b

) 

c

(

a

,

b

))  min

b

(

S

(

k

,

b

) 

c

(

a

,

b

))

i a j b a



b k

14

Weighted Version of Fitch’s Algorithm – Step 2

D o a pre-order (from root to leaves) traversal of tree Select minimal cost character for root For each internal node

j

, select character that produced minimal cost at parent

i

15

Weighted Parsimony Scores

Weighted Parsimony score: Each change is weighted by a score c(a, b).

The weighted parsimony score reduces to the parsimony score when c(a,a)=0 and c(a,b)=1 for all b



a.

16

Evaluating Weighted Parsimony Scores Each position is independent and computed by itself.

Use Dynamic Programming on a given tree.

 If

k

is a node with children

i

and

j

, then

S(i, a) = min x

(

S

(

j, x

)

+c

(

a, x

))

i + min y

(

S

(

k, y

)

+c

(

a, y

)) S(

i, a

)  the minimum score of subtree rooted at

k

when

k

has S(i,a) character

a

.

k

S(j,x)

j

S(k,y) 17

Evaluating Parsimony Scores

Dynamic programming on a given tree

Initialization:

 For each leaf

i S(i,a) =

 set

S(i,a) = 0

if

i

is labeled by

a

, otherwise

Iteration:

 if

i

is node with children

j

and

k

, then

S(i,a) = min x (S(j,x)+c(a,x)) + min y (S(k,y)+c(a,y))

Termination:

 cost of tree is

min x S(r,x)

where

r

is the root

Comment:

To reconstruct an optimal assignment, we need to keep in each node

i

and for each character

a

the two characters x, y that bring about the minimum when

i

has character

a

.

18

Cost of Evaluating Parsimony for binary trees If there are

n

nodes,

m

characters, and

k

possible values for each character, then complexity is O(nmk 2 ). Of course, we still need to search over ALL possible trees and find the best one. One usually resorts to heuristic search techniques.

19

Exploring the Space of Trees

W e’ve considered how to find the minimum number of changes for a given tree topology Need some search procedure for exploring the space of tree topologies possible rooted trees ( 2

n

 3 )!

!

( 2

n

 3 )!

!

 3  5    ( 2

n

 3 ) 20

Counting Trees

n = 3 One Tree: 1 3 n = 4 3 Trees 2 A rooted tree with n leaves has (2n-1) nodes and (2n-2) edges, discounting the edge to the root; hence an unrooted tree has (2n-3) edges. For each additional leaf we add two edges. Therefore we have 1 • 3 • 5 • … • (2n-5)

unrooted

trees with n leaves. Each of such trees has (2n-3) edges, which can be chosen as a root of the rooted tree. Hence we have 1 • 3 • 5 • … • (2n-5) • (2n-3)

rooted

trees with n leaves 21

Exploring the Space of Trees

taxa (n) # of rooted trees 4 5 6 8 10 15 105 945 135,135 30,405,375 22

Maximum Parsimony

1 2 3 4 5 6 7 8 9 10 Species 1 –

A G G G T A A C T G

Species 2 -

A C G A T T A T T A

Species 3 -

A T A A T T G T C T

Species 4 -

A A T G T T G T C G

How many possible unrooted trees?

23

1

Maximum Parsimony

How many possible unrooted trees?

Species 1 Species 2 Species 3 Species 4 1 2 3 4 5 6 7 8 9 10

A G G G T A A A C G A T T A C T G T T A A T A A T T G A A T G T T G T C T T C G

3 1 2 1 2 4 3 4 4 3 2 24

Maximum Parsimony

1 2 tree How many substitutions?

