In a population that meets the Hardy- Weinberg equilibrium assumptions,81% of the individuals are homozygous for a recessive allele.What percentage of the individuals would be expected to be heterozygous for this locus in the next generation?
A)19%
B)18%
C)72%
D)81%

Question

Quizplus · Accepted Answer

According to Hardy-Weinberg equilibrium, the frequencies of alleles and genotypes should not change from one generation to the next. Therefore, the frequency of the recessive allele in the population is:
q² = 0.81
q = √0.81
q = 0.9
The frequency of the dominant allele (p) would therefore be:
p = 1 - q
p = 1 - 0.9
p = 0.1
The frequency of heterozygous individuals (2pq) can now be calculated as follows:
2pq = 2(0.1)(0.9)
2pq = 0.18 or 18%
Therefore, the expected percentage of individuals that would be heterozygous for this locus in the next generation is 18%, which corresponds to choice B.

In a Population That Meets the Hardy- Weinberg Equilibrium Assumptions,81