Translating Genotype Means Across Codings

Why Translate Genotype Means?

A genetic effect can be described either by listing the mean trait value for each genotype or by using coefficients in a model. The three means are easy to interpret directly, but a coding can summarize their pattern compactly. Translating between these descriptions shows exactly what a coefficient means under a chosen coding—and when that coding cannot represent the observed pattern without error.

Consider one biallelic locus with alleles \(a\) and \(A\). Let \(aa\), \(Aa\), and \(AA\) denote the genotypes, and let their quantitative-trait means be \(m_0\), \(m_1\), and \(m_2\), respectively. The subscripts count the number of copies of \(A\). Throughout, these are hypothetical genotype-specific means on the same trait scale, such as millimetres of a measured phenotype. The calculations describe differences between means; they do not, on their own, establish that genotype causes the differences.

Definition: A genotype-specific mean \(m_g\) is the mean trait value for individuals with genotype \(g\). Under a coding, the model mean is the intercept plus the coding coefficient or coefficients multiplied by that genotype's coded value. An exact translation reproduces all three means; it is possible only if the model has enough flexibility, or if the means satisfy the restrictions imposed by the coding.

The Four Codings and Their Coefficients

For the additive coding, assign \(G=0,1,2\) copies of allele \(A\) to \(aa,Aa,AA\). The model has intercept \(\beta_0\) and per-copy coefficient \(\beta_A\). For the dominant coding, use an indicator \(D\) equal to 0 for \(aa\) and 1 for both \(Aa\) and \(AA\). For the recessive coding, use an indicator \(R\) equal to 1 only for \(AA\). Finally, in the reference-based genotypic coding, take \(aa\) as the reference and use separate indicators for \(Aa\) and \(AA\).

\begin{aligned} \text{Additive:}\quad &E[Y\mid G]=\beta_0+\beta_A G,\\ \text{Dominant:}\quad &E[Y\mid D]=\beta_0+\beta_D D,\\ \text{Recessive:}\quad &E[Y\mid R]=\beta_0+\beta_R R,\\ \text{Reference-based:}\quad &E[Y\mid I_1,I_2]=\beta_0+\beta_1 I_1+\beta_2 I_2. \end{aligned}

Here \(I_1=1\) for \(Aa\) and 0 otherwise; \(I_2=1\) for \(AA\) and 0 otherwise. The genotype-specific means implied by each model are:

\begin{array}{c|ccc} \text{Coding} & aa & Aa & AA\\ \hline \text{Additive} & \beta_0 & \beta_0+\beta_A & \beta_0+2\beta_A\\ \text{Dominant} & \beta_0 & \beta_0+\beta_D & \beta_0+\beta_D\\ \text{Recessive} & \beta_0 & \beta_0 & \beta_0+\beta_R\\ \text{Reference-based} & \beta_0 & \beta_0+\beta_1 & \beta_0+\beta_2 \end{array}

These equations provide the translation rules. The additive model requires equal consecutive steps: \(m_1-m_0=m_2-m_1\). The dominant model requires \(m_1=m_2\), while the recessive model requires \(m_0=m_1\). The reference-based model has three parameters and can reproduce any three means.

Key result: When the model reproduces the three genotype means exactly, the coefficients are \( \beta_0=m_0,\ \beta_A=m_1-m_0=m_2-m_1 \) for additive coding; \( \beta_0=m_0,\ \beta_D=m_1-m_0=m_2-m_0 \) for dominant coding; \( \beta_0=m_0=m_1,\ \beta_R=m_2-m_0 \) for recessive coding; and \( \beta_0=m_0,\ \beta_1=m_1-m_0,\ \beta_2=m_2-m_0 \) for reference-based coding. The equalities shown are conditions for exact fit, not identities that hold for every set of means.

Worked Translations

Worked Example: Means That Follow an Additive Pattern

Suppose a hypothetical trait has genotype means \(m_0=10\), \(m_1=13\), and \(m_2=16\) centimetres for \(aa,Aa,AA\), respectively. Translate these means into all four codings and check which models fit exactly.

Additive coding. The two consecutive differences are \(13-10=3\) cm and \(16-13=3\) cm. They are equal, so the additive model fits exactly. Its intercept is \(\beta_0=m_0=10\) cm and its per-copy coefficient is \(\beta_A=3\) cm. The predictions are \(10\), \(10+3=13\), and \(10+2(3)=16\) cm.

