LD and Pleiotropy Confounding in TWAS

Why a TWAS Association Needs a Causal Interpretation

A transcriptome-wide association study (TWAS) tests whether genetically predicted expression is associated with a trait. A small p-value can provide evidence of that association, but it does not by itself show that changing expression would change the trait. Two genetic mechanisms can produce a TWAS signal even when expression has no causal effect on the trait: linkage disequilibrium (LD) between distinct causal variants and horizontal pleiotropy, in which a genetic variant affects expression and the trait through separate pathways.

The distinction matters because the two mechanisms imply different causal structures. With LD, one variant affects expression and a different, correlated variant affects the trait. With horizontal pleiotropy, one variant affects both outcomes, but its effect on the trait does not operate through expression. Neither structure implies an expression-to-trait causal effect.

Definition: Let \(G\) be a vector of standardized genotypes at variants in a locus, with LD matrix \(R=\operatorname{Cov}(G)\). A TWAS prediction is \(P=w'G\), where \(w\) is a vector of expression-prediction weights. A TWAS association is evidence that \(P\) and the trait \(Y\) are associated in the analyzed population. It is not, by definition, evidence that expression causes \(Y\).

The genotype standardization here gives each variant variance 1, so the entries of \(R\) are correlations between genotypes. In particular, an off-diagonal entry can be nonzero because two variants are in LD. LD describes correlation in inherited alleles; it does not mean that one variant causes the other. The prediction weights \(w\) are treated as fixed for the derivations and examples below.

A Covariance Decomposition

Consider a simple model that separates genetic effects on expression, genetic effects on the trait, and a possible causal effect of expression on the trait:

E=a'G+u,\qquad Y=\beta E+\delta'G+\varepsilon.

Here \(E\) is expression, \(a\) is its vector of genetic effects, \(\beta\) is the effect of expression on the trait under this model, and \(\delta\) contains genetic effects on the trait that do not operate through \(E\). The residuals \(u\) and \(\varepsilon\) represent variation not captured by the genetic terms. For the decomposition, assume \(G\) has mean zero and covariance \(R\), and that \(u\) and \(\varepsilon\) are uncorrelated with \(G\). This model is a useful description of possible paths, not a guarantee that the causal assumptions needed to identify \(\beta\) hold.

Substituting the expression equation into the trait equation gives \(Y=(\beta a+\delta)'G+\beta u+\varepsilon\). Since \(P=w'G\), the covariance between the TWAS prediction and the trait is therefore:

\begin{aligned} \operatorname{Cov}(P,Y) &=\operatorname{Cov}\!\left(w'G,(\beta a+\delta)'G+\beta u+\varepsilon\right)\\ &=w'R(\beta a+\delta). \end{aligned}

The second equality follows because \(\operatorname{Cov}(w'G,(\beta u+\varepsilon))=0\) under the stated assumptions, while \(\operatorname{Cov}(w'G,(\beta a+\delta)'G)=w'R(\beta a+\delta)\). Thus the TWAS covariance has a component associated with the expression-mediated path, \(w'R\beta a\), and a component from genetic effects on the trait outside that path, \(w'R\delta\). When \(\beta=0\), the first component vanishes, but the second need not. This is the central reason a TWAS signal can occur without an expression-mediated effect.

LD Between Distinct Causal Variants

In the LD mechanism, one variant changes expression and another changes the trait. If the variants are correlated, a predictor that tags the expression variant can also be correlated with the trait-associated variant. The resulting TWAS association can be strong even though the trait-associated variant has no effect on expression and expression has no effect on the trait.

Worked Example: A Trait Variant in LD With an Expression Variant

Suppose a hypothetical locus contains standardized genotypes \(G_1\) and \(G_2\), with \(\operatorname{Corr}(G_1,G_2)=0.60\). Let expression be \(E=0.50G_1+u\), and let the prediction be \(P=0.50G_1\). The trait is \(Y=0.40G_2+\varepsilon\). Assume \(u\) and \(\varepsilon\) are independent of both genotypes, with variances 0.75 and 0.84, respectively. There is no \(E\)-to-\(Y\) effect in this model: \(\beta=0\).

The assumptions make the stated scales consistent. Since \(\operatorname{Var}(G_1)=\operatorname{Var}(G_2)=1\), \(\operatorname{Var}(E)=0.50^2+0.75=1\) and \(\operatorname{Var}(Y)=0.40^2+0.84=1\). The predictor variance is \(\operatorname{Var}(P)=0.50^2=0.25\).

The covariance is generated by LD:

\operatorname{Cov}(P,Y) =\operatorname{Cov}(0.50G_1,0.40G_2) =0.50(0.40)(0.60) =0.12.

Consequently, the population correlation between \(P\) and \(Y\) is \(0.12/\sqrt{0.25(1)}=0.24\). For illustration, with \(n=800\) independent individuals and prediction weights treated as fixed, the leading large-sample score approximation is \(Z\approx\sqrt{n}\operatorname{Corr}(P,Y)=\sqrt{800}(0.24)\approx6.79\). This is a strong association signal in the simplified model, despite the absence of an expression effect on the trait. Variant \(G_1\) affects expression, variant \(G_2\) affects the trait, and their correlation connects the prediction to the trait.

A useful causal diagram for this example is \(G_1\rightarrow E\), \(G_2\rightarrow Y\), with \(G_1\leftrightarrow G_2\) denoting LD. The double-headed mark represents correlation, not a causal arrow. The diagram contains no arrow \(E\rightarrow Y\), and the two outcomes do not share a causal variant in this construction.