1 change 5 changes 3 A G A G A G G A 4 A G A G MP 25

1 2 1 3 1 4 4 3 2 3 4 2

Maximum Parsimony

1 2 3 4 5 6 7 8 9 10 1 2 3 4 -

A A A A G G G T A A C T G C G A T T A T T A T A A T T G T C T A T G T T G T C G

0 0 0 26

1 2 1 3 1 4 4 3 2 3 4 2

Maximum Parsimony

1 2 3 4 5 6 7 8 9 10 1 2 3 4 -

A G A C A T A A G G T A A C T G G A T T A T T A A A T T G T C T T G T T G T C G

0 3 0 3 0 3 27

Maximum Parsimony

G

1 3

T C G

2 1

C

4

A

2

C T

3

G

1

C

4

A

3

T

3 3 3

A

4

C

2

C

1 G 2 C 3 T 4 A 28

1 2 1 3 1 4 4 3 2 3 4 2

Maximum Parsimony

1 2 3 4 5 6 7 8 9 10 1 2 3 4 -

A G G A C G A T A A A T G T A A C T G A T T A T T A A T T G T C T G T T G T C G

0 3 2 0 3 2 0 3 2 29

1 2 1 3 1 4 4 3 2 3 4 2

Maximum Parsimony

1 2 3 4 5 6 7 8 9 10 1 2 3 4 -

A G G G A C G A A T A A A A T G T A A C T G T T A T T A T T G T C T T T G T C G

0 3 2 2 0 3 2 2 0 3 2 1 30

Maximum Parsimony

G

1

A

3

G

1

G

1

A

2 3

A A

4

G

2

A A

4

G

3

A G

4

A

2

A

2 2 1

4 1 G 2 A 3 A 4 G 31

Maximum Parsimony

1 2 1 3 1 4 4 3 2 3 4 2 0 3 2 2 0 1 1 1 1 3 14 0 3 2 2 0 1 2 1 2 3 16 0 3 2 1 0 1 2 1 2 3 15 32

1 2

Maximum Parsimony

3 1 2 3 4 5 6 7 8 9 10 1 -

A G G G T A A C T G

2 -

A C G A T T A T T A

3 -

A T A A T T G T C T

4 -

A A T G T T G T C G 0 3 2 2 0 1 1 1 1 3 14

4 33

Finding most parsimonious trees exact solutions



Exact solutions can only be used for small numbers of taxa.



Exhaustive search possible trees. examines all



Typically used for problems with less than 10 taxa.

34

Finding most parsimonious trees - exhaustive search

(1) B C Starting tree, any 3 taxa A Add fourth taxon (D) in each of three possible positions: three trees E (2a) B D C (2b) B D C E (2c) B C D E A A E E A Add fifth taxon (E) in each of the five possible positions on each of the three trees -> 15 trees, and so on 35

Finding most parsimonious trees exact solutions

 Branch and bound saves time by discarding families of trees during tree construction that can not be smaller than the smallest tree found so far.

(Here “smaller” means more parsimonious.)  Can be enhanced by specifying an initial upper bound for tree length.

 Typically used only for problems with less than 20 taxa.

36

C2.1

C2.2

C2.3

C2.4

C2.5

Finding most parsimonious trees:

branch and bound B D C A B C A B2 A B C D B3 A C3.1

C3.2

C3.3

C3.4

C3.5

B C1.1

E D C A B C1.2

D C E A B B1 D C A B C1.5

D E A B C1.3

D E C B C1.4

E A D C A C

37

Finding most parsimonious trees -

heuristics

 The number of possible trees increases exponentially with the number of taxa making exhaustive searches impractical for many data sets (an NP complete problem)  Heuristic methods are used to search tree space for most parsimonious trees  The trees found are not guaranteed to be the most parsimonious - they are best guesses 38

Finding most parsimonious trees - heuristics

 Stepwise addition Asis - the order in the data matrix Closest -starts with shortest 3-taxon tree adds taxa in order that produces the least increase in tree length Simple - the first taxon in the matrix is a taken as a reference - taxa are added to it in the order of their decreasing similarity to the reference Random - taxa are added in a random sequence, many different sequences can be used  Recommend random with as many (e.g. 10-100) addition sequences as practical 39

Finding most parsimonious trees - heuristics Branch Swapping: Nearest neighbor interchange (NNI) Subtree pruning and regrafting (SPR) Tree bisection and reconnection (TBR) 40

Finding most parsimonious trees - heuristics 1 Nearest neighbor interchange (NNI)

C D A E A D C B E F A G C D E B F B F G G

41

Finding most parsimonious trees heuristics 2

Subtree pruning and regrafting (SPR) C D A E B F G C D E C E F G F D B G A 42

Finding most parsimonious trees - heuristics 3 Tree bisection and reconnection (TBR) C D A E F B G B G E A F A C D F D B C G E 43

Finding most parsimonious trees heuristics - summary

 Branch Swapping Nearest neighbor interchange (NNI) Subtree pruning and regrafting (SPR) Tree bisection and reconnection (TBR)  The nature of heuristic searches means we cannot know which method will find the most parsimonious trees or all such trees.

 However, TBR is the most extensive swapping routine and its use with multiple random addition sequences should work well.

44

Tree space may be populated by local minima and islands of most parsimonious trees RANDOM ADDITION SEQUENCE REPLICATES FAILURE SUCCESS Branch Swapping Branch Swapping FAILURE Tree Length Branch Swapping Local Minimum GLOBAL MINIMUM Local Minima

45

Multiple most parsimonious trees

 Many parsimony analyses yield multiple equally optimal trees  Multiple trees are due to either: Alternative equally parsimonious optimizations of homoplastic characters Missing data Or both  We can further select among these trees with additional criteria, but  Most commonly relationships common to all the optimal trees are summarized with consensus trees 46

