Why Imputation Quality Matters in Conditional Tests
A conditional association asks whether a variant is associated with a phenotype after accounting for other variants in the model. When genotypes are imputed, the model generally uses dosages, not directly observed allele counts. Each dosage summarizes uncertainty about an unobserved genotype. That uncertainty can differ across variants, so a joint test may depend on how well each variant—and its relationship to the other modeled variants—was imputed.
This matters especially when interpreting a secondary signal. If the lead variant is poorly imputed, conditioning on its dosage may not fully adjust for the lead genotype. A correlated secondary variant can then retain an association even when it has no independent effect in the underlying model. Conversely, uncertainty in the secondary dosage can reduce the information available to detect a real conditional effect. Neither pattern is diagnosed by a p-value alone.
A common imputation summary is an INFO score or dosage \(R^2\). These metrics are intended to summarize how much information the imputed dosages carry about the underlying genotypes. Implementations differ: they may use different estimators, assumptions, or reference quantities. A useful idealized relationship comes from the law of total variance:
Under this idealization, the fraction \(\operatorname{Var}(D_j)/\operatorname{Var}(G_j)\) measures the proportion of genotype variance represented by the dosages. Real INFO and \(R^2\) estimates are not all calculated as this exact ratio. They should be interpreted according to their software definition, and not as a direct guarantee that a particular individual's genotype is correct.
Dosages in a Conditional Model
For a quantitative phenotype \(Y_i\), a two-variant additive model can be written as
where \(D_{i1}\) and \(D_{i2}\) are imputed dosages, \(\mathbf{C}_i\) contains covariates, and the coefficients describe associations with the dosage predictors conditional on the other terms. The test of a secondary variant is typically \(H_0:\beta_2=0\), with covariates and the other variant included in the model. A Wald test uses the estimated coefficient divided by its standard error; under the null and usual large-sample model conditions, this statistic is approximately standard normal.
This analysis requires more than acceptable marginal INFO values. The phenotype model and covariates must be appropriate, observations must be analyzed consistently, and the genotype information must support the conditional comparison. In particular, the joint dosage relationship matters: marginal quality metrics for two variants do not establish that their imputed linkage disequilibrium is accurate. A marginal INFO score also does not show how well the lead dosage captures genotype variation that is relevant after conditioning.
Worked Example: Interpreting a Variant-Specific INFO Score
Suppose a hypothetical imputed variant has allele frequency \(p=0.25\). Under Hardy–Weinberg proportions, the genotype variance is \(2p(1-p)\). In a hypothetical sample, the dosage variance is estimated as 0.270 and the average posterior genotype variance as 0.105.
The expected genotype variance is
The variance decomposition is consistent with the supplied values:
The idealized dosage-variance fraction is therefore \(0.270/0.375=0.720\), or 72.0%. If the imputation software reports an INFO score near 0.72 under a comparable definition, that indicates appreciable uncertainty relative to a well-observed genotype. It does not mean that exactly 72% of individuals have correct genotypes, nor does it by itself determine whether a conditional association is biased. The score is a warning to inspect the role of this variant in the joint model and perform sensitivity checks.
How an Uncertain Lead Can Leave a Secondary Signal
The key issue can be seen by residualizing the secondary dosage against the lead dosage. For clarity, first omit other covariates; with covariates, the same argument applies after adjusting for them. Let \(R_2\) be the residual from the linear projection of \(D_2\) on \(D_1\). By the Frisch–Waugh–Lovell result, the coefficient of \(D_2\) in the joint regression is the slope from regressing the phenotype residual on \(R_2\).
Suppose the true phenotype model contains only the lead genotype, \(Y=\beta_1G_1+\varepsilon\), and the error has mean zero and is uncorrelated with the dosages. Then the population coefficient of \(D_2\) in a regression on both dosages is
The first equality is the definition of the linear projection slope after residualization. For the second, substitute \(Y=\beta_1G_1+\varepsilon\): then \(\operatorname{Cov}(R_2,Y)=\beta_1\operatorname{Cov}(R_2,G_1)+\operatorname{Cov}(R_2,\varepsilon)\), and the final covariance is zero under the stated condition. Thus, even with no independent secondary effect, the fitted secondary coefficient need not be zero if the secondary dosage retains correlation with the true lead genotype after adjustment for the lead dosage. Imperfect imputation is one way this can occur; linkage disequilibrium and the joint dosage structure also matter.
