How Linkage Disequilibrium Alters Marginal GWAS Effects

Why a GWAS Coefficient Can Reflect More Than One Variant

A genome-wide association study typically tests variants one at a time. At each variant, its reported coefficient describes the association between that variant and the phenotype, without conditioning on every other variant in the region. When nearby genotypes are correlated through linkage disequilibrium (LD), a single-variant coefficient can therefore reflect the effects of other correlated variants as well as any effect at the tested variant itself.

This distinction matters when interpreting GWAS results and when using their coefficients as weights in a genetic score. A marginal association is a property of the tested variant’s relationship with the phenotype in the presence of the observed genotype correlations. It is not necessarily that variant’s distinct, conditional effect.

Definition: A marginal GWAS coefficient is the slope from a regression of the phenotype on one variant’s genotype at a time. A joint coefficient is the slope for a variant in a regression that includes the other specified variants simultaneously. In a correctly specified additive model, joint coefficients describe each variant’s effect conditional on the other included variants. They can be interpreted as causal effects only with additional causal assumptions.

We first work with a quantitative phenotype and standardized genotypes. Let \(X_j\) denote the centered genotype dosage at variant \(j\), scaled to have variance 1, and let \(Y\) be a centered phenotype. A standardized genotype is measured in genotype standard deviations, so its regression coefficient is the change in the expected phenotype per one-standard-deviation increase in that genotype. The phenotype may also be standardized, but that is not required.

Model and assumptions: For \(m\) variants, suppose the population model is \(Y=\sum_{k=1}^{m}X_k\beta_k+\varepsilon\), where \(\beta_k\) is the joint regression coefficient at variant \(k\). Assume the genotypes and phenotype are centered, each genotype has variance 1, the additive linear model is appropriate, and \(E[X_j\varepsilon]=0\) for every \(j\). The last condition says that the model’s residual is uncorrelated with each included genotype. Without further causal assumptions, the \(\beta_k\) are conditional association effects, not necessarily causal effects.

The signed LD matrix \(R\) has entries \(R_{jk}=\operatorname{Corr}(X_j,X_k)\). Its diagonal entries are 1, and its off-diagonal entries retain the sign of the genotype correlation. Let \(b_j\) be the population marginal coefficient from regressing \(Y\) on \(X_j\) alone. The central relationship is

b_j=\sum_{k=1}^{m}R_{jk}\beta_k, \qquad \boldsymbol{b}_{\mathrm{marg}}=R\boldsymbol{\beta}.

In this equation, \(R_{jj}\beta_j=\beta_j\) contributes the joint effect at the tested variant. Every other term \(R_{jk}\beta_k\) is a contribution from a correlated variant’s joint effect. If variants are uncorrelated, the off-diagonal entries are zero and the marginal and joint coefficients agree. Otherwise, they need not.

Deriving the LD-Matrix Relationship

A simple regression with an intercept has population slope equal to the covariance of predictor and outcome divided by the predictor’s variance. Because \(X_j\) is centered and has variance 1, the marginal slope is \(b_j=E[X_jY]\). Substitute the joint model for \(Y\), then use the residual assumption:

\begin{aligned} b_j &=E[X_jY]\\ &=E\left[X_j\left(\sum_{k=1}^{m}X_k\beta_k+\varepsilon\right)\right]\\ &=\sum_{k=1}^{m}\beta_k E[X_jX_k]+E[X_j\varepsilon]\\ &=\sum_{k=1}^{m}\beta_k R_{jk}. \end{aligned}

The last equality uses \(E[X_j\varepsilon]=0\) and \(E[X_jX_k]=\operatorname{Corr}(X_j,X_k)=R_{jk}\), since both genotypes are centered and have variance 1. Applying this result to every variant gives the matrix equation. This is an exact population identity under the stated model; estimated GWAS coefficients also vary because of sampling error and may depart from it.

For two standardized genotypes, the matrix and equations become especially transparent. Write \(r\) for their signed correlation:

R= \begin{pmatrix} 1&r\\ r&1 \end{pmatrix}, \qquad \begin{pmatrix} b_1\\ b_2 \end{pmatrix} = \begin{pmatrix} 1&r\\ r&1 \end{pmatrix} \begin{pmatrix} \beta_1\\ \beta_2 \end{pmatrix}, \qquad b_1=\beta_1+r\beta_2,\quad b_2=r\beta_1+\beta_2.

