Local Ancestry Association Under Linkage Disequilibrium

Why Linkage Disequilibrium Changes the Question

At a genomic region, a typed variant and local ancestry can carry overlapping information. Ancestry-specific allele frequencies can make the allele dosage at a marker correlate with the number of chromosome copies inherited from a particular ancestry. Nearby variants may also be correlated through linkage disequilibrium (LD). Consequently, a phenotype association with local ancestry can partly reflect the typed genotype, and a genotype association can partly reflect ancestry-related differences. The size and direction of either association can change when both predictors enter the model.

The key distinction is between a marginal association, which relates the phenotype to one predictor at a time, and a conditional association, which estimates the relationship with one predictor while holding the other fixed in the model. “Holding fixed” describes the regression comparison; it does not imply that genotype and local ancestry can actually be varied independently in the population.

Definition: Let $Y_i$ be a quantitative phenotype for individual $i$, $G_i$ the typed-variant allele dosage, and $L_i$ the local-ancestry dosage (for example, the number of copies assigned to ancestry A). A two-predictor additive model is $$ Y_i=\beta_0+\beta_G G_i+\beta_L L_i+\varepsilon_i. $$ Here $\beta_G$ is the expected change in phenotype per additional counted allele copy, conditional on $L_i$; $\beta_L$ is the expected change per additional ancestry-A copy, conditional on $G_i$. The model tests $H_0:\beta_L=0$ for conditional local-ancestry association and $H_0:\beta_G=0$ for conditional genotype association.

For the ordinary least-squares calculations below, assume the model is correctly specified, observations are independent, and errors have mean zero and constant variance conditional on the predictors. Exact small-sample t tests additionally assume normally distributed errors. In a related sample or a setting requiring a mixed model, the covariance structure must instead be modeled; the conditional interpretation remains, but ordinary least-squares standard errors and tests need not be valid. Treat the genotype and local-ancestry values as observed for these examples.

With covariates, such as sex or global ancestry, include them in the model as well. The conditional tests then adjust for those covariates too. The algebra below uses centered predictors and phenotype after accounting for any intercept and covariates, so its sums of cross-products refer to the remaining variation.

Conditional Coefficients and Their Uncertainty

Write $S_{gg}=\sum_i(G_i-\bar G)^2$, $S_{ll}=\sum_i(L_i-\bar L)^2$, $S_{gl}=\sum_i(G_i-\bar G)(L_i-\bar L)$, $S_{gy}=\sum_i(G_i-\bar G)(Y_i-\bar Y)$, and $S_{ly}=\sum_i(L_i-\bar L)(Y_i-\bar Y)$. The determinant $D=S_{gg}S_{ll}-S_{gl}^2$ measures how much independent variation remains in the two predictors. If $D>0$, the two coefficients are identifiable from this model.

\hat\beta_G=\frac{S_{ll}S_{gy}-S_{gl}S_{ly}}{D}, \qquad \hat\beta_L=\frac{S_{gg}S_{ly}-S_{gl}S_{gy}}{D}.

The expressions show why marginal and conditional results may differ: each conditional coefficient combines both phenotype cross-products and the predictor cross-product. Equivalently, to estimate $\beta_L$, first remove from $L$ the linear variation predictable from $G$, and remove from $Y$ the linear variation predictable from $G$; the slope between those residuals is the conditional ancestry coefficient. This residualization identity is a way to understand the adjustment, not a claim that the predictors are independent.

If the full model has residual sum of squares $RSS_F$ and residual degrees of freedom $df$, then $s^2=RSS_F/df$ estimates the error variance. Under the stated model assumptions, the standard errors are

SE(\hat\beta_G)=\sqrt{s^2\frac{S_{ll}}{D}}, \qquad SE(\hat\beta_L)=\sqrt{s^2\frac{S_{gg}}{D}}.

Each coefficient can be tested using $t=\hat\beta/SE(\hat\beta)$, with $df$ residual degrees of freedom under the ordinary linear model. The test for local ancestry asks whether its association remains after adjustment for the typed variant; the test for genotype asks the reciprocal question. These are distinct one-degree-of-freedom conditional tests. They do not test whether either predictor has a marginal association.

