PPT - Tandy Warnow

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Transcript PPT - Tandy Warnow

Statistical tree estimation
Tandy Warnow
Topics
• Phylogeny as statistical estimation
problem
• Stochastic models of evolution
• Distance-based estimation
• Maximum parsimony tree estimation
• Maximum likelihood tree estimation
• Bayesian tree estimation
Phylogeny estimation as a
statistical inverse problem
Estimation of evolutionary trees as a
statistical inverse problem
• We can consider characters as properties
that evolve down trees.
• We observe the character states at the
leaves, but the internal nodes of the tree also
have states.
• The challenge is to estimate the tree from the
properties of the taxa at the leaves. This is
enabled by characterizing the evolutionary
process as accurately as we can.
DNA Sequence Evolution
-3 mil yrs
AAGACTT
AAGGCCT
AGGGCAT
AGGGCAT
TAGCCCT
TAGCCCA
-2 mil yrs
TGGACTT
TAGACTT
AGCACTT
AGCACAA
AGCGCTT
-1 mil yrs
today
Phylogeny Problem
U
AGGGCAT
V
W
TAGCCCA
X
TAGACTT
Y
TGCACAA
X
U
Y
V
W
TGCGCTT
Markov Model of Site Evolution
Simplest (Jukes-Cantor, 1969):
• The model tree T is binary and has substitution probabilities p(e) on
each edge e.
• The state at the root is randomly drawn from {A,C,T,G} (nucleotides)
• If a site (position) changes on an edge, it changes with equal probability
to each of the remaining states.
• The evolutionary process is Markovian.
More complex models (such as the General Time Reversible model, or the
General Markov model) are also considered, often with little change to
the theory.
Standard DNA site evolution models
Figure 3.9 from Huson et al., 2010
Questions about model trees
• Is the model tree topology identifiable? –
yes
• Are the branch lengths and other numeric
parameters of the model tree identifiable?
– yes
• Is the root of the model tree identifiable? –
no
Answers about model trees
• Is the model tree topology identifiable? –
yes
• Are the branch lengths and other numeric
parameters of the model tree identifiable?
– yes
• Is the root of the model tree identifiable? –
no
Phylogeny estimation methods
•
•
•
•
Distance-based methods
Maximum parsimony
Maximum likelihood
Bayesian MCMC
And other types that are not as commonly
used
Performance criteria
• Running time
• Space
• Statistical performance issues (e.g., statistical
consistency and sequence length requirements)
• “Topological accuracy” with respect to the underlying
true tree, typically studied in simulation.
• Accuracy with respect to a mathematical score (e.g.
tree length or likelihood score) on real data
FN
FN: false negative
(missing edge)
FP: false positive
(incorrect edge)
FP
50% error rate
Statistical Consistency
error
Data
Statistical models
• Simple example: coin tosses.
• Suppose your coin has probability p of
turning up heads, and you want to
estimate p. How do you do this?
Estimating p
• Toss coin repeatedly
• Let your estimate q be the fraction of the time
you get a head
• Obvious observation: q will approach p as the
number of coin tosses increases
• This algorithm is a statistically consistent
estimator of p. That is, your error |q-p| goes
to 0 (with high probability) as the number of
coin tosses increases.
Another estimation problem
• Suppose your coin is biased either towards
heads or tails (so that p is not 1/2).
• How do you determine which type of coin you
have?
• Same algorithm, but say “heads” if q>1/2,
and “tails” if q<1/2. For large enough
number of coin tosses, your answer will be
correct with high probability.
Markov models of character
evolution down trees
• The character might be binary, indicating absence or presence
of some property at each node in the tree.
• The character might be multi-state, taking on one of a specific
set of possible states. Typical examples in biology: the
nucleotide in a particular position within a multiple sequence
alignment.
• A probabilistic model of character evolution describes a random
process by which a character changes state on each edge of
the tree. Thus it consists of a tree T and associated parameters
that determine these probabilities.
• The “Markov” property assumes that the state a character
attains at a node v is determined only by the state at the
immediate ancestor of v, and not also by states before then.
Binary characters
• Simplest type of character: presence (1)
or absence (0).