The term \(r\beta_2\) is the contribution of variant 2’s joint effect to variant 1’s marginal coefficient; \(r\beta_1\) is the corresponding contribution in the other direction. The sign of the contribution depends on both the signed correlation and the joint effect. A positive correlation transmits the same sign; a negative correlation transmits the opposite sign.

Worked Examples With Two Variants

Worked Example: A Variant Tags an Effect at a Correlated Variant

Suppose two standardized genotypes have signed LD \(r=0.70\). In a hypothetical population, the joint effects on a quantitative phenotype are \(\beta_1=0.20\) and \(\beta_2=0\) phenotype units per genotype standard deviation. Find both marginal coefficients.

The LD relationship gives \(b_1=\beta_1+r\beta_2\) and \(b_2=r\beta_1+\beta_2\). Substituting the hypothetical values,

b_1=0.20+(0.70)(0)=0.20,\qquad b_2=(0.70)(0.20)+0=0.14.

Variant 1 has a joint effect of 0.20 phenotype units per genotype standard deviation. Variant 2 has no joint effect in this model, but its marginal coefficient is 0.14 phenotype units per genotype standard deviation because its genotype is correlated with variant 1. Thus variant 2 tags the association at variant 1. The example does not show that variant 2 itself affects the phenotype.

Worked Example: A Marginal Coefficient Changes Sign

Suppose standardized genotypes have \(r=0.80\), and the hypothetical joint coefficients are \(\beta_1=0.10\) and \(\beta_2=-0.20\) phenotype units per genotype standard deviation. Calculate the marginal coefficients and compare their signs with the joint effects.

For variant 1, the correlated contribution from variant 2 is \(r\beta_2\), which is negative. For variant 2, the contribution from variant 1 is positive:

\begin{aligned} b_1&=\beta_1+r\beta_2 =0.10+(0.80)(-0.20) =0.10-0.16=-0.06,\\ b_2&=r\beta_1+\beta_2 =(0.80)(0.10)-0.20 =0.08-0.20=-0.12. \end{aligned}

Variant 1’s joint coefficient is positive, but its marginal coefficient is negative. The negative contribution from variant 2 is larger in magnitude than variant 1’s own positive contribution. Variant 2’s marginal coefficient remains negative. Each value is measured in phenotype units per genotype standard deviation.

This is a sign reversal in the marginal association, not evidence that the conditional effect at variant 1 is negative. Interpreting a marginal sign as the sign of a variant’s distinct effect can therefore be misleading when correlated variants have effects in opposing directions.

Worked Example: Contributions Cancel in a Marginal Association

Consider a hypothetical pair of standardized genotypes with \(r=0.95\), joint coefficients \(\beta_1=0.20\) and \(\beta_2=-0.19\), and a quantitative phenotype. Calculate the two marginal coefficients.

For variant 1, its own positive effect is nearly offset by the negative contribution from variant 2. For variant 2, the positive contribution from variant 1 is nearly offset by its own negative effect:

\begin{aligned} b_1&=0.20+(0.95)(-0.19) =0.20-0.1805=0.0195,\\ b_2&=(0.95)(0.20)-0.19 =0.19-0.19=0. \end{aligned}

The second variant has a nonzero joint coefficient, \(-0.19\), even though its population marginal coefficient is zero under this model. A marginal null result does not, in general, establish that the joint effect is zero. Here the two-variant LD matrix is valid: its determinant is \(1-r^2=1-0.95^2=0.0975\), which is positive.

Worked Example: Translating the Relationship to Raw Genotype Dosages

Standardization is convenient, but GWAS coefficients may also be reported per allele of an effect-allele dosage. Suppose a hypothetical pair of centered dosages \(G_1,G_2\) has \(\operatorname{Var}(G_1)=0.50\), \(\operatorname{Var}(G_2)=0.32\), and \(\operatorname{Cov}(G_1,G_2)=0.16\). Let the joint dosage-scale coefficients be \(\beta_1=0.30\) and \(\beta_2=0\) phenotype units per allele. Find the marginal dosage-scale slopes.

