1. Lecture 15: Analysis of Selection Experiments

2. Variance in the Response to Selection
R = h2S is just the expected value of
the response, but there is a variance
about this value.
Hence, identically-selected replicate lines are still
expected to show variation is response
The major source of such variation is genetic drift

3. æ ( t ) = + z = π + g d e E ( z ) = π + g d æ ( t ) = M
Consider the mean in generation t
The effect of any major environmental trend in generation t
The mean breeding value in generation t
Mean of the original population
Error in estimating the environmental-corrected
mean breeding value from the mean phenotype of
a sample z t = π + g d e
Under this model, the mean of a replicate series of
lines is
R = h2S E ( z t ) = π + g d
The variance is given by
Variance in the environmental trend
(mean is set to be zero)
Variance in the breeding value at generation t æ 2 z ( t ) = g + e d æ 2 e ( t ) = z M
In generation t, Mt individuals
are measured. An upper bound on
the error variance is

4. Variance in Breeding Values
Two sources of variation
(i) Sampling variance in the founding lines
(ii) Genetic drift (inbreeding) within each line æ 2 g ( t ) = 1 M + f ∂ A h z
Inbreeding in generation t
Size of the founding
population t ∂
" 2 f = 1 N e # ' o r
< ( ) -
The mean breeding values in different generations
of the same replicate line are correlated, ( ) æ g t ; = z 1 M + 2 f ∂ h o r
<

5. Variance-covariance structure within a line
Assume the initial sample is sufficiently large so
that we can ignore 1/M0
Variance: s2(gt) = (t/Ne)h2s2z
Covariance: s2(gt, gx) = (x/Ne)h2s2z for x < t
These expressions (which are often called the
pure-drift approximations) will prove useful in
the statistical analysis of selection response

6. The Realized Heritability
Since R = h2 S, this suggests h2 = R/S, so that
the ratio of the observed response over the
observed differential provides an estimate of
the heritability, the realized heritability
Obvious definition for a single generation of response.
What about for multiple generations of response?
Cumulative selection response = sum of all responses R C ( t = i 1 ) X

7. R ( t ) = h S + e ( T ) R b h = S P b = S ( ) R
Cumulative selection differential = sum of the S’s S C ( t ) = 1 i X
(1) The Ratio Estimator for realized heritability
= total response/total differential, ( T ) R b h 2 r = C S
(2) The Regression Estimator --- the slope of the
regression of cumulative response on cumulative differential P b = t S C ( ) R 2
Regression passes through the
origin (R=0 when S=0). Slope = R C ( t ) = h 2 r S + e

8. Note x axis is differential, NOT generations
Ratio estimator
= 17.4/56.9 = 0.292 60 \
Cumulative Differential 5 10 15 20
Cumulative Response
Slope = 0.270
= Regression
estimator
Note x axis is differential,
NOT generations

9. Standard error of the Ratio Estimator
Ratio Estimator, h2r = RT/ST
Recall that the variance for the mean in generation
t is s2(gt) + s2(e) + s2(d)
Assume M0 >> 1 and that we can ignore the
environmental trend variance, then
s2(RT) = s2(gt) + s2(e) = (T/Ne)h2s2z + s2z /MT
Number of individuals sampled in
generation T
s2(RT/ST ) = s2(RT ) /(ST )2
This follows since s2(ax) = a2s2 (x)
Hence, s2(h2r) = [ (T/Ne)h2s2z + s2z /MT ] /(ST )2

10. SE for (OLS) Regression Estimator
The basic linear model is
X = Sc
b = bc
y = R R C ( t ) = h 2 r S + e b C ( O L S ) = ≥ X T 1 y R -
Under the OLS framework (residuals homoscedastic
and uncorrelated), the linear model has the design
matrix X just the vector Sc of cumulative differential
and y = R P = T i 1 S C ( ) R 2 ( ) - V a r h b C O L S i = æ 2 e ≥ X T 1 b æ 2 e = 1 T X i ≥ R C ( ) h r S -

11. Problems with OLS regression approach
Although the OLS regression estimator for realized
heritability is very widely used, it has fatal problems
OLS assumes the residuals are homoscedastic and
uncorrelated. In reality, the covariance structure is
s2(ei) = (i/Ne)h2s2z + s2z /Mi
s2(ek, ei) = (i/Ne)h2s2z for i < k
Hence, the GLS regression is more appropriate
The OLS gives unbiased estimates of the realized
heritability, but it seriously underestimates its SE

