From Variant Set to Gene-Level Conclusion
A gene-based test asks whether a defined collection of variants is associated with a phenotype. The test is only meaningful once that collection is specified: changing its genomic boundaries, functional annotations, or frequency threshold changes the scientific target. The analysis also depends on how variants are weighted and on whether the test is designed for effects that mostly point in one direction or may differ in direction and magnitude.
An end-to-end workflow therefore links each decision to the next. First define the variant set and the analysis population. Then choose weights and a covariate-adjusted null model. Select a test, or a calibrated omnibus procedure, that reflects plausible genetic architectures. Finally check the model and calibration, account for the number of sets tested, and interpret the result as evidence about the specified set rather than as proof of a causal mechanism.
Specify the Set and the Weights
A reproducible set definition identifies the genome build, genomic boundaries or transcript definition, included variant classes, allele-frequency rule, and variant-level quality filters. It also states how allele frequencies are estimated and in which population. These details matter because a variant may be rare in one ancestry group and more common in another, and because different transcripts or annotation rules can assign different variants to the same gene.
Quality filters should be selected before examining the association results. Depending on the data, relevant checks may include genotype quality, call rate, depth, allele balance, and consistency of allele coding across samples. Excluding a variant because its observed association is inconvenient changes the analysis in a data-dependent way. If a variant has missing genotypes, the analysis must use a defined missing-data procedure; treating an unobserved genotype as a reference genotype without justification can distort both variant frequencies and test statistics.
Weights specify each variant's contribution to the test. A weight may reflect a prespecified frequency function, functional annotation, or both. For example, a frequency-based weight that gives greater emphasis to rarer alleles encodes the assumption that rarer variants are more likely, on average, to have larger effects. That is a modeling choice, not a fact guaranteed by the data. A zero or very small weight effectively excludes or downweights a variant, so the weight rule should be stated alongside the set definition.
Worked Example: Defining a Hypothetical Coding Set
Suppose an analysis uses a stated transcript on a stated genome build and includes high-quality protein-altering variants with minor-allele frequency below 1% in the analysis population. Four candidate variants have frequencies 0.004, 0.008, 0.006, and 0.032. Assume all four pass the prespecified genotype-quality filters and the first three meet the annotation rule.
The first three variants meet both the functional and frequency criteria: \(0.004<0.01\), \(0.008<0.01\), and \(0.006<0.01\). The fourth is excluded because its frequency is \(0.032\), which is not below 0.01; it would also be excluded if it failed the annotation rule. Thus the tested set contains three variants, not all four candidates.
If the analysis uses equal weights, each included variant has weight 1. If it instead uses prespecified weights 1.0, 0.8, and 1.2 for the three variants, the rationale for those weights must be defined independently of their observed phenotype associations. The choice changes the test and its target: the weighted analysis emphasizes the third variant more than the second.
Fit the Null Model Before Testing the Set
Let \(Y\) denote the phenotype, \(G_j\) the genotype vector for variant \(j\), and \(C\) the matrix of covariates, including an intercept. For a quantitative phenotype, a common null model is \(Y=C\gamma+\varepsilon\), with independent errors having mean zero and constant variance under the model. For a binary phenotype, a common null model is logistic regression, with \(P(Y_i=1\mid C_i)=\operatorname{logit}^{-1}(C_i^\mathsf{T}\gamma)\). Covariates should be selected for scientific and design reasons, not because they produce a desired association result.
The score for variant \(j\), denoted \(U_j\), measures the association signal remaining after fitting the null model with covariates. In the binary case, for example, a simple score contribution has the form \(U_j=\sum_i G_{ij}(Y_i-\hat{\mu}_i)\), where \(\hat{\mu}_i\) is the fitted event probability under the null model. Its null covariance with other scores depends on the fitted model and the genotypes. The score and covariance used by a valid test must account for covariate adjustment; raw case-control genotype differences are not a substitute.
The model assumptions include appropriate phenotype modeling, adequate control of population structure and other relevant confounding, and valid genotype data. For standard score-test approximations, the null model must be adequately estimated and the score distribution must be sufficiently well approximated by its reference distribution. Relatedness, ancestry structure, or other dependence may require a model or variance estimator that accounts for that dependence. A nominal p-value from a model that ignores material dependence need not be calibrated.
Burden Tests and Their Score Calculation
A burden test combines variant scores into one weighted score. Let \(U=(U_1,\ldots,U_m)^\mathsf{T}\) be the score vector for the \(m\) variants and \(w=(w_1,\ldots,w_m)^\mathsf{T}\) their prespecified weights. The combined score is \(U_B=w^\mathsf{T}U\). If \(V\) is the null covariance matrix of \(U\), then the null variance of the combined score is \(w^\mathsf{T}Vw\). Under the null and the score-test approximation, the standardized statistic is compared with a standard normal distribution.
This formula also explains the central burden-test assumption: weighted variant effects tend to act in a common direction, so combining their scores does not cause substantial cancellation. If directions vary, a burden score can be small even when several individual variant scores are large.
Worked Example: Covariate-Adjusted Burden Score
In a hypothetical binary-phenotype analysis, a logistic null model includes an intercept, age, and ancestry covariates. After fitting that model, the three variant scores are \(U=(0.7200,-0.3500,1.4995)^\mathsf{T}\). Use equal weights \(w=(1,1,1)^\mathsf{T}\), and suppose the fitted null model gives \(w^\mathsf{T}Vw=3.80\). These scores and their covariance are already adjusted for the stated covariates.
