Why Cross-Population Tables Need an Audit
A table can display different allele frequencies across populations even when the difference is not ready for interpretation. Alleles may be coded inconsistently, denominators may not match the number of called genotypes, and unequal sample sizes may make one estimate much less precise than another. A pooled frequency can also answer a different question from a frequency weighted for a specified population mixture. Finally, a table of selected variants may not represent all variants in the genome.
The aim of a diagnostic audit is not to explain a frequency difference by itself. It is to establish what was counted, what population or mixture the summary describes, and how much uncertainty or selection could affect the comparison. The calculations below use hypothetical data.
Start With Allele Labels and Denominators
The formula makes the first checks explicit. Each homozygote contributes two copies of its allele, each heterozygote contributes one copy of each allele, and each called diploid genotype contributes two copies overall. Therefore, with complete calls, the genotype counts must sum to \(n\), and the allele-copy denominator must be \(2n\). With missing calls, use the number of called individuals—not the number enrolled or the nominal sample size—to obtain the denominator.
Before comparing estimates, verify that the same biological allele is being compared in each population. A column headed “alternate-allele frequency” is not sufficient if the alternate allele differs between datasets or if strand orientation has not been harmonized. A frequency for one allele is \(1\) minus the frequency for its complement at a biallelic locus, but a reversed label can make a real contrast appear to change sign.
Worked Example: Reconstructing Frequencies and Harmonizing Labels
Suppose two hypothetical datasets describe the same biallelic locus. The first reports alleles \(A/G\); the second is on the opposite strand and reports the corresponding alleles \(T/C\). Each contains 50 called diploid individuals.
| Dataset | Genotype counts | Reported alternate allele |
|---|---|---|
| North | 32 \(AA\), 16 \(AG\), 2 \(GG\) | \(G\) |
| South | 8 \(CC\), 24 \(CT\), 18 \(TT\) | \(C\) |
In North, the counts sum to \(32+16+2=50\), so the allele-copy denominator is \(2(50)=100\). The \(G\)-copy count is \(2(2)+16=20\), giving \(\hat{p}_{G,\text{North}}=20/100=0.200\).
In South, the counts also sum to 50, with denominator 100. The \(C\)-copy count is \(2(8)+24=40\), giving \(\hat{p}_{C,\text{South}}=40/100=0.400\). Since \(G\) corresponds to \(C\) under the stated strand reversal, the harmonized comparison is \(0.200\) versus \(0.400\), a difference of \(0.200\) in the frequency of that allele.
If the South estimate had instead been reported for \(T\), its value would be \(1-0.400=0.600\). Comparing North \(G\) to South \(T\) would compare different alleles and would create a misleading contrast. The allele correspondence must be established before interpreting the numbers.
A useful audit also distinguishes the number of individuals from the number of allele copies. For example, if 100 individuals were enrolled but only 92 have called genotypes, the complete-call denominator is 184, not 200. If the called counts are 40 \(AA\), 32 \(Aa\), and 20 \(aa\), then the \(a\)-copy count is \(2(20)+32=72\), and its estimated frequency is \(72/184=0.3913\). Dividing by 200 would give \(0.3600\), which incorrectly treats missing genotypes as observed copies of neither allele.
Uncertainty and Unequal Sample Sizes
A frequency estimate is subject to sampling variation. Under random sampling of unrelated diploid individuals, with independent allele copies at the locus (as under Hardy–Weinberg sampling), the allele-copy count can be modeled as binomial with \(2n\) trials and success probability \(p\). Under these assumptions, the estimated standard error is approximately \(\sqrt{\hat{p}(1-\hat{p})/(2n)}\). This approximation is a diagnostic for sampling precision, not a universal model: relatedness, inbreeding, population structure within a sample, or other departures from independent allele-copy sampling can change the uncertainty.
Different sample sizes do not automatically bias allele-frequency estimates, but they generally produce different precision. A frequency estimated from many more sampled individuals will usually have a smaller standard error, all else equal. A table should therefore be read with its sample sizes and uncertainty in view; decimal places alone do not show how reliably two estimates are distinguished.
Worked Example: Reading a Contrast With Unequal Precision
Consider hypothetical complete genotype data at a harmonized biallelic locus. In North, \(n=120\), with 72 \(AA\), 36 \(Aa\), and 12 \(aa\). In South, \(n=40\), with 9 \(AA\), 18 \(Aa\), and 13 \(aa\). Assume unrelated individuals and independent allele copies for this approximate calculation.
The North \(A\)-copy count is \(2(72)+36=180\), out of \(2(120)=240\), so \(\hat{p}_{A,\text{North}}=180/240=0.750\). The South \(A\)-copy count is \(2(9)+18=36\), out of \(2(40)=80\), so \(\hat{p}_{A,\text{South}}=36/80=0.450\). The estimated North-minus-South difference is \(0.750-0.450=0.300\).
