Question

In humans, albinism is caused by loss-of-function mutations in genes involved in the synthesis of melanin, the dark pigment in skin. only people homozygous for a loss-of-function allele (genotype aa) have the albino phenotype. in americans of northern european ancestry, albino individuals are present at a frequency of about 1 in 10,000 (or 0.0001). assuming that genotypes are in hardy−weinberg equilibrium, what is the predicted frequency of caucasians in the united states who carry a single allele for albinism?

Answer 1

The frequency of carriers of the albino allele in the population is. If

To solve this problem, we need to consider the Hardy-Weinberg equation which is the following:

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

where: p = frequency of the "A" allele

q = frequency of the "a" allele

In other words, p^2 refers to the frequency of the homozygous(dominant) genotype AA, q^2 represents the frequency of the homozygous(recessive) genotype aa, and 2pq represents the frequency of the heterozygous genotype.

In the problem, if q = 0.01, and p = 0.99, then the frequency of carriers is 2pq which then becomes 0.0099 x 2. Thus, the predicted frequency of Caucasians in the U.S. who carry a single albino allele is 0.0198.