Why the Regression Setup Matters
A locus-level association analysis is defined not just by a genomic region, but by a particular outcome, sample, genotype encoding, covariate set, and collection of variants in the model. Changing any of these can change the regression parameter being estimated. A useful setup therefore makes those choices explicit before interpreting a coefficient as an association with a genetic variant.
The focus here is the model specification. For a quantitative phenotype, a linear regression provides a clear starting point. The same bookkeeping applies to other regression families, but their link functions and coefficient interpretations differ. Throughout, genotype effects use an additive dosage: the expected number of copies of a specified effect allele.
Phenotype, Sample, and Genotype Dosages
Let \(Y_i\) denote the quantitative phenotype for individual \(i\), for \(i=1,\ldots,n\). State what was measured, the units, and any transformations or exclusions that define the outcome. For example, “log-transformed concentration in units of log milligrams per litre” is more informative than simply naming a biomarker. The analysis sample must also be stated: the regression uses individuals with usable phenotype, genotype, and covariate information under the chosen missing-data rules.
At variant \(j\), define \(G_{ij}\) as individual \(i\)’s dosage of a specified effect allele. With a confidently called diploid genotype, the dosage is usually 0, 1, or 2. If genotype uncertainty is represented by probabilities \(P(G_{ij}=g)\), the dosage is the expected allele count:
Thus a dosage need not be an integer. Its interpretation depends on which allele is counted: changing the effect allele reverses the dosage scale, so the allele convention must be recorded. An additive dosage model treats the expected change in the phenotype as linear in this count. It does not, by itself, include a separate dominance effect or genotype-by-environment interaction.
Worked Example: Converting Genotype Probabilities to Dosage
Suppose a hypothetical individual at a biallelic variant has genotype probabilities 0.10 for zero copies, 0.60 for one copy, and 0.30 for two copies of the chosen effect allele. Find the dosage.
Calculation. The expected allele count is \(G=0(0.10)+1(0.60)+2(0.30)=0+0.60+0.60=1.20\). The dosage is therefore 1.20 copies of the effect allele.
This is not a claim that the individual carries 1.20 physical copies. It summarizes uncertainty over possible integer genotypes. The effect-allele definition is part of the setup; if the other allele were counted, the corresponding dosage would be \(2-1.20=0.80\).
Covariates and the Variant Set
Write the covariates for individual \(i\) as a vector \(C_i\), including an intercept in the model. Depending on the study design, covariates may include age, sex, ancestry-related variables, study site, or technical variables. Their role is to define the adjusted association parameter, not to guarantee that all confounding has been removed. Each covariate needs a clear coding rule: for example, whether age is measured in years and centered, and which category is the reference for a categorical variable.
Let \(S\) denote the variant set under consideration at the locus. Specify its membership in reproducible terms, such as genomic boundaries and the quality-control rules used to retain variants. For every \(j\in S\), state the variant identifier and effect-allele convention. The set \(S\) is a modeling choice; it does not imply that every included variant has a nonzero effect or that every causal variant in the region has been observed.
For a joint additive model, a quantitative phenotype can be represented as
Here \(\beta_0\) is the intercept, \(\gamma\) contains covariate coefficients, and \(\beta_j^{J}\) is the joint coefficient for variant \(j\). The superscript \(J\) emphasizes that this coefficient is defined in a model containing the other variants in \(S\), as well as the covariates. For a phenotype measured in milligrams per litre, the coefficient is in milligrams per litre per additional copy of the specified effect allele, conditional on the model’s other predictors.
The corresponding covariate-adjusted marginal model for variant \(j\) includes that variant but not the other locus variants:
Under this convention, “marginal” does not mean unadjusted for every covariate. It means that the model represents the association with one variant at a time, adjusting for the specified \(C_i\), and omits the other variants in \(S\). The coefficient \(\beta_j^{M}\) and the joint coefficient \(\beta_j^{J}\) are different model parameters; neither should be described without naming the model that defines it.
Worked Example: Writing the Design for a Locus
Consider a hypothetical study with five individuals. The outcome is a measured concentration in milligrams per litre. The chosen covariates are an intercept, age centered at 40 years, and a binary indicator for one sex category. The locus contains variants \(v_1\) and \(v_2\), each coded as dosage of its stated effect allele. For one individual, age is 50 years, the indicator is 1, and the dosages are 1.2 and 0.
Covariate calculation. The centered age value is \(50-40=10\) years. With column order intercept, centered age, indicator, dosage at \(v_1\), dosage at \(v_2\), this individual’s design row is \((1,10,1,1.2,0)\).
Model specification. For \(S=\{v_1,v_2\}\), the joint model is \(Y_i=\beta_0+\gamma_1(\text{age}_i-40)+\gamma_2(\text{indicator}_i)+\beta_{v_1}^{J}G_{i1}+\beta_{v_2}^{J}G_{i2}+\varepsilon_i\). Its design matrix has five columns and five rows for this sample. In a marginal model for \(v_1\), the \(v_2\) dosage column is omitted; the phenotype, analysis sample, and covariate definitions should otherwise be held fixed when that is the intended comparison.
