Allele Frequencies From Diploid Genotype Counts

Why Count Allele Copies?

A genotype count tells us how many individuals have each genotype. An allele frequency, by contrast, describes how common an allele is among the allele copies carried by those individuals. At a diploid locus, each individual contributes two copies. Consequently, a homozygote contributes two copies of its one allele, while a heterozygote contributes one copy of each allele.

This distinction matters even in a small data set: dividing the number of individuals with a genotype by the number of individuals does not, in general, give an allele frequency. The reliable procedure is to count the copies of each allele, add those counts, and divide by the total number of observed allele copies.

Definition: At a biallelic locus with alleles $A$ and $a$, let $n_{AA}$, $n_{Aa}$, and $n_{aa}$ be the counts of the three diploid genotypes in a sample, and let $N=n_{AA}+n_{Aa}+n_{aa}$ be the number of individuals with observed genotypes. The sample frequency of $A$ is $$ \hat{p}=\frac{2n_{AA}+n_{Aa}}{2N}, $$ and the sample frequency of $a$ is $$ 1-\hat{p}=\frac{n_{Aa}+2n_{aa}}{2N}. $$ The denominator $2N$ is the total number of allele copies contributed by the $N$ diploid individuals.

The same calculation works with other allele labels, such as $B$ and $b$ or $G$ and $g$. The labels do not affect the arithmetic. The formula assumes a single locus with exactly two alleles under consideration, diploid individuals, and a genotype call for each individual counted in $N$. It does not require the genotype counts to follow a particular population model.

From Genotype Counts to Allele Counts

The coefficients in the formula come directly from the alleles contained in each genotype. An $AA$ individual contributes two $A$ copies and no $a$ copies. An $Aa$ individual contributes one copy of each. An $aa$ individual contributes two $a$ copies and no $A$ copies. Thus, the number of observed $A$ copies is $2n_{AA}+n_{Aa}$, and the number of observed $a$ copies is $n_{Aa}+2n_{aa}$.

These copy counts must add to the total number of copies. Each of the $N$ individuals contributes two, so there are $2N$ copies altogether. The frequency is the relevant copy count divided by this total. A quick check is that the two allele frequencies add to 1: every observed copy is either $A$ or $a$.

\begin{aligned} \text{number of }A\text{ copies} &= 2n_{AA}+n_{Aa},\\ \text{number of }a\text{ copies} &= n_{Aa}+2n_{aa},\\ \text{total number of copies} &= 2(n_{AA}+n_{Aa}+n_{aa})=2N. \end{aligned}

Worked Example: Calculating Both Frequencies

Suppose a hypothetical sample contains 18 $AA$, 24 $Aa$, and 8 $aa$ individuals. Calculate the sample frequencies of $A$ and $a$.

First, the number of individuals is $N=18+24+8=50$, so the total number of allele copies is $2N=100$. The $A$-copy count is $$ 2(18)+24=36+24=60. $$ Therefore, $$ \hat{p}=\frac{60}{100}=0.600. $$ The $a$-copy count is $$ 24+2(8)=24+16=40, $$ so the frequency of $a$ is $40/100=0.400$. The check $0.600+0.400=1.000$ confirms that the two copy counts account for all 100 copies.

In this sample, 60% of the observed allele copies are $A$, and 40% are $a$. These are frequencies among copies, not percentages of individuals with a particular genotype.

A Short Justification of the Formula

The formula follows by expanding the count of copies rather than by making a model about how genotypes arose. For the $A$-frequency, each $AA$ contributes two copies and each $Aa$ contributes one; each $aa$ contributes none. That gives $2n_{AA}+n_{Aa}$ copies of $A$. Dividing by the total $2N$ gives the fraction of observed copies that are $A$.

The formula for $a$ follows in the same way. The frequencies must sum to 1 because the two copy counts sum to $2N$:

\begin{aligned} \frac{2n_{AA}+n_{Aa}}{2N}+ \frac{n_{Aa}+2n_{aa}}{2N} &=\frac{2n_{AA}+2n_{Aa}+2n_{aa}}{2N}\\ &=\frac{2N}{2N}=1. \end{aligned}

This identity is also a useful error check. If the two calculated frequencies do not sum to 1, check the genotype counts, the copy multipliers, and the denominator.

Worked Example: Heterozygotes and Homozygotes Contribute Different Copies

Consider a hypothetical sample of 40 diploid individuals with 5 $BB$, 30 $Bb$, and 5 $bb$ genotypes. Find the frequency of each allele.

There are $2(40)=80$ allele copies. The 30 heterozygotes each contribute one $B$ and one $b$, giving 30 copies of each allele. The 5 $BB$ individuals contribute $2(5)=10$ additional $B$ copies, while the 5 $bb$ individuals contribute $2(5)=10$ additional $b$ copies. Therefore, the total $B$-copy count is $30+10=40$, and the total $b$-copy count is $30+10=40$.

