Review Question - QID 107015

The disease in question is cystic fibrosis, an autosomal recessive disease. Since the mother is homozygous for the mutation for the autosomal disease, cystic fibrosis, and the father is a possible carrier, the chance that their child will have cystic fibrosis is 9/100 or 9%.

If all the conditions of Hardy Weinberg equilibrium are met (see topic and below), then one can use the equation p^2 + 2pq + q^2 = 1 (where p is the normal allele and q is the mutant allele) to determine the frequencies of the alleles in the population. Since the prevalence of cystic fibrosis (q^2) is 1/100, the frequency of the mutant allele is the square root of 1/100, which is 1/10 (q). Since p + q = 1, then p + 1/10 = 1; therefore, p = 9/10. To determine the frequency of the carriers in the population, use 2pq which is 2 x 1/10 x 9/10 which is 18/100. Therefore, the father has a 18/100 chance of being a carrier for the cystic fibrosis gene. To calculate whether the child will have the disease we multiply the following probabilities: the chance the mother has the mutant alleles (1) and the chance she will give the child a mutated gene (1); the chance the father will give the child the mutated gene (1/2) and the chance the father is a carrier (18/100); therefore, 1 x 1 x 1/2 x 18/100 = 9/100 is the probability that the child will have cystic fibrosis.

Illustration A depicts the Hardy-Weinberg equation and explains the meanings of the associated variables.

Incorrect Answers:
Answer 2: 1/10 is the frequency of the mutant allele (q).
Answer 3: 18/100 is the carrier frequency in the population (2pq).
Answer 4: 81/100 is the frequency of homozygous individuals for the normal phenotype (p^2).
Answer 5: 9/10 is the frequency of the normal allele (p).