short-term selection response

Short-Term Selection Response

R = h2 S

Response to Selection• Selection can change the distribution of

phenotypes, and we typically measure this by changes in mean– This is a within-generation change

• Selection can also change the distribution of breeding values– This is the response to selection, the change

in the trait in the next generation (the between-generation change)

The Selection Differential and the Response to Selection

• The selection differential S measures the within-generation change in the mean– S = * -

• The response R is the between-generation change in the mean– R(t) = (t+1) - (t)

S

o

*p

R

(A) Parental Generation

(B) Offspring Generation

Truncation selection uppermost fraction

p chosen

Within-generationchange

Between-generationchange

The Breeders’ Equation: Translating S into R

∂µyO = πP + h2 P f + Pm

2 πP

Recall the regression of offspring value on midparent value

Averaging over the selected midparents, E[ (Pf + Pm)/2 ] = *,

E[ yo - ] = h2 ( * - ) = h2 SLikewise, averaging over the regression gives

Since E[ yo - ] is the change in the offspring mean, it represents the response to selection, giving:

R = h2 S The Breeders’ Equation (Jay Lush)

• Note that no matter how strong S, if h2 is small, the response is small

• S is a measure of selection, R the actual response. One can get lots of selection but no response

• If offspring are asexual clones of their parents, the breeders’ equation becomes – R = H2 S

• If males and females subjected to differing amounts of selection,– S = (Sf + Sm)/2– An Example: Selection on seed number in plants -- pollination

(males) is random, so that S = Sf/2

Response over multiple generations• Strictly speaking, the breeders’ equation only holds for

predicting a single generation of response from an unselected base population

• Practically speaking, the breeders’ equation is usually pretty good for 5-10 generations

• The validity for an initial h2 predicting response over several generations depends on:– The reliability of the initial h2 estimate– Absence of environmental change between

generations– The absence of genetic change between the

generation in which h2 was estimated and the generation in which selection is applied

(A)

S S

(B)

(C)

S

50% selectedVp = 4, S = 1.6

20% selectedVp = 4, S = 2.8

20% selectedVp = 1, S = 1.4

The selection differential is a function of boththe phenotypic variance and the fraction selected

The Selection Intensity, iAs the previous example shows, populations with thesame selection differential (S) may experience verydifferent amounts of selection

The selection intensity i provides a suitable measurefor comparisons between populations,

i = SpVP= S

æp

Selection Differential UnderTruncation Selection

R code for i: dnorm(qnorm(1-p))/p

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S =*-

Selection Intensity Versions of the Breeders’ Equation

R = h2S = h2 Sæp

æp = i h2æp

We can also write this as Since h2sP = (s2A/s2

P) sP = sA(sA/sP) h sA

R = i h sA Since h = correlation between phenotypic and breedingvalues, h = rPA

R = i rPAsA

Response = Intensity * Accuracy * spread in Va When we select an individual solely on their phenotype,the accuracy (correlation) between BV and phenotype is h

Accuracy of selectionMore generally, we can express the breedersequation as

R = i ruAsA

Where we select individuals based on the index u (for example, the mean of n of their sibs).

ruA = the accuracy of using the measure u topredict an individual's breeding value = correlation between u and an individual's BV, A

Estimating an individual’s BV withprogeny testing. For n progeny, ruA

2 = n/(n+a), where a = (4-h2)/h2

Class problem: show this

Overlapping Generations

Ry =im + ifLm + Lf

h2sp

Lx = Generation interval for sex x = Average age of parents when progeny are born

The yearly rate of response is

Trade-offs: Generation interval vs. selection intensity:If younger animals are used (decreasing L), i is also lower,as more of the newborn animals are needed as replacements

Computing generation intervalsOFFSPRING Year 2 Year 3 Year 4 Year 5 total

Number (sires)

60 30 0 0 90

Number (dams)

400 600 100 40 1140

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Generalized Breeder’s Equation

Ry =im + ifLm + Lf

ruAsA

Tradeoff between generation length L and accuracy r

The longer we wait to replace an individual, the moreaccurate the selection (i.e., we have time for progenytesting and using the values of its relatives)

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Selection on Threshold Traits

Assume some underlying continuous value z, the liability, maps to a discrete trait.

z < T character state zero (i.e. no disease)

z > T character state one (i.e. disease)

Alternative (but essentially equivalent model) is aprobit (or logistic) model, when p(z) = Prob(state one | z)

Liability scaleMean liability before selection

Mean liability after selection(but before reproduction)

Selection differentialon liability scale

qt* - qt is the selection differential on the phenotypic scale

Mean liability in next generationFrequency of character state onin next generation

Frequency of trait

Steps in Predicting Response to Threshold Selection

i) Compute initial mean 0

We can choose a scale where the liabilityz has variance of one and a threshold T = 0

Hence, z - 0 is a unit normal random variableP(trait) = P(z > 0) = P(z - > -) = P(U > -)U is a unit normalDefine z[q] = P(U < z[q] ) = q. P(U > z[1-q] ) = q

For example, suppose 5% of the pop shows thetrait. P(U > 1.645) = 0.05, hence = -1.645

General result: = - z[1-q]

ii) The frequency qt+1 of the trait in the next generation is just

qt+1 = P(U > - t+1 ) = P(U > - [h2S + t ] ) = P(U > - h2S - z[1-q] )

- - t°tq°St = π πt = '(πt) pt ° qt1 q

* -

iii) Hence, we need to compute S, the selection differential on liability

Let pt = fraction of individuals chosen ingeneration t that display the trait

πt = (1 ° pt) E(zjz < 0;πt) +pt E(zjz ∏ 0;πt)- >

This fraction does not display the trait, hencez < 0

This fraction display the trait, hence z > 0When z is normally distributed, this reduces to

Height of the unit normal density functionAt the point tHence, we start at some initial value given h2 and

0, and iterative to obtain selection response

25201510500.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

2.25

0

10

20

30

40

50

60

70

80

90

100

Generation

Sele

ctio

n di

ffer

entia

l S

q, F

requ

ency

of c

hara

cterS q

Initial frequency of q = 0.05. Selection only on adultsshowing the trait

Class problem• Initial freq(trait) = 10%• h2 liability = 0.35• Select 25%

For 3 generations of selection,Compute q, p, R, S

f(z) -- height of normal curve at z = dnorm(z)Pr(U < z ) = p. p = pnorm(z) z = qnorm(p)z[1-q] = qnorm(1-q)

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Some help: 1st gen, 25% saved = 10% trait, 15% other,p = 10/25 = 0.4

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Finite sample correction for iWhen a finite number of individuals form the next generation, previous expressions overestimate i

Suppose M adults measured, of which N are selectedfor p = N/M

E[i] depends on the expected value of the order statistics

Rank the M phenotypes as z1,M > z2,M . .. > zM,M zk,M = kth order statistic in a sample of M

z0k;M = (zk;M ° π)=æStandardized order statistics

E( i ) = 1æ

√1N

NXk=1

E(zk;M ) ° π!

= 1N

NXk=1

E(z0k;M )(

1.1.01

p = N/M, the proportion selected

Exp

ecte

d Se

lect

ion

inte

nsity

M = 10

M = 20

M = 50

M = 100

M = infinity

Finite population size results in infinite M expression overestimating the actual selection intensity, although the difference is small unless N is very small.

Burrows' approximation (normally-distributed trait)

E( i(M;N ) ) ' i °∑ 1 ° p

2p(M +1)∏ 1

i

Bulmer’s approximation:uses infinite population result, with p replaced by

ep = N +1=2M +N=(2M)