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On board do traits fit B.M. model? can we use model fitting to answer evolutionary questions? pattern vs. process table.

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Presentation on theme: "On board do traits fit B.M. model? can we use model fitting to answer evolutionary questions? pattern vs. process table."— Presentation transcript:

1 on board do traits fit B.M. model? can we use model fitting to answer evolutionary questions? pattern vs. process table

2 on board draw random walk with speciation sketch reconstruction of what that would look like expectation under B.M. is difference between two taxa is function of phenetic distance

3 on board draw phenetic distance vs. pairwise trait distance plot and corresponding traitgrams introduce K statistic visualize K on distance plots draw number line for K illustrating ranges of values discuss tip swap for significance

4 3. Tempo and mode of evolution phylogeny and ages from Renner et al. 2008 Syst. Biol. ~3 Ma ~40 Ma StripedSugarMountainSilverRedAsh-leafDipteronia

5 ChangeStasis AdaptiveDirectional selection Stabilizing selection Fluctuating selection (noise with no trend) Non- adaptive Mutation Genetic drift Lack of genetic variation Constraint (?) Antagonistic correlations among traits under selection Swamping by gene flow Pattern: Evolutionary processes that can lead to change or stasis over time

6 Blomberg’s K – measure of phylogenetic signal Blomberg et al. 2003 Evolution examples from Ackerly 2009 PNAS K = 0.18K ~ 1K = 1.62 low brownianhigh phylogenetic signal Data diagnostics

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8 K > 1

9 Brownian motion – assumptions and interpretations Evolutionary models

10 Brownian motion – assumptions and interpretations Evolutionary models ∞ -∞

11 Ornstein-Uhlenbeck model (OU-1) Evolutionary models the math: brownian motion + ‘rubber band effect’ change is unbounded (in theory), but as rubber band gets stronger, bounds are established in practice repeated movement back towards center erases phylogenetic signal, leading to K << 1 see Hansen 1997 Evolution Butler and King 2004 Amer. Naturalist

12 Ornstein-Uhlenbeck model (OU-1) Evolutionary models the math: brownian motion + ‘rubber band effect’ change is unbounded (in theory), but as rubber band gets stronger, bounds are established in practice repeated movement back towards center erases phylogenetic signal, leading to K << 1 see Hansen 1997 Evolution Butler and King 2004 Amer. Naturalist

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14 Ornstein-Uhlenbeck model (OU-2+) Evolutionary models the math: brownian motion + ‘rubber band effect’ with different optimal trait values for clades in different selective regimes Balance of stabilizing selection within clades vs. how different the optima are can lead to strong or weak phylogenetic signal This example would be VERY strong signal see Hansen 1997 Evolution Butler and King 2004 Amer. Naturalist

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16 Early-burst model Evolutionary models the math: brownian motion with a declining rate parameter change is unbounded (in theory), but divergence happens rapidly at first and then rates decline and lineages change little divergence among major clades creates high signal: K >> 1

17 Harmon et al. 2010

18 Assign proportional weighting of alternative models that best fit data

19 Rates of phenotypic diversification under Brownian motion time var(x) 1 felsen = 1 Var(log e (trait)) million yrs

20 Rates of phenotypic diversification under Brownian motion time var(x) higher ratelower rate

21 Diversification of height in maples, Ceanothus and silverswords ~30 Ma ~45 Ma rate = 0.015 felsens0.10 felsens0.83 felsens Ackerly 2009 PNAS ~5.2 Ma Evolutionary rates

22 Rates of phenotypic diversification (estimated for Brownian motion model) Rate (felsens) Leaf sizeHeight Acer Aesculus Arbutoideae Ceanothus lobelioids silverswords North temperate California Hawai’i Acer Aesculus Arbutoideae Ceanothus lobelioids silverswords ±1 s.e. Ackerly, PNAS in review

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24 time var(x)

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26 0 0 0 0 1 2

27 0 0 0 0 1 2 0.12 0.24 0.08 0.52 1.32 2.44 0.56 0.67 0.096 0.96 1.6 2.54 Linear parsimony Squared change parsimony = ML with BL = 1 ML with BL as shown

28 NodeML estimate Lower 95% CI Upper 95% CI A0.56-0.771.89 B0.67-0.431.78 C0.096-0.610.81 D0.9601.95 E1.60.762.45 F2.541.863.2 A B C D E F

29 Oakley and Cunningham 2000

30 Polly 2001 Am Nat

31 Independent contrasts

32 2828 6 16 1616 11 14 8.5 9 11.5 11 19 18 13 12 a b R = 0.74

33 5 11 13 17 9 14 4 12 6 10 16 15 2828 6 16 1616 11 14 8.5 9 11.5 11 19 18 13 12 4848 10 8 3232 -6 6565 2 -2 8686 a bc R = 0.74R = 0.92

34 Oakley and Cunningham 2000

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36 A21223 Fig. 2

37 1) Assume bivariate normal distribution of variables with  = 0 2) Draw samples of 22 and calculate correlation coefficient 3) Repeat 100,000 times! Distribution of correlation coefficients (R) under null hypothesis Crit(R,  = 0.05, df = 20) is 0.423 N 0.423 = 2551 Type I error = 0.051

38 1) Assume bivariate normal distribution of variables with  = 0.5 2) Draw samples of 22 and calculate correlation coefficient 3) Repeat 100,000 times! Crit(R,  = 0.05, df = 20) is 0.423 N < -0.423 = 5 N > 0.423 = 68858 Power = 0.69

39 1) Assume bivariate normal brownian motion evolution along a phylogeny, with  ~ 0.0 2) Calculate R using normal correlation coefficient 3) Repeat 10,000 times! Crit(R,  = 0.05, df = 20) is 0.423 N < -0.423 = 1050 N > 0.423 = 1044 Type I error = 0.21

40 1) Assume bivariate normal brownian motion evolution along a phylogeny, with  ~ 0.0 2) Calculate R using independent contrasts 3) Repeat 10,000 times! Crit(R,  = 0.05, df = 20) is 0.423 N < -0.423 = 246 N > 0.423 = 236 Type I error = 0.048

41 From Ackerly, 2000, Evolution


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