Why Genotype Checks Come Before Covariate Modeling
A genotype covariate can only be useful if the underlying genotype records are sufficiently complete and credible. A low call rate may leave too few participants with usable data; a difference in call rates across assay batches can make missingness consequential; and genotype counts that depart from Hardy-Weinberg expectations can prompt closer review. These checks do not establish that a genotype affects pharmacokinetics. They assess whether the genotype data warrant further analysis and whether a signal needs investigation before fitting a population PK model.
The earlier material on allele, carrier, and diplotype coding established ways to represent genotype categories. Here, the focus is on checking those records. Each calculation should begin by defining which samples and which locus are included, how missing calls are handled, and what assumptions are needed for interpretation.
For a diploid autosomal locus with two alleles, A and G, let \(p\) be the frequency of A and \(q=1-p\) the frequency of G. Under HWE, the expected genotype proportions are \(p^2\) for AA, \(2pq\) for AG, and \(q^2\) for GG. A call rate is a data-completeness measure; an HWE test is a comparison of genotype proportions. Neither replaces the other.
Call Rates and Missing-Call Patterns
Calculate a call rate for a defined denominator, usually all eligible participants for whom genotyping was attempted at that locus. Exclude samples that were not eligible or were never intended to be genotyped, and state that exclusion. A no-call is missing information, not a fourth genotype and not evidence for either allele. Report the numerator and denominator along with the percentage so that the result is auditable.
Worked Example: Overall and Batch-Specific Call Rates
In a hypothetical dataset, genotyping was attempted for 100 eligible participants at one autosomal locus. Ninety-six participants received a genotype call. The overall call rate is
The laboratory record shows that 50 participants were processed on each of two plates. Plate 1 produced 49 calls, and Plate 2 produced 40. Their call rates are
The plate-specific rates imply 89 calls in total, not 96. Therefore, these hypothetical numbers cannot describe the same complete set of records: the overall rate would be \(89/100=89.0\%\). This arithmetic check matters. In a real review, reconcile denominators, plate assignments, and counts before interpreting the discrepancy. If the verified counts were 89 calls, the marked difference between 98.0% and 80.0% would warrant checking plate records and call-generation procedures; it would not, by itself, identify the reason for the difference.
A high overall call rate can conceal a problem concentrated in one batch or participant subgroup. Check rates by assay plate or batch and, where appropriate, by relevant sample characteristics. The strata and denominator must be reported clearly, especially when a subgroup is small. Patterns of missingness can matter: if the chance of a call differs with genotype or with a characteristic related to genotype, the called samples may not represent all eligible participants. The call rate alone cannot establish whether missingness is random.
Allele Frequencies and Expected Genotype Counts
For \(n\) called diploid individuals with observed counts \(n_{AA}\), \(n_{AG}\), and \(n_{GG}\), the total number of called allele copies is \(2n\). Each AA individual contributes two A copies, each AG individual contributes one A copy, and each GG individual contributes none. Thus the estimated A frequency is
These estimates use called genotypes. If calls are missing, \(n\) here is the number with a genotype call, not the number originally eligible for genotyping. This calculation assumes the locus is diploid and autosomal, the genotype categories are correctly specified, and each individual contributes two allele copies.
Worked Example: Allele Frequencies and HWE Expectations
Suppose a hypothetical set of 100 called genotypes contains 60 AA, 30 AG, and 10 GG individuals. There are 200 allele copies. The estimated frequency of A is
The frequency of G is \(\hat{q}=1-0.750=0.250\). Under HWE, the expected proportions are
Multiplying each proportion by 100 gives expected counts of 56.25 AA, 37.50 AG, and 6.25 GG. The expected counts sum to 100. The observed counts are 60, 30, and 10; the differences are not yet evidence of a departure until compared with sampling variability using a suitable test.
The HWE Chi-Square Test
A standard HWE goodness-of-fit test uses the Pearson chi-square statistic. Its null hypothesis is that the genotype counts follow HWE proportions at this locus in the sampled population. The allele frequency is estimated from those same genotype counts. With two alleles and three genotype categories, the degrees of freedom are \(3-1-1=1\): one degree is used by the total-count constraint, and one more by estimating the allele frequency. This usual one-degree-of-freedom approximation requires independent observations and sufficiently large expected genotype counts. The sample should be an appropriate set of unrelated individuals from a reasonably homogeneous population for the HWE model being tested; the locus should be diploid and autosomal, and the genotype categories should be reliable.
The calculation follows directly from the definition of the statistic. Estimate \(\hat{p}\) and \(\hat{q}\) from the allele counts; form expected counts \(n\hat{p}^{\,2}\), \(2n\hat{p}\hat{q}\), and \(n\hat{q}^{\,2}\); then sum the squared observed-minus-expected differences divided by the expected counts. The test’s p-value is the probability, under the null model and its reference distribution, of a statistic at least as large as the observed statistic. It is not the probability that the null hypothesis is true.
Worked Example: Testing a Departure from HWE
Use the hypothetical counts from the preceding example: 60 AA, 30 AG, and 10 GG among 100 called participants. The HWE expected counts are 56.25, 37.50, and 6.25. Each expected count exceeds 5, so the common chi-square approximation is reasonable as a screening calculation under the other assumptions stated above. The statistic is
With \(df=1\), the upper-tail p-value is approximately 0.0455. At a prespecified \(\alpha=0.05\), this is evidence against HWE for this hypothetical sample. The result is marginal relative to that threshold, and it does not demonstrate a genotyping error or a biological mechanism. It is a reason to review the data and assumptions, including the genotype calls and the composition of the sample.
