Week 8: Paired sites tests, gene frequencies, continuous...

Week 8: Paired sites tests, gene frequencies, continuouscharacters

Genome 570

March, 2012

Week 8: Paired sites tests, gene frequencies, continuous characters – p.1/47

An example – two trees

MouseBovine

GibbonOrang

GorillaChimp

Human

MouseBovine

GibbonOrangGorilla

ChimpHuman

Tree I

Tree II


The differences of log likelihoods

site1 2 3 4 5 6 ln L

Tree

I

II

231 232

−1405.61

−1408.80 ...

Diff ... +3.19

−2.971 −4.483 −5.673 −5.883 −2.691 ...−8.003 −2.971 −2.691

−2.983 −4.494 −5.685 −5.898 −2.700 −7.572 −2.987 −2.705

+0.012 +0.013 +0.010 −0.431+0.015+0.111 +0.012 +0.010


The histogram of differences of log-likelihoods

−0.50 0.0 0.50 1.0 1.5 2.0

Difference in log likelihood at site


Paired sites testsWinning sites test (Prager and Wilson, 1988). Do a sign test on thesigns of the differences.

z test (me, 1993 in PHYLIP documentation). Assume differencesare normal, do z test of whether mean (hence sum) difference issignificant.

t test. Swofford et. al., 1996: do a t test (paired)

Wilcoxon ranked sums test (Templeton, 1983).

RELL test (Kishino and Hasegawa, 1989 per my suggestion).Bootstrap resample sites, get distribution of difference of totals.


In our example ...

Winning sites test. 160 of 232 sites favor tree I. P < 3.279 × 10−9

z test. Difference of log-likeihood totals is 0.948104 standarddeviations from 0, P = 0.343077. Not significant.

t test. Same as z test for this large a number of sites.

Wilcoxon ranked sums test. Rank sum is 4.82805 standarddeviations below its expected value, P = 0.000001378765

RELL test. 8,326 out of 10,000 samples have a positive sum,P = 0.3348 (two-sided)


The Shimodaira-Hasegawa test

Starts with a set of user-specified trees

Gets the sitewise log-likelihoods

adjusts each trees’ log-likelihoods to add up to same value

then resamples columns (sites) from these

asks how often a tree will get more than X worse then the best

for each X (the log-likelihood difference between one tree and thebest one) can get a P value.


An outcome of Brownian motion on a 5-species tree


Brownian motion along a tree

x

x

x2

x3

x4

x5

x6x7

x8

x9

x

x1

x0

x x8

v1v2

v3

v8

v9

v4

v6 v7

v10

v5

v11

v12

x x82

x1− x8x x3 9

x x12

x x11

x x6 10 x x7 10

x x10 11

x x11 12

x x12 0

x x8 9

x x0

1 −

−

−

−

4 −

9 −

5 −

−

−

−

−

−

10

11

12


Distribution of tips on a tree under Brownian Motion

‘

3

1

20root v3

v

v

1

2

Tip 1 is the sum of two independent changes each of which is drawnfrom a normal distribution (with mean 0 and variances v3 and v1)so it is normally distributed with mean 0 and variance v3 + v1.



‘

3

1

20root v3

v

v

1

2


Similarly for tip 2 (variance is v3 + v2).



‘

3

1

20root v3

v

v

1

2



They share branch 3, and the change there affects both randomvariables. So they are not independent or uncorrelated.



‘

3

1

20root v3

v

v

1

2




Variance is the expectation of the square (of deviation from themean), and covariance is the expectation of the product of thosedeviations, for the two variables.



‘

3

1

20root v3

v

v

1

2




Variance is the expectation of the square (of deviation from themean), and covariance is the expectation of the product of thosedeviations, for the two variables.

In fact the covariance of the values at tip 1 and tip 2 is the varianceof the shared term that is the same in both of them, so it is v3.


Covariances of species on the tree

2

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

4

v1 + v8 + v9 v8 + v9 v9 0 0 0 0

v8 + v9 v2 + v8 + v9 v9 0 0 0 0

v9 v9 v3 + v9 0 0 0 0

0 0 0 v4 + v12 v12 v12 v12

0 0 0 v12 v5 + v11 + v12 v11 + v12 v11 + v12

0 0 0 v12 v11 + v12 v6 + v10 + v11 + v12 v10 + v11 + v12

0 0 0 v12 v11 + v12 v10 + v11 + v12 v7 + v10 + v11 + v12

3

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

5


Covariances are of form

a b c 0 0 0 0

b d c 0 0 0 0

c c e 0 0 0 0

0 0 0 f g g g

0 0 0 g h i i

0 0 0 g i j k

0 0 0 g i k l


Likelihood under Brownian motion with two species

f(x; µ, σ2

)=

1

σ√

2πexp

(−

(x − µ)2

2σ2

)

L =

p∏

i=1

1

(2π)√

v1v2

exp

(

−1

2

[(x1i − x0i)

2

v1

+(x2i − x0i)

2

v2

])


Minimizing for each character i

Q =(x1i − x0i)

2

v1

+(x2i − x0i)

2

v2

so:dQ

dx0i

= −2(x1i − x0i)

v1

− 2(x2i − x0i)

v2

= 0

and then:

x0i =1v1

x1i + 1v2

x2i

1v1

+ 1v2

So that we have a maximum likelihood estimate of the starting value x0i foreach character.

