Answer:
0.0177
Explanation:
Cystic fibrosis is an autosomal recessive disease, thereby an individual must have both copies of the CFTR mutant alleles to have this disease. The Hardy-Weinberg equilibrium states that p² + 2pq + q² = 1, where p² represents the frequency of the homo-zygous dominant genotype (normal phenotype), q² represents the frequency of the homo-zygous recessive genotype (cystic fibrosis phenotype), and 2pq represents the frequency of the heterozygous genotype (individuals that carry one copy of the CFTR mutant allele). Moreover, under Hardy-Weinberg equilibrium, the sum of the dominant 'p' allele frequency and the recessive 'q' allele frequency is equal to 1. In this case, we can observe that the frequency of the homo-zygous recessive condition for cystic fibrosis (q²) is 1/3200. In consequence, the frequency of the recessive allele for cystic fibrosis can be calculated as follows:
1/3200 = q² (have two CFTR mutant alleles) >>
q = √ (1/3200) = 1/56.57 >>
- Frequency of the CFTR allele q = 1/56.57 = 0.0177
- Frequency of the dominant 'normal' allele p = 1 - q = 1 - 0.0177 = 0.9823
The condition is documented as jaundice.
