Allele frequency and genotype frequency are related but can change differently.
By the end: Derive random-union genotype proportions and explain why a deviation does not identify its cause.
At a two-allele diploid locus, let p be A frequency in each gamete pool and q = 1 − p be a frequency. Random union gives AA with probability p², aa with q², and Aa with 2pq because A can arrive from either parent. The three probabilities sum to one.
The genotype proportions follow from independent gamete union under the specified allele frequencies. For those frequencies to remain constant across generations, the standard Hardy-Weinberg reference also excludes selection, mutation, and migration and assumes drift is negligible. Random mating alone does not prevent other evolutionary forces from changing p.
Suppose A makes up 30% of copies in each of two large gamete pools. The remaining 70% are a. Choose one gamete from each pool independently. The probability of AA is 0.3 times 0.3, or 0.09. The probability of aa is 0.7 times 0.7, or 0.49. These products describe two independent choices. They do not require A to be dominant, useful, or visible in the resulting organism. None of those properties enters the gamete calculation.
A heterozygote can form in two distinct ways. The first gamete can carry A and the second a, with probability 0.3 times 0.7, or 0.21. Alternatively, the first can carry a and the second A, also with probability 0.21. These outcomes add to 0.42. Writing pq would omit half the heterozygotes. The complete distribution is therefore 9% AA, 42% Aa, and 49% aa. Adding 0.09, 0.42, and 0.49 gives one, which checks that every possible pairing has been included exactly once.
Translate those probabilities into a constructed cohort of 100 offspring for bookkeeping. Its expected counts are nine AA, 42 Aa, and 49 aa. Nine AA organisms supply 18 A copies, and 42 Aa organisms supply another 42. The resulting 60 A copies belong in a denominator of 200 total copies, giving 0.3 again. By contrast, 51 organisms carry A, so the carrier fraction is 51%. If A is completely dominant for a specified phenotype, that phenotype also has expected frequency 51%, not 30%.
Expected counts are not guaranteed counts. Random fertilization in a finite cohort can produce counts other than nine, 42, and 49 even when the gamete probabilities are exactly correct. Observed counts must be whole numbers; expectations need not be. The Hardy-Weinberg proportions describe the reference probabilities. A difference between observed and expected counts may reflect ordinary sampling. Deciding whether a discrepancy is unusually large requires a statistical comparison and attention to how organisms were sampled, neither of which the pairing lab performs.
The assumption that both gamete pools have the same frequency also matters. On paper, let one pool have A frequency 0.2 and the other 0.8. Independent union gives AA probability 0.16 and aa probability 0.16. Heterozygotes have probability 0.2 times 0.2 plus 0.8 times 0.8, or 0.68. The offspring A frequency is still 0.5, the average parental contribution. However, their genotype proportions are not 0.25, 0.50, and 0.25. Random union alone does not justify using one shared p when the two pools differ.
| A in pool 1 / pool 2 | AA | Aa | aa | Offspring A |
|---|---|---|---|---|
| 0.5 / 0.5 | 0.25 | 0.50 | 0.25 | 0.50 |
| 0.2 / 0.8 | 0.16 | 0.68 | 0.16 | 0.50 |
Separate a single fertilization calculation from equilibrium across generations. Given equal gamete frequencies and independent union, the genotype probabilities follow immediately. Keeping those allele frequencies unchanged in later generations requires further assumptions about transmission and reproduction. Selection, mutation, gene flow, and finite sampling can change the gamete pool before the next union. The reference is useful because it states what pairing alone predicts. It does not claim that real populations lack these other processes, or that a population close to the reference has experienced no evolution in its history.
Inbreeding increases the chance that paired copies descend from the same ancestral copy. A simple coefficient F summarizes this departure from random pairing. At fixed p, the model gives AA = p² + Fpq, Aa = 2pq times one minus F, and aa = q² + Fpq. Lost heterozygotes are redistributed equally to the two homozygote classes.
F is a coefficient of identity by descent relative to a reference, not simply the percentage of matings between relatives. These formulas hold p fixed and illustrate genotype redistribution. Inbreeding can expose harmful recessive effects and alter later selection, but it does not mechanically create harmful alleles or necessarily change allele frequency in this pairing step.
