1. Introduction to Bioinformatics Tutorial 4 Multiple Alignment and Phylogeny

2. 2 ClustalW Input Fast alignment? Scoring matrix Alignment format Fast alignment options Gap scoring Phylogenetic trees Input sequences

3. 3 ClustalW Output (1) Input sequences Pairwise alignment scores Building alignment Final score

4. 4 ClustalW Output (2) Sequence namesSequence positions Match strength in decreasing order: * :.

5. 5 Phylogenetic Trees Represent closeness between many entities –In our case, genomic or protein sequences human chimp monkey Observed entity Unobserved commonality Distance representation

6. 6 Rooting Trees A tree can be hung from a root –Adds directional information –Requires addition of ‘outgroup’ humanchimpmonkeypig We know this is furthest So we hang the tree from where it joins

7. 7 Phylogeny and Evolution Evolutionary Time Speciation Number of mutations Common Ancestor

8. 8 Tree Reconstruction Build tree based on organism sequences Distance-based methods –Use pairwise alignment scores to build tree –Ignores sequences after initial alignments Character-based methods –Learn a tree with intermediate sequences that minimizes total number of mutations –Slower but generally better results

9. 9 Distance-based Example (1) 234 1 745 2 22 3 2 12 3 4

10. 10 Distance-based Example (2) 34 12 -3 3 2 34 12

11. 11 Distance-based Example (3) 34 34 12 -4 12

12. 12 Newick Tree Format (CFTR_SHEEP:0.01457, (CFTR_HUMAN:0.16153, (CFTR_MOUSE:0.70599, (CFTR_RABIT:2.76042, (CFTR_SQUAC:1.27192, CFTR_XENLA:0.28818) :3.42183) :0.77076) :0.65873) :0.73937, CFTR_BOVIN:0.00953);

13. 13 Phylodendron Input Graphical style Newick tree description Tree size Orientation

14. 14 Calculation of HIV/SIV Neighbor-joining tree Why phylogenetic analyses? Mutations accumulate in the genomes of pathogens, especially viruses, during a spread of an infection. This can be used to document the history of transmission events. Phylogenetic analysis of these mutations may not only be used to reconstruct the history of a pathogen's spread through host populations but can also be used to make predictions about it's future progress. The unsolved HIV/SIV relationship One interesting case, where phylogenetic treebuilding is useful, is the unsolved HIV/SIV relationship: HIV-1, HIV-2 and SIV. AIDS (acquired immunodeficiency syndrome) is caused by two different human viruses: HIV-1, group M and O HIV-2, subtypes A to E There are many related viruses in a variety of non-human primates. These related viruses are called SIV (simian immunodeficiency viruses).

15. 15 Calculation of HIV/SIV Neighbor-joining tree Phylogenetic studies have shown that primate lentiviruses are all in the same clade. Within this clade there are five major lineages (the subscripts denotes the host) : HIV-1 and SIV CPZ (Chimpanzee) HIV-2, SIV SM (Sooty mangabey) and SIV MAC (Captive macaque) SIV AGM (African green monkey) SIV MND (Mandrill) SIV SYK (Sykes´ monkey) The NJ tree in our example is based on the poly protein sequence from HIV-1, HIV-2 and SIV with HTLV-1 as an outgroup. HTLV-1 (human T-lymphotropic virus type 1) is another human retroviral pathogen that has originated from related simian viruses.

16. 16 Calculation of HIV/SIV Neighbor-joining tree Step by step summary: 1.Define all taxa and calculate all pairwise distances. 2.Pick two nodes in the star (i and j) for which the distance is minimal. 3.Define a new node (x) and calculate r i and r j. 4.Calculate d ix and d jx, thereby joining x to i and j respectively. 5.Remove i and j from the star and insert x instead. 6.Calculate d xm for all m in the star. Continue until the star has been resolved and root the tree in a final step.

