. A large population of laboratory animals has been allowed to breed randomly fo
ID: 7047 • Letter: #
Question
. A large population of laboratory animals has been allowed to breed randomly for a number of generations. After several generations, 36% of the animals display a recessive trait (aa), the same percentage as at the beginning of the breeding program. The rest of the animals show the dominant phenotype, with heterozygotes indistinguishable from the homozygous dominants.a. What is the estimated frequency of allele a in the gene pool?
b. What proportion of the population is probably heterozygous (Aa) for this trait
Explanation / Answer
Given that the percentage of animals display the recessive trait aa is 36% qq = q^2 = 36 / 100 = 0.36 q = 0.6 According to Hardy-Weinberg equilibrium p+ q = 1 p = 1 – 0.6 = 0.4 a). estimated frequency of allele a in gene pool = q = 0.6 b). proportion of population with heterozygous trait = Aa = 2pq = 2* 0.6 * 0.4 = 0.48
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