Lecture 3 of the Quantitative Genetics series. Genotypic values, average effects, breeding values and the genetic variance components.
Alleles of a gene vary; there may be two or many allelic forms, and some of these variants affect quantitative traits. Alleles occur in combinations, the genotypes. With two alleles \(A\) and \(a\) there are three genotypes, \(AA\), \(Aa\) and \(aa\), with frequencies \(f_{AA}\), \(f_{Aa}\) and \(f_{aa}\) summing to one.
Allele frequencies follow from genotype frequencies by counting: each homozygote carries two copies, each heterozygote one.
Under random mating, with no selection, mutation or migration, and in a large population, each allele in one generation has an equal chance of being copied into the next. No alleles are added, removed or changed. Genotypes in the offspring are then formed by combining two independently drawn alleles.
These are reached after a single generation of random mating, and are the default assumption throughout the course. Note what this means in practice: even if selection has severely distorted genotype frequencies among adults, one generation of random mating among those adults restores Hardy-Weinberg proportions among the zygotes, provided mating itself is random with respect to the locus.
Our first goal is to decompose a phenotype into a genetic part and an environmental part. Consider a locus with two alleles, and write the expected phenotype of each genotype as a conditional expectation, exactly the object defined in the previous lecture:
The population mean is the average of these three, weighted by their Hardy-Weinberg frequencies:
and the phenotype of an individual is written
with \(g\) the genotypic deviation and \(e\) the environmental deviation, both measured as departures from \(\mu\).
Define the genotypic value \(g\) of each genotype as its deviation from the population mean:
By construction these deviations average to zero over the population: \(p^2g_{AA}+2pq\,g_{Aa}+q^2g_{aa}=0\).
The following step underpins quantitative genetics. Genotypes are not transmitted to offspring, only alleles are, so the quantity required is the value associated with an allele rather than with a genotype.
The average effect of an allele is the expected deviation from \(\mu\) of an individual known to carry that allele, with the second allele drawn at random from the population:
The logic is direct: given one allele is \(A\), the other is \(A\) with probability \(p\) (giving genotype \(AA\)) and \(a\) with probability \(q\) (giving \(Aa\)). Weight the genotypic values accordingly.
\(\beta\) is the expected change in the trait when one \(a\) allele is replaced by one \(A\). It is the single most useful number attached to a locus, and it is what the \(\alpha_j\) in the many-locus model of Lecture 1 refers to. Note that average effects satisfy \(p\alpha_A+q\alpha_a=0\).
Since alleles are what get transmitted, the part of an individual's genotypic value that it can pass on is the sum of the average effects of the two alleles it carries. That sum is the breeding value.
The breeding value will generally not equal the genotypic value exactly. What is left over is the dominance deviation:
giving the decomposition of the genotypic value at one locus:
A parent passes on one allele, not a genotype, so the dominance deviation, which arises from the specific combination of two alleles, is broken up at meiosis and not transmitted. This is precisely why selection responds to the additive component and not to dominance, and why the breeder's equation later contains \(\sigma_A^2\) rather than \(\sigma_G^2\).
Putting it together, the genetic model for one locus is
Consider the case in which there is no dominance. Write the genotypic values in the symmetric form \(x_{aa}=x-s\), \(x_{Aa}=x\), \(x_{AA}=x+s\), so that the heterozygote sits exactly midway between the homozygotes. Then \(\mu=x+(p-q)s\) and the table becomes:
| Genotype | \(x_{ij}\) | \(g_{ij}\) | \(\alpha_i+\alpha_j\) | \(\delta_{ij}\) |
|---|---|---|---|---|
| \(aa\) | \(x-s\) | \(-2ps\) | \(-2ps\) | 0 |
| \(Aa\) | \(x\) | \((p-q)s\) | \((p-q)s\) | 0 |
| \(AA\) | \(x+s\) | \(2qs\) | \(2qs\) | 0 |
Under purely additive gene action every dominance deviation is zero, the genotypic value equals the breeding value, and \(\beta=\alpha_A-\alpha_a=s\) independently of allele frequency. This is the reference case against which dominance and overdominance are judged.
