2. Consensus

3. Consensus tree A consensus tree summarizes information common to two or more trees. bcdeabcdeabcdea

4. Strict consensus Strict consensus includes only those groups that occur in all the trees being considered. bcdeabcdea bcdea bcdea Strict consensus

5. Problem: the split {ab} is found 2 out of 3 times, but is not shown in the strict consensus. bcdeabcdea bcdea bcdea Strict consensus

6. Majority-rule consensus Majority-rule consensus: splits that are found in the majority of the trees are shown. bcdeabcdea bcdea bcdea Majority-rule consensus

7. The percentage of the trees supporting each split are indicated. bcdeabcdea bcde 100 bcdea Majority-rule consensus a 67

8. Problem with Majority-rule consensus However in both trees if we consider only {b,c,d}, then in both trees b is closer to c than b to d, or c to d. bcde bcdae Majority-rule consensus= Strict consensus = a bcdea

9. Adams consensus Adams consensus will give the subtrees that are common to all trees. Adams consensus is useful where there are one or more sequences with unclear positions but there’s a subset of sequences that are common to all trees. bcdae Adams consensus bcdea cdaeb

11. Our goal is to evaluate the reliability of different clades. Problem with consensus of all the MP trees In other words, we do not want to rely just on one best tree, but rather estimate the support for each split based on many equally likely or highly likely trees.

12. Bootstrap (and jackknife)

13. Now we have a tree, but what is the robustness of this tree? African-1 Dugong African-2 Mam-3 African-ref Mam-4 Mam-6 Asian-2 African-3 Asian-3 Asian-1 Mam-5

14. Jackknife A. We create new data sets by sampling half of the characters. (random samples without replacement). We generate 100 pseudo-data sets. Note: we do not change the number of sequences, just the number of positions!

15. Jackknife B. We reconstruct a tree from each data set. POS: 52316 1 : TATTT 2 : CATTT 3 : CACTT N : AACTT POS: 18745 1 : TTTAT 2 : TAACC 3 : TAACC N : TGGGA POS: 18394 1 : TTGTA 2 : TAGAC 3 : TAAAC N : TGAGG Sp1 Sp2 Sp3 Sp4 Sp1 Sp2 Sp3 Sp4 Sp1 Sp2 Sp3 Sp4

16. C. We compute the majority rule consensus. Sp1 Sp2 Sp3 Sp4 Sp1 Sp2 Sp3 Sp4 Sp1 Sp2 Sp3 Sp4 Sp1 Sp2 Sp3 Sp4 67% 100% In 67% of the data sets, the split between SP1+SP2 and the rest of the tree was found. Jackknife

17. Bootstrap The same as jackknife, but instead of sampling N/2 positions, we sample N positions with replacement.

18. Bootstrap A. Resample (100-1000 time). 12345 N 1 : ATCTG…A 2 : ATCTG…C 3 : ACTTA…C N : ACCTA…T 12345 N 1 : AATTT…T 2 : AATTT…G 3 : AACTT…T N : AACTT…T 11244x 12345 N 1 : TTTAT…T 2 : TAACC…G 3 : TAACC…T N : TGGGA…T 47789…x 12345 N 1 : AGGTA…T 2 : AGGAC…G 3 : AAAAC…A N : AAAGG…C 15578…N

19. Bootstrap B. Reconstruct a tree from each data set. 12345 N 1 : AATTT…T 2 : AATTT…G 3 : AACTT…T N : AACTT…T 11244x 12345 N 1 : TTTAT…T 2 : TAACC…G 3 : TAACC…T N : TGGGA…T 47789…x 12345 N 1 : AGGTA…T 2 : AGGAC…G 3 : AAAAC…A N : AAAGG…C 15578…N Sp1 Sp2 Sp3 Sp4 Sp1 Sp2 Sp3 Sp4 Sp1 Sp2 Sp3 Sp4

20. Bootstrap C. Compute the majority rule consensus. Sp1 Sp2 Sp3 Sp4 Sp1 Sp2 Sp3 Sp4 Sp1 Sp2 Sp3 Sp4 Sp1 Sp2 Sp3 Sp4 67% 100% Remark: in a bootstrap tree branch lengths have no meaning.