Lecture 6A – Introduction to Trees & Optimality Criteria - PowerPoint Presentation

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Lecture 6A – Introduction to Trees & Optimality Criteria

Branches:

n

-taxa -> 2n-3 branches

1, 2, 4, 6, & 7 are external (leaves) 3 & 5 are internal branches (edges)

Nodes A – E are terminals

x, y, & z are internal (vertices)

If we break branch 3, we have two sub-trees: (A,B) and (C,(D,E)).

((A,B),C,(D,E)).

Newick

Format

Rooting – The tree is an unrooted

tree.

Also note that there is free rotation around nodes:

(1 2)

(1 2)

(1 2 3)

(1 2 3)

(1 2 3 4)

(1 2 3 4)

Growth of tree space.

The Scope of the Problem

Taxa

Unrooted Trees 3 1 4 3 5 15 6 105

7 945 8 10,395 9 135,135 10 2.027 X 106 22 3 X 10

23

50 3 X 10

74

100 2 X 10

82

1000 2 X 10

2,860

10 mil 5 X 10

68,667,340

II. Optimality Criteria

A. Parsimony First, the score of a tree (i.e., its length) for the entire data set is given by:

l

i

is the length of character

i

when optimized on tree

t

.

w

i

is the weight we assign to character

i

.

The Fitch Algorithm (1971): state sets and accumulated lengths.

(Unordered states with equal transformation costs)

We erect a state set at each terminal node and assign an accumulated length of zero to terminal nodes. This is the minimum number of changes in the daughter

subtree.

The Fitch Algorithm: state sets and accumulated lengths.

1 – Form the intersection of the state sets of the two daughter nodes. If the

intersection is

non-empty, assign the set for the internal node equal to the intersection. The accumulated length of the internal node is the sum of those of the daughter nodes.

2 – If the intersection is empty, we assign the

union

of the two daughter nodes to

the state set for the internal node. The accumulated length is the

sum

of those

of the daughter nodes

plus one

.

empty

Union:

0+0+1=1

non-empty

Intersection:

0+0+0=0

empty

Union:

1+0+1=2

So

l

i

= 2

Sankoff Algorithm – Character-state vectors and step matrices.

Step Matrix – define ci,j  A C G T A -- 4 1 4 C 4 -- 4 1 G 1 4 -- 4 T 4 1 4 --

Step one: Fill in the character-state vectors for terminal nodes.

Each cell is indexed by

s

k

(

i

),

the cost of having

state

i

at node

k

.

Step two: Fill in vectors for other nodes, descending tree.

Node 1 (k = 1):

Node 2 (

k = 2):

A C G T A -- 4 1 4 C 4 -- 4 1 G 1 4 -- 4

T 4 1 4 --

s

1(A)

=

c

AG

+

c

AA

= 1 + 0 = 1,

s

1(C)

=

c

CG

+

c

CA

= 4 + 4 = 8,

s

1(G)

= cGG + cGA

= 0 + 1 = 1,

s

1(T)

=

c

TG

+

c

TA

= 4 + 4 = 8

s

2(A)

= 4 + 4 = 8

s

2(C)

= 0 + 0 = 0

s

2(G)

= 4 + 4 = 8

s

2(T)

= 1 + 1 = 2

For nodes below, we must calculate the cost for each possible state

assignment for daughter nodes.s3(A) = min[s1A + cAj] + min[s2A +

cAj]

s

3(C)

= min[s

1C

+

c

Cj

] + min[s

2C

+

c

Cj

]

s

3(G)

= min[s

1G

+

cGj] + min[s2G + cGj]

s3(T) = min[s1T + c

Tj] + min[s2T + cTj

]So we fill in the character-state vector for node 3:

From daughter node 1

From step matrix

= min[

1

,12,2,12] + min[8,

4

,9,6] = 1+4 = 5

5

= min [

5

,8,

5

,9] + min[12,

0

,12,3] = 5+0 = 5

5

= min [2,12,

1

,12] + min[9,

4

,8,6] = 1+4 = 5

5

= min [

5

,9,

5

,8] + min[12,

1

,12,2] = 5+1 = 6

6

A C G T

A -- 4 1 4

C 4 -- 4 1

G 1 4 -- 4

T 4 1 4 --

2) One can’t compare tree lengths across weighting schemes. In the first example, with

all transformations having the same cost, the length of the character on this tree was 2.

In the second, with a 4:1 step matrix to weight

transversions, the length was 5.

So, li = 5

Points to note:

1) Two types of weighting are possible: weighting of transformations within characters (which we demonstrated with the step matrix) and weighting among characters, which are reflected in the weighted sum of lengths across characters (

w

i

).