The gametic variance is the variance of the effect transmitted by a single allele. A gamete carries \(A\) with probability \(p\) (contributing \(\alpha_A\)) and \(a\) with probability \(q\) (contributing \(\alpha_a\)). Since \(p\alpha_A+q\alpha_a=0\), the mean is zero and
An individual carries two independently drawn gametes, so the additive variance is twice the gametic variance:
The dominance variance is the variance of the deviations \(\delta_{ij}\), which have mean zero:
and, writing \(d\) for the dominance effect (the deviation of the heterozygote from the midpoint of the homozygotes), this evaluates to the compact form
The additive and dominance components are uncorrelated by construction, which is what allows the variances to be added without a covariance term, exactly the condition flagged in the variance section of Lecture 2. Note also that \(\sigma_D^2=\sigma_G^2-\sigma_A^2\) holds for one locus only; with several loci the difference would also contain epistatic variance.
The following dependence is easily overlooked. From the results derived above:
The genotypic values \(x_{AA},x_{Aa},x_{aa}\) are properties of the biology and do not change. Everything else on that list is a property of the population. The same locus, with the same biology, has a different additive variance in two populations that differ in allele frequency, and \(\sigma_A^2=2pq\beta^2\) necessarily goes to zero as \(p\to0\) or \(p\to1\). A locus fixed for one allele contributes nothing to genetic variance, however large its biological effect.
Sickle-cell anaemia. The locus has two alleles \(C\) and \(c\). Individuals \(cc\) have anaemia; \(Cc\) individuals have partial protection against malaria. In parts of West Africa the frequency of \(c\) is \(p_c=0.2\).
1. With \(q=p_c=0.2\) and \(p=p_C=0.8\): \(CC=p^2=0.64\), \(Cc=2pq=0.32\), \(cc=q^2=0.04\).
2. Differential fertility among genotypes is a form of natural selection, and selection is a violation of the Hardy-Weinberg conditions, so the adult population is not in HW proportions. But the question asks about the zygotes of the next generation. Since mating is random with respect to the \(C/c\) genotype, the zygotes are formed by independently sampling two alleles from the adult gamete pool, so Hardy-Weinberg proportions are restored in a single generation, at the new allele frequency.
The distinction is that selection changes the allele frequencies, while random mating re-establishes Hardy-Weinberg proportions at the new frequencies.
Crossing two populations. One locus with alleles \(A\) and \(a\). In population 1 the frequency of \(A\) is \(p_1=0.7\); in population 2 it is \(p_2=0.4\).
1. Population 1: \(AA=0.49\), \(Aa=0.42\), \(aa=0.09\). Population 2: \(AA=0.16\), \(Aa=0.48\), \(aa=0.36\).
2. Allele frequency in the crossbreds is the average of the parental frequencies, \(\bar p=(0.7+0.4)/2=0.55\). But the genotype frequencies are not \(\bar p^2,2\bar p\bar q,\bar q^2\). Each crossbred receives one gamete from each population, so
HW would predict \(0.3025, 0.495, 0.2025\). The crossbreds have more heterozygotes than HW (0.54 against 0.495), so they are not in Hardy-Weinberg proportions. The reason: the two gametes are drawn from populations with different allele frequencies, so the two draws are not identically distributed, which is one of the HW conditions.
3. Now mating is random within a single population. Allele frequency stays at \(\bar p=0.55\) (no selection), and one generation of random mating restores HW proportions: \(AA=0.3025\), \(Aa=0.495\), \(aa=0.2025\). This is the classic F1-to-F2 loss of heterozygosity.
Compute the means. Set \(p=0.7\). For each of the three models of gene action below, compute \(\mu\), \(\alpha_A\), \(\alpha_a\) and \(\beta\).
| Model | \(x_{aa}\) | \(x_{Aa}\) | \(x_{AA}\) |
|---|---|---|---|
| Additive | 1 | 2 | 3 |
| Dominant | 1 | 2 | 2 |
| Overdominant | 1 | 2 | 1 |
Use \(\mu=p^2x_{AA}+2pq\,x_{Aa}+q^2x_{aa}\), \(g_{ij}=x_{ij}-\mu\), \(\alpha_A=p\,g_{AA}+q\,g_{Aa}\), \(\alpha_a=p\,g_{Aa}+q\,g_{aa}\), with \(p=0.7,q=0.3\).