Worked Example: Residual Lead Association After Conditioning
Consider hypothetical centered, rescaled variables with \(\operatorname{Var}(G_1)=1\), \(\operatorname{Var}(D_1)=0.64\), \(\operatorname{Cov}(G_1,D_1)=0.64\), and \(\operatorname{Var}(D_2)=1\). Let \(\operatorname{Cov}(G_1,D_2)=0.80\) and \(\operatorname{Cov}(D_1,D_2)=0.512\). Suppose the phenotype follows \(Y=0.20G_1+\varepsilon\), with no independent effect of variant 2.
Residualizing \(D_2\) on \(D_1\) gives a projection slope \(0.512/0.64=0.80\), so \(R_2=D_2-0.80D_1\). Its variance is
Its covariance with the true lead genotype is
The conditional secondary coefficient is consequently
This nonzero coefficient arises despite the stipulated absence of an independent secondary effect: the residualized secondary dosage still carries information about \(G_1\). In a hypothetical sample of \(n=1000\), suppose the fitted standard error is 0.0405. The Wald statistic is \(0.09756/0.0405=2.41\), giving a two-sided normal-approximation p-value of about 0.016. That p-value is evidence against a zero dosage-model conditional coefficient under the test assumptions; it does not establish that the secondary variant has an independent biological effect.
Joint Tests and Sensitivity Analysis
When testing \(k\) secondary variants together, the null hypothesis is that all their conditional coefficients are zero. Let \(\hat{\boldsymbol{\beta}}_S\) be their estimated coefficient vector and \(\widehat{V}\) its estimated covariance matrix. The Wald statistic is
Under the joint null and standard large-sample conditions, \(W\) is approximately distributed as a chi-squared random variable with \(k\) degrees of freedom. The covariance terms matter: testing each coefficient separately and informally combining the p-values is not the same as this joint test. Imputation uncertainty may affect the coefficient estimates and their covariance, so a joint p-value should be considered alongside variant-specific quality and sensitivity results.
Worked Example: A Joint Secondary-Signal Test and Its Sensitivity
In a hypothetical conditional model, two secondary dosages have estimated coefficients \(0.10\) and \(0.06\) phenotype units per dosage unit. Their standard errors are 0.05 and 0.04, and their estimated coefficient correlation is 0.30. The covariance matrix is therefore
For \(\hat{\boldsymbol{\beta}}_S=(0.10,0.06)^T\), the determinant is \(0.0025(0.0016)-0.0006^2=0.00000364\). Using the inverse of this two-by-two matrix gives
With \(k=2\), the chi-squared reference distribution gives \(p\approx0.0867\). This joint test does not reject the joint null at a 0.05 significance level. It does not show that both effects are zero; rather, the data and model do not provide sufficient evidence to reject the joint null at that threshold.
Suppose the second secondary variant has INFO 0.66, while the first has INFO 0.98. In a hypothetical sensitivity refit using an alternative imputation run, assume the coefficient vector changes to \((0.04,0.02)^T\) and, for a simple illustration, the estimated covariance matrix remains the same. Then
which gives \(p\approx0.700\) on 2 degrees of freedom. The marked change makes the original joint evidence sensitive to the imputation analysis, especially given the lower INFO for the second variant. It does not prove that this variant is responsible for the change or that either analysis is correct. In a real sensitivity analysis, each refit must use its own estimated covariance matrix; the unchanged matrix here is an explicit simplifying assumption for the arithmetic.
Useful sensitivity checks ask whether the conditional result is stable under reasonable changes that address imputation uncertainty. Depending on the data and software, these can include comparing supported imputation outputs, checking the variant-specific dosage distributions and uncertainty summaries, or using posterior genotype draws with an analysis that correctly accounts for the additional uncertainty. Record whether the same individuals, covariates, phenotype definition, and conditioning set were used; otherwise, changes in results may have more than one explanation.
A high INFO value is reassuring but not a certificate of accurate pairwise dosage correlations. A low value is a reason for caution, not an automatic proof that a result is invalid. Likewise, filtering individuals by genotype certainty can change the analyzed sample and introduce selection effects, so it should not be treated as a routine substitute for a properly specified uncertainty analysis. Sensitivity results are most informative when they show both the estimates and their uncertainty, not just whether a p-value crosses a threshold.
Common Mistake
Check Your Understanding
Answer each question using the definitions and results above.
- What does a dosage represent, and why is it not the same as a best-guess genotype?
- Why can a secondary dosage have a nonzero conditional coefficient when the phenotype depends only on the true lead genotype?
- What does a low INFO score indicate, and what conclusion does it not justify on its own?
- For a joint test of three secondary dosages, what are the null hypothesis and the approximate degrees of freedom for the Wald reference distribution?
- Name two features that should be held consistent or reported when comparing conditional results across imputation sensitivity analyses.