Pooled and Population-Specific Allele-Frequency Estimates

Why the Meaning of “Pooled” Matters

At a locus with two alleles, a study may estimate the frequency of allele \(a\) separately in several populations and also report one estimate after combining their samples. These summaries answer different questions. The separate estimates describe allele frequency within each population; the pooled estimate describes the fraction of sampled chromosome copies that carry \(a\) when all samples are combined.

The difference is important whenever populations contribute different numbers of sampled chromosomes. Pooling does not ordinarily give each population equal influence. A population contributing more chromosome copies contributes more to the combined estimate, even if the scientific question is about the average frequency across populations.

Definition: For population \(j\), let \(m_j\) be the number of sampled chromosome copies at a locus, and let \(A_j\) be the number of those copies carrying allele \(a\). The population-specific estimate is \(\hat{p}_j=A_j/m_j\). Across \(K\) populations, the pooled estimate is \(\hat{p}_{\mathrm{pool}}=(\sum_{j=1}^{K} A_j)/(\sum_{j=1}^{K}m_j)\).

For complete genotype data from \(n_j\) diploid individuals in population \(j\), \(m_j=2n_j\). If the genotypes are \(AA\), \(Aa\), and \(aa\), the count of allele \(a\) is the number of heterozygotes plus twice the number of \(aa\) homozygotes:

A_j=n_{Aa,j}+2n_{aa,j}, \qquad \hat{p}_j=\frac{n_{Aa,j}+2n_{aa,j}}{2n_j}.

This is an allele-count calculation, not a genotype-frequency calculation. It does not require Hardy–Weinberg equilibrium: regardless of how genotypes are distributed, each \(Aa\) individual contributes one copy of \(a\), and each \(aa\) individual contributes two.

The Pooled Estimate and Its Target

Writing the pooled estimate in terms of the population-specific estimates makes the weighting explicit. Since \(A_j=m_j\hat{p}_j\), the total allele count is \(\sum_j m_j\hat{p}_j\). Dividing by the total number of sampled chromosome copies gives:

\hat{p}_{\mathrm{pool}} = \frac{\sum_{j=1}^{K}m_j\hat{p}_j}{\sum_{j=1}^{K}m_j} = \sum_{j=1}^{K}w_j\hat{p}_j, \qquad w_j=\frac{m_j}{\sum_{\ell=1}^{K}m_\ell}.

The weights \(w_j\) are the fractions of all sampled chromosome copies contributed by each population. They sum to 1. Thus, the pooled estimate is a weighted average of the estimated population-specific frequencies, with weights set by the numbers of sampled chromosome copies—not by the number of populations.

Interpretation: The pooled estimate is the observed frequency of allele \(a\) among the combined sampled chromosome copies. When samples are used to estimate underlying population frequencies, its corresponding target is the chromosome-count-weighted average of those frequencies, with weights determined by the sampling counts.

To state the target formally, let \(p_j\) be the true frequency of allele \(a\) in population \(j\), and let \(M=\sum_jm_j\). For fixed sample counts, the target associated with pooling is:

p_{\mathrm{pool}}=\frac{\sum_{j=1}^{K}m_jp_j}{M}.

This is not necessarily the frequency in any one population, nor is it automatically a measure of the frequency in a broader real-world population. It is tied to the chosen population set and the chromosome-count weights. If the intended summary instead gives each of the \(K\) populations equal importance, its target is \((1/K)\sum_jp_j\), estimated by \((1/K)\sum_j\hat{p}_j\). That arithmetic average and the pooled estimate coincide when all populations contribute the same number of chromosome copies, but can differ otherwise.

Why the Weighted Target Follows

The relationship between the pooled estimate and its target follows directly from counting. Suppose the sample counts are fixed and the sampled individuals represent their respective populations for the allele-frequency question. Then the expected allele count from population \(j\) is \(E(A_j)=m_jp_j\): each of its \(m_j\) sampled chromosome copies has probability \(p_j\) of carrying \(a\), so the expected total is the sum of those probabilities.

E(\hat{p}_{\mathrm{pool}}) = E\left(\frac{\sum_j A_j}{M}\right) = \frac{\sum_jE(A_j)}{M} = \frac{\sum_jm_jp_j}{M} = p_{\mathrm{pool}}.

Therefore, under these sampling conditions, the pooled allele-frequency estimate is unbiased for the chromosome-count-weighted target. This expectation argument does not require chromosome copies within an individual to be independent; linearity of expectation is enough. It does require that the sampling and allele calls support the stated population frequencies. The result does not make the estimate unbiased for every possible cross-population summary: in general, it is not an estimate of the equally weighted population average.

Worked Example: Counting Alleles Before Pooling

Suppose two hypothetical populations are sampled at a biallelic locus. The target allele is \(a\). Population 1 has 20 diploid individuals: 8 \(AA\), 8 \(Aa\), and 4 \(aa\). Population 2 has 10 diploid individuals: 2 \(AA\), 4 \(Aa\), and 4 \(aa\). Assume all genotypes are called correctly and each individual contributes both chromosome copies.

Population 1. The number of sampled chromosome copies is \(m_1=2(20)=40\). The count of \(a\) copies is \(A_1=8+2(4)=16\), so:

\hat{p}_1=\frac{16}{40}=0.4000.

Population 2. Here \(m_2=2(10)=20\), and \(A_2=4+2(4)=12\). Thus:

\hat{p}_2=\frac{12}{20}=0.6000.

Pooled sample. Combining the counts gives 28 copies of \(a\) among 60 chromosome copies:

\hat{p}_{\mathrm{pool}} = \frac{16+12}{40+20} = \frac{28}{60} = 0.4667.

