39.2k views
3 votes
In a population of 600 people, if p = 20%, what are the expected numbers of individuals with the genotype aa?

a) 120
b) 240
c) 360
d) 480

1 Answer

0 votes

Final answer:

Using the principles of Hardy-Weinberg equilibrium, the calculation shows that the expected number of individuals with genotype aa in a population of 600 people where the frequency of allele A (p) is 20% would be 384. This result does not match any of the provided options.

Step-by-step explanation:

The subject in question is related to Hardy-Weinberg equilibrium, which is a principle of population genetics that asserts that the genetic variation in a population will remain constant from one generation to the next in the absence of disturbing factors. The genotype frequency of aa individuals in a population where p = 20% (which refers to the frequency of the A allele) can be determined using the Hardy-Weinberg equation: + 2pq + = 1. In this equation, q represents the frequency of the a allele, and represents the frequency of the homozygous recessive genotype aa.

Given that p + q = 1, if p = 0.2, then q = 1 - 0.2 = 0.8. The expected frequency of the genotype aa () is 0.8². Multiplying the expected frequency of the aa genotype by the total population, we obtain (0.8²)(600) = (0.64)(600) = 384. Therefore, the expected number of individuals with the genotype aa is 384, which is not one of the options provided, suggesting either a mistake in the question options or a need to re-evaluate the initial given values and calculations.

User Gilles
by
7.9k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories