Why Additive Coding Implies Equal Genotype Steps

Why Dosage Coding Creates Equal Genotype Steps

A genotype can be represented by how many copies of a chosen allele it contains: zero, one, or two. In a quantitative-trait regression, using that count as a single numerical predictor makes a specific claim about the pattern of genotype means. Moving from zero copies to one copy changes the fitted mean by the same amount as moving from one copy to two copies. This is the sense in which dosage coding gives an additive model equal genotype steps.

The claim concerns the model’s fitted or expected means, not a requirement that every observed individual’s trait value changes by a fixed amount, nor a guarantee that the three sample genotype means will lie exactly on a straight line. The distinction matters: the additive model summarizes the mean pattern with one slope, while individual outcomes and observed group averages can vary around that pattern.

Definition: Let $G_i$ be the dosage for individual $i$ of a specified allele, coded $0$, $1$, or $2$. The one-predictor additive model for a quantitative trait $Y_i$ is $$ Y_i = \beta_0 + \beta_1 G_i + \varepsilon_i, $$ where $\beta_0$ is the intercept, $\beta_1$ is the change in the modeled mean per additional allele copy, and $\varepsilon_i$ is the individual’s deviation from that mean. The mean-model assumption is $E(\varepsilon_i\mid G_i)=0$. Consequently, $E(Y_i\mid G_i=g)=\beta_0+\beta_1g$ for $g=0,1,2$.

This model makes a linearity assumption across the three dosage values. The coefficient $\beta_1$ is a trait-scale change per copy of the coded allele, in the units used to measure $Y$. It can be positive, negative, or zero. The intercept $\beta_0$ is the modeled mean for dosage zero. These statements describe a statistical association in the specified model; by themselves, they do not establish that changing a person’s genotype would cause a change in the trait.

Deriving the Three Genotype Means

To obtain each genotype-specific mean, substitute its dosage into the conditional mean. At dosage zero, the dosage term contributes nothing. At dosage one, it contributes one slope. At dosage two, it contributes two slopes.

\begin{aligned} \mu_0 &= E(Y\mid G=0)=\beta_0,\\ \mu_1 &= E(Y\mid G=1)=\beta_0+\beta_1,\\ \mu_2 &= E(Y\mid G=2)=\beta_0+2\beta_1. \end{aligned}

Here $\mu_g$ denotes the model-implied mean trait value for genotype dosage $g$. Subtracting adjacent means gives the two genotype steps:

\mu_1-\mu_0=\beta_1 \qquad\text{and}\qquad \mu_2-\mu_1=\beta_1.

Because the two differences are equal, the middle mean must be halfway between the endpoint means. The following short derivation makes that implication explicit.

Write the endpoint means.
At dosage zero, $\mu_0=\beta_0$; at dosage two, $\mu_2=\beta_0+2\beta_1$.

Take their midpoint.
$\dfrac{\mu_0+\mu_2}{2}=\dfrac{\beta_0+(\beta_0+2\beta_1)}{2}=\beta_0+\beta_1$.

Identify the heterozygote mean.
At dosage one, $\mu_1=\beta_0+\beta_1$. Therefore, $\mu_1=(\mu_0+\mu_2)/2$.

Theorem: Under the one-predictor additive model with dosage coded $0,1,2$, the genotype-specific fitted means satisfy $$ \mu_1=\frac{\mu_0+\mu_2}{2}. $$ Equivalently, $\mu_1-\mu_0=\mu_2-\mu_1=\beta_1$. The heterozygote mean is constrained to the midpoint of the two homozygote means on the modeled quantitative-trait scale.

The midpoint is on the scale of the outcome in this model. For example, if $Y$ is measured in millimetres, the genotype steps and midpoint relation are in millimetres. The result follows from the chosen linear mean model and dosage values; it does not rely on particular allele frequencies or on Hardy–Weinberg equilibrium. Those features can affect how many individuals are observed in each genotype group, but they do not change the algebra above.