Additive: \(\mu=0.09(1)+0.42(2)+0.49(3)=2.4\). Then \(g_{aa}=-1.4\), \(g_{Aa}=-0.4\), \(g_{AA}=0.6\). \(\alpha_a=0.3(-1.4)+0.7(-0.4)=-0.7\); \(\alpha_A=0.3(-0.4)+0.7(0.6)=0.3\); \(\beta=\alpha_A-\alpha_a=1.0\).
Dominant: \(\mu=0.09(1)+0.42(2)+0.49(2)=1.91\). \(g_{aa}=-0.91\), \(g_{Aa}=0.09\), \(g_{AA}=0.09\). \(\alpha_a=0.3(-0.91)+0.7(0.09)=-0.21\); \(\alpha_A=0.3(0.09)+0.7(0.09)=0.09\); \(\beta=0.30\).
Overdominant: \(\mu=0.09(1)+0.42(2)+0.49(1)=1.42\). \(g_{aa}=-0.42\), \(g_{Aa}=0.58\), \(g_{AA}=-0.42\). \(\alpha_a=0.3(-0.42)+0.7(0.58)=0.28\); \(\alpha_A=0.3(0.58)+0.7(-0.42)=-0.12\); \(\beta=-0.40\).
Under overdominance \(\beta\) is negative at \(p=0.7\), although the \(A\) allele is not inferior in any biological sense. The substitution effect is a population quantity and changes sign with allele frequency.
Compute and plot the variances. For the same three scenarios,
compute \(V_G\), \(V_A\) and \(V_D\), reusing the \(\mu,\alpha_A,\alpha_a\) you already have. Then
plot all three as functions of \(p\) over \((0,1)\) using the course R code (var.R).
Interpret the shapes: why are the maxima and minima where they are, and how would you identify the
allele frequencies at which \(V_A=0\)?
Use \(V_G=p^2g_{AA}^2+2pq\,g_{Aa}^2+q^2g_{aa}^2\), \(V_A=2pq\beta^2\), and \(V_D=V_G-V_A\) (valid for a single locus only; with several loci the remainder would also contain epistatic variance).
Reference implementation:
mu <- function(x_aa,x_Aa,x_AA,p){ q <- 1-p; q^2*x_aa + 2*p*q*x_Aa + p^2*x_AA }
Vg <- function(x_aa,x_Aa,x_AA,p){ q <- 1-p; m <- mu(x_aa,x_Aa,x_AA,p)
q^2*x_aa^2 + 2*p*q*x_Aa^2 + p^2*x_AA^2 - m^2 }
alphaA <- function(x_aa,x_Aa,x_AA,p){ q <- 1-p; p*x_AA + q*x_Aa - mu(x_aa,x_Aa,x_AA,p) }
alphaa <- function(x_aa,x_Aa,x_AA,p){ q <- 1-p; p*x_Aa + q*x_aa - mu(x_aa,x_Aa,x_AA,p) }
Va <- function(x_aa,x_Aa,x_AA,p){ q <- 1-p
beta <- alphaA(x_aa,x_Aa,x_AA,p) - alphaa(x_aa,x_Aa,x_AA,p); 2*p*q*beta^2 }
p <- seq(0.01, 0.99, 0.01)
Interpretation. All variances vanish at \(p=0\) and \(p=1\), because a fixed locus is not variable and therefore contributes no variance whatever its biological effect. In the additive case \(\beta\) is constant, so \(V_A=2pq\beta^2\) is a symmetric parabola peaking at \(p=0.5\). Under dominance the peak shifts away from 0.5, because \(\beta\) itself varies with \(p\). Under overdominance \(\beta\) passes through zero at an intermediate frequency, so \(V_A=0\) there while \(V_G\) and \(V_D\) remain positive, a locus with substantial genetic variance but no additive variance, and therefore no response to selection.
To find where \(V_A=0\), solve \(\beta(p)=\alpha_A-\alpha_a=0\) for \(p\), since \(2pq>0\) strictly inside the interval.
