Why Scan for Residual Association?
A strong association at one variant does not tell us whether it captures all the genetic association in a locus. Nearby variants can be correlated through linkage disequilibrium (LD), so a lead variant may partly tag another associated variant. After fitting a model that includes the lead, a scan of the remaining variants asks a narrower question: does any variant still associate with the phenotype after accounting for the selected lead?
This procedure is called a residual scan or sequential conditional scan. It can reveal evidence for a secondary signal when a remaining variant has a sufficiently strong conditional association. It is a model-based search, however, not a guarantee that the variants selected are causal or that each represents a distinct biological mechanism.
The Model and the Sequential Procedure
For a quantitative phenotype, let \(Y_i\) be the phenotype for person \(i\), \(G_{ij}\) the genotype dosage at variant \(j\), and \(Z_i\) the vector of covariates. A dosage is typically coded as the number of copies of a designated allele, with values from 0 to 2. At a given scan stage, let \(S\) denote the set of variants already selected. The model for testing candidate \(j\notin S\) is
The conditional association test for \(j\) evaluates \(H_0:\beta_j=0\) against an alternative that \(\beta_j\ne 0\), while retaining the covariates and all variants in \(S\). Under the usual linear-model assumptions—appropriate mean specification, independent observations, and correctly estimated standard errors—the coefficient estimate divided by its standard error gives a \(t\)-statistic. The residual scan compares these tests across candidates; it does not test only the lead again.
A practical sequence is:
Use a defined initial association scan and a pre-specified rule, such as selecting the variant with the smallest p-value among those eligible for analysis.
Fit the model with the lead and the covariates, then test every eligible variant not already in the model.
Account for the number of candidates tested, using a specified multiple-testing procedure or threshold. If a variant qualifies, add it to the conditioning set.
Rescan the variants still outside the set. Stop when none meets the rule, or when a pre-specified maximum model size or other stopping criterion is reached.
The model and test must be appropriate for the phenotype and study design. For example, the linear model above is for a quantitative outcome; a binary outcome generally requires an appropriate generalized linear model. Population structure, relatedness, and other design features may also require corresponding adjustment. A conditional scan cannot repair a misspecified analysis simply by adding more variants.
Worked Example: Following a Residual Scan to Its Stopping Rule
Consider hypothetical results from a quantitative-trait analysis of \(n=1{,}000\) unrelated participants. The model includes an intercept and three covariates. An initial scan considered 100 eligible variants and selected variant L as the lead. For this example, use Bonferroni family-wise error control at \(\alpha=0.05\) for the variants tested at each residual-scan stage. The reported test statistics use the corresponding linear-model residual degrees of freedom.
After conditioning on L, there are 99 candidates. The threshold is
Suppose candidate C has a conditional coefficient estimate of \(0.100\) phenotype units per dosage unit and \(SE=0.0250\) phenotype units per dosage unit. Its statistic is
With 994 residual degrees of freedom, its two-sided p-value is about \(6.8\times10^{-5}\). This is below \(0.0005051\), so C meets the stated threshold and is added to the conditioning set. The estimated conditional association is positive: holding L and the covariates fixed, one additional dosage unit at C is associated with an estimated \(0.100\)-unit increase in the phenotype.
The next scan tests 98 remaining candidates, so its Bonferroni threshold is \(0.05/98=0.0005102\). Suppose candidate D now has conditional estimate \(0.0900\), standard error \(0.0250\), and \(t=0.0900/0.0250=3.60\), with a two-sided p-value of about \(0.0003339\). D also meets the threshold and is added. In the following scan, suppose the smallest p-value among the 97 remaining candidates is \(0.0012\). Since \(0.0012>0.05/97=0.0005155\), no candidate meets the rule, and this scan stops.
Under this procedure, C and D each provide evidence of residual association at the stage when they were tested. The result supports a model with additional conditional signals, but does not establish that L, C, and D are causal variants.
Why a Conditional Association Can Differ from a Marginal One
A marginal association tests one variant without including other variants in the model. A conditional association estimates its association after accounting for the selected variants. When genotypes are correlated, these quantities need not agree: the marginal slope can combine information associated with both the candidate and the conditioning variant.
A compact derivation makes the distinction clear for two genotypes. Suppose the phenotype and both genotypes have been adjusted for the covariates, the genotypes are centered and standardized to variance 1, and their correlation is \(r\). Let \(\beta_{j\mid L}\) and \(\beta_{L\mid j}\) be their coefficients in the joint two-genotype regression. The marginal slope for \(j\) is its covariance with the adjusted phenotype. Substituting the joint model into that covariance gives
By the same reasoning, the other marginal slope is
These are simultaneous equations, not an identity that uses the lead’s marginal slope in place of its conditional coefficient. Solving them together gives
The denominator shows why the two-genotype conditional estimate can become unstable as \(|r|\) approaches 1: the two genotypes provide increasingly similar information. This formula is an explanation for a simplified setting, not a substitute for fitting the full model used in an actual residual scan.
