Allele Frequencies in Known Population Mixtures

Why Mixture Proportions Matter

A population assembled from multiple source populations does not generally have the same allele frequency as any one source. Its frequency depends on both the frequency in each source and the proportion of the mixture contributed by that source. A source with a high frequency has little influence if it contributes very few individuals; a source that contributes most of the mixture has correspondingly greater influence.

The calculation is straightforward once the weights are defined correctly. Each weight must describe the source population’s share of the mixture, and the weights must account for the entire mixture. The resulting frequency is a weighted average—not necessarily the ordinary average of the source frequencies.

Definition: Suppose a mixture contains $K$ source populations. Let $w_k$ be the proportion of the mixture contributed by source population $k$, and let $p_k$ be the frequency of a specified allele in that source. The mixture frequency of that allele is $$ p_{\mathrm{mix}}=\sum_{k=1}^{K}w_kp_k, $$ where $w_k\geq 0$ for every $k$ and $\sum_{k=1}^{K}w_k=1$. For a diploid autosomal locus, if the weights are proportions of individuals and each individual contributes two allele copies, the same weights apply to allele copies.

Deriving the Weighted-Average Formula

The formula follows by considering a randomly selected allele copy from the mixture. Let $Z$ identify the source population of the individual contributing that copy, and let $A$ denote the event that the copy carries the specified allele. The chance that the copy comes from source $k$ is $P(Z=k)=w_k$. Within that source, the chance that it carries the allele is $P(A\mid Z=k)=p_k$. The law of total probability therefore gives

\begin{aligned} P(A) &=\sum_{k=1}^{K}P(A\mid Z=k)P(Z=k)\\ &=\sum_{k=1}^{K}p_kw_k\\ &=p_{\mathrm{mix}}. \end{aligned}

This argument also explains the assumptions behind the calculation. The source categories must describe the mixture being summarized, the weights must represent the mixture’s actual composition, and each $p_k$ must refer to the same allele at the same locus. When weights are proportions of individuals, the simple formula also assumes equal allele-copy contribution per individual, as for diploid autosomal genotypes. If source frequencies are estimates rather than known values, substituting them gives a plug-in estimate of the mixture frequency; the formula itself does not quantify the uncertainty in that estimate.

The weights can come from stated mixture proportions or from counts of contributing individuals. If $N_k$ individuals come from source $k$, and these counts describe the composition of the mixture, then

w_k=\frac{N_k}{\sum_{j=1}^{K}N_j}.

These are composition weights, not weights chosen merely because a sample happens to contain a certain number of observations from each source. A sample deliberately collected in different proportions from the target mixture does not, by itself, reveal the target mixture’s weights.

Worked Example: Combining Two Source Frequencies

In a hypothetical mixture, source population 1 contributes 70% of individuals and has frequency $p_1=0.12$ for allele $A$. Source population 2 contributes the remaining 30% and has frequency $p_2=0.28$. Find the mixture frequency.

The weights are $w_1=0.70$ and $w_2=0.30$; they are nonnegative and sum to one. Apply the weighted-average formula:

\begin{aligned} p_{\mathrm{mix}} &=w_1p_1+w_2p_2\\ &=(0.70)(0.12)+(0.30)(0.28)\\ &=0.084+0.084\\ &=0.168. \end{aligned}

The specified allele has frequency 0.168, or 16.8%, in this mixture. This value is closer to 0.12 than to 0.28 because the lower-frequency source contributes the larger share. The mixture frequency describes the combined population under the stated composition; it does not imply that either source population has that frequency.

Several Sources and a Check on the Result

With more than two sources, the same calculation adds one weighted contribution for each source. It is often helpful to write each term separately before summing. This makes it easier to detect an omitted source or a weight that has been paired with the wrong frequency.

Result: Because the weights are nonnegative and sum to one, $p_{\mathrm{mix}}$ is between the smallest and largest source frequencies: $$ \min_k(p_k)\leq p_{\mathrm{mix}}\leq\max_k(p_k). $$ If all source frequencies equal $p$, then $p_{\mathrm{mix}}=p$, regardless of the weights.

To see the bounds, let $p_{\min}=\min_k(p_k)$ and $p_{\max}=\max_k(p_k)$. For every source, $p_{\min}\leq p_k\leq p_{\max}$. Multiplying each inequality by its nonnegative weight preserves the inequalities. Adding across sources gives

p_{\min}\sum_{k=1}^{K}w_k \leq \sum_{k=1}^{K}w_kp_k \leq p_{\max}\sum_{k=1}^{K}w_k.

Since the weights sum to one, the outer expressions reduce to $p_{\min}$ and $p_{\max}$, while the middle expression is $p_{\mathrm{mix}}$. Thus the result must lie within the range of the source frequencies. A calculated value outside that range signals an arithmetic or weighting error.