Assumptions, Diagnostics, and Alternatives
An HWE departure is a screening signal with several possible explanations. Genotyping error is one possibility, but departures may also reflect sample composition, non-random mating, selection, relatedness, or chance. A mixture of groups with different allele frequencies can also yield pooled genotype counts that do not follow the single-population HWE proportions. A pooled test alone cannot distinguish these possibilities or establish that a specific subgroup structure exists.
Check the observed genotype counts, allele-frequency calculation, sample exclusions, call rate, and batch-specific patterns before interpreting a p-value. Confirm that allele labels and genotype coding are consistent. Consider whether the sample is suitable for the test and whether multiple loci or subgroups have been tested; when many tests are run, some small p-values can occur by chance. A threshold chosen for one test should not automatically be treated as a universal decision rule for a large panel.
| Check or method | What it assesses | Main limitation |
|---|---|---|
| Call rate | Completeness of genotype calls for a defined locus and sample | Does not show whether called genotypes are correct or missingness is harmless |
| Allele-frequency summary | Observed proportions of allele copies among called samples | Does not test whether genotype proportions match HWE |
| Pearson HWE chi-square test | Departure of observed counts from HWE expectations | Approximation can be poor with small expected counts; a departure does not identify its cause |
| Exact HWE test | HWE evidence when counts are sparse, conditional on the allele counts | Can be conservative or discrete; it still does not diagnose why counts depart |
When expected counts are small, use an exact HWE test rather than relying on the chi-square approximation. In a common conditional formulation, the test considers possible genotype configurations with the observed allele counts fixed, calculates their probabilities under HWE, and assesses how extreme the observed configuration is. Exact-test conventions for a two-sided p-value can differ, so report the method used. Neither method substitutes for inspecting the original calls and quality-control records.
Worked Example: Choosing an Exact Test for Sparse Counts
Suppose a hypothetical sample has 8 AA, 4 AG, and 0 GG calls. Here \(n=12\), so there are 24 allele copies. The estimated A frequency is
The HWE expected counts are approximately \(12(0.8333^2)=8.333\) AA, \(12[2(0.8333)(0.1667)]=3.333\) AG, and \(12(0.1667^2)=0.333\) GG. Because an expected count is far below 5, the usual chi-square approximation is not appropriate. The correct next step is an exact HWE calculation using the observed allele counts, not a chi-square p-value presented as reliable. The zero observed GG count may motivate a review, but these counts alone do not establish a call failure or other cause.
Using the Checks Before PK Modeling
Genotype checks matter because errors in the covariate record can compromise interpretation of genotype-associated PK differences. A low call rate can reduce the usable sample and may do so unevenly across batches or subgroups. A departure from HWE can signal that the sample, genotype calls, or model assumptions merit review. Passing these checks does not prove that genotypes are error-free, that all participants are represented, or that a genotype effect on a PK parameter exists.
Keep the checks in proportion to what they establish. Call rates describe completeness; allele frequencies summarize the observed allele copies; HWE tests compare genotype counts with a specified population model. They are complementary, not interchangeable. Investigate a concerning result using the underlying records and study context, and document denominators, assumptions, test choice, and any follow-up. Do not convert a screening result directly into a claim about a pharmacokinetic mechanism.
Common Mistakes
- Using all eligible participants in the allele-frequency denominator despite no-calls. Calculate allele frequencies from the called diploid genotypes, and report how many calls were available.
- Treating a high overall call rate as proof of complete quality. Inspect call rates by batch and relevant sample groups; the overall average can hide a concentrated problem.
- Using a chi-square approximation with sparse expected counts. Check expected counts and use an exact HWE test when the approximation is not adequate.
- Interpreting HWE departure as proof of genotyping error. It is a signal for review, not a diagnosis; sample composition and other assumptions also matter.
- Interpreting a non-significant test as proof of HWE. A large p-value means the data did not provide strong evidence against the tested model; it does not prove that all assumptions hold.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- At a locus, 180 of 200 eligible participants receive calls. What is the call rate?
- Among 40 called participants, the counts are 20 AA, 16 AG, and 4 GG. Calculate the frequency of allele A.
- For the counts in Question 2, calculate the HWE expected genotype counts using the estimated allele frequencies.
- One HWE expected genotype count is 2.1. Should the usual chi-square approximation automatically be used? What is the alternative?
- A genotype dataset departs from HWE in a test with a small p-value. Does this establish that genotyping error caused the departure? Explain.
Solutions. (1) \(180/200\times100\%=90.0\%\). (2) There are 80 allele copies, of which \(2(20)+16=56\) are A, so \(\hat{p}=56/80=0.700\). (3) \(\hat{q}=0.300\); expected counts are \(40(0.700^2)=19.60\) AA, \(40[2(0.700)(0.300)]=16.80\) AG, and \(40(0.300^2)=3.600\) GG. (4) No. With a small expected count, use an exact HWE test rather than relying on the usual chi-square approximation. (5) No. The result is evidence against HWE under the test assumptions and calls for investigation; it does not identify genotyping error as the cause.