Final answer:
The correct answer is option c. The most prevalent genotype among the options provided is 'c. heterozygote for two disease alleles' due to the higher likelihood of individuals carrying one copy of a mutation rather than being homozygous for it.
Step-by-step explanation:
When examining a family in which both parents are carriers for two different CFTR mutations, we must consider the Hardy-Weinberg principle to determine the most prevalent genotype in the population. The frequencies of the dad's disease allele and mom's disease allele are given as 0.001 and 0.0001, respectively. Under Hardy-Weinberg equilibrium, the frequency of the carrier state for each allele in the population is 2pq, where 'p' is the frequency of the normal allele and 'q' is the frequency of the disease allele.
Let's use the Hardy-Weinberg equation to calculate these frequencies:
- Dad's disease allele: 2pq = 2 * (approx. 1) * 0.001 = 0.002
- Mom's disease allele: 2pq = 2 * (approx. 1) * 0.0001 = 0.0002
Comparing these frequencies, we find the carrier state for dad's disease allele to be more prevalent. Consequently, the homozygote for the dad's disease allele will also be more common than the homozygote for the mom's disease allele.
However, in this particular scenario, we need to evaluate the probability of offspring being heterozygotes carrying both alleles, rather than homozygotes. The parents are each carriers of a different CFTR mutation, not the same one; thus, their children would most likely be heterozygotes for two different disease alleles.
The most prevalent genotype among the three in this population will be c. heterozygote for two disease alleles, as it is more common for individuals to carry one copy of a mutation rather than being homozygous for one.