Allele-Specific Expression QTL Tests

Why Compare the Two Alleles Within an Individual?

A conventional expression QTL analysis asks whether total gene expression differs across individuals with different genotypes. An allele-specific expression (ASE) analysis asks a related but distinct question: within a heterozygous individual, are more RNA molecules produced from one chromosome than from the other? Because the two alleles share many features of the individual’s cellular environment, an imbalance can provide evidence that a regulatory difference acts in cis, on or near the chromosome carrying that allele.

The comparison requires both informative RNA reads and phase information. A read overlapping a transcribed heterozygous site can be assigned to one of the two haplotypes. If the candidate cis variant is phased to that site, the read can then be assigned to the haplotype carrying either the candidate’s effect allele or its alternative allele. Reads that cannot be assigned reliably are not informative for this comparison.

Definition: For individual \(i\), let \(Y_i\) be the number of informative RNA reads assigned to the haplotype carrying a specified candidate-variant allele, and let \(N_i\) be the total number of informative reads assigned to either haplotype. The allele-specific proportion is \(Y_i/N_i\). Under equal expression of the two haplotypes, its expected value is \(1/2\).

The informative unit is the individual, not each sequencing read considered as an independent biological replicate. Reads from the same individual share biological and technical conditions, and their allele counts can vary more than a simple binomial model allows. This extra-binomial variation is called overdispersion.

A Beta-Binomial Model for Allelic Counts

A binomial model assumes that, conditional on the total number of informative reads, each read is independently assigned to the candidate-allele haplotype with the same probability. In real data, the underlying allelic proportion can vary across individuals and samples. A beta-binomial model allows that variation while retaining a count-based likelihood.

Model: For each independent individual \(i\), let \(q_i\) be the probability that an informative read comes from the candidate-allele haplotype. Assume \(Y_i\mid q_i\sim\operatorname{Binomial}(N_i,q_i)\) and \(q_i\sim\operatorname{Beta}(a,b)\). The population mean allelic proportion is \(p=a/(a+b)\). Under the equal-expression null hypothesis, \(H_0:p=1/2\); the two-sided alternative is \(H_a:p\ne1/2\). This model assumes correct phasing and read assignment, independent individuals, and a beta distribution that adequately represents between-individual variation in allelic proportions.

The beta-binomial probability of observing \(y\) reads out of \(n\) is

P(Y_i=y\mid N_i=n)=\binom{n}{y}\frac{B(y+a,n-y+b)}{B(a,b)}.

Here \(B(\cdot,\cdot)\) is the beta function. In practice, the beta-binomial likelihood can be fitted under \(H_0\) and under the alternative; a likelihood-ratio test compares the two fits. A convenient approximation for a score test uses the model-based variance under \(H_0\). Define \(\rho=1/(a+b+1)\), the within-individual overdispersion parameter. The mean and variance of the count are

E(Y_i)=N_i p,\qquad \operatorname{Var}(Y_i)=N_i p(1-p)\{1+(N_i-1)\rho\}.

At \(p=1/2\), the variance becomes \(N_i\{1+(N_i-1)\rho\}/4\). If \(\rho\) is known or estimated appropriately, a normal-approximation test statistic is

Z=\frac{\sum_i Y_i-\frac12\sum_i N_i} {\sqrt{\sum_i \frac{N_i}{4}\{1+(N_i-1)\rho\}}}.

Under the null and the model assumptions, this statistic is approximately standard normal when the data contain enough independent individuals for the approximation to be reasonable. A two-sided p-value is \(2P(Z_{\mathrm{std}}\ge |z_{\mathrm{obs}}|)\), where \(Z_{\mathrm{std}}\sim N(0,1)\). A fitted beta-binomial likelihood test is preferable when the approximation is doubtful, such as with very few individuals.

Worked Tests and Interpretations

Worked Example: Testing Imbalance After Phase Orientation

Suppose 20 hypothetical donors are heterozygous at a candidate cis variant. RNA reads are assigned to the haplotype carrying the same specified candidate allele in every donor. Ten donors have 58 such reads out of 100 informative reads, and the other ten have 55 out of 100. Assume a beta-binomial analysis gives \(\rho=0.01\). Test equal expression using the normal approximation.

The oriented count is \(10(58)+10(55)=1{,}130\), out of \(20(100)=2{,}000\). Under \(H_0:p=1/2\), the expected count is \(2{,}000/2=1{,}000\). The null variance is

\sum_i \frac{N_i}{4}\{1+(N_i-1)\rho\} =20\cdot\frac{100}{4}\{1+99(0.01)\} =500(1.99)=995.

Thus \(SE=\sqrt{995}=31.54\), and

Z=\frac{1{,}130-1{,}000}{31.54}=4.122.

The two-sided normal-approximation p-value is about \(3.8\times10^{-5}\). Under the stated model, these data provide evidence against equal expression of the two haplotypes. The estimated proportion is \(1{,}130/2{,}000=0.565\), corresponding to an estimated allelic ratio of \(0.565/(1-0.565)=1.30\). In this hypothetical sample, the candidate-allele haplotype has about 1.30 times the observed expression of the other haplotype. The result supports a cis-linked imbalance; it does not by itself identify the candidate variant as causal.

Worked Example: Why Overdispersion Changes the Test

In another hypothetical data set, eight heterozygous donors each contribute 100 informative reads. Their counts on the phased candidate-allele haplotype are 58, 55, 62, 48, 57, 60, 53, and 59. The total is \(452\) of \(800\). Suppose the fitted overdispersion is \(\rho=0.02\). Compare the beta-binomial variance calculation with the binomial calculation that incorrectly treats all 800 reads as independent with a common probability.

