Why Analyze a Case-Control Burden?
Rare variants are often too uncommon to test one at a time with useful precision. A burden analysis instead combines variant information into one prespecified summary for each person, then tests whether that summary is associated with case status. This can provide a more focused test of a gene or region, but the conclusion concerns the combined burden—not any one variant in it.
The analysis has two important parts. First, the burden definition determines which variants contribute and how their genotypes are weighted. Second, logistic regression tests whether the burden is associated with the probability of being a case. The weights and variant set must be fixed for the test being reported; choosing them after inspecting the association results changes the inferential problem.
Model and Hypotheses
Let \(Y_i=1\) denote a case and \(Y_i=0\) a control. A logistic model for the probability of case status is
where \(Z_i\) contains any prespecified covariates, \(\alpha\) is an intercept, and \(\gamma\) contains covariate coefficients. The burden test concerns
Under the null, the model includes the intercept and covariates but not a burden effect. The score test evaluates whether the data show evidence for a nonzero burden coefficient at that null model, without requiring a full-model estimate of \(\beta\). A positive score indicates that larger burdens tend to occur among cases more often than predicted under the null; a negative score indicates the reverse.
For a one-degree-of-freedom burden test, the score statistic is the squared efficient score divided by its null variance. With an intercept only and a common fitted null probability \(\hat p\), it has a particularly simple form. Write \(\bar B\) for the mean burden. Then
For this intercept-only null model, \(\hat p\) is the observed case proportion. The centering of \(B_i\) accounts for estimating the intercept: a burden shared equally by every participant cannot distinguish cases from controls. Under the null and appropriate large-sample conditions, \(Q\) is compared with a chi-square distribution with one degree of freedom. The signed score \(U\) indicates direction, while \(Q\) and its two-sided p-value quantify evidence against \(H_0\).
Why the Score Is Centered
A brief derivation explains why the intercept-only formula uses deviations from the mean burden. Under \(H_0\), the fitted probability is \(\hat p\), and the raw score for \(\beta\) is \(\sum_i B_i(Y_i-\hat p)\). But the intercept is estimated from the same outcomes. Its score is \(\sum_i(Y_i-\hat p)=0\). Subtracting \(\bar B\) times this zero quantity leaves the raw score unchanged:
The information for the burden coefficient must likewise account for estimating the intercept. Removing the intercept direction from the burden leaves the centered values \(B_i-\bar B\); their weighted sum of squares is \(\hat p(1-\hat p)\sum_i(B_i-\bar B)^2\). Dividing the squared score by this information gives \(Q\). With covariates, the same principle applies, but burden is residualized against the null-model covariates using weights \(p_i(1-p_i)\).
Worked Example: Constructing the Prespecified Burden
Suppose a hypothetical analysis includes two variants in a region. Before examining case-control results, investigators specify weights \(w_1=2\) and \(w_2=1\), and define \(B_i=2G_{i1}+G_{i2}\), where each \(G\) is the counted-allele dosage. The following genotypes are illustrative.
| Person | \(G_{i1}\) | \(G_{i2}\) | Burden calculation | \(B_i\) |
|---|---|---|---|---|
| A | 1 | 0 | \(2(1)+0\) | 2 |
| B | 0 | 1 | \(2(0)+1\) | 1 |
| C | 1 | 1 | \(2(1)+1\) | 3 |
| D | 0 | 0 | \(2(0)+0\) | 0 |
For person C, the burden is \(3\), while for person D it is \(0\). These values are the predictor used in the regression. A burden of \(3\) does not mean that a particular variant has a three-unit effect, nor does it identify which genotype component is associated with case status. It is simply the value produced by the stated rule.
Worked Case-Control Score Test
The next example uses an intercept-only null model so that every calculation can be checked from burden summaries by case status. Imagine 100 unrelated participants in a hypothetical case-control sample: 50 cases and 50 controls. The prespecified burden has the following distribution.
| Status | Number with \(B_i=0\) | Number with \(B_i=1\) | Number with \(B_i=2\) |
|---|---|---|---|
| Cases | 35 | 10 | 5 |
| Controls | 40 | 8 | 2 |
Worked Example: Calculating the Logistic Score Statistic
Under the intercept-only null, the fitted case probability is \(\hat p=50/100=0.5\). First calculate the burden totals. Among cases, \(\sum B_i=10(1)+5(2)=20\), and \(\sum B_i^2=10(1^2)+5(2^2)=30\). Among controls, \(\sum B_i=8(1)+2(2)=12\), and \(\sum B_i^2=8(1^2)+2(2^2)=16\). Thus \(\sum_i B_i=32\) and \(\sum_i B_i^2=46\).
