1. Parsimony methods the 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

2. A simple example Characters Species 123456 Alpha100110 Beta001000 Gamma110000 Delta110111 Epsilon001110 Evolutionary changes: 0  1 and 1  0 Root: 0 or 1

3. A simple example AlphaBetaDeltaGammaEpsilon 10110 character 1

4. A simple example AlphaBetaDeltaGammaEpsilon 10110 character 1 0 0 1

5. A simple example AlphaBetaDeltaGammaEpsilon 10110 character 1 1 0 1

6. A simple example AlphaBetaDeltaGammaEpsilon 00110 character 2

7. A simple example AlphaBetaDeltaGammaEpsilon 00110 character 2

8. A simple example AlphaBetaDeltaGammaEpsilon 00110 character 2

9. A simple example AlphaBetaDeltaGammaEpsilon 00110 character 2

10. A simple example AlphaBetaDeltaGammaEpsilon 01001 character 3

11. A simple example AlphaBetaDeltaGammaEpsilon 01001 character 3

12. A simple example AlphaBetaDeltaGammaEpsilon 10011 character 4

13. A simple example AlphaBetaDeltaGammaEpsilon 10011 character 4

14. A simple example AlphaBetaDeltaGammaEpsilon 10011 character 4 10011 character 5

15. A simple example AlphaBetaDeltaGammaEpsilon 00010 character 6

16. A simple example Characters 123456 number of changes required 121221 total number of changes required = 9. this first hypothesis requires a total of 9 evolutionary changes

17. A simple example AlphaBetaDeltaGammaEpsilon 1 5 5 4 3 4 2 6 2 colour indicates derived status ( =0, =1) character number

18. A simple example AlphaBetaDeltaGammaEpsilon 1 55 4 3 6 2 4 this alternative hypothesis requires but 8 evolutionary changes.

19. A simple example AlphaBetaDeltaGammaEpsilon 1 55 4 3 6 2 ² 4 homoplasy: the same status arises more than once on the tree

20. A simple example AlphaBetaDeltaGammaEpsilon 1 55 4 3 6 2 ² 4 homoplasy: the same status arises more than once on the tree

21. Rooted and unrooted trees GammaBetaDeltaAlphaEpsilon 1 55 4 3 6 2 ² 4 yet ‘another’ hypothesis requiring but 8 evolutionary changes

22. A simple example AlphaBetaDeltaGammaEpsilon 1 55 4 3 6 2 ² 4 GammaBetaDeltaAlphaEpsilon 1 55 4 3 6 2 ² 4 the two rooted hypotheses requiring 8 changes yield similar unrooted trees

23. Rooted and unrooted trees Alpha 1 5 4 3 2 Delta Gamma Beta Epsilon 6 5 4

24. Rooted and unrooted trees AlphaBetaDeltaGammaEpsilon 00110 AlphaBetaDeltaGammaEpsilon 00110 unrooting trees reduces the number of alternative solutions character 2

25. Rooted and unrooted trees Characters 123456 number of changes required 121221 # alternative trees (rooted) 232221 # alternative trees (unrooted) 121221 unrooting trees reduces the number of alternative solutions

26. Methods of rooting a tree 1.Use an outgroup 2.Use a molecular clock

27. Methods of rooting a tree 1.Use an outgroup Ape3 Ape2 Ape1 Ape4 Monkey root must be along this lineage

28. Methods of rooting a tree 1.Use an outgroup 2.Use a molecular clock only the root is equidistant to all tips

29. Branch lengths Gamma 1 5 2 3 2 Delta Alpha Beta Epsilon 6 5 4 4 2 2 4 4 5 5 +0.5 +1 Characters 123456 # alternative trees (unrooted) 121221 branch lengths are computed as the sum of all character changes (each divided by # alternatives)

30. Branch lengths Gamma Delta Alpha Beta Epsilon 1.5 2.5 1.0 0.5 1.5 the sum of all branch lengths is called the ‘length’ of the tree

31. Branch lengths Gamma Delta Alpha Beta Epsilon 1.5 2.5 1.0 0.5 1.5

32. But how to… 1.count the number of changes in large datasets 2.reconstruct states at interior nodes 3.search among all possible trees for the most parsimonious one 4.handle DNA sequences (4 states) 5.handle complex morphological characters 6.justify the parsimony criterion 7.evaluate statistically different trees