Author response:

We thank the two anonymous reviewers who took the time and effort to read and evaluate our work. We look forward to submitting a revised version of the manuscript that addresses their comments.

A major concern shared between both reviewers is our use of Bayes factors instead pvalues to measure the strength of association. In revision, we will add a section in Supplementary to compare and constrast Bayes factor and p-values. Very briefly here, p-value is the tail probability under the null. Formally, it is defined as P(T > t|H₀), for a test statistic T with obvserved value t computed from data D. But our interest is P(H₀|D) and P(H₁|D), posterior probabilities of the null and alternative models, about which p-value says nothing. With FDR approach, a q-value, the minium FDR at which a null is rejected, which can be estimated from a collection of p-values, has a Bayesian interpretation as the probability that H₀ is true conditioning on rejecting that H₀. This is not quite P(H₀|D) but nevertheless a useful probabilistic statement. For FDR approach to work, however, the collection of tests need to be reasonably independent, and their effect sizes need to be mixed. Both implicit assumptions can fail for cis eQTL analysis.

On the other hand, with Bayes factors we can compute posterior probability P(H₀|D) and P(H₁|D) after specifying prior odds P(H₁)/P(H₀) (or equivalently P(H₁) since P(H₀)+ P(H₁) = 1). In our manuscript, the prior odds used to determined Bayes factor threshold is 1/1000, or about 1 cis eQTL per gene. Bayes factor also allows us to directly compare two non-nested alternative models P(paternal effect|D) and P(maternal effect|D), which is difficult to do using p-values.

It was suggested (by reviewer 2) that POE eQTL should be defined by testing H₀ : θ₀ = θ₁ against H₁ : θ₀ ̸= θ₁ where θ₀ and θ₁ are maternal and paternal effects respectively. This indeed was our initial approach, as evidenced in Table 1 (last column) and Section 4.5 in Methods. Our final approach is more stringent: H₀ : β₀ = β₁ = 0 against H₁ : β₀ = 0,β₁/= 0, to use test for paternal effect as an example (the test for maternal effect can be obtained in a similar fashion). That is, we not only require that paternal and maternal effects be the same, as suggested by reviewer, but also require that they are both 0 under the null. This is partially motivated by an example in Table 1 (Gene ZNF890P) where both β₀ > 0 and β₁ > 0, and β₀/= β₁. In other words, examples like this where both paternal and maternal effects are significant and their differences are also significant were not included in our downstream classification and further analysis.

Figures and data

Example of eQTL statistics.

Comparison of paternal effects and maternal effects of sentinel eQTL in joint analysis.

An example of gene that whose paternal and maternal eQTL have opposite effect sizes.

Enrichment of drug target genes.