2. Branch lengths

3. Branch lengths (3 characters): A C A A C C A A C A C C 2 0 1 0 0 Sum of branch lengths = total number of changes.

4. C A C A C A C A C C AA C A C A 0.5 0

5. Genes: 0 = missing, 1 = exist speciesg1g2g3g4g5g6 s1100110 s2001000 s3110000 s4110111 s5001110

6. Ex: Find branch lengths of: s1 s3 s2 s4 s5

8. Problems with MP MP has many problems. We will go over a small sample of them.

9. Problems with MP 1. The statistical justification of MP is unclear. Why should a tree with the least number of changes be the most likely one given the data?

10. Problems with MP 2. Different transitions should have different probabilities (e.g., transitions versus transversions). In MP, this can be accounted for using cost matrices. However, there’s no objective way to assign costs.

11. Problems with MP 3. Different characters may have different weights (e.g., having versus not having vertebrates should maybe weight more than having or not having nails). In MP, most of the time all characters are assigned equal weights. This can be accounted for in MP by assigning different weights to the different characters. However, there’s no objective way to assign weights to characters.

12. Problems with MP 4. The chance for a change depends on evolutionary distances. It is more likely for an amino-acid replacement to occur between a cucumber and human, than between a chimp and a human. MP ignores evolutionary distances (branch lengths), i.e., each type of transition is assigned the same cost regardless of the branch in which it is inferred to have occurred.

13. Problems with MP 5. The MP score does not change if we consider a rooted tree versus an unrooted one. However branch lenghs do change.

14. s1s4s3 s2 s5 Gene number 2, Option number 1. 01100 1 0 0 1

15. s1s4s3 s2 s5 Gene number 2, Option number 2. 01100 1 0 1 1

16. s1s4s3 s2 s5 Gene number 2, Option number 3. 01100 0 0 0 0 Number of changes for gene 2 (character 2) = 2

17. Gene number 2, Branch lengths s1s4s3 s2 s5 01100 2/3 1/3

18. Gene number 2, The unrooted version s1 s4 s3 s2 s5 0 1 1 0 0 1 10 s1 s4 s3 s2 s5 0 1 1 0 0 0 00

19. Branch lengths are different if one uses a rooted or unrooted tree s1 s4 s3 s2 s5 0 1 1 0 0 0.5

20. Problems with MP 6. MP ignores the chance of multiple substitutions per position. If we see A in one sequence, and C in another, there’s a chance that in fact the evolution was A->G->C. Similarly, if we have A in two sequences, it may be that the evolution was A->-C>-A. MP ignores these possibilities, which is unrealistic, and as a result, MP also underestimates branch lengths.

21. A A Introduction to problems with MP MP underestimates branch lengths C A 1.0 MP branch lengths 0 0 A A C A 1.05 A more realistic solution 0.05

23. Variants of parsimony

24. The simple parsimony which counts changes is the Wagner parsimony. If different changes have different costs, this is weighted parsimony. Variants of parsimony ACGT A0312 C3021 G1203 T2130

25. This method assumes that 0 is the ancestral state, and thus, we can only observe 0->1 changes, but never a reversal (1->0). Computation is easy. The father node of 0 is always 0. Total number of changes = number of 0- >1 changes. Example: small deletions in DNA (0 = no deletion). We assume that a deletion cannot revert to the original sequence. Camin-Sokal

26. This method is directional: the root position influences the score. This parsimony is rarely used today… Camin-Sokal

27. When 0 can change to 1 and not to 2, etc’... 0  1  2. Or when the states are in a linear continuum, and the distance between states 0.45 and 0.99 is abs(0.45-0.99). For example, this can be used to make phylogeny based on fingers’ length. Algorithm: very similar to Sankoff’s. Ordinal scale

28. Dollo Parsimony Dollo’s law states that a complex character, once attained, cannot be attained in that form again. In 0/1 terms, if 0 is the ancestral state and 1 the complex state, 1 can evolve from 0 only once, but 1 can revert to state 0. This, like the Camin-Sokal parsimony is a directional method: the position of the root is important. This method was used to infer phylogenies from restriction enzyme sites.

30. Some additional remarks regarding MP

31. Monophyletic groups: Human Chimp Chicken Gorilla When an unrooted tree is given, you cannot know which groups are monophyletic. You can only say which are not. For example, Chicken + Rat might be monophyletic if the root was between Chicken + Rat and the rest. In fact, the real root of the tree is between Chicken and the rest, hence Chicken and rat are not monophyletic. But, Human and Gorilla are not monophyletic no matter where is the root… Rat

32. We have 6 characters. In each species both 0 and 1 are present. The minimum number of changes is 6 (each character must change at least once). The reason we have more than 6 changes is that some characters had arisen more than once. This is called homoplasy. HOMOPLASY