Gene-Level Association Scores from Variant Statistics

From Variant Evidence to a Gene-Level Test

A genome-wide association study reports evidence at variants, but interpretation often asks whether a genomic region assigned to a gene shows association overall. Combining variant statistics into one gene-level score can answer that narrower statistical question. The answer depends on which variants are assigned to the gene and how their association statistics are related through linkage disequilibrium (LD).

A gene-level p-value is a test of association for a specified collection of variants under a specified null model. It is not a probability that the gene is causal, nor does it identify which variant or biological mechanism produced the association. Those distinctions matter especially when nearby variants are correlated or when several genes could plausibly be linked to the same locus.

Definition: A gene-level association score is a statistic that combines association evidence from a defined set of variants assigned to a gene. A gene-level p-value is the probability, under the specified no-association null model, of obtaining a score at least as extreme as the observed score. Its interpretation is conditional on the variant-set rule, the score, and the null calibration.

Define the Gene’s Variant Set First

A score cannot be interpreted until the set of variants it summarizes is specified. One positional rule might include variants whose genomic coordinates fall within the gene’s start and end coordinates. Another might extend those coordinates by a fixed flank, such as 10 kilobases on each side. A boundary can instead be based on a defined transcript span or another prespecified annotation rule. The chosen rule should be stated, and overlapping or multiple-transcript genes require a consistent policy.

Let \(G\) denote the variants assigned to one gene by that rule, and let \(m\) be the number of variants in \(G\) with usable association statistics and LD information. For each variant \(j\), let \(Z_j\) be its signed association Z statistic. Collect them in the vector \(Z=(Z_1,\ldots,Z_m)^\mathsf{T}\). The sign depends on the effect-allele convention; for a score based on squared statistics, reversing all or some allele signs does not change the score.

The LD matrix \(R\) is an \(m\) by \(m\) correlation matrix for the variant statistics under the null. Its diagonal entries are 1, and its off-diagonal entries describe their null correlations. In common large-sample settings, these correlations are approximated by correlations among genotype dosages. The variants must be aligned consistently between association results and the LD reference, including effect-allele orientation and variant identity. A reference panel that poorly matches the study’s ancestry or variant coverage can give unreliable calibration.

A Quadratic Gene-Level Score

A transparent way to aggregate evidence is to sum squared variant statistics, optionally using prespecified nonnegative weights. A large value indicates that one or more variants have substantial evidence, regardless of the direction of their effects.

Definition: For fixed nonnegative weights \(w_1,\ldots,w_m\), let \(W\) be the diagonal matrix with those weights. The weighted quadratic gene score is \(Q=Z^\mathsf{T}WZ=\sum_{j=1}^{m}w_jZ_j^2\). The unweighted score uses \(w_j=1\) for every variant.

For this score, the standard null calibration assumes that, when none of the variants is associated, the joint statistic vector is approximately multivariate normal: \(Z\sim N(0,R)\). This is an asymptotic model; its adequacy depends on valid variant-level association statistics, an appropriate LD matrix, and a sufficiently accurate normal approximation. Under the null, the distribution of \(Q\) depends on LD, so it is generally not a chi-square distribution with \(m\) degrees of freedom.

Theorem: Under \(Z\sim N(0,R)\), \(Q\) has the null distribution \(\sum_{k=1}^{m}\lambda_k\chi^2_{1,k}\), where \(\lambda_k\) are the eigenvalues of \(R^{1/2}WR^{1/2}\), and the \(\chi^2_{1,k}\) variables are independent chi-square random variables with 1 degree of freedom. Zero eigenvalues contribute zero. The gene-level p-value is the upper-tail probability of this weighted sum at the observed \(Q\).

In practice, software can evaluate this weighted chi-square tail using numerical methods. If \(R\) is the identity matrix and all weights equal 1, every eigenvalue is 1 and \(Q\) has a chi-square distribution with \(m\) degrees of freedom. When variants are correlated, the eigenvalues change: several highly correlated variants do not supply the same amount of independent evidence as the same number of independent variants.

Why the LD Calibration Works

The null distribution follows by expressing the correlated Z statistics in terms of independent standard normal variables. This also shows why a gene’s variant count alone does not determine the reference distribution.

Represent the null statistics.
Under the stated null model, write \(Z=R^{1/2}U\), where the components of \(U\) are independent standard normal variables.

Substitute into the score.
Then \(Q=U^\mathsf{T}R^{1/2}WR^{1/2}U\), a quadratic form in independent standard normal variables.

Diagonalize the quadratic form.
Diagonalize the symmetric matrix \(R^{1/2}WR^{1/2}\). An orthogonal rotation of \(U\) still has independent standard normal components, so the quadratic form becomes \(\sum_k\lambda_k U_k^2\). Each \(U_k^2\) has a chi-square distribution with 1 degree of freedom, giving the stated mixture.

If LD is estimated from a finite reference sample, \(R\) is itself uncertain. A reference that omits variants, uses inconsistent alleles, or represents a different LD pattern can alter the eigenvalues and therefore the p-value. The formula gives a calibration conditional on the LD matrix used; it does not remove errors in that matrix.

Worked Example: Two Independent Variants

Suppose a hypothetical gene’s prespecified interval contains two variants. Their signed Z statistics are \(Z_1=2.0\) and \(Z_2=2.0\), and assume the variants are independent in the appropriate reference, so \(R\) is the two by two identity matrix. Use the unweighted score.