The observed case burden is \(20\). Under the null, half of the total burden is expected among cases, so the score is
The centered sum of squares of burden is
Therefore, the null information and score statistic are
Compared with a chi-square distribution with one degree of freedom, \(Q=1.790\) gives a two-sided p-value of approximately \(0.181\). The positive score means burdens are higher among cases than expected under the null, but this sample does not provide strong evidence against \(H_0\) at the conventional 0.05 level. This p-value is not the probability that \(H_0\) is true, and the test does not establish that the burden causes disease.
Adjustment for a Covariate
With covariates, the null fitted probability can differ across participants. In matrix notation, let \(X\) be the design matrix containing the intercept and all null-model covariates, let \(\hat p_i\) be the null fitted probabilities, and set \(W\) to the diagonal matrix with entries \(\hat p_i(1-\hat p_i)\). The burden is adjusted for these nuisance predictors by subtracting its weighted projection onto \(X\):
Here the null-model score equations imply \(X^T(Y-\hat p)=0\), so the raw burden score equals the score formed from the residualized burden: \(B^T(Y-\hat p)=\widetilde B^T(Y-\hat p)\). Residualization removes the part of the burden predictable from the intercept and covariates. This is essential: a burden that differs across covariate groups should not be credited with association merely because case proportions also differ across those groups.
Worked Example: Adjusting for a Two-Level Covariate
Consider eight hypothetical participants from two covariate strata, with two cases and two controls in each stratum. The null model includes an intercept and stratum. Each stratum therefore has fitted null probability \(\hat p=0.5\). The burdens are shown below.
| Stratum | Case burdens | Control burdens | Total burden |
|---|---|---|---|
| A | 2, 1 | 0, 1 | 4 |
| B | 1, 0 | 0, 1 | 2 |
Within stratum A, the mean burden is \(4/4=1\). The centered burdens are \(1,0,-1,0\), so their squared deviations sum to \(2\). The score contribution is the case burden total minus half the stratum total: \(3-(0.5)(4)=1\).
Within stratum B, the mean burden is \(2/4=0.5\). The centered burdens are \(0.5,-0.5,-0.5,0.5\), so their squared deviations sum to \(1\). The score contribution is \(1-(0.5)(2)=0\). Combining strata gives \(U=1+0=1\), and the information is
The one-degree-of-freedom chi-square p-value is approximately \(0.248\). The positive score has the same directional interpretation as before, but the covariate-adjusted test gives little evidence of association in this tiny illustration. Its asymptotic p-value should not be treated as reliable: eight participants cannot support a dependable large-sample approximation.
Assumptions and Interpretation
A score test is meaningful only for the model and burden that were specified. The phenotype must be coded consistently as case or control, and the logistic mean model must be appropriate under the null, including the chosen covariates. The observations should be independent under the analysis model, or their dependence must be handled with an appropriate method. The burden definition, weights, and included variants should be set before inspecting the test result if the reported p-value is to have its usual interpretation.
The usual chi-square reference distribution is an approximation, not a guarantee. It can be inaccurate when there are very few variant carriers, highly influential observations, or little information for estimating a burden effect. A nominal p-value from a sparse sample may therefore be misleading; use a calibration method suitable for the design and data when the large-sample conditions are doubtful.
A significant burden test supports an association between the prespecified aggregate burden and case status under the fitted model. It does not show that every included variant has an effect, that effects are equal, or that any particular variant is responsible. Nor does it establish causation. The burden compresses multiple genotypes into one number, so distinct variant patterns can produce the same burden and the test cannot disentangle their contributions.
Common Mistakes
- Attributing the result to one variant. A significant test of \(B_i\) does not identify a responsible variant. The tested predictor is the weighted sum, not a variant-specific coefficient.
- Ignoring the intercept or covariates. Comparing raw case and control burden totals without accounting for the null model can give the wrong score and variance. Use the centered formula for an intercept-only model or the covariate-adjusted score for a model with covariates.
- Reading \(Q\) as a signed effect. The statistic \(Q=U^2/I\) is nonnegative and contains no direction by itself. Check the sign of \(U\) to describe whether burden is higher or lower among cases.
- Treating a small-sample chi-square p-value as exact. When carriers are scarce, the reference approximation may be poor. Check whether the information and calibration are adequate before interpreting the p-value.
- Changing weights after seeing the result. Choosing weights to maximize association and reporting the ordinary prespecified-weight p-value ignores that selection. The test described here assumes weights were fixed independently of the outcome association being tested.
Check Your Understanding
Use these questions to check the logic of the analysis.
- For a burden defined as \(B_i=3G_{i1}+G_{i2}\), what is the burden for a person with dosages \(G_{i1}=1\) and \(G_{i2}=2\)?
- In an intercept-only case-control score test, why is burden centered around its sample mean?
- If \(U\) is negative, what does that indicate about the direction of the burden association, and does it by itself establish significance?
- What is the appropriate interpretation of a significant test of a weighted burden containing several variants?
- Why might the usual chi-square p-value be unreliable when only a handful of participants carry any included variant?