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.
1. Population carrier frequency (using Hardy-Weinberg principle in perinatology.com), with the incidence of 1:40000, where:
p = 0.995
q = 0.005
Thus, using the formula p2+2pq+q2=1:
(0.995)2 + 2 x 0.995 x 0.005 + (0.005)2 = 1 (in 100.5 or 1 %)
2.
| Carrier | Non Carrier | |
| Prior | 2/3 | 1/3 |
| Conditional | 1/4 | 1 |
| Joint | 1/6 | 1/3 |
| Posterior | 1/3 | 2/3 |
Sibling's probability of having Pompe Disease gene = 1/3
1/100 (population carrier risk) x 1/3 = 1/300 (probability of the healthy sibling of having a child with Pompe disease)