Question

43. A gene has two alleles a and b. In the gene pool, the a allele accounts for 70 percent of the alleles at this locus. If the population is in Hardy-Weinberg equilibrium, what percentage of the population is heterozygotes with respect to this locus?

A. 21 percent

B. 30 percent

C. 49 percent

D. 70 percent

Answer 1

Hardy-Weinberg equilibrium:

· Hardy-Weinberg equation describes: The most possible distribution of the population genotype of the population, if the frequencies of the alleles are known.

· If allele frequencies are p and q, then the genotype frequencies are p2, q2, 2pq.

· A population is considered to be in Hardy-Weinberg equilibrium when: In a randomly mating population, allele frequency and genotype frequency remain constant.

· Thus, under Hardy-Weinberg equilibrium:

p+ q=1 and

p2+2pq + q2 = 1.

Given:

• Percentage of allele a in the population= 70%
• Frequency of allele a = 70/ 100 = 0.7 = p
• Frequency of allele b = 1-p = 1-0.7 = 0.3 =q
• Frequency of heterozygous = 2pq = 2 x 0.7 x 0.3 = 0.42
• Thus, percentage of heterozygous in the population = 0.42 x 100 = 42 % (Answer)