Question

In achondroplastic dwarfism, an autosomal dominant disease, the gene is lethal in homozygotes. If the frequency of the normal allele is 0.99, what is the heterozygote frequency?

(A) 0.98
(B) 0.01
(C) 0.02
(D) 0.99

Answer 1

The frequency of heterozygote individuals in a population for achondroplastic dwarfism, an autosomal dominant disease with lethal homozygotes and a normal allele frequency of 0.99, is 0.02.

Step-by-step explanation:

In achondroplastic dwarfism, which is an autosomal dominant disease, the gene is lethal in homozygotes. Given that the frequency of the normal allele (a) is 0.99, we can find the frequency of the heterozygote (Aa) individuals in the population using the Hardy-Weinberg equilibrium principle. The formula for Hardy-Weinberg equilibrium is p2 + 2pq + q2 = 1, with p being the frequency of the dominant allele (A) and q being the frequency of the recessive allele (a).

Since q (the frequency of the normal allele) is given as 0.99, p would be 1 - q which equals 0.01. The frequency of heterozygotes is represented by the term 2pq in the formula, which equates to 2(0.01)(0.99).

Thus, the frequency of heterozygotes (Aa) in the population is 2 x 0.01 x 0.99 = 0.0198, which we can round to 0.02. Therefore, the answer to the student's question is (C) 0.02.