• How do we model the presence or
absence of a property?
Cavender-Farris-Neyman (CFN)
• Models binary sequence evolution
• For each edge e, there is a probability
p(e) of the property “changing state”
(going from 0 to 1, or vice-versa), with
0<p(e)<0.5 (to ensure that unrooted
CFN tree topologies are identifiable).
• Every position evolves under the same
process, independently of the others.
Estimating trees under
statistical models…
• Instead of directly estimating the tree,
we try to estimate the process itself.
• For example, we try to estimate the
probability that two leaves will have
different states for a random character.
CFN pattern probabilities
• Let x and y denote nodes in the tree,
and pxy denote the probability that x and
y exhibit different states.
• Theorem: Let pi be the substitution
probability for edge ei, and let x and y
be connected by path e1e2e3…ek. Then
1-2pxy = (1-2p1)(1-2p2)…(1-2pk)
And then take logarithms
• The theorem gave us:
1-2pxy = (1-2p1)(1-2p2)…(1-2pk)
• If we take logarithms, we obtain
ln(1-2pxy) = ln(1-2p1) + ln(1-2p2)+…+ln(1-2pk)
• Since these probabilities lie between 0 and 0.5, these
logarithms are all negative. So let’s multiply by -1 to
get positive numbers.
An additive matrix!
• Consider a matrix D(x,y) = -ln(1-2pxy)
• This matrix is additive (i.e., fits a tree exactly)!
• Can we estimate this additive matrix from what we
observe at the leaves of the tree?
• Key issue: how to estimate pxy.
• (Recall how to estimate the probability of a head…)
Distance-based Methods
Estimating CFN distances
• Consider
dij= -1/2 ln(1-2H(i,j)/k),
where k is the number of characters, and
H(i,j) is the Hamming distance between
sequences si and sj.
• Theorem: as k increases,
dij converges to Dij = -1/2 ln(1-2pij),
which is an additive matrix.
CFN tree estimation
• Step 1: Compute Hamming distances
• Step 2: Correct the Hamming distances,
using the CFN distance calculation
• Step 3: Use distance-based method
(neighbor joining, naïve quartet method,
etc.)
Four Point Method
• Task: Given 4x4 dissimilarity matrix,
compute a tree on four leaves
• Solution: Compute the three pairwise
sums, and take the split ij|kl that gives
the minimum!
• When is this guaranteed accurate?
Error tolerance for FPM
• Suppose every pairwise distance is
estimated well enough (within f/2, for f
the minimum length of any edge).
• Then the Four Point Method returns the
correct tree (i.e., ij+kl remains the
minimum)
Naïve Quartet Method
• Compute the tree on each quartet using
the four-point condition
• Merge them into a tree on the entire set
if they are compatible:
– Find a sibling pair A,B
– Recurse on S-{A}
– If S-{A} has a tree T, insert A into T by
making A a sibling to B, and return the tree
Error tolerance for NQM
• Suppose every pairwise distance is
estimated well enough (within f/2, for f
the minimum length of any edge).
• Then the Four Point Method returns the
correct tree on every quartet.
• And so all quartet trees are compatible,
and NQM returns the true tree.
In other words:
• The NQM method is statistically
consistent methods for estimating CFN
trees!
• Plus it is polynomial time!
Can we use it on DNA sequences?
Standard DNA site evolution models
Figure 3.9 from Huson et al., 2010
Jukes-Cantor DNA model
• Character states are A,C,T,G (nucleotides).
• All substitutions have equal probability.
• On each edge e, there is a value p(e) indicating the
probability of change from one nucleotide to another
on the edge, with 0<p(e)<0.75 (to ensure that JC
trees are identifiable).
• The state (nucleotide) at the root is random (all
nucleotides occur with equal probability).
• All the positions in the sequence evolve identically
and independently.
