Why Use Dosages in Association Tests?
An imputed genotype is uncertain: for an individual, the imputation output may assign probabilities to having zero, one, or two copies of an allele. An association test needs a predictor for each person. Replacing the posterior probabilities with one selected genotype—a hard call—discards uncertainty. A dosage instead summarizes the posterior distribution by its expected allele count. A dosage retains the posterior mean, which a hard call generally discards; both summaries discard information in the full posterior, and neither is uniformly more informative than the other.
The introductory material on genotype posteriors and dosage establishes how probabilities over genotypes yield an expected genotype. Here the focus is how that quantity enters association models, what the resulting coefficients mean, and where the usual dosage-regression approach is an approximation.
The effect allele must be defined consistently across individuals and variants. If the allele coding is reversed, the dosage becomes \(2-D_i\), and the direction of the estimated additive effect reverses. A hard call is commonly the genotype with the largest posterior probability; it is not generally equal to the dosage. Two people can also have the same dosage but different posterior uncertainty.
Worked Example: From Posterior Probabilities to Dosage
Suppose a hypothetical person's imputation posterior probabilities for zero, one, and two copies of the effect allele are \(0.20\), \(0.60\), and \(0.20\). The dosage is \(D=0(0.20)+1(0.60)+2(0.20)=1.00\) allele copies. The most probable genotype is one copy, so the hard call is also one in this case.
The dosage does not tell us that the person certainly has one copy. For example, a different posterior \((0,1,0)\) also gives dosage \(0(0)+1(1)+2(0)=1\), but in that case the imputation assigns all probability to one copy. The first posterior is uncertain across three genotypes; the second is not. Dosage regression uses the same predictor value for both people and therefore does not retain this difference in uncertainty.
For contrast, if the posterior is \((0.45,0.40,0.15)\), the dosage is \(0(0.45)+1(0.40)+2(0.15)=0.70\), while the most probable genotype—and thus the hard call—is zero copies. Turning that posterior into a hard call would replace an expected count of 0.70 by 0.
Additive Dosage Models
An additive association model treats each additional copy of the effect allele as having the same change in the outcome's modelled scale. For a quantitative phenotype \(Y_i\), a common model is linear regression with dosage and covariates:
Here \(\mathbf C_i\) is a vector of covariates such as age, sex, or ancestry-related variables; \(\beta_D\) is the change in the conditional mean of the phenotype per one additional effect-allele copy, holding the included covariates fixed. The usual small-sample \(t\)-test for this coefficient assumes independent errors with mean zero, constant variance, and normally distributed errors conditional on dosage and covariates. Large-sample inference can be more robust to nonnormal errors, but does not remove the need for a suitable mean model and valid variance estimates.
For a binary outcome, a standard dosage model is logistic regression:
In this model, \(\exp(\beta_D)\) is the conditional odds ratio for a one-copy increase in dosage, with other covariates held fixed. It is an odds ratio, not a risk ratio or a change in probability. Logistic-model inference requires independent observations and a correctly specified conditional log-odds model; the usual Wald test also relies on an adequate large-sample approximation.
Dosage and Hard-Call Summaries
A short conditional-expectation argument explains why dosage is a natural predictor. Suppose the outcome model satisfies \(E(Y_i\mid G_i,Z_i,\mathbf C_i)=\beta_0+\beta_DG_i+\boldsymbol{\gamma}^{\mathsf T}\mathbf C_i\), for example because the outcome error has conditional mean zero given \(G_i,Z_i,\mathbf C_i\). If the imputation posterior is appropriate conditional on \(Z_i,\mathbf C_i\), then averaging over the uncertain genotype gives:
The equality follows because expectation is linear. Thus, under those conditions, dosage gives the correct conditional mean for a linear additive model. A hard call \(H_i\), by contrast, is a selected genotype, not the posterior expectation; generally \(H_i\ne E(G_i\mid Z_i)\). Replacing \(D_i\) with \(H_i\) can therefore misrepresent the expected allele count and reduce association-test information.
The same argument does not make ordinary logistic regression on dosage an exact integration over genotype uncertainty. The logistic function is nonlinear, so in general the expected outcome probability across possible genotypes is not the logistic probability evaluated at the expected genotype. For instance, with no covariates, an intercept of zero, and a genotype effect of one, suppose a posterior places probability \(0.5\) on \(G=0\) and \(0.5\) on \(G=2\). Its dosage is 1. The average of the two genotype-specific probabilities is \((0.5000+0.8808)/2=0.6904\), whereas the probability from plugging dosage 1 into the logistic model is \(1/(1+e^{-1})=0.7311\). Dosage-based logistic regression is widely used, but this calculation shows it is not the same as integrating the likelihood over the posterior genotype probabilities.
Worked Example: Testing a Quantitative Association
Consider a hypothetical four-person data set with phenotype values in arbitrary measurement units and effect-allele dosages as follows: \((D_i,Y_i)=(0,1),(1,2),(1,2),(2,4)\). For illustration, fit a linear model with an intercept and dosage only. This tiny data set is not a realistic genomic study; it makes the calculation transparent.
The sample means are \(\bar D=1\) and \(\bar Y=2.25\). The dosage slope is \(\hat\beta_D=\sum_i(D_i-\bar D)(Y_i-\bar Y)/\sum_i(D_i-\bar D)^2\). The numerator is \(1.25+0+0+1.75=3.00\), and the denominator is \(1+0+0+1=2.00\). Therefore \(\hat\beta_D=3.00/2.00=1.50\) phenotype units per additional effect-allele copy. The intercept is \(2.25-(1.50)(1)=0.75\).