Worked Example: Deriving Means From Model Coefficients

Suppose a hypothetical additive regression predicts a quantitative trait measured in centimetres using $G=0,1,2$, with fitted intercept $\hat{\beta}_0=12$ cm and fitted slope $\hat{\beta}_1=1.8$ cm per allele copy. Find the fitted mean for each dosage and check the equal-step condition.

The fitted mean at dosage $g$ is $\hat{\mu}_g=\hat{\beta}_0+\hat{\beta}_1g$. Substituting the coefficients:

\begin{aligned} \hat{\mu}_0 &= 12+1.8(0)=12.0\text{ cm},\\ \hat{\mu}_1 &= 12+1.8(1)=13.8\text{ cm},\\ \hat{\mu}_2 &= 12+1.8(2)=15.6\text{ cm}. \end{aligned}

The first step is $13.8-12.0=1.8$ cm, and the second is $15.6-13.8=1.8$ cm. The midpoint check gives $(12.0+15.6)/2=13.8$ cm. Thus, this fitted additive model predicts a mean 1.8 cm higher per additional copy of the coded allele. This is a model-based association, not by itself a causal conclusion.

Fitted Means Versus Observed Group Means

In an actual sample, the observed mean trait value among dosage-zero individuals need not equal the observed mean among dosage-one individuals minus a constant, and the three observed group means need not satisfy the midpoint relation. The additive regression still produces fitted values on a straight line. When the model is fit by ordinary least squares using dosage alone, it chooses the intercept and slope that minimize the sum of squared differences between observed trait values and their fitted values.

A useful consequence is that the fitted group means have the same midpoint relation even if the observed group means do not. If the sample contains individuals at all three dosage values, the fitted value for any individual in a given dosage group is the same, because the model uses the same $G$ for everyone in that group. The model’s predicted group value is therefore $\hat{\beta}_0+\hat{\beta}_1g$. The fitted values remain equally spaced by construction; the observed group means are allowed to depart from them.

Worked Example: Fitting an Additive Line to Unequal Group Means

Consider hypothetical data on a quantitative trait measured in score units. There are two individuals with $G=0$, three with $G=1$, and two with $G=2$. Their observed group means are 10, 13, and 18 score units, respectively. These observed means are not equally spaced: the steps are $13-10=3$ and $18-13=5$. Fit the ordinary least-squares regression of the trait on dosage and find its three fitted means.

For a regression with an intercept and one numerical predictor, the sample mean dosage is $\bar{G}=(2(0)+3(1)+2(2))/7=7/7=1$. The overall trait mean, calculated by weighting each group mean by its group size, is $\bar{Y}=(2(10)+3(13)+2(18))/7=95/7\approx13.5714$ score units.

Because dosage is constant within each group, the slope can be calculated from the group counts and group means. The numerator is the weighted sum of dosage deviations times mean-trait deviations; the denominator is the weighted sum of squared dosage deviations:

\begin{aligned} \hat{\beta}_1 &= \frac{\sum_g n_g(g-\bar{G})(\bar{Y}_g-\bar{Y})} {\sum_g n_g(g-\bar{G})^2}\\ &= \frac{ 2(0-1)(10-95/7)+3(1-1)(13-95/7)+2(2-1)(18-95/7) }{ 2(0-1)^2+3(1-1)^2+2(2-1)^2 }\\ &=\frac{16}{4}=4.0\text{ score units per copy}. \end{aligned}

The intercept is $\hat{\beta}_0=\bar{Y}-\hat{\beta}_1\bar{G}=95/7-4(1)=67/7\approx9.5714$ score units. The fitted means are therefore:

\begin{aligned} \hat{\mu}_0&=67/7+4(0)=9.5714,\\ \hat{\mu}_1&=67/7+4(1)=95/7\approx13.5714,\\ \hat{\mu}_2&=67/7+4(2)=123/7\approx17.5714. \end{aligned}

Their adjacent differences are both $4.0$ score units, and their midpoint check is $(9.5714+17.5714)/2=13.5714$ score units. The observed group means remain 10, 13, and 18; the fitted values are the additive model’s best-fitting equally spaced predictions, not replacements for the observed data. The unequal group counts affect the fitted line because the least-squares criterion gives each individual equal weight, so a group with more individuals contributes more to the fit.

