short-term selection response: breeder Õs...
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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)*
2520151050
0.00
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
Sel
ecti
on
dif
fere
nti
al
S
q,
Fre
qu
ency
of
cha
ract
erS q
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
109876543210
-0.15
-0.10
-0.05
0.00
0.05
0.10
Generation
Cu
mu
lati
ve
Res
po
nse
to
Sel
ecti
on
(in
ter
ms
of
S)
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)
20151050
-1.0
-0.5
0.0
0.5
1.0
1.5
Generation
Cu
mu
lati
ve
Res
po
nse
(in
un
its
of
S)
m = -0.25
m = -0.5
m = -0.75
h2 = 0.35
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.
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