The frequencies are $$ \hat{p}_B=\frac{2(5)+30}{80}=\frac{40}{80}=0.500, \qquad \hat{p}_b=\frac{30+2(5)}{80}=\frac{40}{80}=0.500. $$ Here the 10 homozygotes contribute 20 copies in total: 10 $B$ copies from the $BB$ individuals and 10 $b$ copies from the $bb$ individuals. They do not contribute 20 copies of each allele. Together, the heterozygotes contribute 60 copies and the homozygotes contribute 20, accounting for all 80 copies.

A General Counting Procedure

For any set of complete diploid genotype counts at a biallelic locus, use the same sequence. Keeping the copy-count step explicit makes it easier to spot mistakes before converting counts to frequencies.

Check the counts.
Confirm that the three genotype counts refer to the same locus and sample, and add them to obtain $N$.

Find the total number of copies.
For $N$ diploid individuals with observed genotypes, use $2N$.

Count copies of the first allele.
For $A$, calculate $2n_{AA}+n_{Aa}$; double the homozygote count and add the heterozygote count.

Divide and check.
Divide the copy count by $2N$. Calculate the other allele's count directly or subtract the first frequency from 1, then verify the frequencies sum to 1.

Worked Example: Unequal Counts and a Frequency Check

A hypothetical data set has 11 $GG$, 29 $Gg$, and 33 $gg$ individuals. Calculate both allele frequencies, reporting each to four significant figures.

The number of individuals is $N=11+29+33=73$, giving $2N=146$ copies. The $G$-copy count is $2(11)+29=22+29=51$. Thus, $$ \hat{p}_G=\frac{51}{146}=0.3493. $$ The $g$-copy count is $29+2(33)=29+66=95$, so $$ \hat{p}_g=\frac{95}{146}=0.6507. $$ The counts add to $51+95=146$, and the rounded frequencies add to $0.3493+0.6507=1.0000$. The sample therefore has a $G$-copy frequency of 0.3493 and a $g$-copy frequency of 0.6507.

What the Calculation Does—and Does Not—Assume

These frequencies summarize the observed allele copies in the individuals whose genotype calls are included. The calculation itself does not assume random mating, Hardy–Weinberg equilibrium, or any particular pattern of genotype occurrence. It is simply a count divided by the total number of counted copies. Those population models and questions about departures from them are separate from this arithmetic.

The denominator must correspond to the individuals actually included. If some individuals have no genotype call at the locus, do not count them in $N$ while also omitting their allele copies from the numerator. Instead, define $N$ as the number of individuals with observed genotypes and use $2N$ as the total number of copies represented by those genotypes. This produces the frequency among the called copies; it does not by itself establish that missing calls are unrelated to genotype.

Likewise, the result is an observed sample frequency. If the sample is being used to describe a larger population, whether and how well it represents that population is a separate question. The copy-count calculation alone does not measure sampling uncertainty or correct for a biased sample.

Common Mistakes

Using individuals as the denominator. Dividing an allele count by $N$ is wrong for diploid individuals; there are $2N$ allele copies. Use the copy count divided by $2N$.
Counting a homozygote once. An $AA$ individual carries two $A$ copies, so the $AA$ count is multiplied by 2 when counting $A$. Apply the same rule to $aa$ when counting $a$.
Treating a heterozygote as two copies of either allele. Each $Aa$ individual contributes one $A$ and one $a$, not two copies of one allele.
Confusing genotype proportions with allele frequencies. The proportion of individuals that are $AA$ is $n_{AA}/N$; the $A$-allele frequency is $(2n_{AA}+n_{Aa})/(2N)$. They answer different questions.
Miscounting total contributions. The homozygotes contribute two copies per homozygous individual in total, allocated to the allele each individual carries. For example, 5 $BB$ and 5 $bb$ individuals contribute 10 $B$ and 10 $b$ copies, or 20 copies altogether.

Key takeaway: For diploid genotype counts, count two copies from every individual. Each homozygote contributes two copies of its allele, each heterozygote contributes one copy of each allele, and each allele frequency is its copy count divided by $2N$.

Check Your Understanding

For each question, keep track of individuals and allele copies separately.

A sample has 12 $AA$, 16 $Aa$, and 12 $aa$ individuals. What are $N$, the total number of copies, and the frequency of $A$?
In a sample with 8 $CC$, 14 $Cc$, and 3 $cc$ individuals, how many copies of $C$ and $c$ are observed?
Why does an $Aa$ individual contribute one copy to each allele count rather than two copies to one of them?
A researcher divides the number of $A$ copies by the number of individuals. What is wrong with the denominator for a diploid sample?
If 10 diploid individuals have complete genotype calls, how many allele copies are represented, regardless of their genotype composition?