Worked Example: Solving for a Conditional Slope
In a hypothetical, covariate-adjusted two-variant analysis, suppose the standardized genotypes have correlation \(r=0.60\), the marginal slope for candidate \(j\) is \(0.220\) phenotype units per standardized genotype unit, and the marginal slope for L is \(0.260\) phenotype units per standardized genotype unit. The two simultaneous equations are
Solving for the candidate’s conditional slope,
Thus the candidate’s conditional slope is \(0.100\) phenotype units per standardized genotype unit. Simply subtracting \(r\) times the lead’s marginal slope would give \(0.220-(0.60)(0.260)=0.0640\), which is not the conditional slope. The simultaneous equations show why the denominator \(1-r^2\) is required.
Lead Selection and Interpretation
The residual scan depends on the variant chosen for conditioning. Conditioning on different variants changes the model and therefore the conditional tests. This is especially relevant when nearby variants are correlated but do not tag the phenotype’s association equally well. The scan’s conclusions should consequently name the conditioning variant or set, rather than report a candidate’s “association” without specifying the model.
Worked Example: A Residual Result Depends on the Lead
Suppose a hypothetical study measures the same quantitative phenotype in the same participants and compares two plausible lead choices, A and B. A was selected by the pre-specified initial scan because it had the stronger marginal association. In the model conditioning on A, candidate C has estimate \(0.0800\) phenotype units per dosage unit and \(SE=0.0200\), giving
Its two-sided p-value is about \(6.334\times10^{-5}\) with large residual degrees of freedom. In a separate conditional model that uses B instead, suppose C has estimate \(0.0300\), the same standard error, and
That test has a two-sided p-value of about \(0.134\). These hypothetical results do not contradict each other: they answer different conditional questions. They indicate that C’s estimated association is more apparent after accounting for A than after accounting for B in these models. They do not, by themselves, establish which variant is causal or explain why the results differ.
There is an additional issue when the lead and residual candidates are selected using the same data in which their p-values are reported. The lead is chosen partly because its observed association is large, and the residual scan searches many candidates for unusually strong results. Ordinary p-values from a model that treats the selected set as fixed do not automatically account for this entire selection process. A multiple-testing correction for the candidates in one residual scan addresses that scan’s stated family of tests, but it does not necessarily correct every consequence of selecting the lead from the same data. Interpretation should therefore distinguish a useful discovery procedure from a fully selection-adjusted inferential procedure.
A Bonferroni threshold is one simple option for controlling the family-wise probability of at least one false rejection across a specified set of tests: for \(m\) tests and target level \(\alpha\), compare each p-value with \(\alpha/m\). Other pre-specified multiplicity procedures may be appropriate. Whatever method is used, report the number of variants tested at each stage and the threshold or procedure applied. When feasible, independent replication can help assess whether selected conditional signals persist.
Common Mistakes and Practical Reporting
- Calling every residual hit a causal secondary variant. A conditional association is evidence under a specified model. LD, unmeasured or imperfectly measured variants, and model limitations can affect which variant appears as a hit.
- Using marginal results as if they were conditional results. The coefficient and p-value from a scan that omits the lead do not answer the question tested by a model that includes it.
- Stopping at the first added variant. If a candidate passes the pre-specified rule, the sequential procedure calls for another scan of the remaining candidates, unless the protocol specifies a different stopping rule.
- Treating a per-test threshold as a scan-wide threshold. A p-value of \(0.01\) may look small, but it does not meet a Bonferroni threshold of \(0.05/99\).
- Assuming selection has no inferential effect. Ordinary conditional test results do not automatically account for choosing the lead and searching the same data for residual hits.
A clear report states the phenotype and model, the covariates, the lead variant and how it was selected, the number of candidates tested at each stage, the multiplicity rule, and the conditional estimates, standard errors, and p-values for selected signals. It should also say when the scan stopped. This makes the result interpretable as a sequence of conditional tests rather than as an unqualified list of associated variants.
Check Your Understanding
Try these questions before reviewing the explanations in the tutorial.
- In a residual scan that conditions on a lead variant, what hypothesis is tested for each remaining candidate?
- A scan tests 49 candidates using Bonferroni family-wise error control at \(\alpha=0.05\). What per-test threshold does this give?
- Why must the marginal-slope equations for two correlated standardized genotypes be solved together to obtain a conditional slope?
- What does it mean if a candidate’s conditional p-value meets the scan threshold after conditioning on one lead, but not after conditioning on another?
- Why does applying a multiple-testing correction within the residual scan not necessarily account for selecting the lead from the same data?