week 10: consensus trees, tree distances, tests of tree...

Week 10: Consensus trees, tree distances, tests of treeshape

Genome 570

March, 2012

Week 10: Consensus trees, tree distances, tests of tree shape – p.1/42

Trees we will use for consensus trees

DA C B E D FG A CG F B E D A CG F B E



A C B E D FG A CG F B E D A CG F B E D

(for unrooted trees we would use partitions induced by branches insteadof clades)




(Do we count this one if the trees are considered rooted? unrooted?)


Their strict consensus tree

A CG F B E D


A distressing case for the strict consensus tree

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

Only one species moves ...


A distressing case for the strict consensus tree

A B C D E F G

... but the strict consensus tree becomes totally unresolved.


Majority-rule consensus tree

A CG F B E D

100

100

67

67100


The Adams consensus tree

For rooted trees, Adams (1972, 1986) suggested:

1. Take all rooted triples on each tree.

2. Retain those that are not contradicted, where lack of resolution doesnot count as contradiction.

3. Construct a tree of these.


Two of the possible triples to examine

DA C B E D FG A CG F B E D A CG F B E

The green triple shows the same rooted topology on all three trees. Thered triple is contradicted and does not get used in the Adams ConsensusTree.


The Adams consensus tree

A CG F B E D


Steel, Böcker, and Dress’s shocking disproof

Steel, M., S. Böcker, and A. W. M. Dress. 2000. Simple but fundamentallimits for supertree and consensus tree methods. Systematic Biology 49(2):363-368.

They put forward three minimal requirements for an unrooted Adams-likeconsensus tree based on observations of quartets, rather than triples.Note that a quartet, like a triple, has three possible topologies, butunrooted ones: ((A,B),(C,D)) and ((A,C),(B,D)) and ((A,D),(B,C)).

The result shouldn’t be altered by relabelling all the species in aconsistent way.

The result should not depend on the order in which the trees areinput.

If a quartet appears in all trees, it should appear in the consensus.

Alas, they then show there is no consensus tree method for unrootedtrees that can satisfy all of these!


A consensus subtree

F C A B G DFCA BDE F AB D E


A consensus subtree



A consensus subtree


B DFA


The symmetric difference metric

AB

C

D

F

E G E

F

AD

B

C

G

Partitions

{ADF | BCEG}

{BC | ADEFG}

Partitions

{ADF | BCEG}

{EG | ABCDF}

{DF | ABCEG} {AD | BCEFG}

{BC | ADEFG}

The symmetric difference is the number of partitions that are in one butnot both of these lists, in this case 3.


The Robinson-Foulds and Branch Length distances

D

E0.1 0.1

G

B

C0.2

0.2 0.15

A0.1

0.2F

0.30.05

0.4

0.2 E

F0.1

0.2

0.3

B

C0.2

0.150.1 0.3

D0.2

0.20.1

A

G

0.4

0.2

0.3

0.1

0.05

0.2

none

0.15

0.1

0.1

0.2

0.2

{ADF | BCEG}

{AD | BCEFG}

{A | BCDEFG}

{B | ACDEFG}

{C | ABDEFG}

{D | ABCEFG}

{E | ABCDFG}

{F | ABCDEG}

{G | ABCDEF}

{DF | ABCEG}

{EG | ABCDF}

Partitions Branch lengths

{BC | ADEFG}

0.15

0.1

0.1

0.2

0.2

0.3

none

none

0.2

0.2

0.1

0.3

difference

−0.2

0.1

−0.1

0.3

0.1

−0.15

0.0

0.0

0.0

0.0

0.1

0.0

RF: Σ | difference | = 1.05

BLD: Σ (difference)2

= 0.1925Week 10: Consensus trees, tree distances, tests of tree shape – p.18/42

Asymmetry at base of a tree

A tree with n tips can, at the bottom fork, have clades whose ultimatesizes are 1 : n − 1, 2 : n − 2, 3 : n − 3, . . . n − 1 : 1.

Under a random branching model, these possibilities are all equiprobable(this is true when there is random death of lineages too). As we can’t tellwhich clade is on the left, we can symmetrize this, always putting thesmaller clade on the left.

So with 7 tips we have equal probabilities of 1 : 6, 2 : 5, and 3 : 4. With aneven number of tips (say 8) we have to be more careful, as 4 : 4 is thenonly half as probable as as each of the others 1 : 7, 2 : 6, and 3 : 5..


Harding’s probability of a tree shape

2:2

1:1 1:1 1:11:11:2

1:3

2:4

4:629

25

23

1 1 111

31


Birth-death process tree

ab

c

d

efghij


If we remove the extinct lineages

ab

c

d

efghij


We get ...

ab

c

d

efghij


Reconstructed lineages from that tree

abcdefghij


Birth-death processes and tree shapes

With branching rate λ and death rate µ,

probability of survival of a lineage for t is

s(t) = Prob (n > 0 | t) =(λ − µ)e(λ−µ)t

λ e(λ−µ)t − µ


Number of reconstructed and actual lineages

Nee, May, and Harvey (1994) argue that the number of lineages at time t

ago who have descendants in the present, given that they started T timeago, is expected to be

E[NR(t, T)] =e(λ−µ)t

s(T − t)

s(T)

and the expected number that actually were there then, given that at leastone is still alive now, is

E[N(t, T)] =e(λ−µ)t

s(T)−

s(t) − s(T)

s(t)s(T − t)


Reconstructed and actual lineages

0 101

10

100

20

2

5

50

155

Actual

Reconstructed

Time

Spe

cies


Rooted tree with ancestors in “intermediate” position

AB

CD

EFG

HIJ

KL

M


Rooted tree with ancestors in “centered” position

AB

CD

EFG

HIJ

KL

M


Rooted tree with ancestors in “weighted” position

AB

CD

EFG

HIJ

KL

M


Rooted tree with ancestors in “inner” position

AB

CD

EFG

HIJ

KL

M


Drawn with diagonal lines

AB

CD

EFG

HIJ

KL

M


Drawn so as to be V-shaped

AB

CD

EFG

HIJ

KL

M


A curvogram

AB

CD

EFG

HIJ

KL

M


A swoopogram

AB

CD

EFG

HIJ

KL

M


A eurogram

AB

CD

EFG

HIJ

KL

M


A circular drawing of a rooted tree

A

B

CD E

F

G

H

I

JK

L

M


The equal-angle algorithm

A

B

I

J

C D

E

F

GH

o72

o72

o144

o72

o216

o36

o108 o

72

o36


The equal-angle algorithm tree for the example

AB

C

D

E

F

G

H

I

J

KL

M


The n-body algorithm for the example

A

B

C

D

E

F

G

H

I

J

K

L

M


The equal-daylight algorithm

F

H

AB

I

J

C D

E

G


Equal-daylight tree for the example

ABCD

E

F

G

HI

J

K

L

MWeek 10: Consensus trees, tree distances, tests of tree shape – p.42/42

week 10: consensus trees, tree distances, tests of tree...

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