Horizontal Pleiotropy at a Shared Variant

Horizontal pleiotropy has a different diagram. A variant affects expression and also affects the trait through a separate route: \(G_1\rightarrow E\) and \(G_1\rightarrow Y\). There is still no arrow \(E\rightarrow Y\). In the covariance decomposition, this is represented by a nonzero \(\delta\) at the expression-associated variant, even when \(\beta=0\).

Worked Example: One Variant With Separate Effects on Expression and Trait

Consider a hypothetical locus with a single standardized genotype \(G_1\). Let \(E=0.50G_1+u\), \(P=0.50G_1\), and \(Y=0.30G_1+\varepsilon\). Assume the residuals are independent of \(G_1\) and each other, with \(\operatorname{Var}(u)=0.75\) and \(\operatorname{Var}(\varepsilon)=0.91\). Expression does not cause the trait in this model; the \(0.30G_1\) term is a direct genetic effect outside the expression pathway.

The variances are \(\operatorname{Var}(E)=0.50^2+0.75=1\), \(\operatorname{Var}(Y)=0.30^2+0.91=1\), and \(\operatorname{Var}(P)=0.50^2=0.25\). The shared genotype creates the following covariance:

\operatorname{Cov}(P,Y) =\operatorname{Cov}(0.50G_1,0.30G_1) =0.50(0.30)(1) =0.15.

The population correlation is \(0.15/\sqrt{0.25(1)}=0.30\). With \(n=400\) independent individuals and fixed weights, the same leading approximation gives \(Z\approx\sqrt{400}(0.30)=6.00\). Again, the association can be strong when \(\beta=0\). Unlike the previous example, expression and trait share a causal variant, but sharing a variant does not establish that the effect on the trait is mediated by expression.

These two cases should not be conflated. LD can produce a TWAS association when different variants underlie expression and trait associations. Horizontal pleiotropy can produce one when the same variant affects both outcomes by separate pathways. A shared genetic signal is therefore not the same claim as a shared causal pathway from expression to trait.

When Genetic Effects Reinforce or Cancel

The non-mediated component \(w'R\delta\) can be positive, negative, or close to zero. Its size depends on the prediction weights, the trait effects, and LD. As a result, a TWAS association is not a simple count of variants that affect both expression and the trait. Contributions can reinforce one another or cancel.

Worked Example: LD Can Also Produce Cancellation

Suppose two standardized genotypes have correlation \(0.50\), so \(R=\begin{pmatrix}1&0.50\\0.50&1\end{pmatrix}\). In a hypothetical locus, let \(E=0.70G_1+u\) and \(P=0.70G_1\). Let the trait be \(Y=0.20G_1-0.40G_2+\varepsilon\), with \(\beta=0\). Assume \(u\) and \(\varepsilon\) are independent of the genotypes, with variances 0.51 and 0.88. Then \(\operatorname{Var}(E)=0.70^2+0.51=1\), and \(\operatorname{Var}(Y)=0.20^2+(-0.40)^2+2(0.20)(-0.40)(0.50)+0.88=1\).

The predictor is associated with each trait component through a different covariance:

\begin{aligned} \operatorname{Cov}(P,Y) &=\operatorname{Cov}(0.70G_1,0.20G_1-0.40G_2)\\ &=0.70\{0.20(1)-0.40(0.50)\}\\ &=0.70(0.20-0.20)=0. \end{aligned}

Here the two contributions cancel exactly. Both variants have nonzero effects on the trait, and \(G_1\) predicts expression, but the population TWAS covariance is zero in this construction. A null TWAS result therefore does not prove that there are no genetic effects at the locus or no causal effect of expression; it means the particular weighted prediction has no net association under this model.

What a TWAS Result Does and Does Not Establish

A TWAS statistic summarizes evidence for association between a genetically weighted expression prediction and a trait. It can prioritize genes or loci for further investigation, but the statistic alone does not determine which causal structure produced the signal. In particular, it does not distinguish an expression-mediated effect from LD tagging or horizontal pleiotropy.

Causal diagrams help make the alternatives explicit. In the LD case, separate genetic causes are connected by genotype correlation. In the pleiotropy case, one genetic cause has distinct paths to expression and trait. The simple equations above show why both structures can generate the same kind of predictor–trait covariance. Interpreting a TWAS signal therefore requires evidence that addresses the causal variants and pathways, rather than relying on the association statistic alone.

Key takeaway: A TWAS association is compatible with a causal effect of expression, but it is also compatible with LD between distinct causal variants or horizontal pleiotropy at a shared variant. Association of genetically predicted expression with a trait is not, by itself, proof that expression mediates the genetic effect.

Common Mistake

Wrong reasoning: “The TWAS association is significant, so the gene’s expression must cause the trait.” This treats an association between \(P\) and \(Y\) as though it identified the causal path from \(E\) to \(Y\).

Correction: The prediction \(P\) is built from genotypes, and those genotypes can be associated with the trait through routes other than expression. Check whether the signal could arise from distinct variants in LD or from a variant with separate effects on expression and trait. A significant statistic supports a genetic association with the prediction; it does not, on its own, establish mediation or causation.

Check Your Understanding

For each question, distinguish the observed association from the causal interpretation.

In the LD example, which variant affects expression, which affects the trait, and what creates the TWAS covariance?
How does the causal diagram for horizontal pleiotropy differ from the diagram with two distinct variants in LD?
In the covariance decomposition, which term can produce a TWAS association when \(\beta=0\)?
Why does a shared causal variant for expression and trait not, by itself, show that expression mediates the trait effect?
In the cancellation example, how can both variants affect the trait while the population TWAS covariance is zero?