Consensus methods - 1

 A consensus tree is a summary of the agreement among a set of fundamental trees  There are many different consensus methods that differ in: 1. the kind of agreement 2. the level of agreement  Consensus methods can be used with any types of tree - not just parsimony 47

Strict consensus methods - 1

 Strict consensus methods require agreement across all the fundamental trees  They show only those relationships that are unambiguously supported by the parsimonious interpretation of the data  The commonest method (strict component consensus) focuses on clades  This method produces a consensus tree that includes all and only those clades found in all the fundamental trees  Other relationships (those in which the fundamental trees disagree) are shown as unresolved polytomies 48

Strict consensus methods - 2

A B

TWO FUNDAMENTAL TREES

C D E F G A B C E D F G A B C D E F G

STRICT COMPONENT CONSENSUS TREE 49

Majority-rule consensus methods

 Majority-rule consensus methods require agreement across a majority of the fundamental trees  May include relationships that are not supported by the most parsimonious interpretation of the data  The commonest method focuses on clades  This method produces a consensus tree that includes all and only those clades found in a majority (>50%) of the fundamental trees  Other relationships are shown as unresolved polytomies  Of particular use in bootstrapping 50

Majority rule consensus

THREE FUNDAMENTAL TREES

A B C D E F G A B C E F D G A B C E D F G Numbers indicate frequency of clades in the fundamental trees A B C E D F G 100 66 66 66 66

MAJORITY-RULE COMPONENT CONSENSUS TREE 51

Reduced consensus methods - 1

 Focuses upon any cladistic relationships (statements that some taxa are more closely related to each other than to some other taxa)  Reduced consensus methods occur in strict and majority-rule varieties  Other relationships are shown as unresolved polytomies  May be more sensitive than methods focusing only on clades 52

Reduced consensus methods - 2

A B C

TWO FUNDAMENTAL TREES

D E F G A G B C D E F A B C D E F G A B C D E F

Strict component consensus completely unresolved STRICT REDUCED CLADISTIC CONSENSUS TREE Taxon G is excluded 53

Consensus methods - 2

Three fundamental trees

Ochromonas Symbiodinium Prorocentrum Loxodes Tetrahymena Spirostomumum Tracheloraphis Euplotes Gruberia Ochromonas Symbiodinium Prorocentrum Loxodes Tetrahymena Spirostomumum Euplotes Tracheloraphis Gruberia Ochromonas Symbiodinium Prorocentrum Loxodes Tetrahymena Euplotes Spirostomumum Tracheloraphis Gruberia

strict (component)

Ochromonas Symbiodinium Prorocentrum Loxodes Tetrahymena Tracheloraphis Spirostomum Euplotes Gruberia

strict reduced cladistic Euplotes excluded

Symbiodinium Prorocentrum Loxodes Tetrahymena Spirostomum Tracheloraphis Gruberia Ochromonas

10 0 majority-rule 66 66 100 100 100

Ochromonas Symbiodinium Prorocentrum Loxodes Tetrahymena Spirostomum Euplotes Tracheloraphis Gruberia

54

Consensus methods - 3

 Use strict methods to identify those relationships unambiguously supported by parsimonious interpretation of the data  Use reduced methods where consensus trees are poorly resolved  Use majority-rule methods in bootstrapping  Avoid other methods which have ambiguous interpretations 55

Parsimony - advantages

 a simple method - easily understood operation  does not seem to depend on an explicit model of evolution  gives both trees and associated hypotheses of character evolution  should give reliable results if the data is well structured and homoplasy is either rare or randomly distributed on the tree 56

Parsimony - disadvantages

 May give misleading results if homoplasy is common or concentrated in particular parts of the tree, e.g: thermophilic convergence base composition biases long branch attraction  Underestimates branch lengths  Model of evolution is implicit - behaviour of method not well understood  Parsimony often justified on purely philosophical grounds - we must prefer simplest hypotheses - particularly by morphologists  For most molecular systematists this is uncompelling 57

Parsimony can be inconsistent

  Felsenstein (1978) developed a simple model phylogeny including four taxa and a mixture of short and long branches Under this model parsimony will give the wrong tree A Model tree p C q q q D p B Rates or Branch lengths p >> q A Parsimony tree C Wrong B D Long branches are attracted but the similarity is homoplastic • • •

With more data the certainty that parsimony will give the wrong tree increases - so that parsimony is statistically inconsistent .

Advocates of parsimony initially responded by claiming that Felsenstein’s result showed only that his model was unrealistic.

It is now recognized that the long-branch attraction (the Felsenstein Zone ) is one of the most serious problems in phylogenetic inference.

58

2. Perfect Phylogeny

Data on species is given by a

Character State Matrix

.

Cell (

p, i

) has value

j

iff character

i

of object (species)

p

has state

j

.

Goal: constructing evolution tree for the species.

