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In a population of 1000 individuals, the frequency of the recessive allele for a particular trait is 0.2. What is the expected frequency of homozygous recessive individuals in the population, assuming Hardy-Weinberg equilibrium?
0.04
0.08
0.16
0.32
Answers (1)
In a population under Hardy-Weinberg equilibrium, the genotype frequencies can be determined using the Hardy-Weinberg equation:
p2 + 2pq + q2 = 1,
where:
p is the frequency of the dominant allele,
q is the frequency of the recessive allele, and
p^2, 2pq, and q^2 represent the frequencies of the three possible genotypes: homozygous dominant, heterozygous, and homozygous recessive, respectively.
Given that the frequency of the recessive allele (q) is 0.2, we can calculate the frequency of the dominant allele (p) as:
p = 1 - q = 1 - 0.2 = 0.8.
Using the Hardy-Weinberg equation, we can now calculate the expected frequency of homozygous recessive individuals (q2) as:
q2 = (0.2)2 = 0.04.
To obtain the expected count of homozygous recessive individuals in a population of 1000 individuals, we multiply the expected frequency (0.04) by the population size:
Expected count = q2 * population size = 0.04 * 1000 = 40.
Therefore, the expected frequency of homozygous recessive individuals in the population is 40/1000 or 0.04.
Option 1 is the correct answer
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