SNP Ascertainment Bias in Allele-Frequency Summaries

Why the Set of Variants Matters

An allele-frequency estimate describes a particular variant in a particular population. But a summary across many variants also depends on which variants were included. If a catalog was built by looking first for common variants in one discovery sample, rare variants in that sample were less likely to enter the catalog. A frequency distribution calculated from the catalog can therefore differ from the distribution across all variants, even if every frequency at every included site was estimated correctly.

This selection effect is called SNP ascertainment bias. It is a variant-level issue: the inclusion of a site depends on its observed or estimated variation in a discovery sample, or on a frequency filter. It is distinct from errors in counting alleles at a selected site. In particular, accurate genotyping and an unbiased frequency estimate among included sites do not, by themselves, make a summary representative of sites that were never included.

Definition: Variant ascertainment is the process by which some variants are selected for inclusion in a data set or analysis. Let \(Q\) denote a variant’s true minor-allele frequency in a specified population, and let \(S=1\) indicate that the variant is selected. The ascertainment probability is \(a(q)=P(S=1\mid Q=q)\). A frequency summary calculated among selected variants describes the distribution conditional on \(S=1\), not automatically the distribution among all candidate variants.

Frequency Filters and Discovery Samples

A simple ascertainment rule keeps a variant only when its true minor-allele frequency \(q\) is at least a threshold \(t\). Under this idealized rule, \(a(q)=1\) for \(q\geq t\) and \(a(q)=0\) otherwise. Real discovery usually uses a finite sample, so inclusion is probabilistic: two variants with the same true frequency can have different observed allele counts, and one can pass a threshold while the other does not.

For example, suppose \(m\) diploid individuals are sampled independently from a population and the minor allele has frequency \(q\leq 0.5\). Under random sampling, the number \(K\) of sampled minor-allele copies has a binomial distribution with \(2m\) trials and success probability \(q\). If the rule requires at least \(c\) copies of each allele, the variant is selected when \(c\leq K\leq 2m-c\). Thus its probability of selection is:

a(q)=P(c\leq K\leq 2m-c\mid Q=q) =\sum_{k=c}^{2m-c}{2m\choose k}q^k(1-q)^{2m-k}.

This formula assumes independent allele-copy sampling from the population and a selection rule based on the observed allele count. Relatedness, population structure in the discovery sample, or other selection criteria can change the inclusion probabilities. The formula is a model for understanding the basic selection mechanism, not a substitute for describing how a particular variant set was constructed.

Worked Example: A Minor-Allele-Frequency Filter Changes the Mean

Consider a hypothetical set of 100 candidate variants. Their true minor-allele frequencies are: 40 variants at 0.02, 30 at 0.10, 20 at 0.25, and 10 at 0.50. Suppose an idealized filter retains a variant only if its minor-allele frequency is at least 0.20. Compare the mean minor-allele frequency before and after filtering.

Before filtering: the sum of the 100 frequencies is \(40(0.02)+30(0.10)+20(0.25)+10(0.50)=0.8+3+5+5=13.8\). The mean across all candidates is \(13.8/100=0.138\).

After filtering: only the 20 variants at 0.25 and the 10 at 0.50 remain. Their frequency sum is \(20(0.25)+10(0.50)=5+5=10\), across 30 variants. The selected-set mean is \(10/30=0.3333\).

The selected-set mean is higher because the rule removed all variants below 0.20. It is a correct summary of the retained variants, but it is not a representative estimate of the mean across the original candidate set.

Worked Example: Finite-Sample Selection Probability

Suppose a hypothetical discovery panel contains five unrelated diploid individuals. A variant is included only when each allele is observed at least twice among the 10 allele copies. Under independent sampling, calculate the inclusion probability for a variant with true minor-allele frequency 0.10, then compare it with the probability for a variant at frequency 0.50.

