Genotype-By-Local-Ancestry Interaction Tests

Why Test a Genotype-by-Local-Ancestry Interaction?

In an admixed individual, the ancestry of the chromosome carrying a particular allele can vary across genomic regions. A genetic association model that allows the effect of genotype to differ with local ancestry asks whether the conditional genotype–outcome relationship changes across local ancestry states. This is an effect-modification question: it is not, by itself, evidence that ancestry causes a change in the biological effect.

The earlier material on decomposing genotype into ancestry-specific components provides a related way to represent genetic dosage. Here the focus is the regression interaction: what it estimates, how to test it, and what conclusions the test does and does not support. The interaction model is an aggregate description based on total genotype and local ancestry; it should not automatically be interpreted as a direct estimate of separate effects for alleles carried on each ancestral background.

Definition: A genotype-by-local-ancestry interaction is a term in a regression model that allows the conditional genotype effect to vary with local ancestry. If \(G\) is the dosage of an effect allele and \(L\) is local ancestry dosage, a product term such as \(G L\), or genotype terms multiplied by indicators for ancestry categories, represents this variation.

A Categorical Local-Ancestry Model

For a diploid locus with two ancestral source populations, let \(L\) be the number of local ancestry copies from source A, so \(L\in\{0,1,2\}\). Let \(G\in\{0,1,2\}\) be the total dosage of a designated effect allele. For a quantitative outcome, one useful model is

E(Y\mid G,L,C)=\beta_0+\beta_G G+\beta_{L1}I(L=1)+\beta_{L2}I(L=2) +\delta_1G I(L=1)+\delta_2G I(L=2)+\gamma^\mathsf{T}C.

Here \(C\) denotes measured covariates, \(I(\cdot)\) is an indicator, and \(L=0\) is the reference category. The model includes genotype and local-ancestry main effects as well as interaction terms. Including the main effects is important: without them, the interaction coefficients generally do not represent the intended differences in genotype slopes.

At each local-ancestry dosage, the conditional change in mean outcome per additional effect allele is:

\begin{aligned} L=0 &: \quad \beta_G,\\ L=1 &: \quad \beta_G+\delta_1,\\ L=2 &: \quad \beta_G+\delta_2. \end{aligned}

Thus \(\delta_1\) is the difference between the genotype slope at \(L=1\) and that at \(L=0\), and \(\delta_2\) is the corresponding difference for \(L=2\). The local-ancestry coefficients \(\beta_{L1}\) and \(\beta_{L2}\) compare mean outcomes across ancestry categories at \(G=0\), conditional on the covariates. They are not genotype effects.

Worked Example: Interpreting Ancestry-Specific Genotype Slopes

Suppose a hypothetical linear regression models a quantitative biomarker in units of mg/dL. After adjustment for the same covariates in each local-ancestry category, the fitted coefficients are \(\hat{\beta}_G=0.12\), \(\hat{\delta}_1=0.04\), and \(\hat{\delta}_2=-0.03\) mg/dL per effect allele. The model also gives \(\hat{\beta}_{L1}=0.40\) and \(\hat{\beta}_{L2}=0.70\) mg/dL.

For \(L=0\), the estimated genotype slope is \(0.12\) mg/dL per additional effect allele. For \(L=1\), it is \(0.12+0.04=0.16\) mg/dL per allele. For \(L=2\), it is \(0.12-0.03=0.09\) mg/dL per allele. The interaction coefficients therefore describe slope differences: \(0.04\) mg/dL per allele for \(L=1\) versus \(L=0\), and \(-0.03\) mg/dL per allele for \(L=2\) versus \(L=0\).

The ancestry main-effect estimates compare fitted outcomes at \(G=0\): the fitted mean for \(L=1\) is \(0.40\) mg/dL above that for \(L=0\), and the fitted mean for \(L=2\) is \(0.70\) mg/dL above that for \(L=0\), holding covariates fixed. These are conditional comparisons, not causal effects of ancestry.

