In this approach, trees are constructed by comparing the ..

1
Phylogenetic TreesLecture 2
Based on Durbin et al 7.4 Gusfield 17
2
Character-based methodsfor 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

3
1. Maximum Parsimony
Input four nucleotide sequences AAG, AAA, GGA,
AGA taken from four species. Question Which
evolutionary tree best explains these sequences ?
4
Example Continued
There are many trees possible. For example
The left tree is preferred over the right tree.
The total number of changes is called the
parsimony score.
5
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
T
C
T
C
C
C
T
T ? C
6
Extension to Many Letters

• What is the parsimony score of

A CAGGTA B CAGACA C CGGGTA D TGCACT E TGCGTA
We do it character after character each score is
computed independently of the others.
7
Fitchs Algorithm of Evaluating Trees

• Assume that a tree is given.
• 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

8
Fitchs Algorithm Step 1

• of changes union operations

9
Fitchs Algorithm Step 1

• Do a post-order (from leaves to root) traversal
  of tree
• Determine possible states Ri of internal node
  i with children j and k

10
Fitchs Algorithm Step 2
11
Fitchs Algorithm Step 2

• Do a pre-order (from root to leaves) traversal
  of tree
• Select state rj of internal node j with
  parent i

12
Weighted Version of Fitchs Algorithm

• Instead of assuming all state changes are
  equally likely, use different costs c(a, b)
  for different changes
• 1st step of algorithm is to propagate costs up
  through tree

13
Weighted Version of Fitchs Algorithm

• Want to determine minimal cost S(i, a)
• of assigning character a to node i
• For leave nodes i

14
Weighted Version of Fitchs Algorithm

• Want to determine minimal cost S(i, a)
• of assigning character a to node i
• For internal nodes

a
i
j
k
b
15
Weighted Version of Fitchs Algorithm Step 2

• Do 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

16
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.

17
Evaluating Weighted Parsimony Scores

• Each position is independent and computed by
  itself.
• Use Dynamic Programming on a given tree.
• If i is a node with children j and k , then
  S(i, a) minx(S(j, x)c(a, x)) miny(S(k,
  y)c(a, y))

S(i, a)?the minimum score of subtree rooted at k
when k has character a.
S(i,a)
S(j,x)
S(k,y)
18
Evaluating Parsimony Scores

• Dynamic programming on a given tree
• Initialization
• For each leaf i set S(i,a) 0 if i is labeled
  by a, otherwise S(i,a) ?
• Iteration
• if i is node with children j and k, then
• S(i,a) minx(S(j,x)c(a,x))
  miny(S(k,y)c(a,y))
• Termination
• cost of tree is minxS(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.
19
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(nmk2).

• Of course, we still need to search over ALL
  possible trees and find the best one. One usually
  resorts to heuristic search techniques.

20
Exploring the Space of Trees

• Weve 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
• Given n sequences there are
  possible rooted trees

21
Counting Trees
n 3 One Unrooted Tree
n 4 3 Unrooted Trees
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
22
Exploring the Space of Trees

23
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?
24
Maximum Parsimony
How many possible unrooted trees?
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
25
Maximum Parsimony
How many substitutions?
MP
26
Maximum Parsimony
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
27
Maximum Parsimony
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
28
Maximum Parsimony
1 - G 2 - C 3 - T 4 - A
29
Maximum Parsimony
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
30
Maximum Parsimony
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
31
Maximum Parsimony
G
A
4 1 - G 2 - A 3 - A 4 - G
2
A
G
A
G
A
2
A
G
A
1
A
G
A
32
Maximum Parsimony
33
Maximum Parsimony
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
34
Finding most parsimonious trees - exact solutions

• Exact solutions can only be used for small
  numbers of taxa.
• Exhaustive search examines all possible trees.
• Typically used for problems with less than 10
  taxa.

35
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
D
C
D
B
B
C
(2b)
(2a)
(2c)
A
A
Add fifth taxon (E) in each of the five possible
positions on each of the three trees -gt 15
trees, and so on
36
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 smaller score i.e.,
  more parsimonious.)
• Can be enhanced by specifying an initial upper
  bound for tree length (total of changes on the
  tree) e.g., from distance method.
• Typically used only for problems with less than
  20 taxa.

37
Finding most parsimonious trees branch and bound
C2.1
B
C
C
C3.1
D
C
B
C2.2
B
C3.2
D
C2.3
C3.3
A
C2.4
C3.4
B2
B3
A
A
C2.5
C3.5
D
B
B
E
E
B
C
D
C
D
C
B1
C1.1
C1.5
A
A
A
B
E
D
D
B
D
B
E
C
C
C1.3
C
E
C1.2
C1.4
A
A
A
38
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

39
Finding most parsimonious trees - heuristics

• Stepwise addition
• Asis - the order in the distance matrix
• Closest -starts with shortest 3-taxon tree and
  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

40
Finding most parsimonious trees - heuristics

• Branch Swapping
• Nearest neighbor interchange (NNI)
• Subtree pruning and regrafting (SPR)
• Tree bisection and reconnection (TBR)

41
Finding most parsimonious trees - heuristics 1

• Nearest neighbor interchange (NNI)

C
D
E
A
F
B
G
D
C
C
D
E
A
E
A
F
B
F
B
G
G
42
Finding most parsimonious trees - heuristics 2

• Subtree pruning and regrafting (SPR)

C
D
E
A
F
B
G
E
C
D
F
E
G
C
F
B
D
A
G
43
Finding most parsimonious trees - heuristics 3

• Tree bisection and reconnection (TBR)

C
D
E
A
F
B
G
B
G
E
F
A
D
C
F
D
C
E
G
44
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.

