Pompe disease is an autosomal recessive condition with an incidence of 1/40,000. Based on this, calculate the population carrier frequency and calculate the risk for the healthy sibling of an affected individual to have a child with Pompe disease. Note healthy sibling’s partner is healthy and has no family history of Pompe disease. Hint: you must know the healthy sibling’s risk to be carrier.
We calculate the population carrier frequency using the Hardy-Weinberg principle. We apply the formula using the incidence of Pompe disease: 1 in 40,000, the formula is p^2 + 2pq + q^2 = 1, where: (I used the ‘Carrier Frequency Calculator’ at perinatology.com)
p = 0.995 (frequency of wild type allele)
q = 0.005 (frequency of mutant allele)
We substitute the values:
0.99003 (p2) + 0.00995 (2pq) + 0.00003 (q2) = 1
We get the Homozygous wild type (carrier frequency) which is 1/100.5 or 1%
|
|
Carrier |
Non-carrier |
|
Prior |
2/3 |
1/3 |
|
Conditional |
1/4 |
1 |
|
Joint |
1/6 |
1/3 |
|
Posterior |
1/6 / (1/6 + 1/3) = 1/3 |
1/3 / (1/6 + 1/3) = 2/3 |
The probability that the healthy sibling is a carrier of the Pompe disease gene is 1/3
We multiply these values to get:
1/3 x 1/100 = 1/300
(the probability of the healthy sibling of an affected individual to have a child with Pompe disease)