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Parsimony methodsthe evolutionary tree to be preferred involves ‘the minimum amount of evolution’
Edwards & Cavalli-Sforza 1963.
• Reconstruct all evolutionary changes along any possible tree• Find tree with least number of changes
A simple example
Characters
Species 1 2 3 4 5 6
Alpha 1 0 0 1 1 0
Beta 0 0 1 0 0 0
Gamma 1 1 0 0 0 0
Delta 1 1 0 1 1 1
Epsilon 0 0 1 1 1 0
Evolutionary changes: 0 1 and 1 0Root: 0 or 1
A simple example
Alpha BetaDelta Gamma Epsilon
1 011 0 character 1
A simple example
Alpha BetaDelta Gamma Epsilon
1 011 0 character 1
0
01
A simple example
Alpha BetaDelta Gamma Epsilon
1 011 0 character 1
1
01
A simple example
Alpha BetaDelta Gamma Epsilon
0 011 0 character 2
A simple example
Alpha BetaDelta Gamma Epsilon
0 011 0 character 2
A simple example
Alpha BetaDelta Gamma Epsilon
0 011 0 character 2
A simple example
Alpha BetaDelta Gamma Epsilon
0 011 0 character 2
A simple example
Alpha BetaDelta Gamma Epsilon
0 100 1 character 3
A simple example
Alpha BetaDelta Gamma Epsilon
0 100 1 character 3
A simple example
Alpha BetaDelta Gamma Epsilon
1 001 1 character 4
A simple example
Alpha BetaDelta Gamma Epsilon
1 001 1 character 4
A simple example
Alpha BetaDelta Gamma Epsilon
1 001 1 character 41 001 1 character 5
A simple example
Alpha BetaDelta Gamma Epsilon
0 001 0 character 6
A simple example
Characters
1 2 3 4 5 6
number of changes required
1 2 1 2 2 1
total number of changes required = 9.
this first hypothesis requires a total of 9 evolutionary changes
A simple example
Alpha BetaDelta Gamma Epsilon
1
5 5 4
34
26
2
colour indicatesderived status ( =0, =1)
character number
A simple example
Alpha BetaDelta Gamma Epsilon
1
5 54
3
6
2
4
this alternative hypothesis requires but 8 evolutionary changes.
A simple example
Alpha BetaDelta Gamma Epsilon
1
5 54
3
6
2
² 4
homoplasy: the same status arises more than once on the tree
A simple example
Alpha BetaDelta Gamma Epsilon
1
5 54
3
6
2
² 4
homoplasy: the same status arises more than once on the tree
Rooted and unrooted trees
Gamma BetaDelta Alpha Epsilon
1
554
3
6
2
²4
yet ‘another’ hypothesis requiring but 8 evolutionary changes
A simple example
Alpha BetaDelta Gamma Epsilon
1
5 54
3
6
2
² 4
Gamma BetaDelta Alpha Epsilon
1
554
3
6
2
²4
the two rooted hypotheses requiring 8 changes yield similar unrooted trees
Rooted and unrooted trees
Alpha
154
32
Delta
Gamma Beta
Epsilon
6
54
Rooted and unrooted trees
Alpha BetaDelta Gamma Epsilon0 011 0
Alpha BetaDelta Gamma Epsilon0 011 0
unrooting trees reduces the number of alternative solutions
character 2
Rooted and unrooted trees
Characters
1 2 3 4 5 6
number of changes required
1 2 1 2 2 1
# alternative trees(rooted)
2 3 2 2 2 1
# alternative trees(unrooted)
1 2 1 2 2 1
unrooting trees reduces the number of alternative solutions
Methods of rooting a tree
1. Use an outgroup2. Use a molecular clock
Methods of rooting a tree
1. Use an outgroup
Ape3
Ape2
Ape1 Ape4
Monkeyroot must be
along this lineage
Methods of rooting a tree
1. Use an outgroup2. Use a molecular clock
only the root is equidistant to all tips
Branch lengths
Gamma
1
5
2
32
Delta
Alpha Beta
Epsilon
6 5
44 22
4
4
55
+0.5+0.5+0.5
+0.5
+0.5
+0.5
+0.5+0.5
+0.5+0.5
+0.5+0.5
+1 +1 +1
Characters
1 2 3 4 5 6
# alternative trees (unrooted)
1 2 1 2 2 1
branch lengths are computed as the sum of all character changes (each divided by # alternatives)
Branch lengths
Gamma
Delta
Alpha Beta
Epsilon
1.5
2.5 1.0
1.0
1.00.5
1.5
the sum of all branch lengths is called the ‘length’ of the tree
Branch lengths
Gamma
Delta
Alpha
Beta
Epsilon
1.5
2.5 1.0
1.0
1.00.5
1.5
But how to…
1. count the number of changes in large datasets2. reconstruct states at interior nodes3. search among all possible trees for the most parsimonious one4. handle DNA sequences (4 states)5. handle complex morphological characters6. justify the parsimony criterion7. evaluate statistically different trees
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