AP Biology · Unit 7 · Topic 7.5
Hardy–Weinberg Equilibrium
If 60% of the allele copies in a population are A, must 60% of the individuals be AA? No: each individual has two copies, and those copies can be paired in different ways. The Hardy–Weinberg model predicts the pairing pattern expected from random fertilization. Comparing that prediction with observations gives us a baseline for investigating a population.
Start with what p and q actually count
Use one autosomal locus in a diploid population with two alleles, A and a. Let \(p\) be the fraction of allele copies that are A and \(q\) the fraction that are a. Both lie between zero and one. Because these are the only two alleles in this model, \(p+q=1\). That identity describes a complete count; it is not evidence that the population is in equilibrium.
If genotype counts are available for \(N\) individuals, each AA contributes two A copies and each Aa contributes one:
You may use these counting formulas without assuming random mating. By contrast, predicting how many individuals are AA from p alone requires an additional model. Also, A does not have to be dominant: the letters label alleles, while dominance concerns how genotypes produce phenotypes.
Why random fertilization gives two routes to Aa
Assume Mendelian segregation and that both the egg and sperm pools have A frequency p and a frequency q. Random union makes the two gamete draws independent. Multiply their probabilities to obtain the probability of each combination.
| Egg allele | Sperm allele | Offspring genotype | Probability |
|---|---|---|---|
| A, probability p | A, probability p | AA | \(p\times p=p^2\) |
| A, probability p | a, probability q | Aa | \(pq\) |
| a, probability q | A, probability p | Aa | \(qp\) |
| a, probability q | a, probability q | aa | \(q^2\) |
The two Aa routes are mutually exclusive ways to obtain the same genotype. Add them: \(pq+qp=2pq\). The factor of two counts the routes, not the number of alleles inside each individual. Thus the expected genotype frequencies are:
These possibilities exhaust the outcomes, giving \(p^2+2pq+q^2=(p+q)^2=1\). A population-wide gamete table is not necessarily an Aa × Aa family cross. That particular cross has A probability one-half from each parent; a population's gamete frequency can instead be 0.2, 0.6, or another value.
What keeps the pattern stable over generations?
Random fertilization explains the genotype proportions at formation of offspring. Stability across later generations also requires processes that would change allele representation to be absent or negligible.
| Condition | Reason it matters |
|---|---|
| Very large breeding population | Makes random genetic drift negligible; the exact mathematical limit is infinite size. |
| Random mating with respect to this locus | Supports independent pairing of alleles rather than preferred genotype combinations. |
| No selection at this locus | Genotypes do not systematically differ in survival or reproductive contribution. |
| No migration or gene flow | Incoming or departing alleles do not alter the pool. |
| No mutation | Allele identities do not change through mutation. |
For the simple autosomal derivation, the gamete pools also have equal allele frequencies and alleles segregate normally. A real population can approximate the model at a particular locus and sampling stage without meeting every condition perfectly.
Check the stability directly. Among offspring with proportions \(p^2,2pq,q^2\), all copies in AA and half the copies in Aa are A. The next allele frequency is \(p^2+\tfrac12(2pq)=p(p+q)=p\). Random pairing alone has not increased the common allele or eliminated the rare one.
Under the stated assumptions, random mating can establish these autosomal genotype proportions in one generation even if the parental proportions differ. That is different from claiming an arbitrary population has been in equilibrium throughout its history.
Read genotype frequencies as functions of allele frequency
Because \(q=1-p\), the three predictions can be plotted against p. The horizontal axis below is not time: each position represents a different hypothetical allele frequency.
At \(p=q=0.5\), the proportions are 0.25 AA, 0.50 Aa, and 0.25 aa. The heterozygote maximum follows without calculus: \(2p(1-p)=0.5-2(p-0.5)^2\). A square cannot be negative, so this expression cannot exceed 0.5. An observed population can have more than 50% heterozygotes, but it would not have these equilibrium proportions at a two-allele locus.
When a recessive phenotype permits a square root
In a hypothetical population of 500 flowering plants, 20 have white flowers. Suppose white flowers uniquely identify aa, A is completely dominant for the alternative color, phenotype classification is reliable, and Hardy–Weinberg proportions are a reasonable assumption at this locus.
First convert the recessive count into a frequency: \(20/500=0.04\). Under the assumptions, this is \(q^2\), not q. Since frequencies are nonnegative:
Expected AA frequency is \(0.8^2=0.64\), and expected Aa frequency is \(2(0.8)(0.2)=0.32\). Multiplying by 500 gives 320 AA, 160 Aa, and 20 aa. Check both totals: \(320+160+20=500\), and a copies equal \(160+2(20)=200\) out of 1,000, recovering \(q=0.20\).
The 160 heterozygotes are 32% of all plants. Among the 480 plants with the dominant phenotype, their expected fraction is \(160/480=1/3\), about 33.3%. The denominator changes when the question restricts the group.