Worked Example: Combining Three Sources

Suppose a hypothetical mixture consists of 50% from source 1, 30% from source 2, and 20% from source 3. At the same locus, their frequencies for allele $B$ are $p_1=0.04$, $p_2=0.15$, and $p_3=0.40$. Calculate the frequency of $B$ in the mixture.

The proportions sum to $0.50+0.30+0.20=1.00$, so they account for the whole mixture. Weight each source frequency by its contribution:

\begin{aligned} p_{\mathrm{mix}} &=(0.50)(0.04)+(0.30)(0.15)+(0.20)(0.40)\\ &=0.020+0.045+0.080\\ &=0.145. \end{aligned}

The mixture frequency is 0.145, or 14.5%. As a check, the smallest source frequency is 0.04 and the largest is 0.40; 0.145 lies between them. The source with the highest frequency contributes only 20% of the mixture, so its frequency does not determine the combined value on its own.

Obtaining Weights From Mixture Counts

Sometimes the composition is given as counts rather than percentages. Convert the counts to proportions of the total before applying the formula. Dividing every source count by the total makes the weights sum to one, which is essential for interpreting the result as a frequency.

Worked Example: Turning Individual Counts Into Weights

In a hypothetical diploid autosomal population mixture, 240 individuals come from source 1 and 160 from source 2. Their frequencies for allele $C$ are $p_1=0.08$ and $p_2=0.23$, respectively. Assume these counts describe the mixture’s composition and that every individual contributes two allele copies. Find the mixture frequency.

There are $240+160=400$ individuals. Thus the source proportions are

w_1=\frac{240}{400}=0.60, \qquad w_2=\frac{160}{400}=0.40.

Because each individual contributes two copies at this autosomal locus, the proportions of individuals are also the proportions of allele copies contributed by the two sources. Substitution gives

\begin{aligned} p_{\mathrm{mix}} &=(0.60)(0.08)+(0.40)(0.23)\\ &=0.048+0.092\\ &=0.140. \end{aligned}

The mixture frequency is 0.140, or 14.0%. The frequency is not the unweighted mean of 0.08 and 0.23: the sources contribute different proportions of individuals and therefore different proportions of allele copies.

Interpreting and Using the Result

A mixture frequency is a summary for a population with a specified composition. It may be used to describe the expected frequency of an allele in a randomly selected allele copy from that mixture. It does not say that every subgroup has the same frequency, nor does it describe a particular individual’s genotype. The population composition is part of the result: changing the weights can change the mixture frequency even when all source-specific frequencies remain fixed.

The weighted-average relationship also extends to a locus with more than two alleles. Apply it separately to each allele, using the same source weights. For example, if a source has frequencies $p_{k,A}$ and $p_{k,B}$ for alleles $A$ and $B$, then the mixture frequencies are $\sum_k w_kp_{k,A}$ and $\sum_k w_kp_{k,B}$. If each source’s allele frequencies sum to one, the mixture frequencies do too:

\sum_a p_{\mathrm{mix},a} =\sum_a\sum_k w_kp_{k,a} =\sum_k w_k\sum_a p_{k,a} =\sum_k w_k =1.

This provides another useful consistency check when summarizing all alleles at a locus. For a single specified allele, check that its mixture frequency is between the smallest and largest source frequencies. For a full set of allele frequencies, check that the mixture frequencies are nonnegative and sum to one, apart from any rounding.

Common Mistakes

Taking an unweighted average. The reasoning “there are two sources, so average their frequencies” is wrong unless the sources contribute equally. Use $\sum_k w_kp_k$, with each source weighted by its share of the mixture.
Using weights that do not sum to one. Counts such as 240 and 160 are not yet mixture weights. Divide each count by the total, or use an equivalent calculation that normalizes by the total contribution.
Pairing the wrong weight with a frequency. Each $w_k$ must multiply the $p_k$ for the same source. Label the sources consistently before substituting values.
Treating sample composition as mixture composition automatically. A sample collected in proportions different from the target mixture does not give the target weights simply because those are the observed sample counts. Use weights that represent the population mixture the frequency is intended to describe.
Reporting a value outside the source range. A weighted average with valid nonnegative weights cannot be below every source frequency or above every source frequency. Recheck the weights and arithmetic if this occurs.

Key takeaway: For a specified allele, the frequency in a known mixture is the sum of each source frequency multiplied by that source’s share of the mixture. Define the weights from the mixture composition, verify that they sum to one, and check that the result lies within the range of the source frequencies.

Check Your Understanding

Work through these questions using the mixture formula and its assumptions.

A mixture is 25% source 1 and 75% source 2. The allele frequencies are 0.32 and 0.12, respectively. What is the mixture frequency?
Three sources contribute 20, 30, and 50 individuals to a diploid autosomal mixture. What are their mixture weights?
Explain why a mixture frequency must lie between the smallest and largest source frequencies when the weights are valid.
Two source populations have the same frequency for an allele. What is the mixture frequency, and does it depend on their relative weights?
A sample contains equal numbers from two sources, but the target mixture is known to contain 80% from source 1. Which weights describe the target mixture, and why?