Jukes-Cantor distances
• Dij = -3/4 ln(1-4/3 H(i,j)/k)) where k is the
sequence length
• These distances converge to an
additive matrix, just as with CFN
distances
Distance-based Methods
UPGMA
While |S|>2:
find pair x,y of closest taxa;
delete x
Recurse on S-{x}
Insert y as sibling to x
Return tree
a
b
c
d
e
UPGMA
Works when
evolution is
“clocklike”
a
b
c
d
e
UPGMA
Fails to produce
true tree if
evolution
deviates too
much from a
clock!
b
a
c
d
e
Statistical Consistency
error
Data
UPGMA is NOT statistically consistent!
error
Data
Better distance-based methods
(all statistically consistent under JC)
•
•
•
•
•
•
Neighbor Joining
Minimum Evolution
Weighted Neighbor Joining
Bio-NJ
DCM-NJ
And others
Quantifying Error
FN
FN: false negative
(missing edge)
FP: false positive
(incorrect edge)
50% error rate
FP
Neighbor joining has poor performance on large
diameter trees [Nakhleh et al. ISMB 2001]
Error Rate
0.8
NJ
0.6
0.4
0.2
0
0
400
800
No. Taxa
1200
1600
Simulation study
based upon fixed
edge lengths, K2P
model of evolution,
sequence lengths
fixed to 1000
nucleotides.
Error rates reflect
proportion of
incorrect edges in
inferred trees.
Summary so far
• Distance-based methods are generally
polynomial time, and can be statistically
consistent under standard sequence
evolution models.
• Yet they can have high error under high
rates of sequence evolution.
What are the options?
Characters
• A character is a partition of the set of
taxa, defined by the states of the
character
• Morphological examples:
presence/absence of wings,
presence/absence of hair, number of
legs
• Molecular examples: nucleotide or
residue (AA) at a particular site within an
alignment
Maximum Parsimony
• Computational issues
• Statistical consistency issues
• Software and methods to find good
solutions
• Comparison to distance-based methods
on simulated data
Maximum Parsimony
• Input: Set S of n aligned sequences of length k
• Output:
– A phylogenetic tree T leaf-labeled by sequences in S
– additional sequences of length k labeling the internal
nodes of T
such that
å H (i, j )
( i , j )ÎE (T )
is minimized, where H(i,j) denotes the Hamming
distance between sequences at nodes i and j
Hamming Distance Steiner Tree Problem
• Input: Set S of n aligned sequences of length k
• Output:
– A phylogenetic tree T leaf-labeled by sequences in S
– additional sequences of length k labeling the internal
nodes of T
such that
å H (i, j )
( i , j )ÎE (T )
is minimized, where H(i,j) denotes the Hamming
distance between sequences at nodes i and j
Maximum parsimony (example)
• Input: Four sequences
–
–
–
–
ACT
ACA
GTT
GTA
• Question: which of the three trees has the
best MP scores?
Maximum Parsimony
ACT
GTA
ACA
ACT
GTT
ACA
GTT
GTA
ACA
GTA
ACT
GTT
Maximum Parsimony
ACT
GTT
2 GTT GTA
1
2
GTA
ACA
ACA
GTT
ACA ACA 1
3
2
MP score = 6
MP score = 5
ACA
ACT
GTA
ACA GTA
2
1
1
MP score = 4
Optimal MP tree
GTT
ACT
GTA
MP: computational complexity
For four leaves, we can do this by inspection
GTA
ACA
ACA
ACT
1
GTA
2
MP score = 4
1
GTT
MP: computational complexity
Using dynamic programming, the optimal labeling
can be computed in O(r2nk) time
GTA
ACA
ACA
ACT
1
GTA
2
1
GTT
MP score = 4
r = # states (4 for nucleotides, 20 for AA, etc.)
n = # leaves
k = # characters (or sequence length)
DP algorithm
• Dynamic programming algorithms on
trees are common – there is a natural
ordering on the nodes given by the tree.
• Example: computing the longest leaf-toleaf path in a tree can be done in linear
time, using dynamic programming
(bottom-up).
Two variants of MP
• Unweighted MP: all substitutions have the
same cost
• Weighted MP: there is a substitution cost
matrix that allows different substitutions to
have different costs. For example:
transversions and transitions can have
different costs. Even if symmetric, this
complicates the calculation – but not by
much.
