A question says that 16% of a population shows a recessive phenotype. You remember p + q = 1, put q = 0.16 and keep calculating. The arithmetic can be flawless while the answer is wrong, because the supplied number described individuals with a phenotype, not the frequency of an allele.
Before using either equation, label the information. Is it an allele frequency, a genotype frequency, a phenotype frequency or a count? This one decision controls the rest of a Hardy–Weinberg calculation.
Keep two equations in separate jobs
For a diploid, two-allele locus, p and q represent allele frequencies, with p + q = 1. Under Hardy–Weinberg proportions, the genotype frequencies are p², 2pq and q². The model assumes random mating and idealised conditions without forces such as selection, migration or mutation changing the frequencies; a sufficiently large population limits drift in the standard introductory treatment. OpenStax, Biology 2e, Population Evolution
These equations are not interchangeable lists of percentages. The first counts allele types. The second describes genotype proportions. Write the genotype labels AA, Aa and aa beside the terms before substituting numbers.
Also separate dominance from frequency. Calling A dominant does not tell you that it is common. The letters describe the genetic model supplied by the question, not a ranking of how frequently the alleles occur.
Worked example: a recessive phenotype
In this invented problem, a population is assumed to follow Hardy–Weinberg proportions at a single two-allele locus with complete dominance. The recessive phenotype appears in 16% of individuals. Under those assumptions, the recessive phenotype identifies aa, so q² = 0.16.
Take the square root: q = 0.40. Then p = 1 − 0.40 = 0.60. The expected genotype proportions are:
| Genotype | Calculation | Expected proportion |
|---|---|---|
| AA | 0.60² | 0.36 |
| Aa | 2 × 0.60 × 0.40 | 0.48 |
| aa | 0.40² | 0.16 |
Check that 0.36 + 0.48 + 0.16 = 1. In a hypothetical group of 250 individuals, these proportions correspond to expected counts of 90 AA, 120 Aa and 40 aa.
The dominant phenotype includes AA and Aa, so its expected proportion is 0.84. It is not p, which equals 0.60. Notice how many distinct quantities can appear in one problem without any contradiction.
Change the wording, change the starting point
Now imagine the question instead states that the frequency of allele a is 0.16. This time q is already supplied. You do not take its square root. Under the same model, p = 0.84 and q² = 0.0256.
The two questions contain the same number, but it refers to different objects. A helpful annotation is “16% of individuals are aa” in the first case and “16% of allele copies are a” in the second.
When checking your work, read the original sentence again before reviewing the multiplication. If you misidentified the starting quantity, repairing the arithmetic alone cannot fix the solution.
When genotype counts are given directly
Consider an original dataset of 100 diploid individuals: 30 AA, 50 Aa and 20 aa. There are 200 allele copies at this locus. Count A copies as 2 × 30 + 50 = 110, so p = 110/200 = 0.55. Count a copies as 2 × 20 + 50 = 90, so q = 0.45.
This direct counting does not require assuming Hardy–Weinberg proportions. If you then want expected genotype proportions under that model, calculate 0.55² = 0.3025, 2 × 0.55 × 0.45 = 0.495 and 0.45² = 0.2025.
Keep observed and expected columns separate. A close visual match is not a substitute for whatever formal assessment your course requires. The original Hardy–Weinberg model is a mathematical baseline, not a declaration that every sampled population follows it exactly. Hardy, 1908
Common traps worth turning into questions
The first trap is using a recessive phenotype frequency as q rather than q². The second is reporting p² when asked for the dominant phenotype. The third is forgetting that heterozygotes contribute one copy of each allele when counting allele frequencies.
A fourth trap is giving a rounded expected count as though it were an observation. Expected values are model quantities and can be fractional. State whether you are reporting a proportion, a percentage or an expected number, and round only as requested.
Our active recall guide can help structure short revision prompts. For this topic, make the first prompt about interpreting the wording, not merely recalling the equations.
Self-test
1. Under the stated two-allele complete-dominance model, 9% have the recessive phenotype. What is q?
Answer: 0.30, because q² = 0.09. The allele frequency is the square root of the genotype proportion here.
2. If p = 0.70 and q = 0.30, what proportion is heterozygous?
Answer: 2pq = 2 × 0.70 × 0.30 = 0.42.
3. Do 40 heterozygous individuals contribute 40 or 80 copies of allele A?
Answer: 40. Each Aa individual contributes one A and one a.
4. Must you assume equilibrium to count allele frequencies from known genotype counts?
Answer: No. Count copies directly. The equilibrium model enters when you predict genotype proportions from allele frequencies.
Practise one question starting from each information type: allele frequency, recessive phenotype frequency and genotype counts. Explain the starting quantity before calculating. Our active recall guide can help organise those prompts; keep the model’s assumptions visible on each answer.