For raw centered genotypes, the marginal slope at variant \(j\) is \(\operatorname{Cov}(G_j,Y)/\operatorname{Var}(G_j)\). Under the joint model and the residual assumption, the vector form is \(b_{\mathrm{marg}}=D^{-1}\Sigma\beta\), where \(\Sigma\) is the genotype covariance matrix and \(D\) is its diagonal matrix of variances. In this example, the calculation for each variant is

\begin{aligned} b_1 &=\frac{\operatorname{Var}(G_1)\beta_1+ \operatorname{Cov}(G_1,G_2)\beta_2}{\operatorname{Var}(G_1)} =\frac{(0.50)(0.30)+(0.16)(0)}{0.50} =0.30,\\ b_2 &=\frac{\operatorname{Cov}(G_2,G_1)\beta_1+ \operatorname{Var}(G_2)\beta_2}{\operatorname{Var}(G_2)} =\frac{(0.16)(0.30)+(0.32)(0)}{0.32} =0.15. \end{aligned}

The second variant’s marginal slope is 0.15 phenotype units per allele, despite having a joint coefficient of zero. The raw-scale relation uses covariances and divides by each tested genotype’s variance; it is not generally the same as multiplying raw-scale coefficients directly by a correlation matrix. The dosage covariance values are consistent with a valid covariance matrix because \(0.50(0.32)-0.16^2=0.1344>0\).

What the Relationship Does—and Does Not—Say

The matrix equation describes how the joint effects combine with the genotype correlation structure to produce population marginal coefficients. It does not by itself determine which variants are causal, prove that a particular joint model is correct, or quantify the sampling uncertainty of a GWAS estimate. Those questions require additional modeling and evidence.

The LD matrix depends on how alleles are coded. If the allele coding at one variant is reversed, the corresponding standardized genotype changes sign. Correlations involving that genotype and its regression coefficients also change sign, while the underlying fitted relationship is unchanged when coding is handled consistently. This is why allele orientation must be consistent when interpreting signed LD alongside GWAS effects.

A related limitation arises when variants are extremely correlated. For two standardized genotypes, the determinant of \(R\) is \(1-r^2\). If \(|r|=1\), the determinant is zero, so the two genotype columns are perfectly linearly dependent. The data cannot distinguish their separate joint coefficients: different pairs of coefficients can produce the same fitted genetic contribution. Even when \(|r|\) is close to 1, separating joint effects can be unstable. Marginal coefficients can still be calculated, but they do not solve this identification problem.

The derivation here uses a linear model for a quantitative phenotype. It should not be applied as an exact identity for coefficients from every GWAS model. In particular, marginal and joint coefficients from logistic regression do not generally obey this same linear relation on the log-odds scale. The model and scale need to be considered before interpreting a coefficient relationship.

Common Mistake: Treating a Marginal Coefficient as a Variant-Specific Effect

Wrong reasoning: “The marginal GWAS coefficient at variant 2 is positive, so variant 2 has a positive independent effect.” This conclusion does not follow when variant 2 is correlated with other variants. As the tagging example showed, a variant with zero joint effect can have a nonzero marginal coefficient.

Correction: A single-variant coefficient reflects the tested genotype’s covariance with the phenotype. Under the stated standardized linear model, that covariance is the sum of the LD-weighted joint effects, \(b_j=\sum_k R_{jk}\beta_k\). To interpret a coefficient as a distinct conditional effect requires a model that conditions on the relevant variants, along with the assumptions needed for that interpretation.

Key takeaway: With standardized genotypes and a correctly specified additive linear model, the population marginal GWAS coefficients satisfy \(\boldsymbol{b}_{\mathrm{marg}}=R\boldsymbol{\beta}\). LD can transmit another variant’s signal, weaken or cancel a marginal association, and even reverse its sign.

Check Your Understanding

Use the model and notation from this tutorial to answer the following questions.

With two standardized genotypes, \(r=0.50\), \(\beta_1=0.10\), and \(\beta_2=0\), what are \(b_1\) and \(b_2\)? State why \(b_2\) need not be zero.
If \(r\) is negative and \(\beta_2\) is positive, what sign does the contribution \(r\beta_2\) make to \(b_1\)?
For \(r=0.70\), \(\beta_1=0.10\), and \(\beta_2=-0.20\), calculate \(b_1\). Does its sign match \(\beta_1\)?
Why can perfectly correlated genotype columns prevent separate joint effects from being identified?
For raw genotype dosages, what covariance-based formula replaces the standardized-genotype relation?