Worked Example: Marginal Ancestry Association Disappears After Adjustment

Suppose a hypothetical cohort of $n=198$ has centered cross-products $S_{gg}=100$, $S_{ll}=100$, $S_{gl}=90$, $S_{gy}=30$, $S_{ly}=27$, and $S_{yy}=100$. The phenotype is measured in arbitrary phenotype units, and each predictor is a dosage in copies. The predictors have correlation $90/\sqrt{100(100)}=0.90$, so they are strongly correlated. Assume the ordinary linear-model conditions stated above.

The determinant is $D=100(100)-90^2=1{,}900$. Thus,

\hat\beta_G=\frac{100(30)-90(27)}{1{,}900} =\frac{570}{1{,}900}=0.3000, \qquad \hat\beta_L=\frac{100(27)-90(30)}{1{,}900} =\frac{0}{1{,}900}=0.

The marginal slope for local ancestry alone would be $S_{ly}/S_{ll}=27/100=0.2700$ phenotype units per ancestry copy. In the two-predictor model, however, the ancestry coefficient is zero: in these hypothetical data, the marginal relationship is accounted for by the component of local ancestry shared with genotype.

The full-model explained cross-product is $\hat\beta_GS_{gy}+\hat\beta_LS_{ly}=0.3000(30)+0(27)=9.000$. Hence $RSS_F=S_{yy}-9=100-9=91$. There are three fitted parameters (intercept and two slopes), so $df=198-3=195$, and $s^2=91/195=0.4667$ phenotype-units squared. The standard error of the local-ancestry coefficient is

SE(\hat\beta_L)=\sqrt{0.4667\frac{100}{1{,}900}}=0.1567 \quad\text{phenotype units per ancestry copy}.

The conditional statistic is $t=0/0.1567=0$, with two-sided $p=1.000$. This is no evidence against $H_0:\beta_L=0$ in the adjusted model; it does not establish that local ancestry has no biological relevance in every model or population.

Worked Example: A Conditional Genotype Association

In another hypothetical cohort, let $n=159$, with $S_{gg}=100$, $S_{ll}=100$, $S_{gl}=50$, $S_{gy}=35$, $S_{ly}=20$, and $S_{yy}=120$. Assume independent, homoskedastic normal errors and a correctly specified linear model. The predictor correlation is $0.5000$, and $D=100(100)-50^2=7{,}500$.

The fitted slopes are

\hat\beta_G=\frac{100(35)-50(20)}{7{,}500} =\frac{2{,}500}{7{,}500}=0.3333, \qquad \hat\beta_L=\frac{100(20)-50(35)}{7{,}500} =\frac{250}{7{,}500}=0.03333.

The explained cross-product equals $0.3333(35)+0.03333(20)=12.3333$ (using the exact coefficients, $12.3333$). Therefore $RSS_F=120-12.3333=107.6667$, and $df=159-3=156$. The residual variance estimate is $s^2=107.6667/156=0.6902$. For either slope, the standard error is

SE=\sqrt{0.6902\frac{100}{7{,}500}}=0.09593.

For genotype, $t=0.3333/0.09593=3.475$, giving a two-sided $p$-value of approximately $0.0006621$ on 156 degrees of freedom. The data provide evidence of association between genotype and phenotype conditional on local ancestry. For local ancestry, $t=0.03333/0.09593=0.3475$, with two-sided $p$ approximately $0.7287$; there is not comparable evidence for a conditional ancestry association. Neither result establishes causation.

What Strong LD Does to the Test

When genotype and local ancestry are highly correlated, $D$ becomes small relative to $S_{gg}S_{ll}$. The standard errors then increase because the data contain little variation that distinguishes the two predictors. In the limiting case of perfect linear dependence, $D=0$, the model cannot estimate separate coefficients: many combinations of $\beta_G$ and $\beta_L$ yield the same fitted values. This is non-identifiability, not evidence that the predictors have no association with the phenotype.

A useful measure of this problem for the two-predictor model is the variance inflation factor $1/(1-r_{GL}^2)$, where $r_{GL}=S_{gl}/\sqrt{S_{gg}S_{ll}}$. It describes how much coefficient variance is inflated by the linear relationship between these predictors, relative to an orthogonal design. It is a diagnostic of precision, not a correction that recovers information absent from the data.