Consider four diploid organisms whose genotypes are all Aa. They contain four A copies and four a copies, so A frequency is one half. Compare a second arrangement with two AA organisms and two aa organisms. It contains exactly the same numbers of A and a copies, but no heterozygotes. This is a bookkeeping comparison of allele pairing, not a claim that adult chromosomes physically rearrange themselves. It shows why allele frequency alone cannot tell you genotype frequency when the random-union assumption does not hold.
Identity by descent describes the history of copies, not merely their letters. Two copies are identical by descent when both trace back to the same copy in a reference ancestry. Two A copies can instead share the letter A while coming from different ancestral copies in that reference. A genotype label does not distinguish these histories. The coefficient F summarizes a probability of shared descent under a specified reference. Its value depends on the reference ancestry and population.
The lab implements a simple mixture of two pairing rules. With probability F, draw one ancestral copy from the reference pool and represent both offspring copies as descendants of it. That part produces AA with probability p and aa with probability q, with no Aa. With probability one minus F, draw the two copies independently, giving the usual p squared, 2pq, and q squared. Combining the rules gives AA frequency Fp plus one minus F times p squared. Expanding that expression yields p squared plus Fpq.
Use p = 0.3 and F = 0.5 as a separate worked calculation. Here pq is 0.21, so Fpq is 0.105. Add 0.105 to the random-union AA proportion 0.09 to obtain 0.195. Add the same amount to aa, changing 0.49 to 0.595. The heterozygote fraction becomes 0.42 times 0.5, or 0.21. Their sum is one. Recover A frequency as 0.195 plus half of 0.21, giving 0.3. The change concerns copy associations, not the total share of A in the model.
At these values, the expected fraction carrying A changes from 0.51 to 0.405 because AA plus Aa is now 0.195 plus 0.21. If A determines a completely dominant phenotype in this constructed example, its phenotype frequency changes too, despite unchanged allele frequency. The recessive phenotype fraction rises from 0.49 to 0.595. Neither phenotype change demonstrates that mutation has created new alleles. Nor does F specify that half the observed matings occurred between relatives. It weights ancestry relationships between paired copies in this particular model.
| Pairing rule | AA | Aa | aa | A carriers | A copies |
|---|---|---|---|---|---|
| F = 0 | 0.090 | 0.420 | 0.490 | 51% | 30% |
| F = 0.5 | 0.195 | 0.210 | 0.595 | 40.5% | 30% |
Pairing can nevertheless affect what happens later. If aa has lower reproductive contribution in a specified environment, increasing its proportion can expose more a copies to that selection. Selection then changes the contributions reaching the next generation. This later effect is distinct from the fixed-frequency pairing calculation. The lab stops before that selection step and assigns no genotype-specific fitness values. It cannot estimate individual consequences or infer ancestry behind an observed F. Its purpose is to separate genotype redistribution from the mechanisms that subsequently change allele frequency.
p is A frequency and q = 1 − p is a frequency. The middle term includes both parental orders, A then a and a then A.
Changing the pairing of a fixed pool need not change its allele frequencies.
Selected AA 37.5%, Aa 25%, aa 37.5%. Recovered A frequency: 50%.
| p = 0.5 | AA | Aa | aa | A frequency |
|---|---|---|---|---|
| Random union, F = 0 | 0.25 | 0.50 | 0.25 | 0.50 |
| F = 0.5 | 0.375 | 0.25 | 0.375 | 0.50 |
In the lab, first choose Balanced: 50% for A and Random union: F = 0. The two displayed rows should match because both now use independent pairing. AA occupies one quarter, Aa one half, and aa one quarter of each row. The horizontal extent represents fractions of organisms with each genotype, not fractions of A and a copies. Each heterozygote belongs to one genotype category while contributing one copy to each allele category, which is why the displayed genotype shares need a separate recovery calculation.