The fitted values are \(0.75,2.25,2.25,3.75\); the residuals are \(0.25,-0.25,-0.25,0.25\). Thus the residual sum of squares is \(4(0.25^2)=0.25\). There are \(4-2=2\) residual degrees of freedom, so the estimated residual variance is \(0.25/2=0.125\). The standard error of the slope is \(SE(\hat\beta_D)=\sqrt{0.125/2}=0.250\), because the sum of squared deviations in dosage is 2. The test statistic for \(H_0:\beta_D=0\) is \(t=1.50/0.250=6.00\), with \(df=2\). Under the model's normal-error assumptions, the two-sided p-value is approximately \(0.0267\).
This calculation gives evidence against a zero linear dosage association at the 0.05 level under the stated model. It does not establish a causal effect. The estimate is \(1.50\) phenotype units per allele copy, and the very small sample makes the inference highly imprecise: a 95% \(t\)-interval is \(1.50\pm4.303(0.250)\), or approximately \([0.424,2.576]\) units per copy. The assumed error distribution and variance cannot be meaningfully checked with four observations.
Worked Example: Interpreting a Logistic Dosage Result
In a hypothetical study of 800 people (320 cases and 480 controls), a logistic regression includes imputed dosage, age, sex, and ancestry covariates. Suppose the fitted dosage coefficient is \(0.350\) with standard error \(0.120\). The Wald statistic is \(z=0.350/0.120=2.92\), giving a two-sided p-value of approximately \(0.00354\) under the large-sample standard normal approximation.
The estimated odds ratio per additional effect-allele copy is \(\exp(0.350)=1.42\). Conditional on the included covariates and the model, this corresponds to 1.42 times the odds of case status per additional copy, not a 42-percentage-point increase in disease probability. A 95% Wald interval for the log-odds coefficient is \(0.350\pm1.96(0.120)=[0.115,0.585]\). Exponentiating the endpoints gives an odds-ratio interval of approximately \([1.12,1.80]\). It describes uncertainty in the model-based odds ratio; it is not a probability statement about where the fixed parameter lies.
This interpretation depends on correct allele coding, an adequate logistic mean model, appropriate covariate adjustment, and valid inference. In particular, the calculation treats dosage as the model predictor; it is not a full integration of the binary-outcome likelihood over each person's posterior genotype distribution.
Assumptions, Diagnostics, and Alternatives
Dosage association tests rely on more than the arithmetic of a regression fit. The imputation probabilities should be reasonably calibrated for the people and variants analyzed, allele coding should be harmonized, and genotype information should not be systematically missing in a way that creates spurious association. The additive model should be suitable for the scientific question: it imposes a linear change per allele copy on the quantitative mean, or on the log odds in the logistic model. Covariates must address important sources of confounding, including ancestry structure where appropriate. An association after adjustment is still an association, not proof of causation.
Useful checks include confirming the effect-allele definition and dosage range, inspecting dosage distributions and missingness, and checking that implausible dosage values or coding errors are absent. For quantitative outcomes, inspect residual patterns for nonlinearity and unequal variance, and assess whether influential observations dominate the result. For binary outcomes, check model fit and sparse-data problems. Compare results across sensible covariate specifications and examine whether the effect direction is consistent with the stated allele coding. A surprising result should prompt checks of sample alignment, allele orientation, and imputation quality rather than an automatic biological interpretation.
Dosage regression is often preferable to hard-call regression because it uses the expected allele count rather than making a categorical decision for every person. Hard calls may be useful when a downstream method requires discrete genotypes, but they can discard information, especially when posterior probabilities are diffuse. A more complete alternative is a method that uses the genotype posterior probabilities directly in an uncertainty-integrated likelihood or test. Such methods can distinguish posterior distributions that have the same dosage, but require suitable implementation and assumptions. Dosage regression is a practical summary-based approach, not a guarantee that all genotype uncertainty has been propagated.
Check Your Understanding
Answer each question, then compare with the solutions below.
- A hypothetical posterior over genotypes \(0,1,2\) is \((0.30,0.50,0.20)\). What is the dosage, and what is the hard call?
- In a linear dosage model, \(\hat\beta_D=-0.40\) with standard error \(0.15\). Calculate the Wald \(t\)-statistic and state the estimated effect in context if the outcome is measured in milligrams per litre.
- Why can two people with the same dosage still have different genotype uncertainty?
- In a logistic dosage model, what does \(\exp(\beta_D)\) represent, and why is it not generally a change in probability?
Solutions. (1) \(D=0(0.30)+1(0.50)+2(0.20)=0.90\) copies. The largest posterior probability is 0.50 at genotype 1, so the hard call is one copy. (2) \(t=-0.40/0.15=-2.67\). The estimated conditional mean outcome decreases by \(0.40\) milligrams per litre per additional effect-allele copy, holding model covariates fixed. (3) Dosage records only the posterior mean; different probability distributions can have the same mean but different spreads or shapes. (4) It is the conditional odds ratio for a one-copy increase, holding included covariates fixed. Logistic regression models log odds, and odds do not change by a fixed probability amount across baseline risks.