The observed score is \(Q=Z_1^2+Z_2^2=(2.0)^2+(2.0)^2=4+4=8.0\). Since \(R\) is the identity and \(W\) is also the identity, both eigenvalues are 1. Thus \(Q\) has a chi-square distribution with 2 degrees of freedom under the null.

For a chi-square variable with 2 degrees of freedom, the upper-tail probability at \(q\) is \(\exp(-q/2)\). Therefore, \(p=\exp(-8.0/2)=\exp(-4.0)=0.0183\). Under the specified null model, a score of 8.0 or larger would occur about 1.83% of the time. This is evidence against the no-association null for this two-variant set; it does not show that the gene itself causes the phenotype.

Worked Example: Perfectly Correlated Variants

Now suppose a hypothetical interval contains two variants in perfect LD, with \(Z_1=Z_2=2.0\). The LD matrix is \(R=\begin{pmatrix}1&1\\1&1\end{pmatrix}\), which has eigenvalues 2 and 0. This singular matrix represents one independent direction of variation, not two.

The observed unweighted score is again \(Q=(2.0)^2+(2.0)^2=8.0\). The null distribution is \(2\chi^2_1+0\chi^2_1\), or equivalently twice a chi-square variable with 1 degree of freedom. Its upper-tail probability at 8.0 is \(P(\chi^2_1\geq 8.0/2)=P(\chi^2_1\geq 4.0)\), approximately 0.0455.

The score is numerically the same as in the independent-variant example, but its calibrated p-value differs because the null variation differs. Treating the two statistics as independent would give 0.0183 and overstate the evidence under this perfectly correlated model. In real data, LD is rarely exactly perfect across all variants, but the example shows why counting every correlated statistic as independent is not a valid general calibration.

Boundary Choices Change the Question

Changing a gene boundary changes the variant set, the observed score, and often the LD matrix used for calibration. A wider window may capture additional association evidence, but it also changes what is being tested. Boundary rules should be set before inspecting which choice gives the smallest p-value. A positional assignment is a convenient statistical definition; it does not prove that every included variant acts through that gene.

Worked Example: A Flanking-Window Sensitivity Check

Consider a hypothetical gene spanning coordinates 100,000 to 199,999. Three variants have coordinates and signed Z statistics: 150,000 with \(Z=2.4\), 180,000 with \(Z=1.2\), and 205,000 with \(Z=2.0\). For this illustrative calculation, assume the included variants are independent and use an unweighted score.

With the gene-body boundary alone, the first two variants are included. The score is \(Q=(2.4)^2+(1.2)^2=5.76+1.44=7.20\). Under independence, \(Q\) has a chi-square distribution with 2 degrees of freedom, giving \(p=\exp(-7.20/2)=\exp(-3.60)=0.0273\).

If the prespecified rule instead extends the interval 10,000 bases beyond each end, the variant at 205,000 is also included. The score becomes \(Q=7.20+(2.0)^2=11.20\). With three independent variants, \(Q\) has a chi-square distribution with 3 degrees of freedom; its upper-tail probability at 11.20 is approximately 0.0107. The lower p-value reflects a different variant set and test, not proof that the expanded interval identifies a causal gene. In an actual analysis, LD among all included variants would need to be incorporated rather than assumed away.

Interpreting and Reporting the Result

A gene-level p-value describes how surprising the observed aggregate statistic would be under a specified null model for the defined variant set. It is not \(P(\text{gene is causal}\mid\text{data})\), and it does not distinguish direct gene effects from linkage to a causal variant acting through another gene. A small value can arise because one variant has a strong association, because several variants contribute evidence, or because the locus has complex LD. The quadratic score alone does not tell these explanations apart.

A useful report states the gene-boundary rule, the number of variants tested, the score and weighting scheme, how the LD matrix was obtained, and the resulting p-value. It should also state whether the p-value is nominal or adjusted for testing multiple genes. A nominal p-value describes one gene-level test; testing many genes increases the chance of observing small values by chance. Any claim about a causal gene requires evidence beyond this association score.

Common Mistakes

Assuming a sum of squared Z statistics always has \(m\) degrees of freedom. That is true for the unweighted score only when the variants are independent under the null. With LD, use the eigenvalue-weighted chi-square null distribution.
Expanding a gene window after seeing the results. A window chosen to maximize the score changes the test and can make the reported p-value too optimistic. Prespecify the rule or report the window sensitivity transparently.
Calling a small gene-level p-value proof of a causal gene. The test concerns association in a defined variant set. Correlated variants and positional assignments do not establish the causal gene or mechanism.
Using an incompatible LD reference. Incorrect variant matching, allele alignment, or ancestry mismatch can misrepresent the covariance of the Z statistics and miscalibrate the p-value.

Key takeaway: A gene-level association score is meaningful only together with its variant-boundary rule and LD-aware null calibration. Its p-value quantifies evidence against a specified no-association model; it is not a posterior probability of causality.

Check Your Understanding

Try these questions before reviewing the worked examples.

Why must a gene-level analysis specify a genomic boundary rule before interpreting its score?
Under what condition does the unweighted sum of squared Z statistics have a chi-square distribution with \(m\) degrees of freedom?
What role does the LD matrix play in calibrating a quadratic gene-level score?
Two variants have identical Z statistics. Explain why their combined evidence can differ depending on whether they are independent or in perfect LD.
What does a small gene-level p-value support, and what does it not establish about the gene?