Fitch’s algorithm for
unweighted MP on a fixed tree
• We process the characters independently.
• Let c be the character we are examining, and
let c(v) be the state of leaf v.
• Let A(v) denote the set of optimal nucleotides
at node v (for an MP solution to the subtree
rooted at v). Hence A(v)={c(v)} if v is a leaf.
Fitch’s algorithm for fixed-tree (unweighted) maximum parsimony
Sankoff’s DP algorithm for
weighted MP
Assume a given rooted binary tree T and a
single character.
Root tree T at some internal node. Now, for
every node v in T and every possible letter x,
compute
Cost(v,x) := optimal cost of subtree of T
rooted at v, given that we label v by x.
Base case: easy
General case?
DP algorithm (cont.)
Cost(v,x) =
miny{Cost(v1,y)+cost(x,y)} +
miny{Cost(v2,y)+cost(x,y)}
where v1 and v2 are the children of v, and
y ranges over the possible states (e.g.,
nucleotides), and cost(x,y) is an
arbitrary cost function.
DP algorithm (cont.)
We compute Cost(v,x) for every node v and
every state x, from the “bottom up”.
The optimal cost is
minx{Cost(root,x)}
We can then pick the best states for each node
in a top-down pass. However, here we have
to remember that different substitutions have
different costs.
MP: solvable in polynomial time if
the tree is given
Optimal labeling can be computed in O(r2nk) time
GTA
ACA
ACA
ACT
1
GTA
2
1
GTT
MP score = 4
r = # states (4 for nucleotides, 20 for AA, etc.)
n = # leaves
k = # characters (or sequence length)
But finding the best tree is NP-hard!
Optimal labeling can be computed in O(r2nk) time
GTA
ACA
ACA
ACT
1
GTA
2
MP score = 4
1
GTT
Solving NP-hard problems
exactly is … unlikely
• Number of
(unrooted) binary
trees on n leaves
is (2n-5)!!
• If each tree on
1000 taxa could
be analyzed in
0.001 seconds,
we would find the
best tree in
2890 millennia
#leaves
#trees
4
3
5
15
6
105
7
945
8
10395
9
135135
10
2027025
20
2.2 x 1020
100
4.5 x 10190
1000
2.7 x 102900
Approaches for “solving” MP
1.
2.
3.
Hill-climbing heuristics (which can get stuck in local optima)
Randomized algorithms for getting out of local optima
Approximation algorithms for MP (based upon Steiner Tree
approximation algorithms).
Local optimum
Cost
Global optimum
Phylogenetic trees
NNI moves
TBR moves
Approaches for “solving” MP
1.
2.
3.
Hill-climbing heuristics (which can get stuck in local optima)
Randomized algorithms for getting out of local optima
Approximation algorithms for MP (based upon Steiner Tree
approximation algorithms).
Local optimum
Cost
Global optimum
Phylogenetic trees
Good parsimony codes
• TNT (not easy to use but is very
effective)
• PAUP* (much easier to use, not quite
as effective on large datasets)
Problems with heuristics for MP
(OLD EXPERIMENT)
Shown here is the performance of a TNT heuristic maximum parsimony analysis on a
real dataset of almost 14,000 sequences. (“Optimal” here means best score to date,
using any method for any amount of time.) Acceptable error is below 0.01%.
0.2
0.18
Performance of TNT with time
0.16
0.14
Average MP
0.12
score above
optimal, shown as 0.1
a percentage of
0.08
the optimal
0.06
0.04
0.02
0
0
4
8
12
Hours
16
20
24
Summary (so far)
• Maximum Parsimony is an NP-hard
optimization problem, but can be solved
exactly (using dynamic programming) in
polynomial time on a fixed tree.
• Heuristics for MP are reasonably fast,
but apparent convergence can be
misleading. And some datasets can
take a long time.
Is Maximum Parsimony statistically
consistent under CFN?
• Recall the CFN model of binary
sequence evolution: iid site evolution,
and each site changes with probability
p(e) on edge e, with 0 < p(e) < 0.5.
• Is MP statistically consistent under this
model?
