Why Compare Two Burdens?
A rare-variant burden test compresses several genotype columns into one score per person, then tests whether that score is associated with the phenotype. The compression is consequential: it determines which genetic patterns contribute strongly and which distinctions are discarded. An unweighted carrier burden gives each person a score of zero or one, regardless of how many qualifying variants they carry. A frequency-weighted allele burden instead sums allele dosages after multiplying them by variant-specific weights.
These approaches can produce different results even for the same people and the same phenotype. A comparison is useful for understanding what each score measures, not for selecting whichever gives the smaller p-value after looking at the results. The worked calculations below use a small, hypothetical case-control dataset. Because the sample is deliberately small, the asymptotic p-values are illustrative rather than reliable evidence of association.
Model and Score Calculation
For a transparent comparison, use an intercept-only logistic model under the null hypothesis of no association with the chosen burden. Under the alternative, add one burden score \(B_i\), either \(C_i\) or \(W_i\), as a predictor:
Assume subjects are independent, the case-control labels are correctly recorded, and the burden is fixed once the genotypes and weights are specified. The intercept-only null model gives every subject the same fitted case probability, \(\hat p=\bar Y\), the observed case fraction. With no additional covariates, the score for \(\gamma\) and its null variance, after accounting for the nuisance intercept, are:
The two expressions for \(U\) agree because \(\sum_i(Y_i-\hat p)=0\). When a normal approximation is appropriate, the standardized statistic is \(Z=U/\sqrt{\widehat{\operatorname{Var}}_0(U)}\), and a two-sided p-value is \(2\{1-\Phi(|Z|)\}\), where \(\Phi\) is the standard normal cumulative distribution function. This approximation requires enough information for the score statistic to be close to normal. Sparse genotypes and very small samples can make that condition doubtful; a computed asymptotic p-value does not fix the problem.
For the weighted burden in the examples, define \(q_j\) as the observed minor-allele frequency at variant \(j\), and use \(w_j=1/\sqrt{q_j(1-q_j)}\). This is an explicit inverse-standard-deviation frequency weighting for this exercise; it is not the only possible weighting scheme. Weights must be specified before testing, and their choice affects the burden and therefore the test.
Worked Example: Constructing Both Burdens
Consider six hypothetical cases and six hypothetical controls, typed at three qualifying variants. Each entry is the allele dosage. The case labels and genotypes below are invented for this worked example.
| Person | Status | Variant A | Variant B | Variant C |
|---|---|---|---|---|
| 1 | Case | 1 | 1 | 1 |
| 2 | Case | 0 | 1 | 1 |
| 3 | Case | 0 | 1 | 1 |
| 4 | Case | 0 | 0 | 0 |
| 5 | Case | 0 | 0 | 0 |
| 6 | Case | 0 | 0 | 0 |
| 7 | Control | 1 | 1 | 1 |
| 8 | Control | 0 | 0 | 1 |
| 9 | Control | 0 | 0 | 1 |
| 10 | Control | 0 | 0 | 0 |
| 11 | Control | 0 | 0 | 0 |
| 12 | Control | 0 | 0 | 0 |
The carrier score is one for a person with any nonzero dosage and zero otherwise. Thus the case scores are \(1,1,1,0,0,0\), and the control scores are also \(1,1,1,0,0,0\). This collapsing discards whether a carrier has one or several alternate alleles, and it does not distinguish which variant was carried.
There are 24 called chromosomes. The observed frequencies are \(q_A=2/24=1/12\), \(q_B=4/24=1/6\), and \(q_C=6/24=1/4\). The corresponding weights are \(w_A=1/\sqrt{(1/12)(11/12)}=12/\sqrt{11}=3.618\), \(w_B=1/\sqrt{(1/6)(5/6)}=6/\sqrt{5}=2.683\), and \(w_C=1/\sqrt{(1/4)(3/4)}=4/\sqrt{3}=2.309\). Person 1 therefore has weighted burden \(3.618+2.683+2.309=8.611\), while person 8 has weighted burden \(2.309\). The weighted case scores are \(8.611,4.993,4.993,0,0,0\); the weighted control scores are \(8.611,2.309,2.309,0,0,0\).
Worked Score-Test Comparison
Worked Example: Calculating the Two Score Statistics
In the dataset above, \(n=12\) and \(\hat p=6/12=0.5\). For the carrier burden, there are six scores equal to one and six equal to zero, so \(\bar C=0.5\) and \(\sum_i(C_i-\bar C)^2=6(0.5)^2+6(-0.5)^2=3\). Among cases, three have \(C_i=1\); among controls, three have \(C_i=1\). Therefore:
The resulting \(Z_C=0/\sqrt{0.750}=0\), with nominal two-sided normal p-value \(1.000\). The data provide no difference in the carrier-score sums between cases and controls.
