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
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
The correct option is B 0.42
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.
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.
