1. Testing alternative hypotheses

2. Outline Topology tests: –Templeton test Parametric bootstrapping (briefly) Comparing data sets

3. Topology tests The best tree for your data contradicts a prior hypothesis. This does not mean that the data refute the hypothesis Compare the optimality score of the best tree and the best trees given the hypothesis

4. Tree space Region of tree space satisfying the hypothesis Optimal tree Optimal tree satisfying the hypothesis

5. Does one tree explain the data significantly better than the other? If the data are “significantly” more compatible with the optimal tree than the constrained tree, the hypothesis is rejected Parsimony framework –Constrained tree length = X –Optimal tree length = Y –Is the cost (X-Y) significant?

6. Templeton test A T G T G A A C A A B T G T G A C C A A C T G C G G C C T A D A G C G G C G T A E A A C T A A G T G F A A C T A A G C G L1 1 1 1 1 2 2 1 2 1 = 12 L4 3 2 2 2 2 1 3 3 2 = 20

7. Templeton test A T G T G A A C A A B T G T G A C C A A C T G C G G C C T A D A G C G G C G T A E A A C T A A G T G F A A C T A A G C G L1 1 1 1 1 2 2 1 2 1 = 12 L4 3 2 2 2 2 1 3 3 2 = 20 Diff 2 1 1 1 0-1 2 1 1 =

8. Templeton test ScoreRank 21.5 15.5 -1-5.5 Sum of the negative ranks = 5.5 N (number of chars varying in length) = 8 P-value = ca. 0.045

9. Problems of topology tests The tests compare trees, they don’t compare the competing hypotheses

10. Tree space Region of tree space satisfying the hypothesis Optimal tree Optimal tree satisfying the hypothesis

11. Another problem of topology tests Suppose we had a prior hypothesis that species A-B form a clade We conduct a phylogenetic analysis of 8 species and find that A-B do not form a clade The shortest tree that has them as a clade is 6 steps longer (decay = -6) which is significant under a Templeton test

12. Suppose we had a prior hypothesis that species A-Z form a clade We conduct a phylogenetic analysis of 100 species and find that A-Z do not form a clade The shortest tree that has them as a clade is 6 steps longer (decay = -6) which is significant under a Templeton test Another problem of topology tests

13. Are these results equivalent? The two hypotheses are differently stringent –The former delimits a much larger proportion of tree- space One solution is to reverse the question: If the hypothesis were true, how likely is it that the optimal tree would reject it? –Requires parametric bootstrapping

14. Find the region of tree space that is plausible if the hypothesis is true: Tree space Optimal tree Optimal tree satisfying the hypothesis Hypothesis rejected

15. Find the region of tree space that is plausible if the hypothesis is true: Tree space Optimal tree Optimal tree satisfying the hypothesis Hypothesis not rejected

16. How do you do this? Find the optimal tree under the constraint (not just the optimal topology but also branch lengths, etc.) Simulate data up that tree many times For each data set calculate the cost of the hypothesis If the observed cost was greater than the cost from the simulated data, the hypothesis is rejected.

17. Strepsiptera sister to the Diptera (Whiting et al. 1997)

18. Could be a long-branch problem (Huelsenbeck, 1997)

19. What if this were the true tree? Probability of Strepsiptera being sister to Diptera on the MP tree = 92%

20. Testing hypotheses Topology tests are good ways to test hypotheses Parametric bootstraping tests are powerful but laborious Other approaches are available using likelihood or Bayesian approaches (later)

21. Multiple data sets for the same sets of taxa Analyze each data set separately and then compare the trees (consense) Concatenate the data and conduct a single combined analysis (combine)

22. Argument for consensus If the same clades appear with multiple data sets we can be more confident The method is conservative

23. Is consensus conservative? Barrett et al. 1994. Syst. Zool. 40:486

25. Arguments against combined analysis Some data sets might have strong misleading signals (e.g., due to lab errors) Different partitions might have tracked different histories

26. Conditional combined analysis Assess if the data look like they have tracked different histories –If they do not: combine –If they do: analyze separately Can you do this with topology tests?

27. Optimal tree for data set 2 Optimal tree for data set 1 Do they conflict?

28. But topology tests can be used more carefully Two data sets don’t conflict significantly if there is one tree that neither data set rejects Two data sets do conflict if: –Data set 1 rejects all trees that lack a certain clade –Data set 2 rejects all trees that have that same clade

29. Optimal tree for data set 2 Optimal tree for data set 1 Significantly worse Optimal tree without the constraint for data set 2 Optimal tree with the constraint for data set 1 Significantly worse

30. The Incongruence Length Difference (ILD) test (Farris et al., 1994) Conflict is manifest as longer trees (or lower likelihood) Look to see how length (or likelihood) increases when we combine data Determine significance compared to random partitions