Testing Whether the Genotype Effect Varies

In the categorical model, the null hypothesis of no genotype-by-local-ancestry interaction is

H_0:\delta_1=\delta_2=0 \qquad\text{versus}\qquad H_a:\text{at least one of }\delta_1,\delta_2\text{ is nonzero}.

This is a two-degree-of-freedom test because it tests two restrictions. A Wald test uses the estimated interaction vector \(\hat{\boldsymbol{\delta}}=(\hat{\delta}_1,\hat{\delta}_2)^\mathsf{T}\) and its estimated covariance matrix \(\widehat V\):

W=\hat{\boldsymbol{\delta}}^\mathsf{T}\widehat V^{-1}\hat{\boldsymbol{\delta}}.

When the model is correctly specified, the null holds, and the sample is sufficiently large for the coefficient estimates to be approximately normal, \(W\) is approximately distributed as a chi-square random variable with 2 degrees of freedom. The p-value is the probability, under that reference distribution and \(H_0\), of obtaining a statistic at least as large as the observed \(W\). It is not the probability that \(H_0\) is true.

Worked Example: A Two-Degree-of-Freedom Interaction Test

Consider hypothetical fitted interaction estimates \(\hat{\boldsymbol{\delta}}=(0.10,-0.20)^\mathsf{T}\), in outcome units per effect allele. Suppose the estimated inverse covariance matrix is

\widehat V^{-1}= \begin{pmatrix} 100 & -12.25\\ -12.25 & 50 \end{pmatrix}.

The Wald statistic is

\begin{aligned} W &= \begin{pmatrix}0.10&-0.20\end{pmatrix} \begin{pmatrix}100&-12.25\\-12.25&50\end{pmatrix} \begin{pmatrix}0.10\\-0.20\end{pmatrix}\\ &=100(0.10)^2+2(-12.25)(0.10)(-0.20)+50(-0.20)^2\\ &=1.00+0.49+2.00=3.49. \end{aligned}

Against a chi-square distribution with 2 degrees of freedom, the p-value is \(P(\chi^2_2\ge 3.49)=\exp(-3.49/2)\approx0.1746\). At significance level \(\alpha=0.05\), this test does not provide evidence that the genotype slope varies across the three local-ancestry categories. It does not establish that the slopes are equal; estimates may be imprecise, particularly if some genotype–ancestry combinations are uncommon.

A Linear Trend in Local Ancestry

Instead of estimating a separate slope for each ancestry dosage, a model can impose a linear trend in the genotype slope as \(L\) increases. Centering local ancestry at one copy gives a convenient parameterization:

E(Y\mid G,L,C)=\beta_0+\beta_GG+\beta_L(L-1)+\theta G(L-1)+\gamma^\mathsf{T}C.

In this model, \(\beta_G\) is the genotype slope at \(L=1\), and \(\theta\) is the change in genotype slope per additional local ancestry A copy. The slopes at \(L=0,1,2\) are respectively \(\beta_G-\theta\), \(\beta_G\), and \(\beta_G+\theta\). The model therefore constrains these three slopes to lie on a straight line. The one-degree-of-freedom interaction test is \(H_0:\theta=0\).

Worked Example: Testing a Linear Ancestry Trend

Suppose a hypothetical fitted model gives \(\hat{\theta}=0.08\) biomarker units per effect allele per additional A-ancestry copy, with \(SE(\hat{\theta})=0.04\) in the same units. Under the large-sample Wald approximation, the test statistic is

z=\frac{\hat{\theta}}{SE(\hat{\theta})}=\frac{0.08}{0.04}=2.00.

The two-sided normal-reference p-value is approximately \(0.0455\). At \(\alpha=0.05\), this provides evidence of a linear trend in genotype slope with local ancestry under the model assumptions. If \(\hat{\beta}_G=0.10\) biomarker units per allele at \(L=1\), the fitted slopes are \(0.10-0.08=0.02\) at \(L=0\), \(0.10\) at \(L=1\), and \(0.10+0.08=0.18\) at \(L=2\), all in biomarker units per effect allele.