12. GLS regression EstimateC ( t ) = h 2 r S + e
X = Sc
b = bc
y = R
The variance-covariance matrix V has elements
h2 is what we are trying to estimate. Use an iterative
approach. Try some initial value, use GLS to update this
value, use the new value for next round of updating.
continue until values stabilize
We can directly estimate the phenotypic
variance from the data
Vii = (i/Ne)h2s2z + s2z /Mi
Vji = Vij = (i/Ne)h2s2z for i < j - b C ( G L S ) = T V 1 R - V a r h b C ( G L S ) i = T 1

13. Just how well does the Breeders’ Equation work?
Sheridan (1988) compared realized heritability
estimates with estimates of heritability obtained
from resemblances between relatives in the
base populations
Punch-line: Good, but not great, fit in many settings
Problems with a wider meta-analysis is that standard
errors are often not presented nor is the data presented
in a form that allows their calculation.

14. Comparison of realized and (relative-based) Heritability estimates
Species
Significant Differences
NS difference
Total
Drosophila
14 (23%)
47 (77%) 61
Tribolium
7 (27%)
19 (73%) 26
Mice/Rats
6 (18%)
28 (82%) 34
Poultry/Quail
5 (45%)
6 (55%) 11
Swine/Sheep
8 (53%)
7 (47%) 15

15. Asymmetric Selection Response
Divergent Selection Experiment: Select some replicate
lines for increased trait value, other for decreased
value
Expectation: roughly
equal response in up and
down directions, R = h2S Rc Sc
Often an asymmetric response
is observed, with a significant
difference in the slope of up
vs. down-selection lines

16. Potential Causes: I. Design Defects
Different selection differentials (Plot is Rc vs. t, not Rc vs. Sc)
Drift (sample size not sufficiently large)
Scale effects
Undetected environmental trends
Transient effects from previous selection
Decay of epistatic response
Undetected selection on correlated traits

17. Scale effects Transform to a log scale
When the trait biologically cannot go below
a specific value (i.e., 0), as we down-select towards
zero, expect less response.

18. Potential Causes: II. Nonlinear Parent-Offspring regression
+ h2S
- S
- h2S
+ S
+ h2S

19. Major gene with dominance
What can cause a non-linear parent-offspring
regression?
Major gene with dominance
G x E
Departures from normality

20. Potential Causes: III. Inbreeding depression
True genetic response
in the absence of inbreeding
Change in mean due to
inbreeding depression.
Depresses upward response,
Enhances downward response

21. Potential Causes: IV. Genetic Asymmetry
• Requires changes in allele frequencies.
• The same absolute change in an allele frequency
can result rather different changes in the
variance in the + vs. - change direction.
• This results in departures in the additive genetic
variance in up vs. down-selected lines, and hence
changes in h2 and response.

22. Additive variance, VA, with no dominance (k = 0)
If p =1/2, then VA is
the same for p+d and
p-d d
Allele frequency, p VA d
If p = 1/2, VA
different for p+d
and p-d

23. Additive variance, VA, with complete dominance (k = 1)

24. Additive variance, VA, with overdominance (k = 10)

25. Control Populations z = π + g d e E ( z ° ) = g h S
Until now, we have been ignoring the bias caused by
not accounting for any environmental trend.
One way to deal with this is to include an unselected
control population in the design
Mean of selection population in generation t
Genetic mean of selection population, random effect,
expected value = h2Sc(t), var = drift variance z s ; t = π + g d e c
Shared environmental trend, random
effect, mean 0, var = s2d
Mean of control population in generation t
Genetic mean of control population, random effect
with expected value = 0 and var = drift variance
Hence, E ( z s ; t c ) = g h 2 S C -

26. Estimating trends with a control populationR t = ( z s ; c ) 1 C S -
The use of a control also accounts for inbreeding
depression
Complication 1: If G x E is present, then E ( z s ; t c ) = h 2 S C + d -
Complication 2: Selection inbreeds a population
quicker, so control must to comparatively inbred
to fully account for inbreeding depression

27. Divergent Selection Designs
An alternative experimental design to remove
a common environmental trend is the divergent
selection design
Mean of up-selected line z u ; t = π + g d e
Mean of down-selected line
Response estimated by R t = ( z u ; 1 ) d C S -
Note that this design also accounts for inbreeding
depression (assuming up/down lines equally inbred)

