Frequency of Phenylketonuria (PKU) in a Genetically Constant Population

Phenylketonuria (PKU) is a recessive genetic disorder that affects the metabolism of the amino acid phenylalanine. In a genetically constant population, where the allele frequencies do not change over time, the frequency of PKU can be calculated based on the principles of Hardy-Weinberg equilibrium.

The Hardy-Weinberg equilibrium states that in an ideal population, the frequencies of alleles and genotypes will remain constant from generation to generation if certain conditions are met. These conditions include a large population size, random mating, no migration, no mutation, and no natural selection.

To calculate the frequency of the carrier genotype for PKU, we need to consider the frequency of the recessive allele (q) and apply the Hardy-Weinberg equation:

p^2 + 2pq + q^2 = 1

Where:
p^2 = frequency of the homozygous dominant genotype (AA)
2pq = frequency of the heterozygous genotype (Aa)
q^2 = frequency of the homozygous recessive genotype (aa)

Let's assume that q^2 represents the frequency of PKU in the population, which is given as 1 in 10000 or 0.0001. Therefore, q^2 = 0.0001.

Now we can calculate q by taking the square root of q^2:

q = √0.0001 = 0.01

Since q represents the frequency of the recessive allele, and the allele frequencies must add up to 1, we can calculate the frequency of the dominant allele (p) as:

p = 1 - q = 1 - 0.01 = 0.99

To calculate the frequency of carriers (Aa), we multiply the frequencies of the two alleles:

2pq = 2 * 0.99 * 0.01 = 0.0198

Converting this to a percentage, we get:

0.0198 * 100 = 1.98%

Therefore, the frequency of the carrier genotype (Aa) is approximately 1.98%, which corresponds to option C in the given answer choices.