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Mark and Recapture Distribution of Fitness Effects

Topic 13Method of Moments

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Outline

Mark and Recapture

Distribution of Fitness Effects

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Mark and Recapture

The Lincoln-Peterson method of mark and recapture

• The size of an animal population in a habitat of interest is an important questionin conservation biology.

• In many case, Individuals are often too difficult to find, a census is not feasible.

• One estimation technique is to capture some of the animals, mark them andrelease them back into the wild to mix randomly with the population.

• Some time later, a second capture from the population is made.

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Mark and Recapture

Some of the animals were not in the first capture and some, which are tagged, arerecaptured. Let

• t be the number captured and tagged,

• k be the number in the second capture,

• r be the number in the second capture that are tagged, and let

• N be the total population size.

Thus, t and k is under the control of the experimenter. The value of r is random andthe populations size N is the parameter to be estimated.

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Mark and Recapture

We can guess the the estimate of N by considering two proportions.

the proportion of the tagged fish ≈ the proportion of tagged fish

in the second capture in the population

r

k≈ t

N

This can be solved for N to find

N ≈ kt

r.

The advantage of obtaining this as a method of moments estimator is that we evaluatethe precision of this estimator by determining, for example, its variance.

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Mark and RecaptureTo begin, let

Xi =

{1 if the i-th individual in the second capture has a tag.0 if the i-th individual in the second capture does not have a tag.

The Xi are Bernoulli random variables with success probability P{Xi = 1} = t/N. Thenumber of tagged individuals is X = X1 + X2 + · · ·+ Xk and the expected number oftagged individuals is

µ = EX = EX1 + EX2 + · · ·+ EXk =t

N+

t

N+ · · ·+ t

N=

kt

N.

The proportion of tagged individuals, X̄ = (X1 + · · ·+ Xk)/k, has expected value

EX̄ =µ

k=

t

N. Thus, N =

kt

µ.

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Mark and RecaptureNow in this case, we are estimating µ, the mean number recaptured with r , the actualnumber recaptured. So, to obtain the estimate N̂. we replace µ with the previousequation by r .

N̂ =kt

r.

We simulate the process in a lake having 4500 fish.

> N<-4500;t<-400;k<-500 #population 4500, 400 tagged, recapture 500

> r<-rep(0,2000) #set a vector of zeros for 2000 simulations

> fish<-c(rep(1,t),rep(0,N-t)) #tag t fish

> for (j in 1:2000){r[j]<-sum(sample(fish,k))}

> Nhat<-k*t/r #compute estimate of population

> mean(Nhat);sd(Nhat)

[1] 4606.933

[1] 666.1918

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Mark and Recapture

Histogram of r

r

Frequency

20 30 40 50 60

0100200300400500600

Histogram of Nhat

Nhat

Frequency

3000 5000 7000

0100200300400500600

Exercise. Describe the histograms above. Comment of the mean and standarddeviation of the estimate N̂.

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• Fitness is a central concept in the theory of evolution.

• Relative fitness is quantified as the average number of surviving progeny of aparticular genotype compared with average number of surviving progeny ofcompeting genotypes after a single generation.

• Consequently, the distribution of fitness effects for newly arising mutations is abasic question in evolution.

Even though more fit genes are more likely to fix in a populations, humans have, justby chance, genes that are deleterious. We will consider, in humans, the distribution offitness effects of such mutations

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Our model for this distribution will be the gamma family of random variables.

A Γ(α, β) random variable has mean α/β and variance α/β2. Because we have twoparameters, the method of moments methodology requires us, in step 1, to determinethe first two moments.

µ1 = E(α,β)X1 =α

β

µ2 = E(α,β)X21 = Var(α,β)(X1) + (E(α,β)X1)2

=α

β2+

(α

β

)2

=α

β2+α2

β2=α(1 + α)

β2.

NB. Var(Y ) = EY 2 − (EY )2. So, EY 2 = Var(Y ) + (EY )2.

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Distribution of Fitness EffectsThe first two moments

µ1 =α

βand µ2 =

α

β2+α2

β2=α(1 + α)

β2

For step 2, we solve for α and β. Note that

µ2 − µ21 =

α

β2,

µ1

µ2 − µ21

=α/β

α/β2= β,

and

µ1 ·µ1

µ2 − µ21

=α

β· β = α, or α =

µ21

µ2 − µ21

.

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β =µ1

µ2 − µ21

and α = βµ1 =µ2

1

µ2 − µ21

.

For step 3, set the first two sample moments

x̄ =1

n

n∑i=1

xi and x2 =1

n

n∑i=1

x2i

to obtain estimates

β̂ =x̄

x2 − (x̄)2and α̂ = β̂x̄ =

(x̄)2

x2 − (x̄)2

as required in step 4.

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Distribution of Fitness EffectsIn a recent manuscript based on the exomes of 44 unrelated Yoruban females, wefound estimates of α̂ = 0.19 and β̂ = 5.18. To investigate the method of momentsestimates on simulated data using R, we consider 2000 repetitions of 500 independentobservations of a Γ(0.19, 5.18) random variable.

> xbar <- rep(0,2000); x2bar <- rep(0,2000)

> for (i in 1:2000){x<-rgamma(500,0.19,5.18);

xbar[i]<-mean(x);x2bar[i]<-mean(x^2)}

> betahat <- xbar/(x2bar-(xbar)^2)

> alphahat <- betahat*xbar

> mean(alphahat);sd(alphahat)

[1] 0.1979031

[1] 0.02837945

> mean(betahat);sd(betahat)

[1] 5.472508

[1] 0.97692740.0 0.2 0.4 0.6 0.8 1.0

02

46

810

x

dgam

ma(

x, 0

.19,

5.1

8)

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Histogram of alphahat

alphahat

Frequency

0.10 0.15 0.20 0.25 0.30

0100

200

300

400

500

Histogram of betahat

betahat

Frequency

3 4 5 6 7 8 9

0100

200

300

400

Exercise. Describe the histograms above. Comment of the mean and standarddeviation of the estimates α̂ and β̂. The correlation r(α̂, β̂) = 0.810 for thissimulation. What does this tells us about the estimates? Repeat the simulation andgive a plot of α̂ versus β̂.

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