Genetics & evolution · SL/HL
Allele frequency
Find allele frequencies from diploid genotype counts.
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Number formats & units
Use decimal dots in datasets. Single values also accept decimal commas and scientific notation such as 1.2e-3 or 1.2 × 10⁻³.
Calculated locally. No account or input upload.
Result & working
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Formula & example
The relationship
What the variables mean
| Symbol | Meaning |
|---|---|
| Total number of individuals. |
Worked example
- AA count
- 36
- Aa count
- 48
- aa count
- 16
p = 0.6, q = 0.4
Follow the example steps
- N = 36+48+16 = 100
- p = (2AA + Aa)/(2N) = (2×36+48)/200 = 0.6
- q = (2aa + Aa)/(2N) = (2×16+48)/200 = 0.4
- p + q = 1
Common mistakes
There are two alleles per diploid individual.
Incorrect approach
AA = 30, Aa = 40, aa = 30: p = (30 + 40)/100 = 0.7
Correct approach
p = (2 × 30 + 40)/(2 × 100) = 0.5
More checks before you submit
- Count two alleles per diploid individual.
Method notes
This calculates observed allele frequencies; it does not by itself show whether the population is in Hardy–Weinberg equilibrium.