1. Lecture 15: Reconstruction of Phylogeny Adaptive characters: 1.May indicate derived character (special adaptation) e.g. Raptorial forelegs in mantids 2.May indicate convergent evolution e.g cacti & euforbs

2. Adaptive Characters Must ask: Are characters independent? What constitutes a single character? e.g. paedomorphosis in salamanders many traits → monophyletic single trait → polyphyletic Different characters give different trees!

3. Unrooted Trees no common ancestor no time implied similarity only compatible with 2(s) – 3 rooted trees A BD C A B C D B A C DD C B AC D B A A B C D

4. Topologies total # of topologies: (2n – 3)! / (2 n - 2 (n - 2)!) for 5 spp: 105 topologies for 15 spp: 213,458,046,676,875 topologies!!

5. Rooted Trees Show common ancestor Imply time Time Common Ancestor

6. Making a Rooted Tree Method 1: (morphological characters) 1) Choose independent characters 2) eliminate analogies 3) differentiate ancestral from derived homologies 4) parsimony Eliminate uninformative characters → compatible with all trees

7. Example: Sunfish (Centrarchidae) black crappie warmouth bluegill redear outgroup: largemouth bass

8. 1) Identify independent characters & transitions Transition: aDoubling of pyloric caeca bReduced number of trunk scales cLoss of 2-3 vertebral centra dLoss of 1-2 pectoral fin rays eLoss of subocular bar fEnlarged levator muscle gAddition of dorsal fin spot

9. 2) Construct character matrix: characters f & g are not shared (not informative) abcdefg Bass (outgroup) 0000000 Crappie1100000 Warmouth1111000 Bluegill1111101 Redear1111110

10. Most Parsimonious Tree: bass crappie warmouth bluegill redear a b c d e f g abcdefg Bass (outgroup)0000000 Crappie1100000 Warmouth1111000 Bluegill1111101 Redear1111110

11. Method 2 1) parsimony → unrooted tree 2) determine polarities → root tree (outgroup analysis) crappie warmouth bluegill redear outgroup analysis places root here

12. Using Molecular Sequences for Phylogenies molecular sequences = “characters” Direct: 1) nucleotide sequence of homologous genes 2) a.a. sequence of polypeptide

13. Indirect Molecular Analysis 1)DNA hybridization stability ≈ % similarity 2) allozyme frequency ≈ “genetic distance” 3) immunological distance : affinity of antibody to protein of reference sp.

14. Parsimony Analyses Similar to morphological: Goal: minimum # of changes in sequence can estimate branch lengths (time…)

15. More parsimony Not all substitutions are equal: Transitions: pur→ pur; pyr → pyr i.e. A ↔ G; C ↔ T Transversions: pur ↔ pyr i.e. A, G ↔ C, T Transitions >> Transversions (DNA repair mechanisms) Weight analysis (many options for this) ACGT A0.60.10.20.1 C-0.60.10.2 G--0.60.1 T---0.6 Kimura 2-parameter matrix:

16. Maximum Likelihood Finding the tree mostly likely (highest probability) to give the character set: A C G A G 0.1 0.2 0.6 0.2 probability: 0.25 X 0.1 X 0.2 X 0.2 X 0.6 = 0.0006 (0.25) Assume: each site evolves independently

17. Neutral Theory of Molecular Evolution Most nucleotide substitutions & polymorphisms are result of selectively neutral mutations (silent substitutions) Selection doesn’t have to be neutral But, drift must be more important Controversy resulted but…

18. Molecular Clock Prediction from neutral theory single gene evolves at constant rate Even if neutral theory doesn’t hold: selection coefficient averaged over time → linear rate Neutral theory important null model

19. Tests of Molecular Clocks: 1) Plot molecular divergence against fossil record 2) Relative rate test: independent of fossil record - outgroup comparison: all in group equal distance to outgroup C to E = D to E C to E : c + b + e D to E : d + b + e if rate of substitution constant c = d C D E B A d e b c

20. Substitution Rates Substitution rates may vary if: selection coefficients vary diff’ns in pop’n size diff’ns in generation time diff’ns in exposure to mutagens diff’ns in metabolic rate etc.

21. Sometimes molecular & morphological data give different results! e.g. Chachalaca Neotropical bird Morphology & behaviour : like a chicken 4 proteins : like a duck??