Final answer:
The correct answer is option a. Given the frequencies of 0.001 and 0.0001 for the parents' CFTR mutation alleles, the most prevalent genotype in the population under Hardy-Weinberg equilibrium would be a heterozygote for two disease alleles. This is because it's improbable for children to be homozygous for the lower frequency alleles and heterozygotes might have an advantage in the population.
Step-by-step explanation:
Autosomal Recessive CFTR Mutations and Genotype Prevalence
When examining a family with two different CFTR mutations (for cystic fibrosis) with parents as carriers, it's important to determine the most prevalent genotype within the population. Given the dad's disease allele frequency of 0.001 and the mom's disease allele frequency of 0.0001 under the Hardy-Weinberg equilibrium, we can predict the prevalence of genotypes in their offspring and the population at large.
Under Hardy-Weinberg conditions, the most common genotype would typically be the homozygous dominant or the heterozygous genotype, due to the high frequency of the normal allele in the population. However, between the options provided:
- a. heterozygote for two disease alleles
- b. homozygote for dad's disease allele
- c. homozygote for mom's disease allele
The most prevalent genotype based on the given frequencies would be the heterozygote for two disease alleles because both parents are carriers of the disease, therefore, there is a 0 percent probability that any child will be homozygous for either disease allele due to the very low frequency of these alleles. Moreover, heterozygotes benefit from a potential heterozygote advantage, which could contribute to their prevalence in the population.
The correct option for the most prevalent genotype among those given is a. heterozygote for two disease alleles.