From Allele Probabilities to Genotype Proportions
At a biallelic locus, an individual inherits one allele from each of two gametes. If the probability of allele \(A\) in each gamete is \(p\), and the probability of allele \(a\) is \(q\), pairing the gametes gives probabilities for the three possible genotypes: \(AA\), \(Aa\), and \(aa\). The key detail is that \(Aa\) can be formed in two different ways: \(A\) from the first gamete and \(a\) from the second, or the reverse.
This derivation gives the Hardy–Weinberg genotype proportions from allele frequencies. It depends on how gametes pair, not on a particular cross between two specified parental genotypes. The resulting proportions describe the model's expected genotype distribution; they do not promise that every finite sample will have exactly those proportions.
Assumptions for the Derivation
The calculation requires a biallelic locus and diploid genotypes: each individual receives two alleles at the locus. It also assumes that the probability of drawing \(A\) is \(p\) and of drawing \(a\) is \(q\) for either gamete, and that the two gamete draws are independent. Equivalently, alleles from the same population gamete pool pair at random, without a dependence that changes the probability of one draw given the other.
These assumptions are the relevant part of the Hardy–Weinberg population model for this derivation. If gamete allele frequencies differ between the two contributions, or if the contributions are not independent, the multiplication used below need not apply. For example, in a specified parental cross, the offspring genotype probabilities depend on the parental genotypes and their transmissions. That is a different calculation from pairing two independent draws from a population allele pool.
Because the locus has only \(A\) and \(a\), their gamete probabilities exhaust the possibilities: \(p+q=1\). The derivation uses the multiplication rule for independent events to find each ordered pair of gamete alleles. It then adds probabilities for distinct, mutually exclusive ordered pairs that yield the same unordered genotype.
The Pairing Derivation
First distinguish the order of the two gamete contributions. The first entry below is the allele from one gamete and the second is the allele from the other. For example, \(Aa\) and \(aA\) are different ordered outcomes, but both produce the same diploid genotype \(Aa\).
| First gamete | Second gamete | Probability | Genotype |
|---|---|---|---|
| \(A\) | \(A\) | \(p \times p = p^2\) | \(AA\) |
| \(A\) | \(a\) | \(p \times q = pq\) | \(Aa\) |
| \(a\) | \(A\) | \(q \times p = pq\) | \(Aa\) |
| \(a\) | \(a\) | \(q \times q = q^2\) | \(aa\) |
The \(AA\) genotype occurs only when both gametes carry \(A\), so its probability is \(p \times p=p^2\). Similarly, \(aa\) occurs only when both carry \(a\), giving \(q \times q=q^2\). The heterozygote occurs in either of the two middle rows. Since those ordered outcomes cannot happen simultaneously, their probabilities add:
Thus, the three genotype probabilities are \(p^2\), \(2pq\), and \(q^2\). The factor of 2 is not an extra assumption or adjustment. It counts the two possible orders in which one \(A\) and one \(a\) can arrive. In contrast, there is only one ordered combination for each homozygote.
Why the Proportions Sum to One
The probability calculation provides a direct check on the result. Every pair of gametes must produce exactly one of the three genotypes: \(AA\), \(Aa\), or \(aa\). As these outcomes are mutually exclusive and exhaustive, their probabilities must total one. Algebra confirms that requirement because \(p+q=1\), so \((p+q)^2=1\).
This check also helps detect a common error. If the heterozygote is counted only once as \(pq\), the proposed proportions sum to \(p^2+pq+q^2\), which is generally less than one. The missing probability is the second ordered heterozygote outcome, \(aA\), with probability \(qp\). Since \(pq=qp\), including both gives \(2pq\).
The same identity shows that the three proportions are nonnegative and normalized: allele frequencies satisfy \(0\leq p\leq1\) and \(0\leq q\leq1\), so each genotype proportion is nonnegative, and together they account for the full population under the model. The identity is a check of the derivation, not evidence by itself that observed data follow the model.
Worked Example: Deriving Proportions When \(p=0.70\)
Suppose, hypothetically, that \(A\) has frequency \(p=0.70\) in the gamete pool. The other allele has frequency \(q=1-p=1-0.70=0.30\). Assume the two gamete contributions are independent draws with these same probabilities.
