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 Epsilon1 011 0 character 1

A simple example

Alpha BetaDelta Gamma Epsilon1 011 0 character 1

0

01

A simple example

Alpha BetaDelta Gamma Epsilon1 011 0 character 1

1

01

A simple example

Alpha BetaDelta Gamma Epsilon0 011 0 character 2

A simple example

Alpha BetaDelta Gamma Epsilon0 011 0 character 2

A simple example

Alpha BetaDelta Gamma Epsilon0 011 0 character 2

A simple example

Alpha BetaDelta Gamma Epsilon0 011 0 character 2

A simple example

Alpha BetaDelta Gamma Epsilon0 100 1 character 3

A simple example

Alpha BetaDelta Gamma Epsilon0 100 1 character 3

A simple example

Alpha BetaDelta Gamma Epsilon1 001 1 character 4

A simple example

Alpha BetaDelta Gamma Epsilon1 001 1 character 4

A simple example

Alpha BetaDelta Gamma Epsilon1 001 1 character 41 001 1 character 5

A simple example

Alpha BetaDelta Gamma Epsilon0 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