Statistical Consistency
error
Data
Statistical consistency under CFN
• We will say that a method M is
statistically consistent under the CFN
model if:
– For all CFN model trees (T,Θ) (where Θ
denotes the set of substitution probabilities
on each of the branches of the tree T), as
the number L of sites goes to infinity, the
probability that M(S)=T converges to 1,
where S is a set of sequences of length L.
Is MP statistically consistent?
• We will start with 4-leaf CFN trees, so
the input to MP is a set of four
sequences, A, B, C, D.
• Note that there are only three possible
unrooted trees that MP can return:
– ((A,B),(C,D))
– ((A,C),(B,D))
– ((A,D),(B,C))
Analyzing what MP does on
four leaves
• MP has to pick the tree that has the
least number of changes among the
three possible trees.
• Consider a single site (i.e., all the
sequences have length one).
• Suppose the site is A=B=C=D=0. Can
we distinguish between the three trees?
Analyzing what MP does on
four leaves
•
•
•
•
•
•
•
Suppose the site is A=B=C=D=0.
Suppose the site is A=B=C=D=1
Suppose the site is A=B=C=0, D=1
Suppose the site is A=B=C=1, D=0
Suppose the site is A=B=D=0, C=1
Suppose the site is A=C=D=0, B=1
Suppose the site is B=C=D=0, A=1
Uninformative Site Patterns
Uninformative site patterns are ones that
fit every tree equally well. Note that any
site that is constant (same value for
A,B,C,D) or splits 3/1 is parsimony
uninformative.
On the other hand, all sites that split 2/2
are parsimony informative!
Parsimony Informative Sites
• [A=B=0, C=D=1] or [A=B=1, C=D=0]
– These sites support ((A,B),(C,D))
• [A=C=1, B=D=0] or [A=C=0, B=D=1]
– These sites support ((A,C),(B,D))
• [A=D=0,B=C=1] or [A=D=1, B=C=0]
– These sites support ((A,D),(B,C))
Calculating which tree MP picks
When the input has only four sequences, calculating what MP does
is easy!
1.Remove the parsimony uninformative sites
2.Let I be the number of sites that support ((A,B),(C,D))
3.Let J be the number of sites that support ((A,C),(B,D))
4.Let K be the number of sites that support ((A,D),(B,C))
5.Whichever tree is supported by the largest number of sites, return
that tree. (For example, if I >max{J,K}, then return ((A,B),(C,D).)
6.If there is a tie, return all trees supported by the largest number of
sites.
MP on 4 leaves
• Consider a four-leaf tree CFN model tree
((A,B),(C,D)) with a very high probability of change
(close to ½) on the internal edge (separating AB from
CD) and very small probabilities of change (close to
0) on the four external edges.
• What parsimony informative sites have the highest
probability? What tree will MP return with probability
increasing to 1, as the number of sites increases?
MP on 4 leaves
• Consider a four-leaf tree CFN model tree
((A,B),(C,D)) with a very high probability of change
(close to ½) on the two edges incident with A and B,
and very small probabilities of change (close to 0) on
all other edges.
• What parsimony informative sites have the highest
probability? What tree will MP return with probability
increasing to 1, as the number of sites increases?
MP on 4 leaves
• Consider a four-leaf tree CFN model tree
((A,B),(C,D)) with a very high probability of change
(close to ½) on the two edges incident with A and C,
and very small probabilities of change (close to 0) on
all other edges.
• What parsimony informative sites have the highest
probability? What tree will MP return with probability
increasing to 1, as the number of sites increases?
Summary (updated)
• Maximum Parsimony (MP) is statistically consistent
on some CFN model trees.
• However, there are some other CFN model trees in
which MP is not statistically consistent. Worse, MP is
positively misleading on some CFN model trees.
This phenomenon is called “long branch attraction”,
and the trees for which MP is not consistent are
referred to as “Felsenstein Zone trees” (after the
paper by Felsenstein).
• The problem is not limited to 4-leaf trees…
Performance on data
• Statistical consistency or inconsistency
is an asymptotic statement, and
requires a proof; it really has nothing
much to say about performance on finite
data.
• To evaluate performance on finite data,
we use simulations.
