Orthogonal Dominance Contrasts Under Hardy-Weinberg Equilibrium

Why Genotype Contrasts Depend on Allele Frequency

At a biallelic locus, a heterozygote can depart from the midpoint of the two homozygote values. That departure is dominance. But separating a set of genotype values into an additive part and a dominance part is not just a matter of assigning a fixed contrast to each genotype. The population frequencies matter: the additive contrast is defined relative to the population’s allele frequency, and the dominance deviations that remain after fitting it depend on that frequency.

Under Hardy–Weinberg equilibrium (HWE), genotype probabilities have a simple form. This lets us derive contrasts that have mean zero and are uncorrelated in the population. The resulting variance decomposition is useful for describing the contribution of additive and dominance effects at a locus, while making clear that these are population-specific quantities.

Genotype Values and HWE Probabilities

Consider a biallelic locus with alleles \(A\) and \(a\). Let \(p\) be the frequency of \(A\), and let \(q=1-p\) be the frequency of \(a\). Define \(G\) as the number of \(A\) alleles carried by an individual, so \(G=0,1,2\) for genotypes \(aa,Aa,AA\), respectively. Under HWE, their probabilities are \(q^2,2pq,p^2\).

Use the standard parameterization in which the genotypic values are \(-a\) for \(aa\), \(d\) for \(Aa\), and \(+a\) for \(AA\). Here, \(a\) is half the difference between the homozygote values, and \(d\) is the heterozygote’s deviation from their midpoint. The parameter \(a\) is distinct from the allele label \(a\). These values describe the genetic contribution at this locus; environmental variation is not included in this single-locus decomposition.

Definition: The additive contrast is the centered allele count \(G-2p\). Its regression coefficient for the genotypic value is the average effect of allele substitution, denoted here by \(\alpha\). The dominance deviation \(D\) is the part of the genotypic value left after subtracting its population mean and its additive component. Thus, \(Y=\mu_G+\alpha(G-2p)+D\), where \(Y\) is the genotypic value and \(\mu_G=E[Y]\).

The centering \(G-2p\) is appropriate because \(E[G]=2p\) under HWE. The coefficient \(\alpha\) is not generally the homozygote half-difference \(a\): when \(d\ne0\), the frequency of heterozygotes shifts the average effect of substituting one allele for another.

Deriving the Average Effect and Dominance Deviations

First calculate the mean genotypic value by weighting each genotype value by its HWE probability:

\mu_G=q^2(-a)+2pq\,d+p^2a =a(p-q)+2pqd.

The variance of the allele count is \(2pq\). Regressing \(Y\) on the centered count \(G-2p\) gives a slope equal to \(\operatorname{Cov}(G,Y)/\operatorname{Var}(G)\). Since \(E[G]=2p\), this covariance is \(E[GY]-E[G]E[Y]\). Also, \(E[GY]=2p^2a+2pqd\): only \(Aa\) and \(AA\) contribute, with their \(G\) values of 1 and 2. Therefore,

\begin{aligned} \alpha &=\frac{2p^2a+2pqd-2p\{a(p-q)+2pqd\}}{2pq}\\ &=a+d(q-p). \end{aligned}

The additive component is consequently \(\alpha(G-2p)\). Subtracting this component and the mean from each genotype value gives the dominance deviations:

\begin{aligned} D_{aa}&=-2p^2d,\\ D_{Aa}&=2pqd,\\ D_{AA}&=-2q^2d. \end{aligned}

These expressions show why the dominance deviations are frequency-dependent even though the original genotype values \(-a,d,+a\) are held fixed. They are the residuals after the additive contrast has been fitted for the particular population allele frequency.

Worked Example: Decomposing Genotype Values at an Unequal Allele Frequency

Suppose a hypothetical locus is in HWE, with \(p=0.30\), \(q=0.70\), \(a=2\), and \(d=1\). The genotype values are \(-2,1,2\) for \(aa,Aa,AA\). Calculate the mean, average effect, and genotype-specific additive and dominance components.

The HWE probabilities are \(q^2=0.49\), \(2pq=0.42\), and \(p^2=0.09\). The mean is

\mu_G=(0.49)(-2)+(0.42)(1)+(0.09)(2) =-0.98+0.42+0.18=-0.38.

The average effect is

\alpha=a+d(q-p)=2+1(0.70-0.30)=2.4.

The centered allele counts \(G-2p\) are \(-0.60,0.40,1.40\). Multiplying by \(\alpha\), the additive components are \(-1.44,0.96,3.36\). The dominance deviations are

D_{aa}=-2(0.30)^2(1)=-0.18,\quad D_{Aa}=2(0.30)(0.70)(1)=0.42,\quad D_{AA}=-2(0.70)^2(1)=-0.98.

For each genotype, mean plus additive component plus dominance deviation recovers its original value: \(-0.38-1.44-0.18=-2\), \(-0.38+0.96+0.42=1\), and \(-0.38+3.36-0.98=2\). The decomposition therefore reproduces all three genotype values.

Orthogonality and the Variance Decomposition

The dominance deviations have mean zero under HWE. Substituting their values and probabilities gives

$$ E[D]=q^2(-2p^2d)+2pq(2pqd)+p^2(-2q^2d)=0. $$

They are also orthogonal to the additive contrast: their population covariance with \(G-2p\) is zero. A direct check uses the centered counts \(-2p,1-2p,2q\) for \(aa,Aa,AA\):

\begin{aligned} E[(G-2p)D] &=q^2(-2p)(-2p^2d) +2pq(1-2p)(2pqd) +p^2(2q)(-2q^2d)\\ &=4p^3q^2d+4p^2q^2(q-p)d-4p^2q^3d\\ &=4p^2q^2d\{p+(q-p)-q\}=0. \end{aligned}

The cancellation uses \(p+q=1\) and the HWE probabilities. Thus, the additive component and dominance deviation have zero population covariance in this setting. Because both have mean zero, their covariance is also their expected product.

