Some of these slides have been borrowed from Dr. Paul Lewis, Dr. Joe Felsenstein. Thanks!

Transcription

1 Some of these slides have been borrowed from Dr. Paul Lewis, Dr. Joe Felsenstein. Thanks! Paul has many great tools for teaching phylogenetics at his web site:

2 Correlation of states in a discrete-state model character 1: character 2: species states change in character 1 branch changes #1 #2 change in character 2 #1 #2 Y N 0 6 Y N 0 18 Week 8: Paired sites tests, gene frequencies, continuous characters p.35/44

3 Example from pp of the MacClade 4 manual. Character 1 Q: Does character 2 tend to change from yellow blue only when character 1 is in the blue state? Character by Paul O. Lewis 2

4 Concentrated Changes Test More precisely: how often would three yellow blue changes occur in the blue areas of the cladogram on the left if these 3 changes were thrown down at random on branches of the tree? Answer: % of the time. Thus, we cannot reject the null hypothesis that the observed coincident changes in the two characters were simply the result of chance by Paul O. Lewis 3

5 Data for Two Characters, X and Y A B X Y Var(X) Var(Y) Cov(XY) Correlation = = = = C D The negative correlation is fairly strong, but would it weaken if it were recognized that there are not really 4 independent data points here by Paul O. Lewis 5

6 Y X

7 Fig. 6 from Felsenstein (1985) 2007 by Paul O. Lewis 3

8 Fig. 7 from Felsenstein (1985) 2007 by Paul O. Lewis 4

9 Fig. 5 from Felsenstein (1985) Felsenstein, J Phylogenies and the comparative method. American Naturalist 125: by Paul O. Lewis 2

10 An outcome of Brownian motion on a 5-species tree Week 8: Paired sites tests, gene frequencies, continuous characters p.11/44

11 An outcome of Brownian motion on a 5-species tree Week 8: Paired sites tests, gene frequencies, continuous characters p.12/44

12 An outcome of Brownian motion on a 5-species tree Week 8: Paired sites tests, gene frequencies, continuous characters p.13/44

13 An outcome of Brownian motion on a 5-species tree Week 8: Paired sites tests, gene frequencies, continuous characters p.14/44

14 Likelihood under Brownian motion with two species f ( x; µ, σ 2) = ) 1 ( σ 2π exp (x µ)2 2σ 2 L = p i=1 ( 1 (2π)σi 2 exp 1 v1 v 2 2σi 2 [ (x 1i x 0i ) 2 + (x 2i x 0i ) 2 v 1 v 2 ]) Week 8: Paired sites tests, gene frequencies, continuous characters p.15/44

15 Brownian motion along a tree x 1 x x x v x x v 2 x 8 x x 8 9 v 8 x 3 x x v 3 x 9 x x 9 0 v 9 x 6 x x 5 7 x x 6 10 x x v 7 10 x x x 6 v x 10 v x x v x 4 11 x x v x x x 12 0 v 12 x x v 10 x 0 Week 8: Paired sites tests, gene frequencies, continuous characters p.8/44

16 Covariances of species on the tree v 1 + v 8 + v 9 v 8 + v 9 v v 8 + v 9 v 2 + v 8 + v 9 v v 9 v 9 v 3 + v v 4 + v 12 v 12 v 12 v v 12 v 5 + v 11 + v 12 v 11 + v 12 v 11 + v v 12 v 11 + v 12 v 6 + v 10 + v 11 + v 12 v 10 + v 11 + v v 12 v 11 + v 12 v 10 + v 11 + v 12 v 7 + v 10 + v 11 + v 12 Week 8: Paired sites tests, gene frequencies, continuous characters p.9/44

17 Covariances are of form a b c b d c c c e f g g g g h i i g i j k g i k l Week 8: Paired sites tests, gene frequencies, continuous characters p.10/44

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19 Brownian Motion Model X i Model predicts that the uncertainty associated with the difference in trait values (i.e. D ij = X i -X j ) grows linearly with time. That is, Var(D ij ) = (t i +t j ) σ 2 So the longer you wait, the less you can t i predict about t j D ij X j X k 2007 by Paul O. Lewis 6

20 Brownian Motion Model X i X j t i t j Ancestral trait value estimated to be weighted average of descendants' values. Weights are inverses of branch lengths (i.e. times) X k = X i /t i + X j /t j 1/t i + 1/t j 2007 by Paul O. Lewis 7

21 Brownian Motion Model X i t i X k t j X j X k is not observed, and thus has some extra uncertainty associated with its value. This extra uncertainty can be modeled by adding an extra bit of length (t k ') to the branch subtending X k t k + t k ' The amount of extra uncertainty that should be added is: t k ' = t i t j / (t i + t j ) 2007 by Paul O. Lewis 8

22 Data for Two Characters on Tree X Y A B C D E F by Paul O. Lewis 9

23 Contrasts (left minus right) X Y by Paul O. Lewis 10

24 Scaling Contrasts A-B and C-D contrasts are expected to be on the same scale because the path length associated with both of these is 4 E-F has a path length of 7, which means this contrast is expected to be larger Adding in the extra uncertainty associated with estimating E and F, this path length expands from 7 to 9 Can put all 3 contrasts on same scale by dividing by standard deviation (square root of variance) 2007 by Paul O. Lewis 11

25 Rescaled Contrasts Asterisks indicate that these original trait values have been scaled by dividing by the square root of the path length. X Y Variance proportional to X* Y* A-B /2 = -3-2/2 = -1 C-D /2 = -2-2/2 = -1 E-F /3 = /3 = These path lengths are proportional to the variance, and thus they are all that are needed to place the 3 contrasts for a given trait on the same scale by Paul O. Lewis 12

26 Y contrast X contrast

27 Correlation of Contrasts A-B C-D X -3-2 Y -1-1 Var(X) Var(Y) Cov(XY) Correlation = = = = E-F The CONTRAST program in Joe Felsenstein's PHYLIP package performs independent contrasts Correlation of the raw X and Y trait values was , which is both stronger and of opposite sign. Note that the sample size is now 3 rather than by Paul O. Lewis 13