For the weighted burden, the sum of case scores is \(8.611+4.993+4.993=18.596\), and the sum of control scores is \(8.611+2.309+2.309=13.230\). Since \(Y_i-\hat p\) is \(0.5\) for cases and \(-0.5\) for controls:
For the variance calculation, the total weighted-score sum is \(31.826\), and the sum of squares is approximately \(208.813\). Hence \(\bar W=31.826/12=2.652\), and
Thus \(Z_W=2.683/\sqrt{31.102}=0.481\), with nominal two-sided normal p-value approximately \(0.631\). The weighted score is higher on average among cases in this dataset, but the standardized difference is small. The two nominal p-values are not evidence that the weighting scheme is superior: they test different one-dimensional summaries, and this sample is too small to rely on their normal calibration.
What the Comparison Assumes
Both scores reduce a multi-variant set to one number per person. A burden test using a single coefficient assumes that the variants’ effects contribute in a sufficiently coherent direction for aggregation to be informative. If some variants increase disease risk while others decrease it, their contributions can cancel. Weighting changes how strongly each allele contributes; it does not remove the shared-direction assumption.
The two burdens also encode different biological summaries. Carrier collapsing treats a person with one qualifying alternate allele like a person carrying several, and treats all qualifying variants alike. It can be useful when carrier status is the intended exposure or when dosage distinctions are not the target. A weighted allele burden retains dosage and permits variants to contribute unequally, but the result depends on the frequency estimates, weight formula, and variant set. If the weights are estimated from a small or unrepresentative sample, they may be unstable or poorly suited to the scientific question.
A single-variant analysis preserves variant-specific results rather than combining them, but it does not answer the same set-level question and can have low power for very rare alleles. In practice, burden results should be read alongside the definition of the qualifying set, the score construction, and the assumed direction of effects. Do not compare p-values as though the two tests differed only by a harmless rescaling: changing the burden changes the alternative pattern the test is sensitive to.
Worked Example: Opposing Effects Can Cancel
Suppose a separate hypothetical set contains two equally weighted variants. At the first variant, the case group has a total of three alternate alleles and the control group has one. At the second, the case group has one alternate allele and the control group has three. Use weight one for each variant, so the case weighted-burden sum is \(3+1=4\), and the control sum is \(1+3=4\). With four cases and four controls, \(\hat p=0.5\), and the weighted score is
The aggregate score has no case-control contrast, even though the variant-specific allele counts point in opposite directions. This calculation does not establish that either variant has a causal effect; it shows why a one-direction burden can lose signal when effects oppose one another. A frequency-weighted score can alter the relative contributions, but it still sums them into one direction and can still cancel.
Diagnostics and Common Pitfalls
Before interpreting a comparison, verify the genotype coding, the qualifying-variant rule, and the weights. For each variant, check allele counts and frequencies; confirm that a dosage of two contributes twice the allele weight; and confirm that carrier status uses “any nonzero dosage,” not the dosage sum itself. Then inspect the distribution of each person-level burden by phenotype. A few extreme weighted scores may dominate the result, while a carrier score can hide substantial differences in the number or identity of carried alleles.
In binary-phenotype data, sparse cells and small numbers of carriers can make the normal approximation to a score statistic inaccurate. The example has only 12 subjects, so its nominal normal p-values are calculations for comparison, not well-calibrated inferential conclusions. For sparse binary-phenotype tests, calibration methods such as saddlepoint methods may be needed; study design, covariates, and the null model also matter. The intercept-only calculation here is not a substitute for a covariate-adjusted analysis when confounders or design factors need adjustment.
- Mistake: “The weighted burden has a nonzero score, so the weighted test demonstrates an association.” Correction: A nonzero score is not by itself significant evidence. In this example, the nominal standardized statistic is only \(0.481\), and the small-sample normal calibration is doubtful.
- Mistake: “A carrier burden and an allele burden are interchangeable because both summarize the same variants.” Correction: They retain different information. The carrier score discards dosage and variant identity; the weighted score retains dosage and applies variant-specific weights.
- Mistake: “Frequency weighting corrects for opposing effects.” Correction: Weighting changes magnitudes, not directions. Contributions with opposite signs can still offset in a single burden.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- If a person has dosages \(0,1,2\) at three variants, what is the carrier burden, and what is the unweighted allele-count burden?
- In the 12-person example, why does the carrier score statistic equal zero?
- What does the frequency-weighted score retain that the carrier burden discards?
- Why should the reported normal p-values in the worked comparison not be treated as dependable evidence?
Solutions. (1) The carrier burden is one because at least one dosage is nonzero; the unweighted allele-count burden is \(0+1+2=3\). (2) Exactly three cases and three controls have carrier score one, so the case and control carrier-score sums are equal and \(U_C=0\). (3) It retains allele dosage and permits different variants to have different contributions through their weights. (4) There are only 12 subjects, so sparse data may violate the conditions needed for accurate normal calibration of the score statistic; the p-values are illustrative approximations, not reliable conclusions.