45
Tree space may be populated by local minima and
islands of most parsimonious trees
RANDOM ADDITION SEQUENCE REPLICATES
Tree
SUCCESS
FAILURE
FAILURE
Length
Branch
Swapping
Branch Swapping
Branch Swapping
Local
Minimum
Local
GLOBAL
Minima
MINIMUM
46
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

47
Consensus methods

• 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

48
Strict consensus methods

• 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

49
Strict consensus methods
TWO FUNDAMENTAL TREES
A
B
C
D
E
F
G
B
E
F
G
A
C
D
B
D
F
G
A
C
E
STRICT COMPONENT CONSENSUS TREE
50
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 (gt50) of the fundamental trees
• Other relationships are shown as unresolved
  polytomies
• Of particular use in bootstrapping

51
Majority rule consensus
THREE FUNDAMENTAL TREES
A
B
C
D
E
F
G
B
E
F
G
A
C
D
B
E
D
G
A
C
F
A
B
C
E
D
F
G
66
100
66
66
Numbers indicate frequency of clades in the
fundamental trees
66
MAJORITY-RULE COMPONENT CONSENSUS TREE
52
Reduced consensus methods

• 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

53
Reduced consensus methods
TWO FUNDAMENTAL TREES
B
D
F
G
A
G
B
C
D
E
F
A
C
E
B
D
F
A
C
E
Strict component consensus
completely unresolved
STRICT REDUCED CLADISTIC CONSENSUS TREE
Taxon G is excluded
54
Consensus methods - 2
strict reduced cladistic
Three fundamental trees
strict (component)
Euplotes excluded
Ochromonas
Ochromonas
Symbiodinium
Symbiodinium
Symbiodinium
Prorocentrum
Prorocentrum
Loxodes
Prorocentrum
Loxodes
Tetrahymena
Loxodes
Tetrahymena
Spirostomumum
Tracheloraphis
Tracheloraphis
Tetrahymena
Euplotes
Spirostomum
Spirostomum
Gruberia
Euplotes
Tracheloraphis
Ochromonas
Gruberia
Symbiodinium
Gruberia
Prorocentrum
Ochromonas
Loxodes
majority-rule
Tetrahymena
Spirostomumum
Euplotes
Tracheloraphis
100
Gruberia
Ochromonas
100
Symbiodinium
100
66
Prorocentrum
Loxodes
66
Tetrahymena
Euplotes
Spirostomumum
100
Tracheloraphis
Gruberia
55
Consensus methods

• 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

56
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

57
(No Transcript)
58
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

59
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
B
Parsimony tree
Long branches are attracted but the similarity
is homoplastic
Model tree
C
A
Rates or
p
p
Branch lengths
q
p gtgt q
Wrong
q
q
B
D
C
D

• 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 Felsensteins 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.

60
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.

61
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)
62
(No Transcript)
63
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.

2
3
1
4
E
B
D
5
A
C
64
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 is
  state from 0 to 1).

65
Solution to Perfect Phylogeny Problem

• Definition Given a 0-1 matrix M, Okj Mjk1
  i.e., Ok is the set of objects that have
  character k.
• Theorem M has a perfect phylogenetic tree iff
  the sets Oi are laminar, ie for all i, j,
  either Oi and Oj are disjoint, or one includes
  the other.

Laminar
Not Laminar
66
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).

2
3
1
4
E
D
B
5
A
C
67
Proof (cont.)

• ? Assume for all i, j, either Oi and Oj 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.

68
Proof (cont.)

• ? Induction step Assume correctness for n-1
  characters, and consider a matrix with n
  characters (non-zero columns).
• WLOG assume that O1 is not contained in Oj for j
  gt 1.
• Let S1 be the set of objects that have character
  1, and S2 be the remaining objects. Then each
  character belongs to objects in S1 or S2, but not
  both. By inductive hypothesis there are trees T1
  and T2 for S1 and S2. Combining them as below
  gives the desired tree.

69
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 Oi is not a proper
  subset of Oj.
• Proof Oi Oj gt 0 means the 1s in Oi are not
  covered by the 1s in Oj.

70
Efficient Implementation (2)

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

4
5
3
1
2
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.
0
0
0
1
1
A
0
0
1
0
0
B
0
1
0
1
1
C
1
0
1
0
0
D
0
0
0
0
1
E
71
Examples
laminar
4
5
3
1
2
0
0
0
1
1
A
0
0
1
0
0
B
0
1
0
1
1
C
1
0
1
0
0
D
0
0
0
0
1
E
72
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. Need O(mn) time.
2
3
1
4
E
D
B
5
A
C