Worked Example: A Conditional Ancestry Association Persists

Consider a third hypothetical cohort with $n=99$ and centered cross-products $S_{gg}=100$, $S_{ll}=100$, $S_{gl}=80$, $S_{gy}=20$, $S_{ly}=30$, and $S_{yy}=50$. The phenotype is in arbitrary units. The predictors have correlation $0.8000$, and $D=10{,}000-6{,}400=3{,}600$. Assume the ordinary linear-model conditions, including normal errors for the t test.

The conditional coefficients are

\hat\beta_G=\frac{100(20)-80(30)}{3{,}600} =\frac{-400}{3{,}600}=-0.1111, \qquad \hat\beta_L=\frac{100(30)-80(20)}{3{,}600} =\frac{1{,}400}{3{,}600}=0.3889.

The marginal local-ancestry slope is $30/100=0.3000$, while the adjusted estimate is $0.3889$ phenotype units per ancestry copy. The fitted cross-product is $(-0.1111)(20)+(0.3889)(30)=9.4444$, so $RSS_F=50-9.4444=40.5556$. With $df=99-3=96$, $s^2=40.5556/96=0.4225$, and

SE(\hat\beta_L)=\sqrt{0.4225\frac{100}{3{,}600}}=0.1083, \qquad t=\frac{0.3889}{0.1083}=3.590.

The two-sided $p$-value is approximately $0.00052$ on 96 degrees of freedom. Under this model, local ancestry is associated with the phenotype conditional on the typed genotype. The test does not identify a causal mechanism: an untyped linked variant, model misspecification, or residual confounding could also contribute to the observed conditional association.

Diagnostics, Alternatives, and Interpretation

Before interpreting a conditional test, check that the model has enough independent variation to estimate both slopes. Inspect the genotype–ancestry correlation, the determinant or equivalent condition diagnostics, and the standard errors. A large standard error, unstable coefficient under modest model changes, or near-perfect predictor correlation warns that the data do not cleanly separate the two signals. Also check whether the phenotype–predictor relationship is plausibly additive and linear on the chosen scale, and whether residual patterns or known sampling structure undermine the error assumptions.

A marginal admixture-mapping regression is simpler and can be more precise for detecting a broad ancestry-correlated signal, but it cannot distinguish that signal from a typed variant correlated with ancestry. A SNP-only regression answers a different marginal question. Including both predictors gives conditional tests that attempt this separation, at the cost of precision when LD is strong. Residualizing one predictor and the phenotype on the other is algebraically equivalent to the corresponding conditional slope in ordinary least squares, but it does not remove collinearity or create independent information. Stratifying or fitting separate ancestry groups changes the comparisons and may discard information; it is not automatically a substitute for the joint conditional model.

A conditional coefficient is an association parameter defined by the specified model and study sample. Its interpretation depends on dosage coding, phenotype scale, covariate adjustment, and the available overlap of genotype and ancestry values. It should not be described as the effect of changing ancestry while keeping genotype fixed in a literal intervention sense. When local-ancestry calls are uncertain, treating them as exact can also affect inference; the estimates here condition on the supplied values.

Common pitfall: “Local ancestry has a significant marginal association, so it must have an independent ancestry effect.” This reasoning ignores correlation with genotype and other covariates. The correction is to state which model produced the result, report the conditional coefficient and its uncertainty, and describe it as association conditional on the included predictors—not as proof of an independent causal effect.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

If $r_{GL}=0$, what does the two-predictor model imply about the relationship between a predictor’s marginal slope and its conditional slope?
Why can a local-ancestry coefficient have a larger standard error when genotype and local ancestry are strongly correlated?
In the second worked example, what does the genotype test’s small p-value say, and what does it not say?
What happens to separate genotype and ancestry coefficient estimates if $D=0$?

Solutions. (1) With centered, uncorrelated predictors, $S_{gl}=0$, so $\hat\beta_L=S_{ly}/S_{ll}$ and $\hat\beta_G=S_{gy}/S_{gg}$; the marginal and conditional slopes coincide for these two-predictor fits. (2) Strong correlation leaves little predictor variation unique to local ancestry. Since $SE(\hat\beta_L)=\sqrt{s^2S_{gg}/D}$, a smaller $D$ inflates its standard error. (3) It is evidence against $H_0:\beta_G=0$ in the specified model, conditional on local ancestry. It is not proof of causation or evidence that the ancestry coefficient is nonzero. (4) Perfect linear dependence makes the two slopes non-identifiable from these data; no unique pair of separate estimates exists.

Local Ancestry Association Under Linkage Disequilibrium

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