If the frequency of the disorder in the population is 0.0070, what is the percentage of heterozygous carriers

15.322 %

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 %)

the answer would be 3

answer; /// i believe the correct answer is (reactants);