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Week 10: Consensus trees, tree distances, tests of treeshape
Genome 570
March, 2014
Week 10: Consensus trees, tree distances, tests of tree shape – p.1/44
Trees we will use for consensus trees
DA C B E D FG A CG F B E D A CG F B E
Week 10: Consensus trees, tree distances, tests of tree shape – p.2/44
Trees we will use for consensus trees
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)
Week 10: Consensus trees, tree distances, tests of tree shape – p.3/44
Trees we will use for consensus trees
A C B E D FG A CG F B E D A CG F B E D
Week 10: Consensus trees, tree distances, tests of tree shape – p.4/44
Trees we will use for consensus trees
A C B E D FG A CG F B E D A CG F B E D
(Do we count this one if the trees are considered rooted? unrooted?)
Week 10: Consensus trees, tree distances, tests of tree shape – p.5/44
Trees we will use for consensus trees
A C B E D FG A CG F B E D A CG F B E D
Here is a clade that is found on only two of the trees, so it is not includedin the Strict Consensus Tree.
Week 10: Consensus trees, tree distances, tests of tree shape – p.6/44
Their strict consensus tree
A CG F B E D
Week 10: Consensus trees, tree distances, tests of tree shape – p.7/44
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 ...
Week 10: Consensus trees, tree distances, tests of tree shape – p.8/44
A distressing case for the strict consensus tree
A B C D E F G
... but the strict consensus tree becomes totally unresolved.
Week 10: Consensus trees, tree distances, tests of tree shape – p.9/44
Majority-rule consensus tree
A CG F B E D
100
100
67
67100
Week 10: Consensus trees, tree distances, tests of tree shape – p.10/44
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.
Week 10: Consensus trees, tree distances, tests of tree shape – p.11/44
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. The
red triple is contradicted and does not get used in the Adams ConsensusTree.
Week 10: Consensus trees, tree distances, tests of tree shape – p.12/44
The Adams consensus tree
A CG F B E D
Week 10: Consensus trees, tree distances, tests of tree shape – p.13/44
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, but
unrooted 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!
Week 10: Consensus trees, tree distances, tests of tree shape – p.14/44
A consensus subtree
F C A B G DFCA BDE F AB D E
Week 10: Consensus trees, tree distances, tests of tree shape – p.15/44
A consensus subtree
F C A B G DFCA BDE F AB D E
Week 10: Consensus trees, tree distances, tests of tree shape – p.16/44
A consensus subtree
F C A B G DFCA BDE F AB D E
B DFA
Week 10: Consensus trees, tree distances, tests of tree shape – p.17/44
A supertree
F C A B G D
F C A BA B G DC A B D
Construct a tree with all tips, for which each of the smaller trees is asubtree. What to do if there is conflict? There are various suggestions.
Week 10: Consensus trees, tree distances, tests of tree shape – p.18/44
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.
Week 10: Consensus trees, tree distances, tests of tree shape – p.19/44
The Robinson-Foulds and Branch Length distances
D
E0.1
0.1
G
B
C0.2
0.20.15
A0.1
0.2F
0.30.05
0.4
0.2 E
F
0.1
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
= 0.43748
Week 10: Consensus trees, tree distances, tests of tree shape – p.20/44
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 deathof lineages too).
As we can’t tell which clade is on the left, we can symmetrize this, alwaysputting the smaller 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..
Week 10: Consensus trees, tree distances, tests of tree shape – p.21/44
Harding’s probability of a tree shape
2:2
1:1 1:1 1:11:11:2
1:3
2:4
4:62
9
2
5
2
3
1 1 111
3
1
Week 10: Consensus trees, tree distances, tests of tree shape – p.22/44
Birth-death process tree
ab
c
d
efghij
Week 10: Consensus trees, tree distances, tests of tree shape – p.23/44
If we remove the extinct lineages
ab
c
d
efghij
Week 10: Consensus trees, tree distances, tests of tree shape – p.24/44
We get ...
ab
c
d
efghij
Week 10: Consensus trees, tree distances, tests of tree shape – p.25/44
Reconstructed lineages from that tree
abcdefghij
Week 10: Consensus trees, tree distances, tests of tree shape – p.26/44
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
Week 10: Consensus trees, tree distances, tests of tree shape – p.27/44
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)
Week 10: Consensus trees, tree distances, tests of tree shape – p.28/44
Reconstructed and actual lineages
0 101
10
100
20
2
5
50
155
Actual
Reconstructed
Time
Sp
ecie
s
Week 10: Consensus trees, tree distances, tests of tree shape – p.29/44
Rooted tree with ancestors in “intermediate” position
AB
CD
EFG
HIJ
KL
M
Week 10: Consensus trees, tree distances, tests of tree shape – p.30/44
Rooted tree with ancestors in “centered” position
AB
CD
EFG
HIJ
KL
M
Week 10: Consensus trees, tree distances, tests of tree shape – p.31/44
Rooted tree with ancestors in “weighted” position
AB
CD
EFG
HIJ
KL
M
Week 10: Consensus trees, tree distances, tests of tree shape – p.32/44
Rooted tree with ancestors in “inner” position
AB
CD
EFG
HIJ
KL
M
Week 10: Consensus trees, tree distances, tests of tree shape – p.33/44
Drawn with diagonal lines
AB
CD
EFG
HIJ
KL
M
Week 10: Consensus trees, tree distances, tests of tree shape – p.34/44
Drawn so as to be V-shaped
AB
CD
EFG
HIJ
KL
M
Week 10: Consensus trees, tree distances, tests of tree shape – p.35/44
A curvogram
AB
CD
EFG
HIJ
KL
M
Week 10: Consensus trees, tree distances, tests of tree shape – p.36/44
A swoopogram
AB
CD
EFG
HIJ
KL
M
Week 10: Consensus trees, tree distances, tests of tree shape – p.37/44
A eurogram
AB
CD
EFG
HIJ
KL
M
Week 10: Consensus trees, tree distances, tests of tree shape – p.38/44
A circular drawing of a rooted tree
A
B
CD E
F
G
H
I
JK
L
M
Week 10: Consensus trees, tree distances, tests of tree shape – p.39/44
The equal-angle algorithm
A
B
I
J
C D
E
F
GH
o72
o72
o144
o72
o216
o36
o108 o
72
o36
Week 10: Consensus trees, tree distances, tests of tree shape – p.40/44
The equal-angle algorithm tree for the example
AB
C
D
E
F
G
H
I
J
KL
M
Week 10: Consensus trees, tree distances, tests of tree shape – p.41/44
The n-body algorithm for the example
A
B
C
D
E
F
G
H
I
J
K
L
M
Week 10: Consensus trees, tree distances, tests of tree shape – p.42/44
The equal-daylight algorithm
F
H
AB
I
J
C D
E
G
Week 10: Consensus trees, tree distances, tests of tree shape – p.43/44
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.44/44
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