Keep p fixed at one half and choose Intermediate: F = 0.5. The reference row remains unchanged, while the selected row becomes 0.375 AA, 0.25 Aa, and 0.375 aa. Heterozygotes lose 0.25 of the cohort relative to the reference. Each homozygote class gains 0.125. This equal redistribution preserves allele frequency because the extra AA copies balance the A copies removed with heterozygotes. Recovering p gives 0.375 plus 0.25 divided by two, or 0.5. The lab reports that recovered value alongside the proportions.
Next choose Complete identity: F = 1 without changing p. The model now produces only AA and aa, each with probability 0.5. It still contains both alleles. Complete homozygosity is therefore different from fixation, which means that only one allele remains at the locus. The four-organism pairing diagram makes the distinction visible. Two AA and two aa organisms have no heterozygotes but retain both variants. Calling that population fixed for A would confuse a property of paired genotypes with a property of the whole gene pool.
Now leave F at one and select Rare A: 10%. Predict the outcome before comparing the rows. Independent pairing gives 0.01 AA, 0.18 Aa, and 0.81 aa. Complete identity instead gives 0.10 AA, zero Aa, and 0.90 aa. In both cases A accounts for 10% of copies. Under complete dominance, however, A carriers make up 19% in the reference and 10% in the selected arrangement. A visible phenotype survey would therefore give a different percentage from a gene-copy survey even though both described the same locus.
The largest absolute heterozygote change occurs around balanced allele frequencies because the term pq is largest there. Compare the two products directly. At p = 0.5, pq is 0.25, whereas at p = 0.1 it is 0.09. Increasing F from zero to one removes heterozygote fractions 0.50 and 0.18 respectively. This does not mean that rare variants are unaffected by pairing. Their homozygote probabilities can change by a large factor even when the absolute change occupies a smaller portion of the full genotype row.
| A frequency | Aa at F = 0 | Aa at F = 1 | AA at F = 0 | AA at F = 1 |
|---|---|---|---|---|
| 0.1 | 18% | 0% | 1% | 10% |
| 0.5 | 50% | 0% | 25% | 50% |
Finally, move the A-frequency slider to either endpoint while keeping an F preset selected. At p = 0, every genotype is aa for every allowed F. At p = 1, every genotype is AA. Pairing cannot assemble a missing allele. Controls replace theoretical proportions immediately and preserve the other parameter. They do not advance generations or sample organisms. Intermediate slider positions change the assumed starting frequency, not an observed evolutionary trajectory. The endpoint comparison checks the mechanism because the formulas must agree when only one allele is available.
Observed genotype counts can differ from the reference through nonrandom mating, selection, population mixture, sampling variation, or errors in measurement and classification. Pooling distinct subpopulations can produce a heterozygote deficit even when each subgroup mates randomly.
A departure therefore flags a model-checking problem, not a unique diagnosis. Also specify the life stage sampled: selection between fertilization and adulthood can change genotype proportions. The lab compares theoretical proportions, not a statistical test of an observed sample or an inference about a person’s ancestry.
A population boundary determines which gametes could actually meet. Imagine two equally sized breeding groups, each contributing an expected 100 offspring. A frequency is 0.2 in the first group and 0.8 in the second. Assume independent gamete union within each group, with equal frequencies in its two parental pools. The first group has expected genotype counts four AA, 32 Aa, and 64 aa. The second has 64 AA, 32 Aa, and four aa. Every group's proportions fit its own random-union reference exactly by construction.
Pooling the groups gives 68 AA, 64 Aa, and 68 aa among 200 offspring. There are 136 A copies in AA organisms and 64 in Aa organisms, totaling 200 among 400 copies. Pooled p is therefore 0.5. If you incorrectly treat the collection as one freely mixing breeding group, you predict 50 AA, 100 Aa, and 50 aa. Observed heterozygotes fall short of that prediction by 36. The mismatch arises because cross-group gamete unions were never part of the process that produced these offspring.
