Case 1
- Dominant allele frequency (p)
- 0.6
0.4800
Open with these valuesHardy-Weinberg Calculator
0.48002pq (carriers)
Result: 0.4800 2pq (carriers)Enter the dominant allele frequency p. The headline figure is the heterozygote frequency 2pq — the carriers, the number Hardy-Weinberg problems usually ask for. The two homozygotes follow in one step each: p² and q² with q = 1 − p, and all three add up to 1.
The numbers at a glance
| Dominant allele frequency (p) | Result |
|---|---|
| 0.4000 | 0.4800 |
| 0.5000 | 0.5000 |
| 0.6000Your value | 0.4800 |
| 0.7000 | 0.4200 |
| 0.8000 | 0.3200 |
Worked examples
Case 2
- Dominant allele frequency (p)
- 0.5
0.5000
Open with these valuesCase 3
- Dominant allele frequency (p)
- 0.25
0.3750
Open with these valuesHow it's calculated
2pq = 2 × p × (1 − p)
- StepEnter the dominant allele frequency p, a proportion between 0 and 1.
- StepRead 2pq — the share of the population carrying both alleles.
- ResultFor the homozygotes, square p and square 1 − p.
What this number means
What a population geneticist can hold against reality is genotype frequencies, not allele frequencies, and this page turns the one into the other. Enter p, the frequency of the dominant allele, and the headline figure is 2pq — the share of the population carrying one copy of each allele, the carriers a Hardy-Weinberg question usually asks after. The default p = 0.6 works through in two steps: q = 1 − 0.6 = 0.4, then 2 × 0.6 × 0.4 = 0.48. Beside it sit p² = 0.36 and q² = 0.16, and the three come to exactly 1. The expression is symmetric, so p = 0.4 would give the same 0.48 — the carriers do not reveal which allele is the common one. It also peaks in the middle: at p = 0.5 the value reaches 0.5000 and goes no higher, so under this model no population is more than half heterozygous. What the three figures are not is a description of a real population. They show what a population would look like if nothing were happening to it: random mating, no selection, no migration, no new mutations, and enough individuals for drift not to bite. That is the limitation and the purpose in one. The equation is used as a baseline, and a population that misses these frequencies is pointing at whichever assumption fails for it.
A null model: nothing is happening
The equation describes a population with random mating, no selection, no migration, no new mutations and a size large enough for drift to be negligible. A population that deviates is breaking one of those assumptions, and that deviation is the actual finding.
The three frequencies always add up to 1
p² + 2pq + q² is the full expansion of (p + q)², and p + q = 1. Every individual is one of the three genotypes, so the frequencies are exhaustive.
p fixes q, so one field is enough
The two allele frequencies add up to 1, so q = 1 − p follows from p alone. Square p and square 1 − p for the two homozygote frequencies beside the 2pq shown here.
Commonly misread
A population off the predicted frequencies breaks the calculation.
It breaks one of the model's assumptions, which is exactly how the model is used. The deviation is the finding.
I have to enter both p and q.
p alone is enough, because the two add up to 1 and q = 1 − p.
p = 1 should give the largest share of heterozygotes.
It gives none at all: with p = 1 the recessive allele is missing, so 2pq is 0.
Reference table
| p | p² and q² | Heterozygotes 2pq |
|---|---|---|
| 0 | 0 and 1 | 0.0000 |
| 0.25 | 0.0625 and 0.5625 | 0.3750 |
| 0.3 | 0.09 and 0.49 | 0.4200 |
| 0.5 | 0.25 and 0.25 | 0.5000 |
| 0.6 | 0.36 and 0.16 | 0.4800 |
| 1 | 1 and 0 | 0.0000 |
Questions
How do I calculate the Hardy-Weinberg genotype frequencies?
From the dominant allele frequency p, the recessive one is q = 1 − p. The three genotypes are then p² homozygous dominant, 2pq heterozygous and q² homozygous recessive. With p = 0.6 that gives 0.36, 0.48 and 0.16, which add up to 1.
Why does this page put 2pq first?
Because it is the figure a Hardy-Weinberg question usually asks for: the carriers. p² and q² are one squaring away from numbers you already hold, while 2pq needs both alleles at once. The table lists all three for every p.
What do p and q mean?
p is the frequency of the dominant allele and q that of the recessive allele, for one gene with two alleles. They are proportions between 0 and 1 that add up to 1, so p fixes q = 1 − p.
Why do the three frequencies add up to 1?
Because p² + 2pq + q² is the full expansion of (p + q)², and p + q = 1. Every individual is one of the three genotypes, so the frequencies are exhaustive.
What assumptions does the Hardy-Weinberg model make?
Random mating, no selection, no migration, no new mutations, and a population large enough for drift to be negligible. A population that deviates from the predicted frequencies is breaking one of them — which is exactly how the model is used.
Sources and last check
Information, not professional advice.