Hardy–Weinberg Equilibrium With Multiple Alleles

Why the Number of Alleles Changes the Calculation

At a locus with two alleles, the Hardy–Weinberg model gives three genotype categories. With three or more alleles, the same probability principle applies, but there are more distinct genotypes to keep track of. The key distinction is whether the two allele copies are the same or different: a homozygote has two copies of one allele, while a heterozygote has one copy of each of two different alleles.

For a heterozygote, the allele pair can arrive in either order. An individual with genotype $A_iA_j$ could receive $A_i$ from one gamete and $A_j$ from the other, or receive $A_j$ and $A_i$, respectively. Those are distinct ordered events, but they describe the same unordered genotype. This is why the heterozygote probability has a factor of two, whereas the homozygote probability does not.

Model and Notation

Definition: Consider a locus with $k$ alleles, denoted $A_1,\ldots,A_k$. Let $p_i$ be the population frequency of allele $A_i$, where $p_i\geq 0$ and $\sum_{i=1}^{k}p_i=1$. Under Hardy–Weinberg equilibrium, the two allele copies in an individual are modeled as independent draws from this allele distribution. Genotypes are treated as unordered pairs: $A_iA_i$ is a homozygote, and $A_iA_j$, for $i\neq j$, is a heterozygote.

This model describes random union of gametes at the locus: the allele in one gamete has probability $p_i$ of being $A_i$, and the allele in the other gamete has the same probabilities, independently. For this calculation, assume the two gamete allele distributions are the same and that their allele choices are independent. In the usual population interpretation, random mating with respect to the locus produces this genotype distribution from the allele frequencies. Applying the model does not establish that real data meet its assumptions.

The genotype proportions depend on the allele frequencies, not on how the alleles are named. There are $k$ homozygote categories and one heterozygote category for each distinct pair of alleles. Counting these categories once gives $k+k(k-1)/2=k(k+1)/2$ possible unordered genotypes.

Deriving the Genotype Proportions

First consider an ordered pair of gametes. Independence means the probability of receiving $A_i$ from the first gamete and $A_j$ from the second is $p_i p_j$. A homozygote $A_iA_i$ corresponds to just one ordered outcome, $A_i$ followed by $A_i$. A heterozygote $A_iA_j$, with $i\neq j$, corresponds to two ordered outcomes. Adding those probabilities gives the genotype formulas.

\begin{aligned} P(A_iA_i) &= P(A_i\text{ from the first gamete},\,A_i\text{ from the second})=p_i p_i=p_i^2,\\ P(A_iA_j) &= P(A_i,A_j)+P(A_j,A_i)\\ &=p_i p_j+p_j p_i=2p_i p_j,\qquad i<j. \end{aligned}

Writing the heterozygote formula only for $i<j$ ensures that each unordered genotype is counted once. The formula $2p_i p_j$ is also valid when the two allele labels are written in the opposite order, but $A_iA_j$ and $A_jA_i$ are not separate genotype categories.

Theorem: Under the multiple-allele Hardy–Weinberg model, the expected proportion of genotype $A_iA_i$ is $p_i^2$, and the expected proportion of genotype $A_iA_j$ is $2p_i p_j$ for each distinct pair $i<j$. These genotype proportions sum to one: $$ \sum_{i=1}^{k}p_i^2+\sum_{i<j}2p_i p_j =\left(\sum_{i=1}^{k}p_i\right)^2=1. $$

The identity follows by expanding the square of the sum of allele frequencies: each squared term $p_i^2$ appears once, and each cross-product $p_i p_j$ appears twice. Since the allele frequencies sum to one, the total probability across all unordered genotypes is one. This provides a useful check on a table of calculated proportions.

Worked Example: Three Alleles and Six Genotype Categories

Suppose a hypothetical population has three alleles, $A_1,A_2,A_3$, with frequencies $p_1=0.50$, $p_2=0.30$, and $p_3=0.20$. Assume the model’s independent, equally distributed gamete draws. Calculate the expected proportion of every genotype.

For each homozygote, square its allele frequency:

P(A_1A_1)=0.50^2=0.250,\quad P(A_2A_2)=0.30^2=0.0900,\quad P(A_3A_3)=0.20^2=0.0400.

For each heterozygote, multiply the two allele frequencies and double the product:

\begin{aligned} P(A_1A_2)&=2(0.50)(0.30)=0.300,\\ P(A_1A_3)&=2(0.50)(0.20)=0.200,\\ P(A_2A_3)&=2(0.30)(0.20)=0.120. \end{aligned}

There are three homozygote and three heterozygote categories, for six categories in total. Their proportions sum to $0.250+0.0900+0.0400+0.300+0.200+0.120=1.000$, as required. For example, the model assigns 0.300, or 30.0%, of individuals to genotype $A_1A_2$. These are model-based population proportions, not a claim that a finite sample must contain exactly those percentages.

From Proportions to Expected Counts

If the population model gives a genotype proportion $q$, then a sample of size $n$ has expected count $nq$ for that genotype. This is an expectation, not a guarantee about the realized count. The calculation uses the same genotype proportions; it does not change the probability formula.

Worked Example: Expected Counts With Four Alleles

In a second hypothetical population, suppose a locus has four alleles with frequencies $0.40, 0.30, 0.20,$ and $0.10$, respectively. Assume Hardy–Weinberg proportions and consider a sample of $n=500$ individuals. Calculate the expected count for each genotype.