For \(AA\), both gametes must carry \(A\):
For \(Aa\), either ordering is possible:
For \(aa\), both gametes must carry \(a\):
The check is \(0.49+0.42+0.09=1.00\). Under the stated model, the expected genotype proportions are 0.49 \(AA\), 0.42 \(Aa\), and 0.09 \(aa\). These are proportions, so the corresponding percentages are 49%, 42%, and 9%.
Interpreting the Model Proportions
Each expression is the probability that a randomly formed individual has a particular genotype under the assumptions above. In a population described by this model, the same values are its expected genotype proportions. A proportion is dimensionless: it can be reported as a decimal or as a percentage. If a population has \(N\) individuals, multiplying a model proportion by \(N\) gives the corresponding expected count. An expected count need not be an integer, and observed counts can differ because a finite sample need not match its expectation exactly.
The formulas do not say that an observed group must contain precisely the predicted fractions, nor do they establish that the assumptions hold in any particular population. They translate specified allele frequencies into genotype probabilities under a particular pairing model. Comparing observed counts with model expectations is a separate inferential task.
Worked Example: The Heterozygote Has Two Routes
Consider a hypothetical gamete pool with \(p=0.20\) and \(q=0.80\). We calculate the two ordered ways to form \(Aa\) separately before adding them.
The probability that the first gamete supplies \(A\) and the second supplies \(a\) is:
The reverse ordering has probability:
The two events are distinct and mutually exclusive, so the total heterozygote probability is \(0.16+0.16=0.32\), equivalently \(2pq=2(0.20)(0.80)=0.32\). For completeness, \(P(AA)=(0.20)^2=0.04\) and \(P(aa)=(0.80)^2=0.64\). The sum \(0.04+0.32+0.64=1.00\) confirms that all three genotypes are included.
If only one ordering had been counted, the reported heterozygote proportion would be 0.16 rather than 0.32, and the three proposed proportions would sum to \(0.04+0.16+0.64=0.84\). The shortfall of 0.16 is exactly the omitted reverse ordering.
From Proportions to Expected Counts
To express model proportions as expected counts, multiply each proportion by the number of individuals. This is a conversion of scale, not a new genotype-probability derivation. The genotype proportions still depend on the assumed allele frequencies and independent pairing; the population size determines the expected counts.
Worked Example: Expected Counts in a Hypothetical Population
Suppose a hypothetical group contains \(N=400\) diploid individuals, and the model uses \(p=0.45\) and \(q=0.55\). These allele probabilities sum to one, and assume independent gamete pairing as required by the derivation.
The genotype proportions are:
Multiplying each proportion by \(N=400\) gives the model's expected counts:
The expected counts sum to \(81+198+121=400\), as they must because the proportions sum to one. Under this model, the expected number of heterozygotes is 198 in a group of 400. This statement describes an expectation; an actual group of 400 need not contain exactly 198 heterozygotes.
Common Mistakes and a Practical Check
- Counting only one heterozygote ordering. The reasoning “one \(A\) and one \(a\) has probability \(pq\)” omits the distinct ordering in which \(a\) arrives first. Add the two mutually exclusive probabilities: \(pq+qp=2pq\).
- Multiplying probabilities without the independence assumption. The product \(p^2\) for \(AA\), for example, uses the probability \(p\) for each draw and independence between draws. If the draws are dependent, that product is not justified by the marginal allele frequency alone.
- Treating expected proportions as guaranteed observed proportions. The formulas describe the model distribution. Finite observed counts can differ from their expected values, even when the model is appropriate.
- Forgetting to check the allele frequencies. Before calculating, confirm \(p\) and \(q\) are valid probabilities and that \(p+q=1\). If one frequency is given, calculate the other as its complement.
A reliable calculation therefore has three checks: state the allele frequencies and pairing assumption; calculate both ordered routes to \(Aa\); and verify \(p^2+2pq+q^2=(p+q)^2=1\). These steps make clear both where the factor of 2 comes from and why the results form a complete genotype distribution.
Check Your Understanding
Use the independent pairing model stated in this tutorial for each question.
- If \(p=0.35\), what is \(q\), and what are the model proportions of \(AA\), \(Aa\), and \(aa\)? Verify that they sum to one.
- Write the two ordered gamete combinations that produce \(Aa\). Why are their probabilities added?
- A student calculates \(P(Aa)=pq\). Identify the omitted outcome and correct the calculation.
- For \(p=0.60\), \(q=0.40\), and \(N=500\), calculate the expected count of each genotype under the model.
- In one or two sentences, explain why expected genotype counts do not have to equal observed counts in a finite population.