Dominant coding. Exact fit would require \(m_1=m_2\), but \(13\ne16\). The model therefore cannot reproduce all three means. If we use \(aa\) and \(Aa\) to set its coefficients, \(\beta_0=10\) cm and \(\beta_D=13-10=3\) cm, it predicts \(13\) cm for \(AA\), not the observed mean of \(16\) cm.

Recessive coding. Exact fit would require \(m_0=m_1\), but \(10\ne13\). Setting the intercept and the \(AA\) contrast gives \(\beta_0=10\) cm and \(\beta_R=16-10=6\) cm. This predicts \(10\) cm for \(Aa\), rather than \(13\) cm, so it is not an exact representation.

Reference-based coding. The intercept is \(\beta_0=10\) cm. The heterozygote contrast is \(\beta_1=13-10=3\) cm, and the \(AA\) contrast is \(\beta_2=16-10=6\) cm. The model gives \(10\), \(10+3=13\), and \(10+6=16\) cm. Both contrasts are differences from the \(aa\) reference.

Worked Example: A Dominant Pattern

For a different hypothetical trait, suppose the means are \(m_0=8\), \(m_1=12\), and \(m_2=12\) milligrams. The two genotypes carrying at least one \(A\) allele have the same mean.

Dominant coding. Because \(m_1=m_2=12\), the model fits exactly. Its intercept is \(\beta_0=8\) mg, and the carrier contrast is \(\beta_D=12-8=4\) mg. It predicts \(8\) mg for \(aa\), and \(8+4=12\) mg for each of \(Aa\) and \(AA\). Thus, \(\beta_D\) is the difference between the mean for either carrier genotype and the \(aa\) mean.

Reference-based coding. The same means give \(\beta_0=8\) mg, \(\beta_1=12-8=4\) mg, and \(\beta_2=12-8=4\) mg. These two equal contrasts make explicit that each non-reference genotype has the same mean difference from \(aa\).

Additive and recessive checks. Additive coding would require equal steps, but \(12-8=4\) mg while \(12-12=0\) mg. Recessive coding would require \(m_0=m_1\), but \(8\ne12\). Neither coding reproduces all three means exactly. The pattern fits the dominant model, but this description alone does not demonstrate a biological mechanism.

Worked Example: A Recessive Pattern

Now suppose a hypothetical enzyme concentration has means \(m_0=6\), \(m_1=6\), and \(m_2=11\) units per litre.

Recessive coding. Since \(m_0=m_1=6\), the model fits exactly. The intercept is \(\beta_0=6\) units per litre, and the \(AA\) contrast is \(\beta_R=11-6=5\) units per litre. The predictions are \(6\), \(6\), and \(6+5=11\) units per litre.

Reference-based coding. The coefficients are \(\beta_0=6\) units per litre, \(\beta_1=6-6=0\) units per litre, and \(\beta_2=11-6=5\) units per litre. The zero \(Aa\) contrast means the \(Aa\) mean equals the \(aa\) reference mean in this example.

Additive and dominant checks. The consecutive additive steps are \(6-6=0\) and \(11-6=5\) units per litre, so they are not equal. Dominant coding would require \(m_1=m_2\), but \(6\ne11\). The recessive model is the only one of these two simpler codings that reproduces all three means exactly.

Worked Example: Means That Do Not Fit a Simpler Coding

Suppose a hypothetical trait has means \(m_0=5\), \(m_1=7\), and \(m_2=14\) seconds. Translate the means, and assess whether the simpler models reproduce them exactly.

Reference-based coding. The coefficients are \(\beta_0=5\) seconds, \(\beta_1=7-5=2\) seconds, and \(\beta_2=14-5=9\) seconds. Substitution gives predictions \(5\), \(5+2=7\), and \(5+9=14\) seconds, so this coding represents all three means.

Additive coding. The consecutive steps are \(7-5=2\) seconds and \(14-7=7\) seconds. Since they differ, no single intercept and per-copy coefficient can match all three means. If we choose the intercept and slope to match the two homozygotes, then \(\beta_0=5\) seconds and \(\beta_A=(14-5)/2=4.5\) seconds per copy. This predicts \(5+4.5=9.5\) seconds for \(Aa\), not the given 7 seconds. The endpoint-based value \(4.5\) is a defined contrast here, not an exact additive fit to all three means.

