Final answer:
In this population, the frequency of the disease phenotype is 0.5%, which corresponds to the frequency of the ff genotype. The frequencies of the f and F alleles can be determined by multiplying the frequencies of the ff and FF genotypes by 2, respectively. The proportion of the population that are heterozygous carriers of the disease allele can be calculated by taking the product of the frequencies of the f and F alleles and multiplying it by 2.
Step-by-step explanation:
In this population, the frequency of the disease phenotype is 0.5%, which means that 0.005 of the individuals have the cystic fibrosis disease phenotype. Since the disease is caused by an autosomal recessive allele, the frequency of the homozygous affected individuals (ff) can be determined by taking the square root of the disease phenotype frequency. In this case, the square root of 0.005 is 0.07, which represents the frequency of the ff individuals.
Since the population is in Hardy-Weinberg equilibrium, the sum of the frequencies of the ff and FF individuals must equal 1. Therefore, the frequency of the FF individuals can be calculated by subtracting the frequency of the ff individuals (0.07) from 1, which is 0.93.
The frequencies of the f and F alleles can be determined by multiplying the frequencies of the ff and FF individuals by 2, respectively. Therefore, the frequency of the f allele is 2 * 0.07 = 0.14 (14%) and the frequency of the F allele is 2 * 0.93 = 1.86 (186%).
As for the proportion of the population that are heterozygous carriers of the disease allele, it can be calculated by taking the product of the frequencies of the f and F alleles and multiplying it by 2. Therefore, the proportion of the population that are heterozygous carriers is 2 * 0.14 * 0.93 = 0.26 (26%).