The setup identifies what each column means; this five-person description alone does not establish that the coefficients can be estimated reliably. In particular, the design must have sufficient independent variation, a condition considered below.
Marginal and Joint Parameters in a Small Example
The distinction between the two parameters is easiest to see when the phenotype is associated with two correlated dosage columns. The following hypothetical values are deliberately small so that the regression calculations can be checked directly. There are no additional covariates beyond the intercept.
Worked Example: One-Variant and Two-Variant Coefficients
Suppose four hypothetical individuals have the following phenotype and dosages:
| Individual | \(Y_i\) | \(G_{i1}\) | \(G_{i2}\) |
|---|---|---|---|
| 1 | 10 | 0 | 0 |
| 2 | 13 | 0 | 1 |
| 3 | 15 | 1 | 1 |
| 4 | 18 | 1 | 2 |
Marginal model for variant 1. Regress \(Y\) on an intercept and \(G_1\), omitting \(G_2\). The mean phenotype at \(G_1=0\) is \((10+13)/2=11.5\), and the mean at \(G_1=1\) is \((15+18)/2=16.5\). Since the dosage difference between these groups is 1, the fitted slope is \(\beta_1^{M}=16.5-11.5=5\) phenotype units per copy. The fitted intercept is 11.5, so the marginal fitted values are 11.5, 11.5, 16.5, and 16.5.
Joint model. Now fit \(Y_i=\beta_0+\beta_1^{J}G_{i1}+\beta_2^{J}G_{i2}\). The data satisfy \(Y_i=10+2G_{i1}+3G_{i2}\): for the four rows, the right-hand sides are \(10\), \(10+3=13\), \(10+2+3=15\), and \(10+2+6=18\). Thus the joint fitted coefficients are \(\beta_0=10\), \(\beta_1^{J}=2\), and \(\beta_2^{J}=3\), with zero residual for every row.
The two columns are not perfectly redundant: after centering, \(G_1\) takes values \((-0.5,-0.5,0.5,0.5)\), while \(G_2\) takes values \((-1,0,0,1)\), so they are not proportional. The joint model is therefore identifiable here. The example shows why the coefficient for \(G_1\) in a one-variant model need not equal its coefficient in a model that also includes \(G_2\). It does not establish that either coefficient is a causal effect.
Identifiability and Model Conditions
For the linear model, collect the outcome values into \(Y\), and let \(D\) be the design matrix whose columns contain the intercept, covariates, and the dosage columns specified by the model. The ordinary least-squares coefficient vector is uniquely determined when the columns of \(D\) are linearly independent. If a column is an exact linear combination of others, the data cannot distinguish all the corresponding coefficients.
A short algebraic argument explains this requirement. If \(D\) has full column rank, then for any nonzero coefficient vector \(a\), \(Da\ne 0\), and therefore
Thus \(D^{\mathsf T}D\) is positive definite and has an inverse, so the least-squares solution \((D^{\mathsf T}D)^{-1}D^{\mathsf T}Y\) is unique. Conversely, if the columns are linearly dependent, there is a nonzero \(a\) with \(Da=0\). Adding \(a\) to any coefficient vector leaves its fitted values unchanged, because \(D(\beta+a)=D\beta\). The individual coefficients are then not uniquely identified by this design.
Full rank is not the only practical consideration. Dosage columns that are nearly redundant can produce imprecise estimates even when the coefficients are technically identifiable. Also specify the regression assumptions relevant to the intended analysis. For the linear model, the conditional mean is represented by the stated linear predictor, and the residual has mean zero given the predictors. Independent observations and an appropriate variance model are needed for standard uncertainty calculations; normal residuals are an additional condition for exact small-sample normal-theory inference. These assumptions do not follow merely from writing down the design matrix.
Practical Checks Before Interpretation
A careful setup records the outcome scale, sample definition, effect alleles, covariate coding, and variant set together. When comparing a marginal and a joint parameter, keep the outcome, sample, and covariate definitions consistent unless a deliberate change is part of the question. Otherwise, differences in coefficients may reflect different analysis populations or adjustment sets as well as different variant inclusion.
The fitted coefficient is an association on the chosen phenotype scale, per allele copy, under the specified model. For example, a coefficient of 0.25 for a phenotype in milligrams per litre means a 0.25 milligrams-per-litre difference in the model’s conditional mean per additional dosage unit, holding the other included predictors fixed. It does not by itself show that changing the genotype would cause that difference.
Check Your Understanding
Use the definitions and model setup in this tutorial to answer the following questions.
- A genotype has probabilities 0.20, 0.50, and 0.30 for zero, one, and two copies of the effect allele. What is its dosage, and what does that number represent?
- In the stated convention, which predictors are included in a covariate-adjusted marginal model for one variant? Which additional predictors appear in the corresponding joint model?
- A phenotype is measured in centimetres and a dosage coefficient is 0.4. State its units and the model-based interpretation of a one-unit increase in dosage.
- Why can exact linear dependence among dosage and covariate columns prevent unique estimation of individual regression coefficients?
- Why is it incorrect to call a joint coefficient a causal effect solely because other variants in the locus were included in the model?