Short-Term SelectionResponse: Breeder’s

equationBruce Walsh lecture notes

Uppsala EQG courseversion 31 Jan 2012

Response to Selection

• Selection can change the distribution ofphenotypes, and we typically measure thisby changes in mean– This is a within-generation change

• Selection can also change the distributionof breeding values– This is the response to selection, the change in

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

The Selection Differential andthe Response to Selection

• The selection differential S measures thewithin-generation change in the mean

– S = µ* - µ

• The response R is the between-generationchange in the mean

– R(t) = µ(t+1) - µ(t)

S

µo

µ*µp

R

(A) Parental Generation

(B) Offspring Generation

Truncation selection uppermost fraction

p chosen

Within-generationchange

Between-generationchange

The Breeders’ Equation: Translating S into R

!"yO = µP + h2 Pf + Pm

2µP

Recall the regression of offspring value on midparent value

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

E[ yo - µ ] = h2 ( µ! - µ ) = h2 S

Likewise, averaging over the regression gives

Since E[ yo - µ ] 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 issmall, the response is small

• S is a measure of selection, R the actualresponse. One can get lots of selection butno response

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

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

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

Price-Robertson indentity

• S = cov(w,Z)

• The covariance between relativefitness (w = W/Wbar), scaled to havemean fitness = 1

• VERY! Useful result

Consider 5 individuals, zi = trait valueni= number of offspring

Unweighted S = 7, offspring-weighted S= 4.69

Response over multiple generations

• Strictly speaking, the breeders’ equation only holds forpredicting a single generation of response from anunselected base population

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

• The validity for an initial h2 predicting response over severalgenerations depends on:

– The reliability of the initial h2 estimate

– Absence of environmental change between generations

– The absence of genetic change between the generation inwhich h2 was estimated and the generation in whichselection 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, i

As 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 =S!VP

=S!p

Selection Differential UnderTruncation Selection

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

Likewise,

S =µ*- µ

Selection Intensity Versions of the Breeders’Equation

R = h2S = h2 S!p

!p = i h2 !p

Since h2"P = ("2A/"2

P) "P = "A("A/"P) = h "A

R = i h "A

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

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 ruA"A

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

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

Overlapping Generations

Ry =im + if

Lm + Lf

h2"p

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 intervals

114040100600400Number(dams)

90003060Number(sires)

totalYear 5Year 4Year 3Year 2OFFSPRING

Generalized Breeder’s Equation

Ry =im + if

Lm + Lf

ruA"A

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)

Selection on ThresholdTraits

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)

Frequency of character state onin next generation

Frequency of trait

Observe: trait valuesare either 0,1. Popmean = q (frequencyof the 1 trait)

Want to map fromq unto the underlyingliability scale, whereBreeder’s equationRz = h2Sz holds

Liability scale Mean liability before selection

Selection differentialon liability scale

Mean liability in next generation

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

Mean liability in next generation

Steps in Predicting Response to Threshold Selection

i) Compute initial mean µ0

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

Hence, z - µ0 is a unit normal random variable

P(trait) = P(z > 0) = P(z - µ > -µ) = P(U > -µ)

U is a unit normal

Define z[q] = P(U < z[q] ) = q. P(U > z[1-q] ) = q

For example, suppose 5% of the pop shows the trait. P(U > 1.645) =

0.05, hence µ = -1.645. Note: in R, z[1-q] = qnorm(1-q), with

qnorm(0.95) returning 1.644854

General result: µ = - z[1-q]

Steps in Predicting Response to Threshold Selection

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

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

µt = (1" pt) E(z|z < 0;µt) + pt E(z|z # 0;µt)*

-- ttq

St = µ µt ="(µt) pt" qt

1 q*

This fraction does not display the trait, hence z < 0

When z is normally distributed, this reduces to

Height of the unit normal density functionat the point µt

Hence, we start at some initial value given h2 andµ0, and iterative to obtain selection response

This fraction displays the trait, hence z > 0

µt = (1" pt) E(z|z < 0;µt) + pt E(z|z # 0;µt)*

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

Permanent Versus TransientResponse

Considering epistasis and shared environmental values,the single-generation response follows from the midparent-offspring regression

R = h2 S +S!2

z

"!2

AA

2+

!2AAA

4+ · · · + !(Esire,Eo) + !(Edam,Eo)

!