Object A B C D E c 1 1 2 3 0 1 c 2 1 0 2 3 1 Character c 3 2 1 3 4 0 c 4 0 2 3 1 0 c 5 0 1 1 0 1 59

Motivation: Evolution Tree

Internal nodes correspond to speciation events, where some character (attribute) is acquired.

Assumptions: 1. No reversals (characters are not lost) 2. No convergences (a character is created only once) 60

61

Perfect Phylogeny for a 0-1 Matrix

A 0-1 matrix: Each character is either 0 (non exists) or 1 (exists).

 Each of the

n

objects label exactly one leaf of T  Each of the

m

characters labels exactly one edge of T  Object

p

has exactly the characters labeling the path from

p

to the root.

A perfect phylogeny for the matrix: Tree with no convergence, no reversals. 3 2 A 1 1 2 1 3 0 4 0 5 0 B 0 0 1 0 0 1 C 1 1 0 0 1 4 E D B A 5 C D E 0 0 0 1 1 0 1 0 0 0 62

The (Binary) Perfect Phylogeny Problem

Problem: Given a 0-1 matrix M, determine if it has a perfect phylogeny, and construct one if it does. (Note: edges are labeled by characters: edge labeled by i represent changing character i’s state from 0 to 1).

4 D B 3 E 2 A 1 5 C A B C D E 1 1 0 1 0 0 2 1 0 1 0 1 3 0 1 0 1 0 4 0 0 0 1 0 5 0 0 1 0 0 63

Solution to Perfect Phylogeny Problem

Definition

: Given a 0-1 matrix M, O

k

={

j

: M

jk

=1}; i.e., O

k

set of objects that have character

k

. is the A B C D E

Theorem

: M has a perfect phylogenetic tree iff the sets {O

i

} are

laminar

, ie: for all includes the other.

i

,

j

, either O

i

and O

j

are disjoint, or one 1 1 0 1 0 0 Laminar 2 3 1 0 1 0 1 0 1 0 1 0 4 0 0 0 1 0 5 0 0 1 0 0 A B C D E 1 1 0 1 0 0 2 Not Laminar 1 3 0 4 0 0 1 1 0 0 0 0 1 1 0 1 0 5 0 1 1 0 1 64

Proof

 : Assume M has a perfect phylogeny, and let

i

,

j

be given.

Consider the edges labeled

i

and

j.

Case 1

: There is a root to leaf path containing both. Then one is included in the other (2 and 1 below).

Case 2

: not case 1. Then they are disjoint (2 and 3 below).

4 D B 3 2 E 1 A 5 C 65

Proof (cont.)

 : Assume for all

i

,

j

, either O

i

and O

j

are disjoint, or one includes the other. We prove by induction on the number of characters that it has.

Basis

: one character. Then there are at most two objects, one with and one without this character.

A B 1 1 0 B 1 A 66

Proof (cont.)

 : Induction step: Assume correctness for

n-1

characters, and consider a matrix with

n

characters (non-zero columns). WLOG assume that O 1 is not contained in O

j

for

j

> 1.

Let S 1 be the set of objects that have character 1, and S 2 be the remaining objects. Then each character belongs to objects in S 1 or S 2 , but not both. By induction there are trees T 1 and T Combining them as below gives the desired tree.

2 for S 1 and S 2 . A B C D E 1 1 0 1 0 1 2 1 0 1 0 0 3 0 1 0 1 0 4 0 0 0 1 0 5 0 0 1 0 0 T 1 1 T 2 67

Efficient Implementation

1.

Sort the columns 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 O

i

Proof: O is not a proper subset of O

j

.

i

– O

j >

0 means the 1’s in O

i

are not covered by the 1’s in O

j .

A B C D E 1 1 0 1 0 0 2 1 0 1 0 1 3 0 1 0 1 0 4 0 0 0 1 0 5 0 0 1 0 0 A B C D E 2 1 0 1 0 1 1 1 0 1 0 0 3 0 1 0 1 0 5 0 0 1 0 0 4 0 0 0 1 0 68

Efficient Implementation (2)

2.

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

mn

).

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.

A B C D E 1 0 1 2 1 0 1 0 0 1 1 0 0 1 0 3 0 1 1 0 0 5 0 0 0 1 0 4 0 0 69

Examples

A B C D E 0 1 1 0 1 0 0 1 0 1 1 1 0 1 0 Not laminar 0 1 0 0 1 1 0 0 0 0 A B C D E 1 0 1 2 1 0 1 0 0 1 laminar 3 5 1 0 0 0 1 0 0 1 0 1 0 0 0 1 0 4 0 0 70

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.

4 D B 3 E 2 A 1 5 C A B C D E 2 1 0 1 0 1 1 1 0 1 0 0 3 0 1 0 1 0 1 0 0 5 0 0 0 1 0 4 0 0 71