Remember that
R = h2S
R=
VA
V
S, since h 2 = A
VP
VP
R=
VA
cov(wi , zi ), since S = cov(wi , zi )
VP
rearranging, we get
R = VA
cov(wi , zi )
cov(wi , zi )
= VA
VP
Vzi
R = VA b wi zi , since b wi zi =
cov(wi , zi )
Vzi
Note the difference between S, the selection differential, and
bw z the selection gradient.
i i
Clearly the response to selection, R, depends on the additive
genetic variance, VA. But how do we understand VA in
terms of allele frequencies and their “additive” effects on the
phenotype?
Note: for the proof that S = cov(wi, zi) see page 310, Freeman
&Heron, third edition.
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First, a graphical representation of partitioning the variance
(acknowledgements to F. Bashey for the visuals).
Fortunately, variances can be partitioned in to different
components. Below
VP= total phenotypic variance
VG=total genetic variance
VE=total environmental variance
VGxE=total variance due to the interaction between genotypes
and environment.
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The total genetic variance can be portioned into
VA=Additive genetic variance
VD=Dominance variation
VEPI=Epistatic variation
Our main goal today is to understand the partitioning of VA
and VD in a one locus, two-allele system (i.e., no possible
epistasis).
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The environmental variance can also be partitioned into
different components.
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h2, heritability in the narrow sense is then:
H2, heritability in the broad sense is:
The total genetic variance divided by the total phenotypic
variance. Only h2 correctly scales the response to selection.
Why?
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Now, what is VA? How can we calculate it from first
principles? What are the underlying population genetics?
We now take the simplest possible case, and consider one
locus, with two alleles, 1 and 2. Hence there is no epistasis.
Why? We also assume that there is no environmental
variance, VE=0, and as such VGxE is also zero. Why? As
previously, p gives the frequency of allele 1.
In the drawing above, let a be the phenotype of genotype 11,
and -a be the phenotype of genotype of genotype 22. Hence
the distance between 11 and 22 is 2 times a. (See Falconer &
Mackay, 3rd edition.) d is the dominance deviation. In the
figure above, allele 1 is partially dominant. d/a is degree of
dominance. In the figure below, allele 1 is dominant.
Note: our previous dominant coefficient, h = (1 - d/a)/2.
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In the figure below, d is “overdominant” (i.e., the
heterozygote is more extreme than the homozygotes.
Now, we calculate average effects.
The average effect (mean deviation from the mean
phenotype) of allele 1 is equal to
Why? Think about the probability of allele 1 finding itself
with allele 1, and the probability of finding itself with allele
2.
The average effect of allele 2 is equal to
Now we calculate the breeding values for each genotype.
Geno
11
12
22
Pheno
a
d
-a
Breeding value
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freq
p2
2pq
q2
The additive genetic variation is equal to the variance in
breeding values. In the space below, show how you would
calculate the variance in breeding values. Hint see breeding
values worksheet on web site.
With some algebra (see web site for solution), it can be
shown that the variance in breeding values is equal to
var(BV) = VA =
.
The dominance variation is equal to
VD=
Here are some things to note:
1. the dominance deviation, d, can contribute to VA if q>p.
2. The additive genetic variation can be derived in terms of
gene frequencies (p and q), the additive effects of alleles (a),
and the dominance deviation (d). Thus there is a population
genetic base to quantitative genetics.
3. The average effects are somewhat abstract quantities, but
the breeding values can be measured as 2 times the mean
difference between the progeny and the population mean.
The mean difference is double because the parent only
contributes half the genes. (see Falconer & MacKay, p114)
Now, and this is non-trivial, extend your excel worksheet to
calculate and graph, h2, VA, VD
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