Genotype Dosages in Conditional Models

Why Dosage Matters in a Joint Model

At a locus, a joint regression estimates the association of each included variant with an outcome while accounting for the other included variants and covariates. This conditional comparison is sensitive to how genotypes are represented. A hard call assigns each person one genotype, usually the genotype with the highest imputation probability. A dosage instead retains the expected allele count implied by the genotype probabilities. Replacing dosages with hard calls can therefore change both the predictors and the conditional estimates.

For a biallelic variant, the allele count $G$ is 0, 1, or 2. Imputation provides probabilities for those three possible genotypes. The dosage summarizes these probabilities as an expected allele count. It is not a claim that an individual literally has a fractional number of alleles.

Definition: If $P(G=g\mid I)$ is the imputation probability of genotype $g$, given the available imputation information $I$, the allele-count dosage is $$ D=E(G\mid I)=\sum_{g=0}^{2}gP(G=g\mid I) =P(G=1\mid I)+2P(G=2\mid I). $$ A hard call is a single assigned genotype, commonly the $g$ with the largest probability. Dosages lie between 0 and 2; hard calls take only the values 0, 1, and 2.

Dosage in Conditional Regression

For a quantitative outcome $Y_i$, a joint linear regression with two variants can be written

Y_i=\beta_0+\beta_1D_{i1}+\beta_2D_{i2}+\boldsymbol{\gamma}^{\mathsf T}\mathbf{C}_i+\varepsilon_i.

Here $D_{i1}$ and $D_{i2}$ are the allele-count dosages for person $i$, $\mathbf{C}_i$ is a vector of included covariates, and $\varepsilon_i$ is the model error. Under the additive dosage specification, $\beta_1$ is the expected change in the outcome per one-unit increase in variant 1 dosage, conditional on variant 2 dosage and the covariates. The same interpretation applies to $\beta_2$, conditioning on variant 1 dosage and the covariates.

Ordinary least squares is appropriate for this example when the conditional mean is adequately represented by the stated linear model, observations are independent, and the error variance assumptions used for the reported standard errors are appropriate. In particular, the usual homoskedastic standard errors assume errors have a common variance conditional on the predictors. Dosages are predictors in this regression; the fitted model does not turn them into observed genotypes. With a binary outcome, a corresponding generalized linear model can include dosages in its linear predictor, but the coefficient scale and interpretation differ.

The regression estimates association conditional on the variables included in the model. It does not establish that a variant causes the outcome. Also, conventional regression standard errors treat the supplied dosages as predictor values. They should not be described as automatically incorporating every source of uncertainty in genotype imputation.

Worked Example: Calculating an Allele-Count Dosage

Suppose a hypothetical participant has imputation probabilities $P(G=0\mid I)=0.25$, $P(G=1\mid I)=0.60$, and $P(G=2\mid I)=0.15$. Calculate the dosage and the most-probable-genotype hard call.

Dosage. The probabilities sum to $0.25+0.60+0.15=1.00$, and the expected allele count is

$$ D=0(0.25)+1(0.60)+2(0.15)=0+0.60+0.30=0.90. $$

Hard call. The largest probability is 0.60, for $G=1$, so the most-probable-genotype hard call is 1. The dosage, 0.90, preserves the probability-weighted allele count; the hard call replaces the probability distribution with one integer.

For another hypothetical participant with probabilities $0.45, 0.10, 0.45$ for genotypes 0, 1, and 2, respectively, the dosage is $0(0.45)+1(0.10)+2(0.45)=1.00$. The two most-probable genotypes tie, so a hard-call rule would need a tie-breaking convention. The dosage is still well defined.

Worked Comparison in a Two-Variant Model

The following hypothetical data make the change from dosage to hard call explicit. The outcome $Y$ is a quantitative measurement in arbitrary units. Variant 1 is represented by genotype values $X_1$, assumed to be observed for this illustration. Variant 2 has dosage $D_2$, as well as the most-probable hard call $H_2$. The hard calls are compatible with the dosages: dosages 0.2 and 0.4 have call 0; 1.8 and 1.6 have call 2; and 0.6, 1.4, 0.8, and 1.2 have call 1 under possible genotype-probability distributions.

Person	$X_1$	$D_2$	$H_2$	$Y$
1	-1	0.2	0	-2.2
2	1	1.8	2	4.2
3	-1	0.4	0	-3.9
4	1	1.6	2	1.9
5	-1	0.6	1	-2.6
6	1	1.4	1	2.6
7	-1	0.8	1	-2.3
8	1	1.2	1	2.3

Worked Example: Estimating the Conditional Dosage Effect

Fit the hypothetical linear model $Y_i=\beta_0+\beta_1X_{1i}+\beta_2D_{2i}+\varepsilon_i$, with an intercept. The predictors and outcome each have mean zero in this dataset, so the fitted intercept is 0. The model assumes a linear conditional mean and independent errors; the standard errors below additionally use the usual common-variance OLS formula.

The centered cross-products are $S_{11}=\sum X_1^2=8$, $S_{22}=\sum(D_2-\bar D_2)^2=2.4$, and $S_{12}=\sum X_1(D_2-\bar D_2)=4$. The outcome cross-products are $S_{1Y}=22$ and $S_{2Y}=11.6$. For two centered predictors, the coefficient estimates solve the cross-product normal equations:

\begin{pmatrix}8&4\\4&2.4\end{pmatrix} \begin{pmatrix}\hat\beta_1\\\hat\beta_2\end{pmatrix} = \begin{pmatrix}22\\11.6\end{pmatrix}.