First calculate the four homozygote proportions by squaring each frequency. Then calculate the six heterozygote proportions by doubling each distinct pairwise product:

Genotype	Proportion calculation	Expected proportion	Expected count, $500q$
$A_1A_1$	$(0.40)^2$	0.160	80
$A_2A_2$	$(0.30)^2$	0.0900	45
$A_3A_3$	$(0.20)^2$	0.0400	20
$A_4A_4$	$(0.10)^2$	0.0100	5
$A_1A_2$	$2(0.40)(0.30)$	0.240	120
$A_1A_3$	$2(0.40)(0.20)$	0.160	80
$A_1A_4$	$2(0.40)(0.10)$	0.0800	40
$A_2A_3$	$2(0.30)(0.20)$	0.120	60
$A_2A_4$	$2(0.30)(0.10)$	0.0600	30
$A_3A_4$	$2(0.20)(0.10)$	0.0400	20

The proportions sum to $0.160+0.0900+0.0400+0.0100+0.240+0.160+0.0800+0.120+0.0600+0.0400=1.000$. The expected counts sum to $80+45+20+5+120+80+40+60+30+20=500$. Thus, under the stated model, the expected count for $A_1A_2$ is 120 in a sample of 500, and the expected count for $A_4A_4$ is 5. Actual observed counts can differ from these expectations.

Checking the Formula for a Particular Genotype

A reliable way to apply the rule is to identify the number of ordered gamete outcomes represented by the genotype. If both copies match, there is one outcome. If they differ, there are two. The next example makes that distinction explicit for a particular allele pair and checks the full set of proportions.

Worked Example: Why the Heterozygote Probability Is Doubled

Suppose a hypothetical three-allele population has $p_1=0.60$, $p_2=0.25$, and $p_3=0.15$. Under independent gamete draws, find the probability of genotype $A_2A_3$, and compare it with the probability of genotype $A_2A_2$.

For $A_2A_3$, one gamete order has probability $(0.25)(0.15)=0.0375$. The reverse order has the same probability, so the unordered genotype probability is $0.0375+0.0375=0.0750$, or $2(0.25)(0.15)=0.0750$. For $A_2A_2$, there is only one matching ordered outcome, so its probability is $(0.25)(0.25)=0.0625$. It is not doubled.

As a check, the other genotype proportions are $P(A_1A_1)=0.60^2=0.360$, $P(A_3A_3)=0.15^2=0.0225$, $P(A_1A_2)=2(0.60)(0.25)=0.300$, and $P(A_1A_3)=2(0.60)(0.15)=0.180$. Including $A_2A_3$ and $A_2A_2$, the sum is $0.360+0.0625+0.0225+0.300+0.180+0.0750=1.000$.

Common Mistake and Why the Model Matters

A common mistake is to double every genotype probability because each individual has two allele copies. That reasoning confuses an allele copy with an unordered genotype. A homozygote such as $A_2A_2$ can arise through only one allele-label order, $A_2$ then $A_2$; its probability is $p_2^2$. A heterozygote such as $A_2A_3$ can arise through two orders, $A_2$ then $A_3$ or $A_3$ then $A_2$; its probability is $2p_2p_3$.

A related error is to list both $A_iA_j$ and $A_jA_i$ as separate genotypes and also use $2p_ip_j$ for each. The genotype is an unordered pair, so those are one category. Either combine the two ordered outcomes into one probability $2p_ip_j$, or retain the ordered outcomes separately as $p_ip_j$ each—but do not do both.

The model matters because it turns a set of allele frequencies into a complete, internally consistent set of genotype proportions. It is useful for describing the expected genotype distribution at a multi-allelic locus under random union of gametes. When applying the model to data, its assumptions must be appropriate to the population and setting; a calculated expected proportion alone does not show that the assumptions hold.

Key takeaway: For $k$ alleles with frequencies summing to one, assign $p_i^2$ to each homozygote $A_iA_i$, and $2p_ip_j$ to each distinct heterozygote $A_iA_j$. Count each unordered genotype once, and check that all proportions sum to one.

Check Your Understanding

Use the independent-gamete Hardy–Weinberg model in answering these questions.

A locus has three alleles with frequencies $0.50, 0.30,$ and $0.20$. What is the expected proportion of the heterozygote formed by the first and third alleles?
Explain why the probability of a homozygote is not doubled, even though an individual receives one allele from each of two gametes.
For four alleles, how many homozygote categories and how many distinct heterozygote categories are there?
If a genotype has expected proportion $0.08$ in a sample of 250 individuals, what is its expected count? Does that mean the observed count must equal it?
State one assumption about the two gamete allele choices that is needed to use the genotype formulas in this tutorial.

Genotype	Proportion calculation	Expected proportion	Expected count, \(500q\)
\(A_1A_1\)	\((0.40)^2\)	0.160	80
\(A_2A_2\)	\((0.30)^2\)	0.0900	45
\(A_3A_3\)	\((0.20)^2\)	0.0400	20
\(A_4A_4\)	\((0.10)^2\)	0.0100	5
\(A_1A_2\)	\(2(0.40)(0.30)\)	0.240	120
\(A_1A_3\)	\(2(0.40)(0.20)\)	0.160	80
\(A_1A_4\)	\(2(0.40)(0.10)\)	0.0800	40
\(A_2A_3\)	\(2(0.30)(0.20)\)	0.120	60
\(A_2A_4\)	\(2(0.30)(0.10)\)	0.0600	30
\(A_3A_4\)	\(2(0.20)(0.10)\)	0.0400	20