28. Variance in Response R ( t ) = z ° g + e R ( t ) = z ° π + g d e
We have been assuming that we can ignore s2d.
With a control line and/or divergent selection, don’t have
to worry about this.
Control: R C ( t ) = z s ; c π + g d e -
The common dt term cancels R C ( t ) = z s u ; d g + e -
Divergent design
Again, common dt term cancels

29. æ [ R ( t ) ] = f + B h ' A Design ft A B (t > t’) æ [ R ( t ) ; ]
The resulting variance and covariances in response become æ 2 [ R C ( t ) ] = f + B h z ' A æ [ R C ( t ) ; ] = 2 f + B h z ' A o r
<
Design ft A
B (t > t’)
Selection in a single direction, no control
fs,t
1/Ns
1/Ms,t
Selection in a single direction, with control
fs,t + fc,t
1/Ns + + 1/Nc
1/Ms,t + 1/Mc,t
Divergent Selection, no control
fu,t + fd,t
1/Nu + + 1/Nd
1/Mu,t + 1/Md,t

30. Variance with a Control
Control populations are not without a cost.
When does the use of a control population result
in a reduced variance?
Variance w/ control - variance without control = æ 2 ( R C t ) = N + 1 M ∂ h z d - t æ 2 z h N
> d
Hence (ignoring M terms),
However, this approach runs the risk of an
undetected directional environmental trend
compromising the estimated heritability.
Regardless of the value of s2d, if sufficient generations are used, the
optimal design (in terms of giving the smallest expected variance in
response) is not to use a control.

31. Optimal Experimental Design
The coefficient of variance (CV) provides one measure
for comparing different designs C V [ R ( t ) ] = æ E
Design
E [ R(t) ]
CV [ R(t) ]
Selection in one direction, with control
th2isz
(2/Nt)1/2/hi
Selection in one direction, no control
(1/Nt)1/2/hi
Divergent Selection,
no control
2th2isz
(1/2Nt)1/2 /hi
CV scales with
Nt = total #
over the entire
experiment

32. Example Suppose we plan to select the upper 5% of the population
on a trait with h2 = 0.25
How large must N be to give a CV of 0.01 when no
control is used?
p = > i = Assuming drift variance dominates
s2d, then
CV = 0.01 = (1/Nt)1/2/hi = (1/Nt)1/2/(0.5*2.06)
or Nt = 1/(0.01*0.5*2.06)^2 = 9426
Hence, we must have at least 9,426 selected parents
over the course of the experiment

33. Nicholas' Criterion
Alternative criterion for choosing Nt suggested by Nicholas
Suppose we wish a certain probability that the ACTUAL
response will be at least b of the expected response
This is just a unit normal P r ( R C t )
> Ø E [ ] = æ 1 ∂ U V -
= 1/CV
Add E [Rc(t)] to each side, divide each by s[Rc(t)]
Solve for Nt to give desired probability
Note for b = 1, that Pr(U > 0) = 1/2, so that 50% of the time
the actual response exceeds the expected response.

34. Example Again suppose i = 2.05 and h2 =0.25. What value
of NT is required for a 95% probability of the
observed response is at least 90% of its expected
value?
Here b = 0.9 and since Pr( U > -1.65) = 0.95
We have (b -1)/CV = -0.01/CV = -1.65, or
CV = 0.01/1.65
Here CV = (1/0.5*2.06)/(Nt)1/2
Solving CV = 0.01/1.65 gives Nt = 257

35. Mixed-Model estimation
PROVIDED that we have the full pedigree of
individuals in the selection experiment, we can
use mixed-model methodology (e.g., BLUP & REML)
Power: Mixed-model accounts for ALL the
covariances in the sample, not just those
between means in different generations, but
also ALL of the covariances between related
individuals.

36. Basic model: the so-called animal model
Trait value of jth individual from generation i y i j = π + a e
Additive genetic value
Vectorize the data as y = B @ 1 2 . t C A ; w h e r i n y = 1 π + a e
The (simple) model becomes
Here, Aii = (1+fj), Aij = 2Qij
With additional fixed effects, y = X b + a e

37. b a = 1 n X The estimated mean in generation k is the average
of the estimated breeding values in generation k, b a k = 1 n X j
Interesting complication: The BLUP estimate of a
requires a prior estimate of the heritability h2.
The relationship matrix A fully accounts for the effects
of drift and the generation of linkage disequilibrium
(assuming the infinitesimal model holds).