17. 17 Step1 minimum

18. 18 Step1 (cont.) The calculation starts with the star: The branch lengths between node 5 and 10 and between 6 and 10 are calculated with these formulas: In this case L = 9 New node x = 10 r i =r 5 =Σd 5k /(L-2) = 3.22406/(9-2) = 0.46058 r j =r 6 =Σd 6k /(L-2) = 3.22758/(9-2) = 0.461083 d ix =d 5 10 =(d 5 6 + r 5 - r 6 )/2 = (0.06088 + 0.46058 - 0.461083)/2 = 0.0301886 d jx =d 6 10 = d 5 6 - d 5 10 = 0.06088 - 0.0301886 = 0.0306914

19. 19 Step1(cont.)

20. 20 Step2 minimum

21. 21 Step2 (cont.) Calculation of the new branches: In this case L = 8 New node x = 11 r i =r 3 =Σd 3k /(L-2) = 2.715455/(8-2) = 0.452576 r j = r 4 =Σd 4k /(L-2) = 2.50096/(8-2) = 0.416827 d ix =d 3 11 =(d 3 4 + r 3 - r 4 )/2 = (0.125 + 0.452576 - 0.416827)/2 = 0.080375 d jx =d 4 11 = d 3 4 - d 3 11 = 0.125 - 0.080375 = 0.044625

22. 22 Step1(cont.)

23. 23 Step3 minimum

24. 24 Step3 (cont.) Calculation of the new branches: In this case L = 7 New node x = 12 r i =r 2 =Σd 2k /(L-2) = 2.252265/(7-2) = 0.450453 r j =r 11 =Σd 11k /(L-2) = 2.108208/(7-2) = 0.4216415 d ix =d 2 12 =(d 2 11 + r 2 - r 11 )/2 = (0.109705 + 0.450453 - 0.4216415)/2 = 0.069258 d jx =d 11 12 = d 2 11 - d 2 12 = 0.109705 - 0.069258 = 0.040447

25. 25 Step1(cont.)

26. 26 Step 7 In this case L = 3 New node x = 16: r 13 = 0.843684 r 15 =0.728574 d 13 16 = 0.131758 d 15 16 = 0.016648

27. 27 Step 7 (cont.) Because node 9 is the outgroup, the root will be placed between node 9 and the other nodes. The distance between node 9 and the first internal node is 0.563519.

28. 28 Conclusions HIV-2 (H2) is more closely related to SIV (S) from sooty mangabey than to HIV-1 (H1). HIV-1 seems to be more closely related to SIV from chimpanzee. This means that HIV- 1 and HIV-2 have originated independently from two different SIV strains. There must have been a cross-species transmission from chimpanzee SIV to human HIV-1. There also seems to have been a cross- species transmission from human to MAN/MAC.

29. 29 Conclusions As one can see the branch between the H2-ROD A and the to SIV taxa has a low support. Only 56% of the trees have this topology. Therefore the transmission events from human to non-human primates are very uncertain.

30. 30 Exercise In this exercise you will perform a phylogenetic analysis of the human globin sequences. You will compare your results to current prevalent knowledge on the globin family, according to the following summary on the globin sequences: Myoglobin and hemoglobins diverged from one another before the emergence of worms, about 800 million year ago. The hemoglobins diverged into two families (the α-family and β-family) following a gene duplication, about 450 million years ago, which is before the emergence of mammals. The α-family diverged into the zeta, teta and alpha genes, and the β-family diverged into the beta, gamma_G, gamma_A, delta and epsilon genes, all following a series of gene duplications. The most recent duplication was that gamma_G from gamma_A, which occurred around the separation of the simians (humans, chimp, gorilla, etc.) from the pro- siminas (such as lemurs and lorises), about 55 million years ago. (adapted from Graur and Li, 1999)

31. 31 Exercise (cont.) 1.Reconstruct the phylogenetic tree of the human globins using Neighbor joining. Make sure tree is properly rooted (by defining an outgroup) according to the information in the above summary. Point out where the hemoglobins and myoglobin diverged, and where the α-family and β-family diverged. 2.Which of the following groups are monophyletic according to the tree you obtained: (i) alpha, beta, delta, (ii) alpha, teta, zeta, (iii) epsilon, beta, delta 3.Bootstrap the tree you built with 1000 bootstrap iterations. Display the tree with the bootstrap values displayed. On which branch was the lowest bootstrap value obtained? Explain what this means.