Dominant and Recessive Genetic Contrasts

Why Group Genotypes?

For a biallelic variant, let \(G\) count copies of a designated effect allele: \(G=0\), \(G=1\), or \(G=2\). A genotypic model can estimate a separate mean trait value for each of these three groups. A dominant or recessive contrast instead combines two groups by imposing that their mean trait values are equal. This can make the model simpler and can improve precision when that equality is a reasonable approximation. It can also hide a real difference when the grouping assumption is wrong.

The labels dominant and recessive describe the genotype-mean pattern being modeled for the designated allele. They do not, by themselves, establish a biological mechanism. Always state which allele is counted and which genotype groups are being compared.

Model Codings and Fitted Effects

For the dominant model, define \(D=I(G\geq 1)\), which is 1 for either carrier genotype and 0 for \(G=0\). A two-group mean model is

Here \(\beta_0\) is the mean for \(G=0\), while \(\beta_D\) is the difference between the pooled carrier mean and the \(G=0\) mean. The model predicts the same mean, \(\beta_0+\beta_D\), for \(G=1\) and \(G=2\).

For the recessive model, define \(R=I(G=2)\). Then

Now \(\beta_0\) is the pooled mean for \(G=0\) and \(G=1\), and \(\beta_R\) is the difference between the \(G=2\) mean and that pooled mean. The model predicts a shared mean for \(G=0\) and \(G=1\).

These are different one-parameter contrasts, not interchangeable names for a general association. A positive coefficient means the group coded 1 has a higher mean trait value than the reference group; a negative coefficient means it has a lower mean. The coefficient is expressed in the trait's units. For example, if \(Y\) is measured in millimeters of mercury, the coefficient is a difference in millimeters of mercury.

Connection to the Unrestricted Genotypic Model

An unrestricted genotypic model allows all three genotype means to differ. One convenient coding uses \(G=0\) as the reference and two indicators, \(I(G=1)\) and \(I(G=2)\):

In this coding, \(\beta_1=\mu_1-\mu_0\) and \(\beta_2=\mu_2-\mu_0\). The dominant model is the restriction \(\beta_1=\beta_2\); the recessive model is the restriction \(\beta_1=0\). Thus, each contrast replaces the unrestricted model's three potentially distinct means with two fitted means. The restriction saves one parameter, but it is an assumption about the pattern of means.

The pooling calculation follows directly from least squares. If a set of genotypes is constrained to share a fitted mean \(m\), the sum of squared deviations for those observations is minimized by their combined sample mean. To see this, let the observed trait values in the pooled set be \(y_1,\ldots,y_N\). The derivative of the sum of squared deviations with respect to \(m\) is

Setting the derivative to zero gives \(Nm=\sum_i y_i\), hence \(m=\bar{y}\). If the observations are summarized by genotype-specific sample sizes and means, this combined mean is the weighted average of those means. This explains why the two carrier groups in the dominant model are pooled by sample size, rather than by taking an unweighted average of their two means.

Assumptions and Precision

For the simple group-mean calculations and standard errors below, suppose observations are independent, the conditional mean follows the chosen genotype model, and the residual variance is common across genotype groups. Write that common variance as \(\sigma^2\). Normal residuals are additionally required for exact small-sample \(t\)-based inference; they are not needed just to define or calculate the group-mean contrasts. In an association analysis, covariates or relatedness may require a more appropriate fitted model, but the meaning of a dominant or recessive contrast remains the stated comparison of fitted means.

Under these assumptions, the variance of a sample mean from a group of size \(n_g\) is \(\sigma^2/n_g\). For two disjoint groups with sizes \(n_A\) and \(n_B\), the variance of their difference is the sum of those variances:

A dominant contrast uses the combined carrier count for one group and the \(G=0\) count for the other. A recessive contrast uses the \(G=2\) count and the combined \(G=0,1\) count. Pooling can increase the size of a comparison group and reduce its contribution to the standard error. But it does not guarantee a more precise or more useful estimate of every possible genotype difference: it estimates a different, restricted contrast.

Choosing and Reporting a Contrast

A dominant or recessive model can be useful when the corresponding grouping is scientifically motivated or when it was specified in advance. It reduces the number of genotype-specific means that must be estimated. This may be valuable when one genotype group is small, especially when the contrast pools that group with a biologically and statistically defensible comparison group.

The same grouping can be a liability. If the two genotypes forced to share a mean truly differ, pooling can obscure that difference and produce a misleading summary. A recessive contrast can be particularly imprecise when \(G=2\) is rare, because its effect is estimated from that small group. A dominant contrast may have more carriers, but it cannot reveal a difference between one and two effect-allele copies. Genotype counts, the allele being counted, and the contrast should therefore be reported clearly.

A concise interpretation names both groups and the outcome units. For example: “Under the dominant model, carriers of at least one copy of the designated allele had an estimated mean trait value 2.33 \(U\) higher than individuals with no copies.” This reports an association under a specified model; it does not establish that the allele caused the trait difference.

Common Mistake

A common error is to calculate the dominant pooled mean as the unweighted average of the \(G=1\) and \(G=2\) sample means, even when the genotype groups have different sizes. In the first example, that incorrect calculation would be \((12+13)/2=12.5\ U\). It gives each genotype mean equal weight, not each individual. The correct carrier mean is \(\{60(12)+30(13)\}/90=12.33\ U\), because there are twice as many \(G=1\) observations as \(G=2\) observations.

Another error is to interpret a dominant coefficient as the effect of each additional allele copy. It is a carrier-versus-noncarrier difference under a model that assigns the same fitted mean to one-copy and two-copy genotypes. An additive per-allele slope is a different contrast.