Under the null, the expected count is \(800/2=400\), so the observed excess is 52. For the beta-binomial model,

\operatorname{Var}_0\Big(\sum_iY_i\Big) =8\cdot\frac{100}{4}\{1+99(0.02)\} =200(2.98)=596.

The standard error is \(\sqrt{596}=24.41\), giving \(Z=52/24.41=2.130\), with a two-sided normal-approximation p-value of about \(0.0332\). Under the naive binomial calculation, the variance would be \(800(1/2)(1/2)=200\), the standard error would be \(\sqrt{200}=14.14\), and \(Z=52/14.14=3.677\), with a two-sided p-value of about \(0.000236\).

The naive calculation produces much stronger evidence because it ignores within-donor dependence and between-donor variation. The beta-binomial result is the appropriate one under the stated overdispersed model. The difference is not a minor technical adjustment: treating reads as independent replicates can make uncertainty far too small.

Worked Example: Orienting Reads to the Candidate Allele

Consider two hypothetical heterozygous donors, each with 100 informative reads. In donor A, the candidate allele is phased to the reference base at the reporter site, and 60 reads carry that base. In donor B, the candidate allele is phased to the alternate base, and 40 reads carry the reference base. Determine the candidate-oriented count and explain what happens if the counts are instead combined by reporter-site reference status.

For donor A, 60 reads come from the candidate-allele haplotype. For donor B, the candidate allele is on the alternate-base haplotype, so its count is \(100-40=60\). The oriented total is therefore \(60+60=120\) of \(200\), giving an estimated candidate-haplotype proportion of \(120/200=0.60\) and an allelic ratio of \(0.60/0.40=1.50\).

If the reference-base counts were combined without accounting for phase, the total would be \(60+40=100\) of 200, apparently exactly balanced. That calculation compares reporter alleles, not the candidate-variant haplotypes, because the phase orientation reverses between donors. With only two donors, this example is a phase-orientation demonstration, not a reliable basis for an asymptotic significance test.

Justification for the Overdispersion Adjustment

The variance formula follows from conditioning on the donor-specific proportion \(q_i\). For a beta distribution with mean \(p\), write \(\operatorname{Var}(q_i)=p(1-p)\rho\). Conditional on \(q_i\), the binomial mean is \(N_iq_i\) and variance is \(N_iq_i(1-q_i)\). The law of total variance gives

\begin{aligned} \operatorname{Var}(Y_i) &=E\{\operatorname{Var}(Y_i\mid q_i)\} +\operatorname{Var}\{E(Y_i\mid q_i)\}\\ &=N_i\{p(1-p)-\operatorname{Var}(q_i)\} +N_i^2\operatorname{Var}(q_i)\\ &=N_i p(1-p)\{1+(N_i-1)\rho\}. \end{aligned}

When \(\rho=0\), this reduces to the binomial variance \(N_i p(1-p)\). When \(\rho>0\), the variance is larger, with the difference increasing as the read depth per individual grows. This explains why adding many reads from a small number of donors cannot substitute for observing more independent individuals.

What ASE Evidence Can Establish

An allelic imbalance test asks whether expression differs between phased chromosomes among individuals heterozygous at the candidate variant. In the natural log scale, the allelic effect can be summarized as

\theta=\log\left(\frac{p}{1-p}\right),

where \(\theta=0\) corresponds to equal expression. A positive value means the candidate-allele haplotype is more highly expressed on average; a negative value means it is less highly expressed. This is a within-individual contrast, not a comparison of total expression between genotype groups. Individuals homozygous at the candidate variant do not provide this particular allelic contrast, though they may contribute to other QTL analyses.

A significant imbalance is consistent with a cis-regulatory difference on the chromosome carrying the associated allele. However, nearby variants in linkage disequilibrium may explain the same pattern, so the result does not prove that the tested variant itself is causal. Nor does a non-significant result establish equal expression: low read depth, few heterozygotes, or substantial overdispersion can leave an effect imprecisely estimated.

The interpretation also depends on valid read assignment. Reference-mapping preference can make one haplotype appear more abundant even when expression is balanced. Personalized or bias-aware alignment, reliable phasing, and filtering of ambiguous reads help address this problem. Other sources of imbalance—including copy-number changes, imprinting, RNA editing, and transcript-specific effects—should be considered in context. Bulk-tissue cell composition and sample-specific technical factors can also complicate interpretation.

Common Mistakes

Counting reads as independent biological replicates. A large read total does not remove variation between individuals. Model overdispersion and base inference on independent donors.
Ignoring phase orientation. A reference allele is not always on the same candidate-variant haplotype across donors. Orient counts to the candidate allele before combining them.
Calling allelic imbalance proof of causality. The signal can arise from another linked cis variant or a technical artifact. Describe it as evidence for a cis-linked difference, not definitive identification of a causal variant.
Interpreting a non-significant test as equality. A large p-value does not demonstrate that the alleles are equally expressed; it may reflect limited information or high overdispersion.

Key takeaway: ASE-QTL tests use phase-oriented counts from heterozygous individuals to test whether one candidate-linked haplotype is expressed more than the other. A beta-binomial model accounts for extra-binomial variation, while careful phasing and mapping-bias checks are essential for a credible cis interpretation.

Check Your Understanding

Try these questions before reviewing the worked examples.

Why are homozygous individuals at the candidate variant uninformative for the within-individual allelic contrast defined here?
What does the overdispersion parameter \(\rho\) represent, and how does it change the variance of the read count?
A candidate allele is phased to the reference reporter base in one donor and to the alternate reporter base in another. What must be done before combining the donors’ counts?
What does a significant test of \(H_0:p=1/2\) support, and what does it not establish about causality?
Why can a very large number of reads from a few donors still yield less reliable inference than reads from many independent donors?