Breeder’s Equation

Response from shared environmental effects

Permanent component of response

Transient component of response --- contributesto short-term response. Decays away to zero

over the long-term

Response from epistasis

Permanent Versus TransientResponse

The reason for the focus on h2S is that thiscomponent is permanent in a random-matingpopulation, while the other components aretransient, initially contributing to response, butthis contribution decays away under random mating

Why? Under HW, changes in allele frequenciesare permanent (don’t decay under random-mating),while LD (epistasis) does, and environmentalvalues also become randomized

Response with Epistasis

R = S

"h2 +

!2AA

2!2z

!

R(1 + #) = S

"h2 +(1 c)! !2

AA

2!2z

!

The response after one generation of selection froman unselected base population with A x A epistasis is

The contribution to response from this single generationafter # generations of no selection is

c is the average (pairwise) recombination between lociinvolved in A x A

Response with Epistasis

Contribution to response from epistasis decays to zero aslinkage disequilibrium decays to zero

Response from additive effects (h2 S) is due to changes in allele frequencies and hence is permanent. Contribution from A x A due to linkage disequilibrium

R(1 + #) = S

"h2 +(1 c)! !2

AA

2!2z

!

R(t + # ) = t h2 S + (1 " c)! RAA(t)

Why unselected base population? If history of previousselection, linkage disequilibrium may be present andthe mean can change as the disequilibrium decays

More generally, for t generation of selection followed by# generations of no selection (but recombination)

RAA has a limitingvalue given by

Time to equilibrium afunction of c

What about response with higher-order epistasis?

Fixed incremental differencethat decays when selection

stops

Maternal Effects:Falconer’s dilution model

z = G + m zdam + e

G = Direct genetic effect on characterG = A + D + I. E[A] = (Asire + Adam)/2

maternal effect passed from dam to offspring m zdam is

just a fraction m of the dam’s phenotypic value

m can be negative --- results in the potential for a reversed response

The presence of the maternal effects means that responseis not necessarily linear and time lags can occur in response

Parent-offspring regression under the dilution model

In terms of parental breeding values,

E(zo | Adam, Asire, zdam) =Adam

2+

Asire

2+ mzdam

Regression of BV on phenotype

A = µA + bAz ( z " µz ) + e

With no maternal effects, baz = h2

The resulting slope becomes bAz = h2 2/(2-m)

Parent-offspring regression under the dilution model

!A,M = m !2A / ( 2"m)

With maternal effects, a covariance between BV and maternal effect arises, with

∆µz = Sdam

"h2

2"m+m

!+ Ssire

h2

2"m

The response thus becomes

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Response to a single generation of selection

Reversed response in 1st generation largely due to negative maternal correlation masking genetic gain

Recovery of genetic response after initial maternal correlation decays

h2 = 0.11, m = -0.13 (litter size in mice)

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Selection occurs for 10 generations and then stops

Ancestral RegressionsWhen regressions on relatives are linear, we can think of theresponse as the sum over all previous contributions

For example, consider the response after 3 gens:

8 great-grand parentsS0 is there selectiondifferential$3,0 is the regressioncoefficient for an offspring at time 3on a great-grandparentFrom time 0

4 grandparentsSelection diff S1

$3,1 is the regression

of relative in generation3 on their gen 1 relatives

2 parents

Ancestral Regressions

$T,t = cov(zT,zt)

More generally,

The general expression cov(zT,zt), where we keep track of theactual generation, as oppose to cov(z, zT-t ) -- how many generationsSeparate the relatives, allows us to handle inbreeding, where the(say) P-O regression slope changes over generations of inbreeding.