The combined score is \(U_B=0.7200-0.3500+1.4995=1.8695\). Its standard error under the null is \(\sqrt{3.80}=1.9494\) to four significant figures. Therefore,
For a two-sided standard normal score test, the p-value is \(2P(Z\geq 0.9590)\approx0.338\), where \(Z\sim N(0,1)\). At a significance level of 0.05, this result is not statistically significant. It does not establish that the set has no effect; it indicates that this test does not provide strong evidence against its null hypothesis. The negative middle score also illustrates how opposing directions can partly cancel in a burden sum.
Choose a Test for the Plausible Genetic Architecture
A variance-component test such as SKAT uses a quadratic form in variant scores, conventionally \(Q=\sum_j w_j^2U_j^2\). Squaring the scores avoids cancellation between positive and negative contributions. Its reference distribution is generally a weighted sum of chi-squared variables, with weights determined by the null covariance structure; it is not generally a single chi-squared distribution. A burden test is often more powerful when effects are mostly aligned, while a variance-component test can be more appropriate when effects may differ in direction or only a subset of variants is associated.
An adaptive or omnibus test addresses uncertainty about the architecture by combining evidence across component tests or burden-variance mixtures. Examples include SKAT-O and ACAT-O. The omnibus procedure must use its own valid calibration: choosing the smallest p-value among several component tests and reporting it as if it had been the only test does not account for that selection. A calibrated omnibus p-value accounts for the procedure's combination rule under its assumptions.
| Scientific expectation | Reasonable starting test | Key consideration |
|---|---|---|
| Most included effects point in the same direction | Burden test | Opposing effects can cancel in the weighted sum |
| Effects may have different directions, or only some variants may contribute | Variance-component test such as SKAT | Calibration depends on the null covariance and score distribution |
| Direction and sparsity are uncertain | Calibrated adaptive or omnibus procedure | Use the omnibus p-value, not an unadjusted minimum component p-value |
The choice should be driven by the scientific question and a defensible view of plausible effects, not by whichever test gives the smallest p-value in the observed data. If many genes or sets are tested, the analysis must also account for that multiplicity. The threshold or adjustment should reflect the number and structure of the hypotheses being tested.
Worked Example: Choosing and Interpreting an Omnibus Result
Suppose investigators analyze eight prespecified genes for a binary outcome. For one gene, the burden and SKAT component p-values are 0.021 and 0.0042. A prespecified calibrated ACAT-O procedure, combining the planned component evidence, returns p=0.0038. The familywise significance level is 0.05 and a Bonferroni threshold is used for the eight genes.
The per-gene threshold is \(0.05/8=0.00625\). The omnibus p-value satisfies \(0.0038<0.00625\), so this gene meets the stated Bonferroni criterion. The result is evidence against the null of no association for the defined set, under the model and calibration assumptions. It does not show which variant drives the evidence, nor does it establish that the gene causes the outcome. The component p-values provide context, but the inference in this example is based on the calibrated omnibus result and the prespecified multiplicity threshold.
Diagnostics, Limitations, and a Justification
Before interpreting a result, check that the analyzed variant list matches the written set definition, that allele coding and frequency calculations are consistent, and that sample inclusion and missingness are handled as planned. Examine whether covariates and ancestry structure are represented adequately, whether the null model converged, and whether the test's null calibration is credible. The earlier material on null calibration diagnostics is relevant when evaluating empirical or model-based calibration. Sparse counts, separation in a binary model, and extreme influence from one variant can make an ordinary approximation unreliable; the procedure and its calibration need to match the data conditions.
A short derivation verifies the burden variance used above. Since \(U\) has null covariance \(V\), the variance of a linear combination follows from the covariance rule for linear transformations:
Under the null, the score vector has mean zero, so \(U_B=w^\mathsf{T}U\) is centered at zero and has variance \(w^\mathsf{T}Vw\). Standardizing it gives \(Z_B\); when the score approximation is valid, \(Z_B\) is approximately standard normal. This justifies the denominator and reference distribution in the burden calculation, and shows why ignoring covariance among variant scores can give an incorrect standard error.
No single gene-based procedure is uniformly best. A burden test can lose power under mixed directions; a variance-component test can be less efficient when effects are genuinely aligned; and an omnibus test may pay a power cost for covering multiple architectures. All depend on the specified set, model, and calibration. An association signal is not a causal estimate, and a nonsignificant result is not evidence that every variant in the set has zero effect.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- A set definition includes variants with frequency below 1%. Is a variant with frequency exactly 0.010 included under a strict “below 1%” rule? Why?
- For weights \(w=(1,2)^\mathsf{T}\) and scores \(U=(0.5,-0.1)^\mathsf{T}\), calculate the burden score.
- Why can a burden score be small when a set contains associated variants with effects in opposite directions?
- When is a calibrated omnibus test preferable to selecting the smallest component p-value?
- Eight genes are tested with a Bonferroni familywise level of 0.05. Does a gene p-value of 0.007 meet the threshold?
Solutions. (1) No. A strict below-0.010 rule excludes a value equal to 0.010. (2) \(w^\mathsf{T}U=(1)(0.5)+(2)(-0.1)=0.30\). (3) Positive and negative variant scores can cancel in the weighted sum. (4) It is preferable when the plausible effect architecture is uncertain and evidence across planned component tests is to be combined while accounting for the selection or combination rule. (5) No. The threshold is \(0.05/8=0.00625\), and \(0.007>0.00625\).