For \(q=0.10\), \(K\sim\operatorname{Binomial}(10,0.10)\), and selection means \(2\leq K\leq 8\). Therefore, \(P(S=1)=1-P(K=0)-P(K=1)-P(K=9)-P(K=10)\). The excluded probabilities are \(P(K=0)=0.9^{10}=0.3486784\), \(P(K=1)=10(0.1)(0.9)^9=0.3874205\), \(P(K=9)=10(0.1)^9(0.9)=0.000000009\), and \(P(K=10)=(0.1)^{10}=0.0000000001\). Thus the inclusion probability is approximately \(1-0.7360989=0.2639\).

For \(q=0.50\), the binomial distribution is symmetric. The excluded counts are 0, 1, 9, and 10. Here \(P(K=0)=P(K=10)=1/1024\), and \(P(K=1)=P(K=9)=10/1024\). Hence \(P(S=1)=1-22/1024=0.9785\). Under this rule, a variant at frequency 0.50 is much more likely to be discovered than one at frequency 0.10, even though both are drawn from the same panel design.

How Ascertainment Changes a Frequency Distribution

Let \(f(q)\) describe the distribution of minor-allele frequencies among all candidate variants. A selected variant must both have frequency \(q\) and pass the ascertainment rule. By the definition of conditional probability, the frequency distribution among selected variants is proportional to the original distribution multiplied by the chance of selection:

f(q\mid S=1)=\frac{a(q)f(q)}{\int a(u)f(u)\,du},

where the denominator is the overall probability that a candidate variant is selected. In a finite collection of variants, the same logic applies with sums and weights in place of densities and integrals. Variants with larger \(a(q)\) make up a larger share of the selected set than of the original set.

A useful special case follows when selection probability is a nondecreasing function of minor-allele frequency. The mean frequency among selected variants is then at least as large as the mean among all candidate variants. This statement concerns the mean of the minor-allele frequency under the stated condition; it does not imply that every ascertainment process raises every possible frequency summary.

Justification: The selected mean is \(E[Q\mid S=1]=E[Q\,a(Q)]/E[a(Q)]\), so its difference from the original mean \(E[Q]\) is \(\operatorname{Cov}(Q,a(Q))/E[a(Q)]\). If \(a(q)\) is nondecreasing, take two independent copies \(Q\) and \(Q'\). Then \(2\operatorname{Cov}(Q,a(Q))=E[(Q-Q')(a(Q)-a(Q'))]\geq 0\), because the two differences have the same sign. Since \(E[a(Q)]>0\) when any variants are selected, the selected mean cannot be smaller under these conditions.

The result is about the distribution of variants, not necessarily the accuracy of an estimate at one site. Also, a threshold can have more complicated effects when selection depends on several populations, annotations, or quality criteria. The relevant question is always how the inclusion probability relates to the quantity being summarized.

Discovery in One Population, Summaries in Another

Cross-population summaries are especially sensitive to where variants were discovered. Suppose the frequency in a discovery population is \(p_A\), and the frequency of the same labelled allele in a target population is \(p_B\). If selection depends on the discovery sample, the target population’s selected frequency distribution is not generally its full distribution. For a joint distribution \(f(p_A,p_B)\), it is:

f(p_B\mid S=1) = \frac{\int a(p_A)f(p_A,p_B)\,dp_A} {\int\int a(p_A)f(p_A,p_B)\,dp_A\,dp_B}.

This expression shows why a discovery rule can affect a target-population summary even when the target population played no role in selection. Variants common or readily detected in population A are preferentially represented. Variants rare in A but common in B, including some population-specific variants, may be underrepresented or absent. The direction and size of the resulting distortion depend on how frequencies in A and B are related across the candidate variants.

Worked Example: A Discovery Panel Changes a Target-Population Mean

Consider four hypothetical variants, each given equal weight. Their frequencies refer to the same labelled allele in populations A and B. Suppose discovery in A retains variants with frequency at least 0.10. The frequencies are: variant 1, \(p_A=0.40\), \(p_B=0.40\); variant 2, \(p_A=0.30\), \(p_B=0.10\); variant 3, \(p_A=0.02\), \(p_B=0.30\); and variant 4, \(p_A=0.01\), \(p_B=0.40\).