The additive variance is the variance of \(\alpha(G-2p)\). The dominance variance is the variance of \(D\). Orthogonality means there is no cross-product term when the variance of their sum is expanded:

\begin{aligned} V_A&=\alpha^2\operatorname{Var}(G)=2pq\alpha^2,\\ V_D&=E[D^2]=4p^2q^2d^2,\\ \operatorname{Var}(Y)&=V_A+V_D. \end{aligned}

For the dominance result, weighting the squared deviations gives

\begin{aligned} E[D^2] &=q^2(4p^4d^2)+2pq(4p^2q^2d^2)+p^2(4q^4d^2)\\ &=4p^2q^2d^2(p^2+2pq+q^2)\\ &=4p^2q^2d^2, \end{aligned}

because \(p^2+2pq+q^2=(p+q)^2=1\). This is the dominance variance for the stated genotype-value parameterization and HWE population, not a frequency-free property of \(d\).

Worked Example: Checking Orthogonality Numerically

For a hypothetical HWE population, let \(p=0.25\), \(q=0.75\), and \(d=2\). The dominance deviations are \(-0.25,0.75,-2.25\) for \(aa,Aa,AA\). The centered allele counts are \(-0.50,0.50,1.50\), and genotype probabilities are \(0.5625,0.375,0.0625\).

Calculate the weighted cross-product:

\begin{aligned} E[(G-2p)D] &=(0.5625)(-0.50)(-0.25) +(0.375)(0.50)(0.75) +(0.0625)(1.50)(-2.25)\\ &=0.0703125+0.140625-0.2109375=0. \end{aligned}

The dominance variance is

$$ V_D=4(0.25)^2(0.75)^2(2)^2=0.5625. $$

The zero cross-product verifies orthogonality for these frequencies and deviations; the variance calculation agrees with the general formula. Both calculations use the HWE genotype probabilities given above.

Allele Frequency Changes the Variance Attributed to Dominance

For fixed \(d\), \(V_D=4p^2q^2d^2\) is largest at \(p=q=0.5\), where it equals \(d^2/4\). It approaches zero as either allele becomes rare and is zero at fixation. This follows because \(pq\) is largest at equal frequencies and vanishes when either frequency is zero. The formula describes variance across genotypes, so a heterozygote deviation contributes little to population variance when heterozygotes are uncommon.

Worked Example: Comparing Two Allele Frequencies

Consider the same hypothetical genotype parameters \(a=1\) and \(d=1\) in two HWE populations. Compare one with \(p=0.10,q=0.90\) and one with \(p=q=0.50\). The genotype values are \(-1,1,1\) in both populations.

In the first population, \(\alpha=1+1(0.90-0.10)=1.8\). Thus,

V_A=2(0.10)(0.90)(1.8)^2=0.5832,\qquad V_D=4(0.10)^2(0.90)^2(1)^2=0.0324.

In the second population, \(\alpha=1+1(0.50-0.50)=1\), giving

V_A=2(0.50)(0.50)(1)^2=0.50,\qquad V_D=4(0.50)^2(0.50)^2(1)^2=0.25.

The homozygote and heterozygote values did not change, but the variance attributed to dominance is much larger in the balanced-frequency population, where HWE predicts more heterozygotes. The additive variance also changes because \(\alpha\) depends on frequency. In each population the two components sum to the genotypic variance: \(0.5832+0.0324=0.6156\) in the first and \(0.50+0.25=0.75\) in the second.

Assumptions, Scope, and a Common Mistake

The formulas here assume a single biallelic locus, the stated \(-a,d,+a\) genotype values, and HWE probabilities \(q^2,2pq,p^2\) in the population whose variance is being described. For \(0<p<1\), the additive contrast has nonzero variance and its regression coefficient is defined. At fixation, there is no allele-count variation from which to estimate that coefficient, although the dominance variance formula correctly gives zero. The decomposition concerns genotypic values at this locus; it does not by itself describe total phenotypic variance or establish that genetic effects cause an observed trait difference.

Common mistake: A reader may treat \(d\) as the dominance variance, or assume that \(d\) and the dominance variance are interchangeable across populations. But \(d\) is the heterozygote’s deviation in the specified genotype-value scale, whereas \(V_D=4p^2q^2d^2\) also depends on allele frequency. The correction is to report the genotype-value parameterization and the population frequency assumptions alongside the variance.

Key takeaway: Under HWE, the average allele-substitution effect is \(\alpha=a+d(q-p)\), and the residual dominance deviations are orthogonal to the centered allele count. Their variance is \(4p^2q^2d^2\), so dominance variance is a property of both the genotype values and the allele frequencies.

Check Your Understanding

Use the definitions and assumptions in this tutorial to answer the following questions.

For \(p=0.40\), what are the HWE probabilities of \(aa,Aa,AA\), and what is the expected allele count \(E[G]\)?
If \(a=1.5\), \(d=0.5\), \(p=0.25\), and \(q=0.75\), calculate \(\alpha\).
Why is the average effect of allele substitution generally not equal to the homozygote half-difference \(a\) when \(d\ne0\)?
For fixed \(d\), at what allele frequencies is dominance variance maximized, and what is its value there?
Explain why a nonzero heterozygote deviation does not guarantee substantial dominance variance in a population.