Question

In a study of the Hopi, a Native American tribe of central Arizona, Woolf and Dukepoo (1959) found 26 albino individuals in a total population of 6000. This form of albinism is controlled by a single gene with two alleles: albinism is recessive to normal skin coloration. Calculate the expected allele frequencies and genotype frequencies if the population were in Hardy-Weinberg equilibrium. What is the allele frequency of the normal allele

Answer 1

Answer:

0.9342

Explanation:

The Hardy-Weinberg equation states that p² + 2pq + q² = 1,

where p is the frequency of the dominant 'normal' (n) allele and q is the frequency of the recessive 'albino' (a) allele in the population, while q² represents the frequency of the homo-zygous albino genotype (aa), p² represents the frequency of the homo-zygous normal genotype (nn) and 2pq represents the frequency of the heterozygous genotype (na).

In this case, the frequency of individuals in the population that have the genotype aa (q²) is equal to 26/6000 = 0.004333. In consequence, q is equal to √ 0.004333 = 0.0658.

Moreover, the allele frequency of the normal (n) allele p is equal to 1 - q = 1 - 0.0658 = 0.9342, so p² (nn) = (0.9342)² = 0.8727.  

Finally, the frequency of the heterozygous genotype (na) is 2pq = 2 x 0.9342 x 0.0658 = 0.123.