mutation/selection balance, mutation rate calculation

Question

If the incidence at birth before selection of a recessive mutation is given (ex. 1 in 90,000), how do you calculate the mutation rate if mutation and selection are balanced in maintaining the allele? I understand that at an equilibrium between mutation and selection, p = μ/s, but how do you calculate any of those terms from just the incidence at birth?

 

How do you calculate the frequency of the mutation at equilibrium if there's a selective disadvantage for heterozygotes (ex. 2%)?

Answer 1

Suppose a rare autosomal recessive disorder has a birth incidence of 1 in 90,000. This means 1 in 90,000 individuals is born homozygous recessive for the disease-causing mutation.

We can use the Hardy-Weinberg equilibrium to estimate the allele frequency and carrier rate.

Let:

1. p = frequency of the normal allele (B)
2. q = frequency of the mutant allele (b)
3. p² = frequency of homozygous dominant individuals (BB)
4. 2pq = frequency of heterozygous carriers (Bb)
5. q² = frequency of affected individuals (bb)

Given:

q² = 1/90,000 ≈ 0.0000111

Therefore:

q = √0.0000111 ≈ 0.00333

p = 1 - q = 0.99667

p² = (0.99667)² ≈ 0.9934

2pq = 2 × 0.99667 × 0.00333 ≈ 0.00664

Conclusion: Approximately 0.66% of the population, or 1 in 150 individuals, are carriers of the mutation (heterozygous but unaffected).