Hardy-Weinberg carrier frequency
This calculator uses the Hardy-Weinberg equation (p² + 2pq + q² = 1) to estimate the carrier frequency (2pq) for an autosomal recessive trait.
For a trait with two alleles: p is the frequency of the wild-type (normal) allele and q is the frequency of the mutant allele. The three genotype frequencies in the population are:
- p² — homozygous wild type (AA)
- 2pq — heterozygous carriers (Aa)
- q² — homozygous affected (aa)
The Hardy–Weinberg principle states that genotype frequencies p², 2pq, and q² remain constant from generation to generation (are in equilibrium) if allele frequencies remain constant and the following assumptions are met:
-
No natural selection:
- Matings are at random
- All individuals reproduce equally
- All offspring survive
- No appreciable rate of mutation
- No migration
- No genetic drift
Expressed as:
p² + 2pq + q² = 1
Example: Cystic fibrosis (CF)
Consider cystic fibrosis (CF) in northern Europeans, where the prevalence of affected individuals (aa) is approximately 1 in 2500.
Let:
- A = wild-type allele
- a = cystic fibrosis mutation
For CF:
q² = 1/2500
q = 1/50
Since p + q = 1:
p = 1 − q = 49/50
The carrier frequency is:
2pq = 2 × (49/50) × (1/50) = 98/2500 = 0.04
| Maternal | |||
|---|---|---|---|
| A (p) | a (q) | ||
| Paternal | A (p) | AA (p²) | Aa (pq) |
| a (q) | Aa (pq) | aa (q²) | |
All calculations must be confirmed before clinical use. The suggested results are not a substitute for clinical judgment. Neither Perinatology.com nor any other party involved in the preparation or publication of this site shall be liable for any special, consequential, or exemplary damages resulting from the use of this material.
Carrier Frequency Calculator
Enter the frequency of affected individuals (q²) as either a percent or a proportion (1 in N), then select Calculate.