Quantifying Error
FN
FN: false negative
(missing edge)
FP: false positive
(incorrect edge)
50% error rate
FP
Performance in practice
From Nakhleh et al., PSB 2002
Summary so far
• Maximum Parsimony is not statistically consistent under
standard sequence evolution models, but it can be consistent on
some model trees.
• Maximum parsimony is NP-hard and computationally intensive
in practice.
• In contrast, distance-based methods can be statistically
consistent and polynomial time.
• Yet MP is sometimes more accurate than the leading distancebased methods such as neighbor joining.
• MP remains one of the popular techniques for phylogeny
estimation.
What are the options?
Statistical Methods
• Maximum Likelihood: find model tree
most likely to have generated the
observed data
• Bayesian Estimation: output distribution
of tree topologies in proportion to their
likelihood for having generated the
observed data (marginalizing over all
the possible numeric parameters) R
(General Time Reversible) model
Maximum Likelihood
• Input: sequence data S,
• Output: the model tree (tree T and
parameters theta) s.t. Pr(S|T,theta) is
maximized.
NP-hard to find best tree.
Important in practice.
Good heuristics (RAxML, FastTree,
IQTree, PhyML, and others)
Computing the probability of
the data
• Given a model tree (with all the parameters
set) and character data at the leaves, you can
compute the probability of the data.
• Small trees can be done by hand.
• Large examples are computationally intensive
- but still polynomial time (using dynamic
programming, similar to Sankoff’s algorithm
for MP on a fixed tree).
Cavender-Farris model
calculations
• Consider an unrooted tree with topology
((a,b),(c,d)) with p(e)=0.1 for all edges.
• What is the probability of all leaves
having state 0?
We show the brute-force technique.
Brute-force calculation
Let E and F be the two internal nodes in the tree
((A,B),(C,D)).
Then Pr(A=B=C=D=0) =
• Pr(A=B=C=D=0|E=F=0) +
• Pr(A=B=C=D=0|E=1, F=0) +
• Pr(A=B=C=D=0|E=0, F=1) +
• Pr(A=B=C=D=0|E=F=1)
The notation “Pr(X|Y)” denotes the probability of X given Y.
Calculation, cont.
Technique:
• Set one leaf to be the root
• Set the internal nodes to have some specific
assignment of states (e.g., all 1)
• Compute the probability of that specific
pattern
• Add up all the values you get, across all the
ways of assigning states to internal nodes
Calculation, cont.
Calculating Pr(A=B=C=D=0|E=F=0)
• There are 5 edges, and thus no change on any edge.
• Since p(e)=0.1, then the probability of no change is
0.9. So the probability of this pattern, given that the
root is a particular leaf and has value 0, is (0.9)5.
• Then we multiply by 0.5 (the probability of the root A
having state 0).
• So the probability is (0.5)x (0.9)5.
Maximum likelihood under
Cavender-Farris
• Given a set S of binary sequences, find the
Cavender-Farris model tree (tree topology and edge
parameters) that maximizes the probability of
producing the input data S.
ML, if solved exactly, is statistically consistent under
Cavender-Farris (and under the DNA sequence
models, and more complex models as well).
The problem is that ML is hard to solve.
“Solving ML”
• Technique 1: compute the probability of the
data under each model tree, and return the
best solution.
• Problem: Exponentially many trees on n
sequences, and infinitely many ways of
setting the parameters on each of these
trees!
“Solving ML”
• Technique 2: For each of the tree topologies,
find the best parameter settings.
• Problem: Exponentially many trees on n
sequences, and calculating the best setting of
the parameters on any given tree is hard!
Even so, there are hill-climbing heuristics
for both of these calculations (finding
parameter settings, and finding trees).
Approaches for “solving” ML
1.
2.
Hill-climbing heuristics (which can get stuck in local optima)
Randomized algorithms for getting out of local optima
Local optimum
Cost
Global optimum
Phylogenetic trees
Bayesian analyses
• Algorithm is a random walk through space of all possible model
trees (trees with substitution matrices on edges, etc.).
• From your current model tree, you perturb the tree topology and
numerical parameters to obtain a new model tree.