Without the equilibrium assumption, the 20 white plants do not determine how many of the other 480 are AA versus Aa. For example, 480 AA plus 20 aa would give \(q=40/1000=0.04\), while 480 Aa plus 20 aa would give \(q=520/1000=0.52\). The same recessive count can therefore accompany very different allele frequencies.
When genotypes are known, count first and predict second
For a separate invented sample, researchers identify 90 AA, 60 Aa, and 50 aa individuals. Here \(N=200\). Do not take the square root of the observed aa fraction to estimate q: we can count the actual alleles directly.
Now use these estimated allele frequencies to predict genotype counts under the equilibrium hypothesis: \(200p^2=72\), \(200(2pq)=96\), and \(200q^2=32\).
| Genotype | Observed, O | Expected, E | O − E |
|---|---|---|---|
| AA | 90 | 72 | 18 |
| Aa | 60 | 96 | −36 |
| aa | 50 | 32 | 18 |
The observed heterozygote frequency is 0.30, while the expected frequency is 0.48. The expected counts still contain 240 A copies, matching the observed sample. Allele frequencies can agree while genotype proportions differ.
Calculating \(0.6^2+2(0.6)(0.4)+0.4^2=1\) does not test the sample: the algebra holds for any p and q summing to one. The testable comparison is between the observed counts and the predicted counts.
Optional extension: quantify the mismatch
A Pearson goodness-of-fit statistic sums squared deviations scaled by expected counts. For the table:
All three expected counts are well above five. For an appropriately sampled population, the usual large-sample comparison uses one degree of freedom here: three categories, minus one constraint for their total, minus one independently estimated allele frequency. At a 5% significance level, the critical value is about 3.84, so 28.125 rejects the Hardy–Weinberg-proportions hypothesis for this dataset.
This is not a diagnosis of natural selection. Population mixing, nonrandom mating, selection, or genotype errors could contribute to a mismatch. Small expected counts call for an appropriate exact test instead. Failing to reject in another dataset would mean insufficient evidence of a departure, not proof that every equilibrium condition holds.
Keep a snapshot separate from an evolutionary history
A departure from expected genotype proportions and a change in allele frequency over time answer different questions. Suppose a hypothetical parental population has 100 AA and 100 aa, with no Aa. It has \(p=q=0.5\). Random mating under the ideal assumptions predicts 50 AA, 100 Aa, and 50 aa among 200 offspring: genotype proportions change, but allele frequencies stay at 0.5.
Conversely, a locus may show approximately equilibrium genotype proportions at successive censuses while its allele frequency changes between generations. Random mating can produce the pairing pattern again after another process changes the pool. A match at one locus also says nothing conclusive about every other locus in the genome.
Always identify the locus, population boundary, and life stage sampled. A mixture of separately mating populations is not necessarily one randomly mating population. Use the simple equations here for a diploid autosomal two-allele model; haploid, sex-linked, or multi-allele cases require the corresponding model.
Check the assumption before the arithmetic
1. If p = 0.3, what genotype frequencies does the model predict?
First find \(q=0.7\). Then AA is \(0.3^2=0.09\), Aa is \(2(0.3)(0.7)=0.42\), and aa is \(0.7^2=0.49\). The total is 1. A can be dominant and still be the less frequent allele.
2. Why not use q = 0.04 in the white-flower example?
The 0.04 is a fraction of aa individuals, not a fraction of a allele copies. Under equilibrium it equals \(q^2\), giving \(q=0.2\). Without equilibrium, count genotypes rather than applying that square root.
3. If 36% express a dominant phenotype under equilibrium, what is p?
The recessive phenotype accounts for \(1-0.36=0.64=q^2\). Thus \(q=0.8\) and \(p=0.2\). Taking the square root of the dominant fraction would wrongly treat AA and Aa together as AA alone.
4. Can 80% of a real population be heterozygous?
Yes. The 50% maximum applies to equilibrium expectations at a two-allele locus, not to every possible observed population. Nonrandom crosses or other processes can produce a heterozygote excess.
5. What can we conclude from the chi-square example?
The observed genotype counts disagree with the tested proportions beyond what the specified large-sample test would usually attribute to sampling. Investigate population structure, mating, selection, and data quality. The statistic alone cannot choose the cause or reconstruct past allele frequencies.
Concept references: OpenStax Biology 2e: Population Evolution; Iowa State University Digital Press: Gene Frequencies; University of Maryland MathBench: Hardy–Weinberg chi-square reasoning; University of Washington: Chi-square critical values. Scope reference: College Board AP Biology Course and Exam Description, Topic 7.5. Explanations, hypothetical datasets, worked calculations, graph, and questions were independently composed; no source figures, slides, or worksheets are reproduced.
Official curriculum reference: College Board AP Biology course page. The explanation and worked example are independently written for this study site.