The result is that

Q =(x1i − x2i)

2

v1 + v2


Likelihood after estimating initial coordinates

Substituting in our estimates of x0i, we end up with

L =1

(2π)p (v1v2)12p

exp

(−

1

2

p∑

i=1

(x1i − x2i)2

v1 + v2

)

and this finally turns into:

ln L = −p ln(2π) −1

2p ln (v1v2) −

1

2

p∑

i=1

(x1i − x2i)2

v1 + v2

This actually goes to infinity as either v1 or v2 goes to zero! This is relatedto the problem that Edwards and Cavalli-Sforza had with their maximumlikelihood method in 1964.


If there is a clock ...If instead we constrain v1 = v2 because assume a clock:

ln L = K′ − p ln(v1 + v2) −1

2

D2

(v1 + v2)

which leads tov1 = v2 = D2/(4p)

(which is half as big as it should be!)

The number of parameters being estimated is p + 1, which rises as weconsider more characters. The fact that the ratio of data to parametersdoes not rise without limit is the reason why likelihood misbehaves in thiscase.


The difference between ML and REML

Information we use for ML inference:

1.0 2.0 3.0 4.0

species 1species 2 species 3species 4

species 1species 2 species 3species 4

Information we use for REML inference:

1.0+x 2.0+x 3.0+x 4.0+x

Does it matter that we don’t know x ? It makes it unnecessary to estimatethe starting value x0, and that eliminates p parameters. It means that theratio of data to parameters does then rise as we add characters.


Using only differences between populations (REML)

We assume that we have observed only the differences x1i − x2i, and notthe actual locations on the phenotype scale. Then

L =

p∏

i=1

1√

2π√

v1 + v2

exp

(−

1

2

(x1i − x2i)2

v1 + v2

)

ln L = K −p

2ln (v1 + v2) +

1

2 (v1 + v2)

n∑

i=1

(xi1 − xi2)2


Likelihood with two species using REML

ln L = K −p

2ln (v1 + v2) +

D2

2 (v1 + v2)

ln L = K −p

2ln (vT) +

D2

2 vT

vT = D2/p

The number of parameters being estimated is 1 (it is the sum v1 + v2).The number of parameters does not rise as we consider more characters.


“Pruning” a tree in the Brownian motion case

+v1 v2

v3 v4v5

v6

x1 x

2x 3

x 4

v1 v2

x1 x

2

v3 v4v5

v6

δ

x 3x 4

x12

v1 v2δ =v2v1

+

x1

x2x

12 v2v1+

=

v1v2+


What about quantitative characters?

For neutral mutation and genetic drift, can show that for a quantitativecharacter with additive genetic variance VA and population size N thegenetic (additive) value of the population mean is:

Var(∆g) = VA/N

If mutation and drift are at equilibrium:

E[V

(t+1)A

]= V

(t)A

(1 −

1

2N

)+ VM


In neutral traits additive genetic variance rules

so thatE [VA] = 2NVM

wherebyVar[∆g] = (2NVM) /N = 2VM

an analogue of Kimura’s result for neutral mutation.There is a precise analogue of this for multiple characters.

Thus to transform characters to independent Brownian motions of equalevolutionary variance, we could use the additive genetic variance VA.


With selection ... life is harder

There is the quantitative genetics formula of Wright and Fisher (1920’s)

∆z = h2S

and Russ Lande’s (1976) recasting of that in terms of slopes of meanfitness surfaces:

S = VP

d log (w)

dx

∆z = (VA/VP) VP

d log (w)

dx= VA

d log (w)

dx


Selection towards an optimum

P

Vs

Fit

nes

sPhenotype

If fitness as a function of phenotype is:

w(x) = exp

[−

(x − p)2

2Vs

]

Then the change of mean phenotype “chases” the optimum:

m′ − m =VA

Vs + VP

(p − m)


A character changing by “chasing” an adaptive peak

time

The course of change of the population mean is expected to be somewhatsmoother than the changes of the peak of the fitness surface.


Sources of evolutionary correlation among characters

Variation (and covariation) in change of characters occurs for two reasons:

1. Genetic drift, with the covariances being proportional to the additivegenetic covariances

2. Selection, with the covariances being affected by both the additivegenetic covariances and the covariation of the selection pressures.