• Compute the probability of the data (character states at the
leaves) for the new model tree.
– If the probability increases, accept the new model tree.
– If the probability is lower, then accept with some probability (that
depends upon the algorithm design and the new probability).
• Run for a long time…
Bayesian estimation
After the random walk has been run for a very long time…
• Gather a random sample of the trees you visit
• Return:
– Statistics about the random sample (e.g., how many trees
have a particular bipartition), OR
– Consensus tree of the random sample, OR
– The tree that is visited most frequently
Bayesian methods, if run long enough, are statistically consistent
methods (the tree that appears the most often will be the true
tree with high probability).
MrBayes is standard software for Bayesian analyses in biology.
170
Maximum Likelihood
Estimation
Statistical vs.
gene Bayesian
tree estimationTree
methods
L
Tree 1
Tree 2
Numeric parameters (branch lengths, subs tu on matrix)
Figure 8.3 The x-axis indicates the numeric parameters for the GTR model trees that can equip
different tree topologies, Tree 1 and Tree 2; the y-axis indicates the likelihood score. Note that
Summary
There are many statistically consistent methods:
•Maximum Likelihood
•Bayesian MCMC methods
•Distance-based methods (like Neighbor Joining and the
Naïve Quartet Method)
But not maximum parsimony, not maximum
compatibility, and not UPGMA (a distance-based
method)
But statistical consistency is not the only important thing
– performance on data matters at least as much.
Also, the model assumptions under which methods are
statistically consistent are not particularly realistic!
Classical Sequence Evolution
-3 mil yrs
AAGACTT
AAGGCCT
AGGGCAT
AGGGCAT
TAGCCCT
TAGCCCA
-2 mil yrs
TGGACTT
TAGACTT
AGCACTT
AGCACAA
AGCGCTT
-1 mil yrs
today
Standard DNA site evolution models
Figure 3.9 from Huson et al., 2010
DNA sequence evolution models
• The models in this figure are the standard ones:
– Every edge has a substitution probability
– The model also allows 4x4 substitution matrices on the
edges:
• Simplest model: Jukes-Cantor (JC) assumes that all
substitutions are equiprobable
• General Time Reversible (GTR) Model: one 4x4
substitution matrix for all edges
• Not in this figure:
– General Markov (GM) model: different 4x4 matrices allowed
on each edge
– No Common Mechanism model: different substitution
probabilities for each combination of edge and site
• All of these models assume substitution-only
evolution
DNA sequence evolution models
• The models in this figure are the standard ones:
– Every edge has a substitution probability
– The model also allows 4x4 substitution matrices on the
edges:
• Simplest model: Jukes-Cantor (JC) assumes that all
substitutions are equiprobable
• General Time Reversible (GTR) Model: one 4x4
substitution matrix for all edges
• Not in this figure:
– General Markov (GM) model: different 4x4 matrices allowed
on each edge
– No Common Mechanism model: different substitution
probabilities for each combination of edge and site
• All of these models assume substitution-only
evolution
DNA sequence evolution models
• The models in this figure are the standard ones:
– Every edge has a substitution probability
– The model also allows 4x4 substitution matrices on the
edges:
• Simplest model: Jukes-Cantor (JC) assumes that all
substitutions are equiprobable
• General Time Reversible (GTR) Model: one 4x4
substitution matrix for all edges
• Not in this figure:
– General Markov (GM) model: different 4x4 matrices allowed
on each edge
– No Common Mechanism model: different substitution
probabilities for each combination of edge and site
• All of these models assume substitution-only
evolution
The Classical Phylogeny
Problem
U
AGGGCAT
V
W
TAGCCCA
X
TAGACTT
Y
TGCACAA
X
U
Y
V
W
TGCGCTT
Much is known about this problem from a mathematical
and empirical viewpoint
U
AGGGCAT
V
W
TAGCCCA
X
TAGACTT
Y
TGCACAA
X
U
Y
V
W
TGCGCTT
However…
U
V
W
AGGGCATGA
AGAT
X
TAGACTT
Y
TGCACAA
X
U
Y
V
W
TGCGCTT