Question

Assume a population is in Hardy-Weinberg equilibrium. If there are two alleles at a locus, R and r, and the frequency of R is known to be .3. Which statement below is accurate?

A. Most individuals are homozygous
B. Most individuals are heterozygous
C. There must be equal numbers of RR and rr homozygotes
D. b & c are correct
E. Cannot tell from information provided

Answer 1

In a population in Hardy-Weinberg equilibrium where the frequency of allele R is 0.3, the most accurate statement is that most individuals are heterozygous. Therefore, the accurate statement is B. Most individuals are heterozygous.

Step-by-step explanation:

In a population that is in Hardy-Weinberg equilibrium, the frequencies of alleles remain constant over generations. The Hardy-Weinberg equation allows us to calculate the expected frequencies of genotypes based on the allele frequencies. In this case, if the frequency of allele R is 0.3, we can use the equation: p² + 2pq+q² = 1 to determine the frequencies of genotypes.

Let's use p to represent the frequency of allele R and q to represent the frequency of allele r. Since there are only two alleles, p + q = 1. Given that the frequency of allele R (p) is 0.3, the frequency of allele r (q) can be calculated as 1 - 0.3 = 0.7.

Using the equation, we can calculate the expected frequencies of genotypes:

1. RR genotype frequency = p² = 0.3² = 0.09
2. Rr genotype frequency = 2pq = 2 * 0.3 * 0.7 = 0.42
3. rr genotype frequency = q² = 0.7² = 0.49

Based on these calculations, we can see that the most frequent genotype is rr (homozygous recessive) with a frequency of 0.49. Therefore, the accurate statement is B. Most individuals are heterozygous.