The approximate standard errors are
The South estimate is less precise in this calculation, consistent with its smaller sample. If the two samples are independent, the estimated standard error of the difference is
A rough normal-approximation 95% interval for the difference is \(0.300\pm1.96(0.06225)\), or approximately \(0.178\) to \(0.422\). Under the stated sampling assumptions, this describes uncertainty in the North-minus-South frequency contrast; it does not identify why the frequencies differ. In particular, the unequal sample sizes show up in precision, not as an automatic correction to either sample’s estimated frequency.
Pooled Frequencies and Mixture Weights
“Pooled frequency” can refer to more than one target. If genotype counts from sampled individuals are combined, the resulting estimate weights populations according to their numbers of observed allele copies. This describes the combined sampled individuals. A frequency for a specified target mixture instead weights each population frequency according to that mixture’s population proportions. Those weights need not match the sample proportions.
For a mixture with population-specific frequencies \(p_1,\ldots,p_K\) and target proportions \(w_1,\ldots,w_K\), where each \(w_k\geq0\) and \(\sum_k w_k=1\), the mixture frequency is \(\sum_k w_kp_k\). The weights encode the target being described; changing them changes the estimand, even when the underlying population-specific estimates stay the same.
Worked Example: Sample Pooling Is Not Always Target Weighting
Suppose a hypothetical sample contains 80 individuals from population 1, where the \(A\)-allele frequency is \(0.100\), and 20 individuals from population 2, where the frequency is \(0.500\). Combining the sampled allele copies weights the two groups by 80% and 20%, respectively:
Now suppose the scientific target is a mixture with equal proportions from the two populations. Its frequency estimate is instead
The estimates differ because they describe different mixtures, not because one calculation is arithmetically wrong. Reporting \(0.180\) as the frequency in an equal-proportion target would silently substitute the sample composition for the target composition.
Ascertainment and the Scope of a Table
A table may include only variants that passed a discovery or quality-control screen. This is ascertainment: the observed set of variants has been selected according to a rule, rather than sampled as a representative set of all genomic sites. The selection rule might depend on allele frequency, whether a variant was detected in a discovery population, or other properties relevant to the comparison.
For example, imagine that a hypothetical variant panel includes only loci with an estimated minor-allele frequency of at least 0.10 in a discovery sample from population A. The mean frequency across the listed loci cannot automatically be interpreted as the mean across all genomic loci, and the set may be especially unrepresentative for population A. If the same discovery observations were used both to select variants and to estimate their frequencies, the selection rule can also affect the apparent frequency distribution. A separate evaluation sample can reduce that reuse problem, but it does not make the selected loci representative of all variants.
Thus, ask what variants could have entered the table, in which samples the selection occurred, and whether the analysis concerns those selected loci or a broader genomic target. A valid frequency calculation at each listed locus does not by itself remove ascertainment from a summary across loci.
A Practical Diagnostic Sequence
A consistent sequence makes common table problems easier to find before interpreting population contrasts.
Confirm that each column refers to the same biological allele across populations, accounting for reference/alternate changes and strand orientation.
Check that genotype counts sum to the number of called individuals and calculate allele-copy counts explicitly. Use the called-genotype denominator when data are missing.
Read sample sizes alongside estimates, state the assumptions behind any standard error, and distinguish sample proportions from target mixture weights.
Identify discovery filters or ascertainment criteria and limit summaries to the population of loci the selection process can support.
Common Mistake: Treating a Difference as Its Own Explanation
A frequent error is to see two different entries in a table and conclude that the populations differ biologically in a way caused by a particular process. The entries establish only estimated frequencies for the specified allele, samples, and loci. Differences can be affected by coding errors, sampling variability, nonrepresentative sampling, mixture composition, or variant selection; even after those issues are checked, a frequency contrast alone does not establish its cause.
The correction is to make each claim match what the table supports. First report the harmonized allele and frequency estimates. Then describe the uncertainty and sampling assumptions, identify the target population or mixture, and state any ascertainment limits. Only after these checks should the contrast be interpreted in the context of a separate scientific question.
Check Your Understanding
For each question, state the check or calculation that would resolve the issue.
- A population has 25 called individuals: 10 \(AA\), 10 \(Aa\), and 5 \(aa\). What is the estimated frequency of \(A\), and what denominator should be used?
- Two datasets report alternate-allele frequencies at the same locus, but their alternate alleles differ. What must be verified before comparing the values?
- Why can a frequency estimate from a smaller population sample be less precise without necessarily being biased?
- Two population frequencies are 0.15 and 0.55. Calculate the mixture frequency when the target proportions are 0.75 and 0.25, respectively. Why might that differ from the pooled sample frequency?
- A panel contains only variants detected in one discovery population. What conclusion about a genome-wide mean frequency would require caution, and why?