Answer:
15.322 %
Step-by-step explanation:
Supposing that this is a recessive disorder and the population is in Hardy-Weinberg equilibrium:
p² + 2pq + q² = 1, where
1- the genotypic frequency of the homo-zygous dominant allele is p²,
2- the genotypic frequency homo-zygous recessive allele is q², and
3- the genotypic frequency heterozygous allele is 2pq
Therefore:
q² = 0.007 >> q = √ 0.007 = 0,0836
p = 1 - q = 1 - 0,0836 = 0.9164
2pq = 2 x 0.9164 x 0,0836 = 0.15322 (or 15.322 %)