Population 1 supplied twice as many chromosome copies as population 2, so its estimate has twice the weight. The pooled estimate is also \((40/60)(0.4000)+(20/60)(0.6000)=0.4667\). The arithmetic mean of the two population-specific estimates is \((0.4000+0.6000)/2=0.5000\), a different summary that gives each population equal weight.

Three Ways to Check What an Estimate Represents

Before reporting a combined frequency, name the unit being weighted. In the pooled estimator, the basic units are chromosome copies. The total number of individuals alone does not determine the weight if populations contribute different numbers of called chromosome copies; with complete diploid genotypes, each individual contributes two. Also name the populations included: changing that set changes the target.

A useful check is to write the weights explicitly. If population 1 contributes 80 chromosome copies and population 2 contributes 20, the weights are \(80/100=0.8\) and \(20/100=0.2\). If both estimated frequencies are between 0 and 1, their weighted average must also lie between them. A pooled estimate outside the range of its population-specific estimates signals an arithmetic or data-handling error.

For an equally weighted average, the weights are instead \(1/K\) for all populations. This may be appropriate when the question is specifically about the average among a defined set of populations, with each population counting equally. It is not the same estimand as the frequency among all sampled chromosome copies. Neither summary is universally preferable: the scientific question determines which weighting is relevant.

Worked Example: Unequal Contributions Change the Pooled Target

Consider two hypothetical populations with true frequencies \(p_1=0.20\) and \(p_2=0.80\). For illustration, suppose the samples have exactly those frequencies: 90 diploid individuals are sampled from population 1, and 10 from population 2. Genotypes are complete and correctly called.

Population 1 contributes \(m_1=180\) chromosome copies, of which \(A_1=180(0.20)=36\) carry \(a\). Population 2 contributes \(m_2=20\) copies, of which \(A_2=20(0.80)=16\) carry \(a\). The pooled estimate is:

\hat{p}_{\mathrm{pool}} = \frac{36+16}{180+20} = \frac{52}{200} = 0.2600.

The equal-population average is \((0.20+0.80)/2=0.5000\). The values differ because 90% of the sampled chromosome copies came from population 1, whose frequency is lower. The pooled result, 0.2600, describes the combined sample’s allele frequency; 0.5000 describes the average of the two population frequencies when the populations receive equal weight.

Assumptions and Scope

The arithmetic definition of the pooled sample frequency only requires correctly counted allele copies and a clear denominator. Interpreting that statistic as an estimate of underlying population frequencies requires additional care. The following conditions make the interpretation explicit:

Population labels are meaningful. Each sampled individual is assigned to the population whose frequency is being estimated.
Alleles are called correctly. Genotype errors can alter both the allele counts and the resulting frequencies.
The denominator matches the data. With complete calls for diploid individuals, each contributes two chromosome copies. The formulas here assume that setting.
Sampling represents the stated populations. To generalize a population-specific estimate beyond the observed sample, the sample must provide an appropriate basis for estimating that population’s frequency.
The target weights are intentional. Pooling uses the observed chromosome-count weights. If the scientific target uses different weights, the pooled estimate does not automatically estimate it.

No Hardy–Weinberg assumption is needed for the allele-count formula. Nor does the pooled estimate by itself establish why populations have different frequencies or what the frequency would be in a different population set. It is a descriptive summary of counts, with an inferential target determined by its weights.

Worked Example: Choosing the Summary for the Question

Suppose three hypothetical populations have estimated frequencies \(\hat{p}_1=0.10\), \(\hat{p}_2=0.40\), and \(\hat{p}_3=0.70\). Their samples contribute 20, 20, and 60 chromosome copies, respectively. Compare the pooled estimate with an equally weighted population average.

For the pooled estimate, the chromosome-count weights are \(20/100=0.20\), \(20/100=0.20\), and \(60/100=0.60\). Therefore:

\hat{p}_{\mathrm{pool}} = 0.20(0.10)+0.20(0.40)+0.60(0.70) = 0.020+0.080+0.420 = 0.5200.

For the equally weighted average, each population receives weight \(1/3\):

\hat{p}_{\mathrm{equal}} = \frac{0.10+0.40+0.70}{3} = \frac{1.20}{3} = 0.4000.

If the question concerns the allele frequency among these 100 sampled chromosome copies, 0.5200 is the relevant estimate. If it asks for the average frequency across these three populations with each population counting equally, 0.4000 is the relevant estimate. The calculations use the same population-specific estimates but answer different questions.

Common Mistake

Wrong reasoning: “There are three populations, so the pooled frequency is the average of their three estimated frequencies.”

Correction: That calculation gives an equally weighted population average. A pooled estimate combines allele counts and divides by the total number of chromosome copies. Equivalently, it weights each population-specific estimate by its share of those copies. Before calculating, state whether the target is a combined chromosome-copy frequency or an average that gives populations equal weight.

Key takeaway: Population-specific estimates are \(\hat{p}_j=A_j/m_j\). Pooling gives \(\hat{p}_{\mathrm{pool}}=\sum_jm_j\hat{p}_j/\sum_jm_j\), so populations are weighted by their sampled chromosome counts. An equal-population average is a distinct summary and should be used only when that is the intended target.

Check Your Understanding

Use allele-copy counts and state which target each summary represents.

A population has 12 \(AA\), 10 \(Aa\), and 3 \(aa\) individuals. How many chromosome copies and how many copies of \(a\) were sampled? What is \(\hat{p}\)?
Two populations contribute 30 and 70 chromosome copies. What weights do they receive in the pooled estimate?
Two populations have estimated frequencies 0.25 and 0.65 and contribute equal numbers of chromosome copies. What is the pooled estimate? What is the equally weighted population average?
Explain why pooling genotype counts does not require Hardy–Weinberg equilibrium.
A pooled estimate is reported as the “average frequency across populations.” What question should you ask to determine whether that interpretation is justified?