1. Lecture 8 Short-Term Selection Response
R = h2 S

2. Applications of Artificial Selection
Applications in agriculture and forestry
Creation of model systems of human diseases and disorders
Construction of genetically divergent lines for QTL mapping and gene expression (microarray) analysis
Inferences about numbers of loci, effects and frequencies
Evolutionary inferences: correlated characters, effects on fitness, long-term response, effect of mutations

3. 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)

4. The Selection Differential and the Response to Selection
The selection differential S measures the within-generation change in the mean
S = m* - m
The response R is the between-generation change in the mean
R(t) = m(t+1) - m(t)

5. Truncation selection
uppermost fraction
p chosen
Within-generation
change
Between-generation
change

6. The Breeders’ Equation: Translating S into R
Recall the regression of offspring value on midparent value ( )
Averaging over the selected midparents,
E[ (Pf + Pm)/2 ] = m*,
E[ yo - m ] = h2 ( m* - m ) = h2 S
Likewise, averaging over the regression gives
Since E[ yo - m ] is the change in the offspring mean, it
represents the response to selection, giving:
R = h2 S
The Breeders’ Equation

7. If males and females subjected to differing amounts of selection,
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

8. Break for Problem 1

9. 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

10. The selection differential is a function of both
the phenotypic variance and the fraction selected
20% selected
Vp = 1, S = 1.4
50% selected
Vp = 4, S = 1.6
20% selected
Vp = 4, S = 2.8

11. The Selection Intensity, i
As the previous example shows, populations with the
same selection differential (S) may experience very
different amounts of selection
The selection intensity i provided a suitable measure
for comparisons between populations,

12. Selection Differential Under Truncation Selection
Mean of truncated distribution ( )
Phenotypic standard deviations
Height of the unit normal density function
at the truncation point T
P ( z > T)
Hence, S = s pT/pT
Likewise, i = S/ s = pT/pT ( )
Change in the Variance

13. Selection Intensity Versions of the Breeders’ Equation
Since h2sP = (s2A/s2P) sP = sA(sA/sP) = h sA
We can also write this as
R = i h sA
Since h = correlation between phenotypic and breeding
values, 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

14. Accuracy of selection R = i ruAsA
More generally, we can express the breeders
equation 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 to
predict an individual's breeding value =
correlation between u and an individual's BV
Example: Estimating an individual’s BV with
progeny testing. For n progeny,
ruA2 = n/(n+a), where a = (4-h2)/h2
Mathematica program

15. Overlapping Generations
Lx = Generation interval for sex x
= Average age of parents when progeny are born
The yearly rate of response is
Ry =
im + if
Lm + Lf
h2sp
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

16. Generalized Breeder’s Equation
Ry =
im + if
Lm + Lf
ruAsA
Tradeoff between generation length L and
accuracy r
The longer we wait to replace an individual, the more
accurate the selection (i.e., we have time for progeny
testing and using the values of its relatives)

17. Finite sample correction for i
When a finite number of individuals form the next
generation, previous expressions overestimate i
Suppose M adults measured, of which N are selected
for 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 (
Standardized order statistics

18. p = N/M, the proportion selected Expected Selection intensity1 .1
.01
p = N/M, the proportion selected
Expected Selection intensity
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.

19. Burrows' approximation (normally-distributed trait)
Bulmer’s approximation:
uses infinite population result, with p replaced by

20. 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 a
probit (or logistic) model, when p(z) = Prob(state one | z)

21. Frequency of trait
qt* - qt is the
selection differential
on the phenotypic scale
Liability scale
Mean liability before selection
Selection differential
on liability scale
Mean liability after selection
(but before reproduction)
Frequency of character state on
in next generation
Mean liability in next generation

22. Steps in Predicting Response to Threshold Selection
i) Compute initial mean m0
Hence, z - m0 is a unit normal random variable
U is a unit normal
P(trait) = P(z > 0) = P(z - m > -m) = P(U > -m)
We can choose a scale where the liability
z has variance of one and a threshold T = 0
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 m =
General result: m = - z[1-q]
ii) The frequency qt+1 of the trait in the next
generation is just
qt+1 = P(U > - mt+1 ) = P(U > - [h2S + mt ] )
= P(U > - h2S - z[1-q] )

23. iii) Hence, we need to compute S, the selection
differential on liability
Let pt = fraction of individuals chosen in
generation t that display the trait
This fraction does not display the trait, hence
z < 0
This fraction display the trait, hence z > 0 -
>
When z is normally distributed, this reduces to
Height of the unit normal density function
At the point mt * -
Hence, we start at some initial value given h2 and
m0, and iterative to obtain selection response

24. Initial frequency of q = 0.05. Selection only on adults
showing the trait
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25 10 20 30 40 50 60 70 80 90
100 S q
q, Frequency of character
Selection differential S 25 20 15 10 5
Generation

