Change Stasis

Adaptive Directional selection

• Stabilizing selection• Fluctuating selection (noise with no trend)

Non-adaptive

MutationGenetic 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

Blomberg’s K – measure of phylogenetic signal

Blomberg et al. 2003 Evolutionexamples from Ackerly 2009 PNAS

K = 0.18 K ~ 1 K = 1.62

low brownian high

phylogenetic signal

Data diagnostics

K > 1

Brownian motion – assumptions and interpretations

Evolutionary models

Brownian motion – assumptions and interpretations

Evolutionary models

∞

-∞

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 EvolutionButler and King 2004 Amer.

Naturalist

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 EvolutionButler and King 2004 Amer.

Naturalist

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 EvolutionButler and King 2004 Amer.

Naturalist

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

Harmon et al. 2010

Harmon et al. 2010

Assign proportional weighting of alternative models that best fit

data

Rates of phenotypic diversification under Brownian motion

time

var(x)

1 felsen = 1 Var(loge(trait))

million yrs

Rates of phenotypic diversification under Brownian motion

time

var(x)

higher rate lower rate

Diversification of height in maples, Ceanothus and silverswords

~30 Ma

~45 Ma

rate = 0.015 felsens 0.10 felsens 0.83 felsens

Ackerly 2009 PNAS

~5.2 Ma

Evolutionary rates

Rates of phenotypic diversification (estimated for Brownian motion model)

Rate

(fe

lsens) Leaf sizeHeight

Ace

rAes

culu

sArb

utoi

deae

Cea

noth

uslo

belio

ids

silv

ersw

ords

North temperateCaliforniaHawai’i

Ace

rAes

culu

sArb

utoi

deae

Cea

noth

uslo

belio

ids

silv

ersw

ords

±1 s.e.

Ackerly, PNAS in review

time

var(x)

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

ML with BL as shown

Node ML estimate

Lower 95% CI

Upper 95% CI

A 0.56 -0.77 1.89

B 0.67 -0.43 1.78

C 0.096 -0.61 0.81

D 0.96 0 1.95

E 1.6 0.76 2.45

F 2.54 1.86 3.2

A

B

C

D

E

F

Oakley and Cunningham 2000

Polly 2001 Am Nat

Independent contrasts

28

616

16

1114

8.59

11.511

1918

1312a

b

R = 0.74

511

1317

914

412

610

1010

1615

28

616

16

1114

8.59

11.511

1918

1312

48

108

32

-6-6

65

2-2

86

a

b c

R = 0.74 R = 0.92

Oakley and Cunningham 2000

Oakley and Cunningham 2000

A21223Fig. 2

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 = 2519; N > 0.423 = 2551

Type I error = 0.051

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 = 5N > 0.423 = 68858

Power = 0.69

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 = 1050N > 0.423 = 1044

Type I error = 0.21

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 = 246N > 0.423 = 236

Type I error = 0.048

From Ackerly, 2000, Evolution