31. ILD test ( = Partition Homogeneity Test in PAUP*) One TACATAAACAAGCCTAAAATGCGACACTACGTTCACTGTTACGCTCTCCACTGCCTAGACGAAGAAGCTTCA Two TACATAAACAAGCCCAAAATGCGACACTACGTCCACTGTTATGCTCTCCACTGCCTAGACGAAGACGCTTCA Three TACATAAACAAGCCCAAAATGCGACACTACGTCCACTGTTACGCTCTTCACTGCCTAGACGAGGATGCCTCG Four TACATAAATAAGCCAAAAATGCGACACTACGTTCATTGTTACGCACTCCATTGCCTCGACGAAGAAGCTTCA Five TACATAAACAAACCCAAAATGCGACACTACGTCCACTGTTATGCTCTCCACTGTCTAGACGAAGACGCTTCG Six TACATAAACAAGCCCAAGATGCGTCACTACGTCCACTGCTACGCCCTCCACTGTCTCGACGAGGAGGCCTCG Seven TACATAAACAAACCAAAAATGCGACACTACGTCCATTGTTACGCCCTACACTGCCTAGACGAAGACGCTTCA Eight TACATAAACAAACCAAAAATGCGACACTACGTCCATTGTTACGCCCTACACTGCCTAGACGAAGACGCTTCA Partition 1 Length = 12 Partition 2 Length = 9 Combined L = 21

32. ILD test ( = Partition Homogeneity Test in PAUP*) One TACATAAACAAGCCTAAAATGCGACACTACGTTCACTGTTACGCTCTCCACTGCCTAGACGAAGAAGCTTCA Two TACATAAACAAGCCCAAAATGCGACACTACGTCCACTGTTATGCTCTCCACTGCCTAGACGAAGACGCTTCA Three TACATAAACAAGCCCAAAATGCGACACTACGTCCACTGTTACGCTCTTCACTGCCTAGACGAGGATGCCTCG Four TACATAAATAAGCCAAAAATGCGACACTACGTTCATTGTTACGCACTCCATTGCCTCGACGAAGAAGCTTCA Five TACATAAACAAACCCAAAATGCGACACTACGTCCACTGTTATGCTCTCCACTGTCTAGACGAAGACGCTTCG Six TACATAAACAAGCCCAAGATGCGTCACTACGTCCACTGCTACGCCCTCCACTGTCTCGACGAGGAGGCCTCG Seven TACATAAACAAACCAAAAATGCGACACTACGTCCATTGTTACGCCCTACACTGCCTAGACGAAGACGCTTCA Eight TACATAAACAAACCAAAAATGCGACACTACGTCCATTGTTACGCCCTACACTGCCTAGACGAAGACGCTTCA

33. ILD test ( = Partition Homogeneity Test in PAUP*) Combined L = 25 One TACATAAACAAGCCTAAAATGCGACACTACGTTCACTGTTACGCTCTCCACTGCCTAGACGAAGAAGCTTCA Two TACATAAACAAGCCCAAAATGCGACACTACGTCCACTGTTATGCTCTCCACTGCCTAGACGAAGACGCTTCA Three TACATAAACAAGCCCAAAATGCGACACTACGTCCACTGTTACGCTCTTCACTGCCTAGACGAGGATGCCTCG Four TACATAAATAAGCCAAAAATGCGACACTACGTTCATTGTTACGCACTCCATTGCCTCGACGAAGAAGCTTCA Five TACATAAACAAACCCAAAATGCGACACTACGTCCACTGTTATGCTCTCCACTGTCTAGACGAAGACGCTTCG Six TACATAAACAAGCCCAAGATGCGTCACTACGTCCACTGCTACGCCCTCCACTGTCTCGACGAGGAGGCCTCG Seven TACATAAACAAACCAAAAATGCGACACTACGTCCATTGTTACGCCCTACACTGCCTAGACGAAGACGCTTCA Eight TACATAAACAAACCAAAAATGCGACACTACGTCCATTGTTACGCCCTACACTGCCTAGACGAAGACGCTTCA Partition 1 Length = 14 Partition 2 Length =11

34. Results Sum of Number of tree lengths replicates ----------------------------- 1661 1 1662 2 1663 1 1665* 9 1666 8 1667 9 1668 5 1669 11 1670 10 1671 9 * = sum of lengths for original partition P value = 1 - (87/100) = 0.130000 Sum of Number of tree lengths replicates -------------------------------- 1672 10 1673 7 1674 4 1675 4 1676 1 1677 4 1678 2 1679 1 1680 1 1683 1

35. What does a positive result mean? The data sets have tracked different histories? The original partition is non-random Does not even look at topology

36. Option if you find conflict Conduct separate analyses only Delete taxa until conflict disappears - then combine Combine anyway

37. Conditional conditional combined analysis You believe that conflict reflects data partitions tracking different histories –Keep the data separate and find ways to summarize the discrepancy You believe that conflict reflects artifactual signals (noise) in one or both data sets –Combine anyway in the hope that the real signal will come to dominate