A simple example of selective covariance

a simple example:

(temperate) (arctic) (arctic)(temperate) (temperate)

sizecolorlimblength

size

color

limblength

covariation due not to genetic correlationbut to covariation of the selection pressure

These are Bergmann’s, Allen’s and Glogler’s Rules

not They are presumably the result of genetic correlationsbut result from patterns of selection

Variation and evolutionin plants. Columbia Univ. Press, New York.page 121

G. L. Stebbins. 1950.


A simulated example with two characters

After 100 generations:

−30 −20 −10 0 10 20 30−30

−20

−10

0

10

20

30

Genetic covariances are negative, but the wanderings of the adaptivepeak in the two characters is positively correlated.



After 1000 generations:

−30 −20 −10 0 10 20 30−30

−20

−10

0

10

20

30




After 10,000 generations:

−30 −20 −10 0 10 20 30−30

−20

−10

0

10

20

30



Correcting for correlations among characters

Can we transform the set of characters to remove their correlations andthus end up with independent Brownian motions of equal variance?

We might hope to infer additive genetic covariances by doingquantitive genetics breeding experiments to infer them fromcovariances among relatives.

There is little or no hope of inferring “selective correlations” without acomplete understanding of the functional ecology.

If we are given the tree from molecular data (and are willing toassume that the branch lengths are proportional to those that applyto the morphological characters), we can hope to use the tree toinfer the covariation of the characters.


Correlation of states in a discrete-state model

#2

#1

#2

#1

species states branch changes

change incharacter 2

change incharacter 1

0 6

4 0

Y N

Y

N

1 0

0 18

character 1:

character 2:


A simple case to show effects of phylogeny


Two uncorrelated characters evolving on that tree


Identifying the two clades


A tree on which we are to observe two characters

0.3

0.1

0.25

0.65

0.1 0.1a

b

cd e

(0.7)

(0.2) 0.9


Decomposing it into two-species contrasts ...

0.25

0.65

0.3

0.1

a

b

c0.1 0.1

d e

(0.7)

(0.2) 0.9

(de)(ab)0.075

0.05

(abc)0.1666


Contrasts on that tree

Varianceproportional

Contrast to

y1 = xa − xb 0.4

y2 = 14

xa + 34

xb − xc 0.975

y3 = xd − xe 0.2

y4 = 16

xa + 12

xb + 13

xc − 12

xd − 12

xe 1.11666


Contrasts for the 20-species two-clade example

−3 −2 −1 0 1 2 3−3

−2

−1

0

1

2

3


An example: Smith and Cheverud 2002Smith, R. J. and J. M. Cheverud. 2002. Scaling of sexual dimorphism in body mass: Aphylogenetic analysis of Rensch’s Rule in primates. International Journal of Primatology 23(5):1095-1135.

Fig. 1. The interspecific allometric equation (specific regression, identified as IA) and theindependent contrasts equation (identified as IC) plotted for 105 primate species in raw dataspace, transformed to natural logarithms. The interspecific allometric equation islny = 0.139 + 0.080(lnx), with r = 0.53. The phylogenetically corrected form of thisequation, taken from the independent contrasts analysis, is lny = 0.160 + 0.056(lnx), withr = 0.26. The two equations are not significantly different from each other. The identifiedspecies are Mandrillus sphinx (M), Pongo pygmaeus (O), Gorilla gorilla (G), Pan troglodytes (P),and Homo sapiens (H).


A tree with punctuated equilibrium

Y

GA

FA

I

R

G

E

U

L

LA

V

N

KA

O

MA

CA

B

T

D

C

Z

X

JA

DA

BA

J

HA

A

K

F

M

P

OA

EA

IA

NA

W

H

Q

S


The punctuated tree when we sample 10 species

I

G

E

B

D

C

J

A

F

HWeek 8: Paired sites tests, gene frequencies, continuous characters – p.45/47

Two-species paired comparisons

AB CD EF G H


Pagel’s (1994) test for correlation with discrete 0/1 trait s

When character 1 has state Rates of change incharacter 2 are:

0 1α

β0 0

0

0 1α

β1 1

1

When character 2 has state Rates of change incharacter 1 are:

0 100

0

0 1

γ

δ

γ1

1

1δ


Pagel’s (1994) test for correlation with discrete 0/1 trait s

To : 00 01 10 11

From :

00 −− α0 γ0 0

01 β0 −− 0 γ1

10 δ0 0 −− α1

11 0 δ1 β1 −−

This can be set up as a 4 × 4 model of change with four states, 00, 01,10, and 11, and likelihood ratio tests used.

Complete independence of the changes in the two characters involvesrestricting the parameters so that α1 = α0, β1 = β0, γ1 = γ0, and δ1 = δ0.


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