Why Compare Missingness Across Phenotype Groups?
A genotype call can be missing for technical reasons, such as weak signal or low-quality reads. If missing calls occur at different rates in cases and controls, however, a variant may appear to have a phenotype-related pattern simply because the groups were measured differently. Comparing missingness is therefore a quality-control check: it can flag a potential problem, but it does not establish a genetic association.
The comparison is made separately for each variant. For that variant, classify each sample’s genotype as either missing or called, then compare the proportions missing across phenotype groups. The earlier material on sample and variant call rates provides the basic missingness summaries; here the focus is whether those summaries differ by phenotype.
A called genotype may be any observed genotype category; for this comparison, the categories are combined. The test asks whether missing-versus-called status is associated with group membership. It does not compare allele counts or genotype frequencies among called samples, and it does not recover information about the unobserved genotypes.
The Two-by-Two Table and Its Tests
For cases and controls, arrange the counts for one variant in a two-by-two contingency table. Let \(a\) and \(b\) be the numbers missing in cases and controls, and let \(c\) and \(d\) be the numbers called in cases and controls. The group-specific rates are \(a/(a+c)\) and \(b/(b+d)\). Reporting both rates and their difference gives the size and direction of the imbalance; a test statistic alone does not.
Pearson’s chi-square test compares observed table counts with counts expected under independence. For cell \(i,j\), the expected count is \(E_{ij}=(\text{row total})(\text{column total})/N\), where \(N\) is the table total. The test statistic is the sum of squared observed-minus-expected differences divided by expected counts. For a two-by-two table, its reference distribution has \(df=1\).
The usual chi-square approximation requires independent observations and adequate expected cell counts. A common practical guideline is that expected counts should not be very small; if they are sparse, Fisher’s exact test is a suitable alternative for a two-by-two table. Fisher’s test calculates probabilities under the null while conditioning on the observed row and column totals. Neither test fixes dependence between related samples or confounding by technical variables; those require attention to study design and analysis.
Worked Example: A Difference With Adequate Expected Counts
Suppose a hypothetical study has 300 cases and 300 controls at one variant. Genotypes are missing for 18 cases and 6 controls. The table is:
| Group | Missing | Called | Total |
|---|---|---|---|
| Cases | 18 | 282 | 300 |
| Controls | 6 | 294 | 300 |
| Total | 24 | 576 | 600 |
The missingness rates are \(18/300=0.0600\), or 6.00%, in cases and \(6/300=0.0200\), or 2.00%, in controls. The difference is 4.00 percentage points, and the case rate is three times the control rate. These are descriptive comparisons, not by themselves evidence of a biological effect.
Under the null of equal rates, each group is expected to have \(300(24/600)=12\) missing and \(300(576/600)=288\) called. All expected counts are adequate for the usual chi-square approximation. Thus:
With \(df=1\), the two-sided chi-square p-value is approximately 0.0124. At \(\alpha=0.05\), this is evidence against equal missingness rates for this variant in these groups. It is not the probability that the null hypothesis is true, nor does it show that the variant is associated with the phenotype. The 4.00-percentage-point difference is a reason to investigate data quality and group composition.
Why the Chi-Square Statistic Has This Form
Under the null hypothesis of independence, the proportion missing is estimated from the pooled table. If the missing total is \(m\), the called total is \(N-m\), and the two group sizes are \(n_1\) and \(n_0\), the expected missing counts are \(n_1m/N\) and \(n_0m/N\). The expected called counts are \(n_1(N-m)/N\) and \(n_0(N-m)/N\).
For a two-by-two table, write the observed missing counts as \(a\) and \(b\). The deviations from their expected counts are equal in magnitude and opposite in sign: \(a-n_1m/N=-(b-n_0m/N)\). Substituting the four expected counts into the Pearson sum and simplifying gives the familiar closed form below. This identity is useful for checking a calculation; the sum over cells remains the general definition.
The formula makes clear that the test reflects both the difference in proportions and the amount of data. It does not report the difference in percentage points, so that effect size should be stated separately. A small p-value can accompany a modest difference when sample sizes are large; a large p-value does not prove that the rates are identical.
Worked Example: Sparse Counts and Fisher’s Exact Test
In another hypothetical dataset, 2 of 20 cases and 0 of 20 controls have a missing call at a variant. The rates are \(2/20=10.0\%\) and \(0/20=0\%\), a difference of 10.0 percentage points. The pooled missing rate is \(2/40=0.05\), so each group has an expected missing count of \(20(0.05)=1\). Because these expected counts are small, use Fisher’s exact test rather than relying on the chi-square approximation.