Dominant and recessive coding. Dominant coding would predict the same mean for \(Aa\) and \(AA\), contrary to \(7\ne14\). Recessive coding would predict the same mean for \(aa\) and \(Aa\), contrary to \(5\ne7\). Neither fits exactly.

Why the Translation Works

The reference-based formulas follow by substituting each genotype's indicator values into the model. For \(aa\), both indicators are 0, so its model mean is \(\beta_0\). To match the observed mean, set \(\beta_0=m_0\). For \(Aa\), \(I_1=1\) and \(I_2=0\), so the model mean is \(\beta_0+\beta_1\); matching \(m_1\) requires \(\beta_1=m_1-m_0\). For \(AA\), \(I_1=0\) and \(I_2=1\), giving \(\beta_2=m_2-m_0\).

The same substitution explains why the additive model imposes a restriction. Its means for \(aa,Aa,AA\) are \(\beta_0,\beta_0+\beta_A,\beta_0+2\beta_A\). Subtracting adjacent means gives \(\beta_A\) in both cases:

(\beta_0+\beta_A)-\beta_0 = (\beta_0+2\beta_A)-(\beta_0+\beta_A) = \beta_A.

Therefore, exact additive fit requires \(m_1-m_0=m_2-m_1\). The dominant and recessive conditions follow just as directly: their coded values assign the same predictor value to two genotypes, so the model must give those genotypes the same mean.

Interpreting Coefficients Carefully

A coefficient's meaning depends on its coding. In an additive model, \(\beta_A\) is the change in modelled mean per additional \(A\) allele, under the equal-step assumption. In a dominant model, \(\beta_D\) is the mean difference between the carrier category and \(aa\), when the two carrier genotypes share a mean. In a recessive model, \(\beta_R\) is the difference between \(AA\) and the shared \(aa/Aa\) mean. In the reference-based model, \(\beta_1\) and \(\beta_2\) are separate differences from the \(aa\) mean.

These coefficients have the units of the trait, because they are differences between trait means. The intercept is also a trait mean: in these codings, it is the modelled mean for \(aa\). Changing the reference genotype or the coding changes coefficient values and interpretations, even though the underlying three means need not change.

There is an important distinction between translating supplied genotype means and fitting a regression to individual-level data. If the means do not satisfy a simpler model's restrictions, the model cannot reproduce all three exactly. A fitted regression may still estimate a coefficient, but its value can depend on genotype counts or weights and on the fitting procedure. The three means alone do not determine every such regression estimate. The exact translations above concern matching the stated genotype means, not an unspecified weighted approximation.

Common Mistake

Common mistake: “The additive coefficient is always half the difference between the two homozygote means, so the additive model fits all three genotype means.” The endpoint contrast \((m_2-m_0)/2\) sets a per-copy slope that matches the homozygotes, but it predicts the heterozygote mean \((m_0+m_2)/2\). Unless \(m_1\) equals that midpoint, all three means are not represented by the additive model. Check the two consecutive differences before claiming an exact additive fit.

A second version of the same error is to treat a contrast from one coding as if it had the same meaning under another. For example, the dominant contrast compares carriers with \(aa\), whereas the additive coefficient describes a per-copy step. Keep the model equation beside each coefficient so its interpretation stays tied to the assigned genotype values.

Key takeaway: Start with the three means, write the genotype values for the chosen coding, and equate the model predictions to those means. The reference-based coding always represents three means exactly; additive, dominant, and recessive codings do so only when their respective equality conditions hold.

Check Your Understanding

For each question, state the relevant condition or show the coefficient calculation.

For means \(m_0=4\), \(m_1=7\), and \(m_2=10\) units, what are the exact additive intercept and per-copy coefficient?
For means \(m_0=3\), \(m_1=8\), and \(m_2=8\) milligrams, calculate the dominant-model coefficients and explain why the model fits exactly.
For means \(m_0=9\), \(m_1=9\), and \(m_2=13\) centimetres, calculate the recessive intercept and contrast.
With \(aa\) as the reference, what are the reference-based coefficients for means \(m_0=20\), \(m_1=18\), and \(m_2=25\) units?
Why does knowing three genotype means not always determine the coefficient from a fitted additive regression to individual-level data?