25. 15 Minute Break

26. Permanent Versus Transient Response
Considering epistasis and shared environmental values,
the single-generation response follows from the
midparent-offspring regression ( )
Response from shared
environmental effects
Permanent component of response
Transient component of response --- contributes
to short-term response. Decays away to zero
over the long-term
Breeder’s Equation
Response from epistasis

27. Response with Epistasis
The response after one generation of selection from
an unselected base population with A x A epistasis is ( )
The contribution to response from this single generation
after t generations of no selection is ( )
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
c is the average (pairwise) recombination between loci
involved in A x A
Contribution to response from epistasis decays to zero as
linkage disequilibrium decays to zero

28. Why unselected base population? If history of previous
selection, linkage disequilibrium may be present and
the mean can change as the disequilibrium decays
More generally, for t generation of selection followed by
t generations of no selection (but recombination)

29. Maternal Effects: Falconer’s dilution model z = G + m zdam + e
Maternal effect passed from dam to offspring is just
A fraction m of the dam’s phenotypic value
z = G + m zdam + e
Direct genetic effect on character
G = A + D + I. E[A] = (Asire + Adam)/2
The presence of the maternal effects means that response
is not necessarily linear and time lags can occur in response
m can be negative --- results in the potential for
a reversed response

30. ) ( Parent-offspring regression under the dilution model
In terms of parental breeding values,
Regression of BV on phenotype
With no maternal effects, baz = h2
With maternal effects, a covariance between BV
and maternal effect arises, with
The resulting slope becomes bAz = h2 2/(2-m)
The response thus becomes ( )

31. Response to a single generation of selection
h2 = 0.11, m = (litter size in mice)
Recovery of genetic response after
initial maternal correlation decays
-0.15
-0.10
-0.05
0.00
0.05
0.10
Cumulative Response to Selection
Reversed response in 1st
generation largely due to
negative maternal correlation
masking genetic gain
(in terms of S) 10 9 8 7 6 5 4 3 2 1
Generation

32. Selection occurs for 10 generations and then stops
-1.0
-0.5
0.0
0.5
1.0
1.5
m = -0.25
m = -0.5
m = -0.75
Cumulative Response (in units of S)
h2 = 0.35 20 15 10 5
Generation

33. Gene frequency changes under selection
Genotype
A1A1
A1A2
A2A2
Fitnesses 1
1+s
1+2s
Additive fitnesses
Let q = freq(A2). The change in q from one
generation of selection is: D
In finite population, genetic drift can overpower
selection. In particular, when
drift overpowers the effects of selection

34. Strength of selection on a QTL
Have to translate from the effects on a trait under
selection to fitnesses on an underlying locus (or QTL)
Suppose the contributions to the trait are additive:
Genotype
A1A1
A1A2
A2A2
Contribution to
Character a 2a
For a trait under selection (with intensity i) and
phenotypic variance sP2, the induced fitnesses
are additive with s = i (a /sP )
Thus, drift overpowers
selection on the QTL when

35. D More generally Change in allele frequency:
Genotype
A1A1
A1A2
A2A2
Contribution to trait
a(1+k) 2a
Fitness 1
1+s(1+h)
1+2s D
Change in allele frequency:
s = i (a /sP )
Selection coefficients for a QTL
h = k

36. 15 Minute Break

37. Changes in the Variance under Selection
The infinitesimal model --- each locus has a very small
effect on the trait.
Under the infinitesimal, require many generations
for significant change in allele frequencies
However, can have significant change in genetic
variances due to selection creating linkage disequilibrium
With positive linkage disequilibrium, f(AB) > f(A)f(B),
so that AB gametes are more frequent
With negative linkage disequilibrium, f(AB) < f(A)f(B),
so that AB gametes are less frequent
Under linkage equilibrium,
freq(AB gamete) = freq(A)freq(B)

38. Changes in VA with disequilibrium
Starting from an unselected base population, a single
generation of selection generates a disequilibrium
contribution d to the additive variance
Under the infinitesimal model, disequilibrium only
changes the additive variance.
Additive genetic variance after
one generation of selection
Additive genetic variance in the
unselected base population. Often
called the additive genic variance
disequilibrium
Changes in VA changes the phenotypic variance
The amount disequilibrium generated by a single generation
of selection is
Changes in VA and VP change the heritability
Within-generation change
in the variance
An increase in the variance
generates d > 0 and hence
positive disequilibrium
A decrease in the variance
generates d < 0 and hence
negative disequilibrium *

39. A “Breeders’ Equation” for Changes in Variance
d(0) = 0 (starting with an unselected base population)
Within-generation
change in the variance
-- akin to S
New disequilibrium generated by
selection that is passed onto the
next generation
Decay in previous disequilibrium from recombination
d(t+1) - d(t) measures the response in selection
on the variance (akin to R measuring the mean)
Many forms of selection (e.g., truncation) satisfy
k > 0. Within-generation
reduction in variance.
negative disequilibrium, d < 0
k < 0. Within-generation
increase in variance.
positive disequilibrium, d > 0

40. Break for Problem 2