In a population that is in Hardy-Weinberg equilibrium, the frequency of the recessive homozygous genotype is 0.49.The percentage of the population that is heterozygous is
A)51.
B)49.
C)42.
D)7.
E)3.

Question

Quizplus · Accepted Answer

In Hardy-Weinberg equilibrium, the frequency of the recessive homozygous genotype (aa) is given by q2, where q is the frequency of the recessive allele. So, q2 = 0.49, which means q = 0.7 (because q is the square root of 0.49). 
Since p + q = 1, the frequency of the dominant allele (p) is 0.3. 
The frequency of heterozygotes can be calculated as 2pq (where p and q are the frequencies of the two alleles), which is 2 x 0.3 x 0.7 = 0.42 or 42%. Therefore, the answer is C.

In a Population That Is in Hardy-Weinberg Equilibrium, the Frequency