Justification of the Wald Reference Distribution

The interaction test follows from the approximate sampling distribution of the coefficient estimates. Under standard large-sample regression conditions, under \(H_0\), and with a consistently estimated covariance matrix, the standardized interaction vector is approximately multivariate normal with mean zero and covariance equal to the identity matrix. Its squared length therefore has an approximate chi-square distribution.

\widehat V^{-1/2}\hat{\boldsymbol{\delta}} \ \dot{\sim}\ N(\boldsymbol{0},I_2) \quad\Longrightarrow\quad \left(\widehat V^{-1/2}\hat{\boldsymbol{\delta}}\right)^\mathsf{T} \left(\widehat V^{-1/2}\hat{\boldsymbol{\delta}}\right) = \hat{\boldsymbol{\delta}}^\mathsf{T}\widehat V^{-1}\hat{\boldsymbol{\delta}} \ \dot{\sim}\ \chi^2_2.

The approximation can be poor in small samples, with sparse genotype-by-ancestry combinations, or when the fitted model or covariance estimate is inadequate. A p-value from the formula does not repair those problems.

Assumptions, Diagnostics, and Alternative Models

For a quantitative outcome, the displayed linear model assumes the conditional mean is correctly represented by the specified terms. Standard errors and Wald inference also require an appropriate error covariance model. Independent, equal-variance errors are a simple working assumption; if observations are related or have nonconstant variance, the covariance estimator or model must account for that structure. For binary or count outcomes, an appropriate generalized linear model can use the same interaction structure on its link scale. In logistic regression, for example, the interaction concerns changes in log odds per allele, not changes in probability.

Before interpreting an interaction test, check whether the data contain information to estimate it. Tabulate or otherwise inspect genotype counts across local-ancestry categories. If a category has few individuals or little genotype variation, its slope and the slope differences may be unstable. Check model fit and influential observations, and verify that covariate adjustment and the standard-error method reflect the study design. Local-ancestry calls are treated as observed in these examples; uncertainty in those calls requires additional methods.

The categorical interaction estimates two differences and allows the three genotype slopes to vary freely. The linear-trend interaction estimates one difference per ancestry copy, gaining precision if the trend restriction is reasonable but potentially missing nonlinearity. The ancestry-specific component model, using genotype dosage carried on each ancestral background, can target contrasts between those components more directly. These parameterizations answer related but not identical questions: a total-genotype-by-local-ancestry interaction is not automatically interchangeable with separate ancestry-specific allele effects.

Common pitfall: A significant interaction p-value does not mean that the genotype effect is significant in every ancestry category, nor does a nonsignificant result prove the effects are identical. The interaction test addresses differences between slopes. Estimate and report the category-specific slopes and their uncertainty if those slopes are scientifically relevant.

Interaction estimates can also be difficult to interpret when genotype and ancestry combinations have limited support. Extrapolating a fitted slope into a combination that is absent or nearly absent in the data relies heavily on the model form. Moreover, local ancestry is not randomly assigned: residual population structure, environmental differences, or technical artifacts can contribute to apparent effect modification. The interaction alone cannot distinguish these explanations from genuine ancestry-dependent genetic effects.

Check Your Understanding

Answer each question using the model definitions above. Short solutions follow.

In the categorical model, what is the genotype slope when \(L=2\)?
How many degrees of freedom are used to test \(H_0:\delta_1=\delta_2=0\), and what feature of the model determines this number?
In the centered linear-trend model, what does \(\theta\) measure?
Why can a nonsignificant interaction test not establish that all genotype slopes are equal?
What should be checked if few individuals have \(L=2\) and \(G=2\)?

Solutions. (1) The slope is \(\beta_G+\delta_2\). (2) The test has 2 degrees of freedom because it tests two coefficient restrictions. (3) \(\theta\) is the change in genotype slope for each additional A-ancestry copy under the linear-trend model. (4) Failure to reject may reflect low precision or limited power; it is not proof of equality. (5) Check genotype-by-ancestry support and the stability of the fitted slope and standard error; sparse support can make the interaction estimate unreliable.