The determinant is $8(2.4)-4^2=3.2$. Thus

\hat\beta_1=\frac{2.4(22)-4(11.6)}{3.2}=2.00,\qquad \hat\beta_2=\frac{8(11.6)-4(22)}{3.2}=1.50.

The fitted values follow $Y=2X_1+1.5(D_2-\bar D_2)$. The residuals are $1,1,-1,-1,0,0,0,0$, so the residual sum of squares is $4$. There are $8-3=5$ residual degrees of freedom, accounting for the intercept and two slopes; hence $MSE=4/5=0.800$.

The inverse cross-product matrix has diagonal entries $2.4/3.2=0.750$ for variant 1 and $8/3.2=2.50$ for variant 2. Therefore, the dosage coefficient's standard error is

SE(\hat\beta_2)=\sqrt{MSE\left(\frac{8}{3.2}\right)} =\sqrt{0.800(2.50)}=\sqrt{2.00}=1.414.

In this hypothetical dataset, the estimated conditional association for variant 2 is 1.50 outcome units per one-unit increase in dosage, conditional on $X_1$, with standard error 1.414 outcome units per dosage unit. The estimate describes this fitted association, not a causal effect.

Worked Example: Replacing Dosages With Hard Calls

Now fit the same model but replace $D_2$ with the hard calls $H_2$. The mean hard call is 1, so use the centered predictor $H_2-1$. From the table, its cross-products are $\sum(H_2-1)^2=4$, $\sum X_1(H_2-1)=4$, and $\sum(H_2-1)Y=12.2$. The outcome cross-product with $X_1$ remains 22. Thus the coefficient equations are

\begin{pmatrix}8&4\\4&4\end{pmatrix} \begin{pmatrix}\hat\beta_{1,H}\\\hat\beta_{2,H}\end{pmatrix} = \begin{pmatrix}22\\12.2\end{pmatrix}.

The determinant is $8(4)-4^2=16$, giving

\hat\beta_{2,H} =\frac{8(12.2)-4(22)}{16} =\frac{9.6}{16}=0.600.

The fitted hard-call model's explained sum of squares is $2.45(22)+0.600(12.2)=61.22$. The total centered outcome sum of squares is 65.40, so the residual sum of squares is $65.40-61.22=4.18$. With 5 residual degrees of freedom, $MSE=4.18/5=0.836$. The hard-call slope's standard error is

SE(\hat\beta_{2,H}) =\sqrt{0.836\left(\frac{8}{16}\right)} =\sqrt{0.418}=0.6465.

The hard-call estimate is 0.600 per one-unit increase in hard-call genotype, rather than the dosage estimate of 1.50 per one-unit increase in dosage. Its standard error is also smaller in this particular dataset. Neither difference shows that hard calls are generally preferable: changing the predictor changes its scale, its association with the other variant, and the residual fit. In particular, a smaller standard error alone does not establish that an estimate better represents the association captured by the imputation probabilities.

Why the Conditional Estimate Can Change

A conditional coefficient compares outcome differences associated with a predictor after accounting for the other predictors. If two variants are correlated, their joint regression estimates depend on how each predictor varies after the other has been accounted for. Dosages and hard calls can have different correlations with the other variant, and the hard-call conversion discards distinctions between people who share a call but have different dosages.

The cross-products in the dosage example make this dependence visible. The dosage predictor has $S_{22}=2.4$ and $S_{12}=4$. Its variation remaining after linear adjustment for $X_1$ is

S_{22}-\frac{S_{12}^2}{S_{11}} =2.4-\frac{4^2}{8}=0.400.

The corresponding adjusted outcome cross-product is

S_{2Y}-\frac{S_{12}S_{1Y}}{S_{11}} =11.6-\frac{4(22)}{8}=0.600.

Dividing the adjusted outcome cross-product by the adjusted dosage variation gives $0.600/0.400=1.50$, the same conditional dosage coefficient calculated from the joint equations. This illustrates why a marginal association is not a substitute for a joint-model coefficient: the latter uses the variation left after accounting for the other included predictors.

Key takeaway: In a joint model, use each variant's dosage as a probability-weighted allele count when the aim is to retain imputation information in the predictor. Interpret its coefficient per dosage unit, conditional on the other model terms. A hard-call analysis is a different predictor specification, so its estimate and standard error need not match the dosage analysis.

Common Mistakes and Practical Checks

Treating a dosage as an observed fractional genotype. A dosage of 0.90 is an expected allele count given the imputation information, not a literal genotype. Keep the distinction clear when describing the analysis.
Assuming hard calls and dosages are interchangeable. A hard call discards probability detail. Check which representation was used and report it; do not substitute one silently.
Assuming the dosage standard error must be smaller. Precision in a conditional model depends on residual predictor variation, correlation with the other terms, and residual variance. The worked example has a larger dosage standard error, which is possible and does not by itself establish that the hard-call estimate is better.
Comparing coefficients without their predictor units. State whether a coefficient is per dosage unit or per hard-call allele-count unit, and state the outcome units.
Interpreting a conditional association as causal. Conditioning on included variants changes the association being estimated; it does not by itself identify a causal variant or prove a biological mechanism.

Check Your Understanding

Use the definitions and calculations above to answer the following questions.

If genotype probabilities for 0, 1, and 2 alleles are 0.50, 0.30, and 0.20, what is the dosage, and what is the most-probable-genotype hard call?
In a joint model containing two variant dosages and age, what does the coefficient of the first dosage describe?
Why can replacing dosages with hard calls change a conditional coefficient even when both analyses include the same variants?
Does a smaller standard error in a hard-call analysis prove that hard calls retain more imputation information? Explain.
What additional assumption is needed to use the usual homoskedastic OLS standard error, beyond specifying a linear conditional mean and independent observations?