Conditioning on the margins, the two missing calls can both fall in the case group, both fall in the control group, or be split one per group. Under the null, the probability that both fall in the case group is:
The equally extreme table with both missing calls in controls also has probability \(190/780\). For the conventional two-sided Fisher test, the probabilities of tables no greater than the observed table’s probability are summed. Here those are the two extreme tables, so \(p=2(190/780)\approx 0.4872\). The observed rate difference is notable descriptively, but this small dataset does not provide strong evidence against equal rates. The result does not establish equality; its uncertainty is substantial.
Separate Phenotype Imbalance From Technical Structure
A marginal comparison pools samples within each phenotype group. If cases and controls have different proportions from genotyping batches, sequencing runs, plates, or other technical strata, a pooled difference can reflect that composition rather than a within-stratum difference. Earlier material on batch effects explains why processing structure can influence call rates. For differential missingness, inspect the group rates within relevant technical strata as well as overall.
Worked Example: An Overall Difference Explained by Batch Composition
Consider a hypothetical study with 100 cases and 100 controls processed in two batches. Batch A has a 10% missingness rate in both groups; batch B has a 30% rate in both groups. Cases are mostly in A, while controls are mostly in B:
| Batch | Group | Missing | Called | Total |
|---|---|---|---|---|
| A | Cases | 9 | 81 | 90 |
| A | Controls | 1 | 9 | 10 |
| B | Cases | 3 | 7 | 10 |
| B | Controls | 27 | 63 | 90 |
Pooling across batches gives \(12/100=12.0\%\) missing in cases and \(28/100=28.0\%\) in controls. For the pooled two-by-two table, the total is 200, with 40 missing and 160 called. Under equal missingness, the expected counts are 20 missing and 80 called in each group. Therefore:
The pooled comparison gives \(df=1\) and \(p\approx0.00468\), but within batch A both groups have a 10% rate, and within batch B both have a 30% rate. In this constructed example, the pooled imbalance arises entirely because cases and controls have different batch compositions. It should prompt a technical investigation, not a claim of biological evidence. Stratified comparisons or a model that accounts for batch can assess whether a phenotype difference remains after accounting for this structure.
Interpretation and Practical Follow-Up
A differential-missingness result is a diagnostic signal. It says that, for the tested variant and analyzed samples, the observed proportion of missing genotypes differs across the specified phenotype groups under the assumptions of the test. It does not say which genotypes are missing, because those genotypes were not observed. In particular, a missingness test is not a test of allele-frequency difference or a substitute for a genotype–phenotype association analysis.
When a variant shows a notable imbalance, check the number of missing calls and the rates by group, then inspect whether the pattern tracks batch, DNA or read quality, plate, or other technical features. Confirm that phenotype labels and sample inclusion are correct. If there are additional phenotype strata, compare the relevant group-specific rates rather than collapsing distinct groups without justification. If samples are related or otherwise dependent, the ordinary independent-observation test may not be valid without an analysis that handles that dependence.
Genome-wide testing creates an additional issue: testing many variants increases the chance of small p-values under null conditions. A per-variant p-value should therefore be interpreted in light of the number of tests and an appropriate multiple-testing strategy. Report the rates and their difference alongside the test result, and use the finding as a quality-control lead. A low p-value does not by itself identify the cause of missingness or justify removing a variant or sample.
Common Mistakes
- Wrong reasoning: “Cases have more missing calls, so the variant is associated with disease.” Correction: Missingness is not an observed genotype. The difference flags a potential measurement or sampling problem and is not evidence of an allele or genotype effect.
- Wrong reasoning: “The p-value is large, so the groups have equal call rates.” Correction: A large p-value means the data do not provide strong evidence against the null under the test assumptions; it does not prove equality.
- Wrong reasoning: “A significant pooled test must reflect phenotype.” Correction: Different batch composition can produce a pooled imbalance even when the rates match within each batch. Inspect technical strata.
- Wrong reasoning: “Use Pearson’s chi-square regardless of cell counts.” Correction: Check expected counts and independence. For sparse two-by-two tables, Fisher’s exact test avoids the chi-square approximation.
Check Your Understanding
Use the definitions and examples above to answer these questions.
- At a variant, 15 of 250 cases and 5 of 250 controls have missing calls. What are the two missingness rates, and what is their difference in percentage points?
- What null hypothesis does a two-sided test of differential missingness assess?
- Why might Fisher’s exact test be preferred when expected cell counts are small?
- How can cases and controls have different overall missingness rates when their rates are equal within every batch?
- Why does a small p-value for differential missingness not establish a genetic association with the phenotype?