Across all four variants, the mean frequency in B is \((0.40+0.10+0.30+0.40)/4=0.30\). The discovery rule keeps only variants 1 and 2, so the mean frequency in B among selected variants is \((0.40+0.10)/2=0.25\). In this hypothetical set, discovery in A leads to a lower mean frequency for the selected variants in B because the omitted variants have relatively high frequencies in B.

For comparison, the mean absolute difference \(|p_A-p_B|\) across all variants is \((0+0.20+0.28+0.39)/4=0.2175\). Among the two selected variants it is \((0+0.20)/2=0.10\). Thus this particular ascertainment rule also makes the selected set appear less differentiated by this summary. Neither direction is universal: different relationships between \(p_A\) and \(p_B\) can raise or lower a target mean or a measure of cross-population difference.

Keep Allele Labels Consistent

Minor-allele frequency is useful for describing variation within a population, but the identity of the minor allele can change across populations. A cross-population comparison should distinguish the frequency of a consistently labelled allele, such as the reference or alternate allele, from the minor-allele frequency calculated separately in each population. Otherwise, a change in which allele is called “minor” can hide the direction of the frequency difference.

Worked Example: The Minor Allele Can Switch

At a hypothetical site, the alternate allele has frequency 0.08 in population A and 0.62 in population B. The alternate-allele frequency difference, defined as \(p_B-p_A\), is \(0.62-0.08=0.54\). The minor-allele frequencies are 0.08 in A and \(1-0.62=0.38\) in B, a difference of \(0.30\). These are different summaries: the first tracks the same allele in both populations, while the second reports whichever allele is less common in each population.

If discovery used a minor-allele-frequency threshold of 0.10 in population A, this site would fail the threshold despite its alternate allele being common in B. A catalog discovered only in A could therefore omit a site that is informative about variation in B.

Interpretation and Common Pitfalls

An allele-frequency table from an ascertained SNP set is not automatically wrong. Its interpretation must match its sampling frame: it describes the included variants, under the stated discovery and filtering rules. Trouble arises when a result is presented as though the variants were a representative sample of all sites in the genome, or when a cross-population comparison ignores that the populations may not have had equal roles in variant discovery.

Mistake: “The selected variants have a high mean frequency, so the population contains few rare variants.” The mean may be high because the discovery rule favored common variants. It cannot establish the frequency distribution among unascertained sites.
Mistake: “The frequency estimates are unbiased at each included SNP, so the genome-wide summary is unbiased.” Site-level estimation and variant-level selection are different processes. A summary across selected sites can be distorted even when each included-site estimate is sound.
Mistake: “A SNP discovered in one population provides a neutral comparison with another.” Discovery in the first population affects which variants are available for comparison. State the discovery panel and selection criteria, and interpret the comparison conditional on that ascertainment.
Mistake: “Minor-allele frequencies identify the same allele in both populations.” The minor allele can switch. Use a consistently labelled allele when the direction of a cross-population frequency difference matters.

When assessing a frequency summary, first identify the candidate variant set, the discovery sample, and the inclusion rule. Then ask which variants that rule makes more likely to retain and whether the same selection process applies across the populations being compared. These checks clarify what the reported summary describes, without conflating ascertainment with uncertainty in estimating frequencies at observed sites.

Check Your Understanding

Answer each question using the distinction between a variant’s frequency and its chance of being included.

A catalog retains only variants with minor-allele frequency at least 0.05 in a discovery panel. What distribution does a mean frequency calculated from the catalog summarize?
Why can two variants with the same true frequency have different inclusion outcomes when discovery uses a finite sample?
Under what condition does increasing selection probability with minor-allele frequency imply a selected-set mean that is at least as high as the original mean?
A variant is rare in the discovery population but common in the target population. Explain how discovery-based ascertainment can affect its representation in a target-population summary.
Why can comparing minor-allele frequencies across two populations conceal a change in the frequency of a consistently labelled allele?