Lecture 3 Molecular Evolution and Phylogeny

1. Lecture 3Molecular Evolution and Phylogeny

2. Facts on the molecular basis of life • Every life forms is genome based • Genomes evolves • There are large numbers of apparently homlogous intra-genomic (paralog) and inter-genomic (ortholog) genes • Some genes, especially those related to the function of transcription and translation, are common to ALL life forms • The closer two organisms seem to be phylogenetically, the more similar their genomes and corresponding genes are

3. Central dogma of molecular biology DNA RNA Protein

4. Basic assumptions of molecular evolution • Closer related organisms have more similar genomes • Highly similar genes are homologs (have the same ancestor) • A universal ancestor exists for all life forms • Molecular difference in homologous genes (or protein sequences) are positively correlated with evolution time • Phylogenetic relation can be expressed by a dendrogram (a “tree”)

5. The five steps in phylogenetics dancing 1 Sequence data 2 Align Sequences Phylogenetic signal? Patterns—>evolutionary processes? 3 Distances methods Characters based methods Distance calculation (which model?) 4 Choose a method MB ML MP Wheighting? (sites, changes)? Model? Model? Optimality criterion Single tree LS ME NJ Calculate or estimate best fit tree 5 Test phylogenetic reliability Modified from Hillis et al., (1993). Methods in Enzymology 224, 456-487

6. Why protein phylogenies? • For historical reasons - first sequences... • Most genes encode proteins... • To study protein structure, function and • evolution • Comparing DNA and protein based • phylogenies can be useful • Different genes - e.g. 18S rRNA versus EF-2 protein • Protein encoding gene - codons versus amino acids

7. Protein were the first molecular sequences to be used for phylogenetic inference Fitch and Margoliash (1967) Construction of phylogenetic trees. Science 155, 279-284.

8. Most of what follows taken from: Statistical Physics and Biological Information Institute of Theoretical Physics University of California at Santa Barbara 2001 May 7

9. Understanding trees Root 30 Mya Time 22 Mya 7 Mya same as

10. Understanding trees #2

11. Understanding trees #3

12. Difference in homologous sequences is a measure of evolution time Part of multiple sequence alignment of Mitochondrial Small Sub-Unit rRNA Full length is ~ 950 11 primate species with mouse as outgroup 靈長目 Change similarity matrix to distance matrix: d = 1 - S

14. From alignment construct pairwise distance* *Note: Alignment is not the only way to compute distance

15. Models of sequence evolution

16. Jukes-Cantor (minimal) Model All substitution rates = a all base frequency = 1/4 = 3 Pij(2t) A C

17. Derivation of Jukes-Cantor formula • Let probability of site being a base at time t be P(t) • After elapse time Dt • mutate to other three bases is –3aDt P(t) • Gain from other bases is aDt (1 - P(t)) • Hence • P(t + Dt) = P(t) –3aDt P(t) + aDt (1 - P(t)) • dP(t)/dt = a - 4a P(t) • Write P(t) = a exp(-bt) +c, solution is b= 4a, c=1/4 • P(t) = a exp(- 4a t) +1/4 • If P(0) = 1, then a = ¾. If P(0) = 0, then a = -1/4 • Finally Psame(t) =1/4 +3/4 exp(- 4a t) Pchange(t) =1/4 - 1/4 exp(- 4a t)

18. Hasegawa-Kishino-Yano model Has a more general substitution rate Transition A G or C T Transversion A T or C G

20. Part of Jukes-Cantor distance matrix for primate examples (is much larger; for outgroup) Matrix will be used for clustering methods

21. Clustering

22. UPGMA

23. Neighbor-Joining Method

24. N-J Method produces an Unrooted, Additive tree

25. Neighbor-Joining Method An Example What is required for the Neighbour joining method? 0. Distance Matrix Distance matrix

26. 1. First Step PAM distance 3.3 (Human - Monkey) is the minimum. So we'll join Human and Monkey to MonHum and we'll calculate the new distances. Mon-Hum Mosquito Spinach Rice Human Monkey

27. 2. Calculation of New Distances After we have joined two species in a subtree we have to compute the distances from every other node to the new subtree. We do this with a simple average of distances: Dist[Spinach, MonHum] = (Dist[Spinach, Monkey] + Dist[Spinach, Human])/2 = (90.8 + 86.3)/2 = 88.55 Mon-Hum Spinach Human Monkey

28. 3. Next Cycle Mos-(Mon-Hum) Mon-Hum Rice Spinach Mosquito Human Monkey

29. 4. Penultimate Cycle Mos-(Mon-Hum) Spin-Rice Mon-Hum Rice Spinach Mosquito Human Monkey

30. 5. Last Joining (Spin-Rice)-(Mos-(Mon-Hum)) Mos-(Mon-Hum) Spin-Rice Mon-Hum Rice Spinach Mosquito Human Monkey

31. The result:Unrooted Neighbor-Joining Tree Human Spinach Monkey Mosquito Rice

33. Bootstrapping

34. Why are trees not exact?

35. Pairwise distances usually not tree-like

37. Searching tree space

38. Maximum likelihood criterion

39. Parsimony criterion

40. Parsimony with molecular data

41. Parsimony criterion Paul Higgs:

43. Is the best tree much better than others? L: likelihood at nodes

44. Use Maximum Likelihood to rank alternate trees NJ tree is 2nd best same topology yes yes

45. Use Parsimony to rank alternate trees different topology ; parsimony differentiates weakly

46. Quartet puzzling

48. MCMC: Markov chain with Monte Carlo

49. Topology probabilities according to MCMC

50. Clade probability compared from tree methods NJ method is very fast and close to being the best