Reading the Slope and the Constraint

The sign of $\beta_1$ determines the direction of the fitted steps. If $\beta_1>0$, each additional copy of the coded allele corresponds to an increase in the modeled trait mean. If $\beta_1<0$, each additional copy corresponds to a decrease. If $\beta_1=0$, all three fitted genotype means are equal. The amount $\beta_1$, rather than the genotype label alone, specifies the size of each step.

Worked Example: Interpreting a Negative Dosage Slope

Suppose a hypothetical model for a quantitative trait measured in milligrams has fitted coefficients $\hat{\beta}_0=42$ mg and $\hat{\beta}_1=-3.5$ mg per copy of a specified allele. Calculate and interpret the genotype means.

Substituting each dosage into the fitted mean gives:

\begin{aligned} \hat{\mu}_0&=42+(-3.5)(0)=42.0\text{ mg},\\ \hat{\mu}_1&=42+(-3.5)(1)=38.5\text{ mg},\\ \hat{\mu}_2&=42+(-3.5)(2)=35.0\text{ mg}. \end{aligned}

The first step is $38.5-42.0=-3.5$ mg, and the second is $35.0-38.5=-3.5$ mg. The heterozygote fitted mean is the midpoint: $(42.0+35.0)/2=38.5$ mg. In this model, each additional allele copy is associated with a 3.5 mg lower mean trait value. That interpretation is about the fitted mean per copy, not a claim that every individual with an additional copy will have a trait value exactly 3.5 mg lower.

Equal steps are a substantive restriction on the three genotype means. If the observed pattern is markedly non-linear, a one-slope dosage model may not describe it well, even though the fitted values will always be equally spaced. A model with separate genotype parameters can represent a wider range of mean patterns, but that is a different specification. The important point here is to recognize what a single dosage coefficient asserts: there is one common increment from one dosage category to the next.

Common Mistake and Practical Check

A common mistake is to treat the midpoint result as a fact that must hold exactly for the three observed sample means. For example, a student might see observed means of 10, 13, and 18, note that 13 is not the midpoint of 10 and 18, and conclude that an additive regression cannot be fit. That conclusion confuses observed group averages with model-fitted means. A regression can be fit; its fitted genotype means lie on the best-fitting dosage line and obey the midpoint constraint, while the observed means can differ.

The reverse error is also possible: because fitted means are equally spaced, one might assume that dosage coding has demonstrated that the biological effect is truly additive. It has not. The equality follows from the model’s linear predictor and coding choice. Whether that restriction is a reasonable summary of the data is a scientific and statistical modeling question, not an algebraic consequence of observing a genotype.

Key takeaway: With $G=0,1,2$ and a single dosage slope, the fitted means are $\hat{\beta}_0$, $\hat{\beta}_0+\hat{\beta}_1$, and $\hat{\beta}_0+2\hat{\beta}_1$. Thus, the heterozygote fitted mean is halfway between the two homozygote fitted means, and each adjacent fitted step equals the slope.

Check Your Understanding

Answer each question using the dosage model and its mean interpretation.

A fitted model has intercept 7 trait units and slope 2 trait units per allele copy. What are the fitted means for dosages zero, one, and two?
Show in one equation why the dosage-one fitted mean is the midpoint of the dosage-zero and dosage-two fitted means.
Can observed genotype-specific sample means fail to be equally spaced even when an additive dosage regression has been fit? Explain the difference between observed and fitted means.
A fitted slope is $-1.2$ score units per allele copy. State what the sign and units mean for the modeled genotype steps.
Does the midpoint relation by itself prove that the allele causally changes the trait? Explain what the relation establishes and what it does not establish.