The same pooled proportions can be reproduced by the lab's pairing formula at p = 0.5 and F = 0.36. Heterozygotes would be 0.5 times 0.64, or 0.32, and each homozygote class would be 0.25 plus 0.36 times 0.25, or 0.34. This numerical agreement does not prove shared-copy ancestry within either group. It shows that one set of genotype proportions can fit different causal accounts. The coefficient needed to describe a deficit against the pooled reference need not identify how that deficit arose.
A life-stage comparison supplies a different contrast. Suppose a constructed cohort begins with 25 AA, 50 Aa, and 25 aa offspring. Before the adult census, half the Aa organisms survive while all homozygotes survive. Adults then number 25 in each genotype class, for 75 total. A frequency remains one half, but the heterozygote fraction becomes one third rather than one half. This is genotype-dependent survival, not a change in mating. It produces a deficit at adulthood even though the original fertilization proportions matched the reference.
| Stage | AA organisms | Aa organisms | aa organisms | Aa fraction |
|---|---|---|---|---|
| Initial cohort | 25 | 50 | 25 | 50/100 = 0.50 |
| Adult cohort | 25 | 25 | 25 | 25/75 = 1/3 |
To interpret a real deficit, ask which comparisons distinguish these possibilities. Recording subgroup membership allows separate reference calculations rather than only a pooled one. Comparing life stages can reveal whether genotype frequencies change after fertilization. Checking classification can detect an assay that mistakes heterozygotes for homozygotes. Sampling uncertainty remains relevant even after these checks. No single departure tells you that selection, ancestry, subdivision, or measurement error is the cause. Conversely, agreement at one locus does not exclude selection elsewhere or earlier allele-frequency changes at the same locus.
There is also no contradiction between random pairing and drift. In a finite population, the realized gamete contribution can differ from its expectation by chance. Independent pairing can then apply to that newly sampled pool. The deterministic lab suppresses this sampling so that p and F have separate, readable effects. When transferring the reasoning to observations, state the locus, sampling stage, breeding boundary, and counted unit first. Then recover allele frequency from genotype counts before proposing a mechanism. Those choices determine what the reference predicts and which departures need an explanation.
With A frequency 0.2 in both gamete pools, calculate AA, Aa, and aa proportions. Recover A frequency from them.
Aa has two possible parental orders.
| Genotype | Frequency |
|---|---|
| AA | 4% |
| Aa | 32% |
| aa | 64% |
AA is 0.04, Aa is 2 × 0.2 × 0.8 = 0.32, and aa is 0.64. They sum to one. Counting one A from each Aa and two from each AA gives p = 0.04 + 0.32/2 = 0.20, as required.
Two equally sized populations mate randomly within themselves. Their A frequencies are 0.2 and 0.8. Compare their pooled heterozygote fraction with a single random-mating population at the pooled p.
Calculate heterozygotes within each group before averaging.
| Population model | A frequency | Aa fraction |
|---|---|---|
| First subgroup | 0.2 | 0.32 |
| Second subgroup | 0.8 | 0.32 |
| Pooled observations | 0.5 | 0.32 |
| Random union across pooled group | 0.5 | 0.50 |
Each group has heterozygote fraction 2 × 0.2 × 0.8 = 0.32. The pooled fraction remains 0.32, while pooled A frequency is 0.5. A single randomly mixing population at p = 0.5 would have 0.50 heterozygotes. The deficit reflects subdivision here, not proof of inbreeding within either group.
The explanations and exercises are complete without these optional links. The references cover inheritance, population mechanisms, and evidence for ancestry. Source annotations distinguish textbook explanations, secondary history, and a primary paper whose bibliographic identity alone was checked. The examples are constructed teaching models, not empirical estimates or professional decision advice.
Genetic variation, allele frequencies, and the Hardy-Weinberg reference model. Its closing claim about any genotype deviation implying evolution is too broad: sampling, subdivision, and nonrandom pairing can produce departures without an allele-frequency change. Our lessons distinguish the random-union calculation from stability across generations.
Drift, gene flow, and other contributors to population change. The labs isolate mechanisms rather than estimate real populations.
The Measures subsection defines the coefficient through identity by descent. The lab uses a bounded fixed-frequency mixture model, not a pedigree estimator or clinical risk calculator.