Question

1. The frequency of two alleles in a gene pool is 0.13 (B) and 0.87(b). Assume that the population is in Hardy-Weinberg equilibrium.

Calculate the percentage of heterozygous individuals in the population?

Answer 1

Answer:

1

Explanation:

12th

Biology

Evolution

Hardy-Weinberg Principle

A population is in Hardy - ...

BIOLOGY

A population is in Hardy-Weinberg equilibrium for a gene with only two alleles. If the gene frequency of an allele A is 0.7, the genotype frequency of Aa is

HARD

Share

Study later

ANSWER

The Hardy-Weinberg equation is expressed as p2 + 2pq + q2 = 1, where p is the frequency of the "A" allele and q is the frequency of the "a" allele in the population. In the equation, p2 represents the frequency of the homozygous genotype AA, q2 represents the frequency of the homozygous genotype aa, and 2pq represents the frequency of the heterozygous genotype Aa. In addition, the sum of the allele frequencies for all the alleles at the locus must be 1, so p + q = 1. So if the frequency of an allele A is 0.7, then the frequency of allele a is 0.3. Therefore according to formula: (0.7)2 + 2Aa + (0.3)2 = 1

0.49 + 0.9 + 2Aa = 1

Aa = 0.42 (2Aa represents genotype frequency of genotype Aa). Hence option B is correct.