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Answer :
According to Hardy-Weinberg equilibrium, we can assume that there is no evolution occurring in the population over the generations.
In a population under Hardy-Weinberg equilibrium, if p is the frequency of allele A and q is the frequency of allele a, the expected genotype frequencies are as follows:
AA: p^2
Aa: 2pq
aa: q^2
In the provided information, we have the genotype frequencies for the first generation. We can use these frequencies to estimate the allele frequencies (p and q). According to the Hardy-Weinberg equilibrium, these allele frequencies should remain the same in the second generation.
Let's calculate p and q:
p^2 (frequency of AA) = 0.36
q^2 (frequency of aa) = 0.16
From this, we can calculate p and q:
p^2 = 0.36 -> p ≈ √0.36 ≈ 0.6
q^2 = 0.16 -> q ≈ √0.16 ≈ 0.4
Given that p ≈ 0.6 and q ≈ 0.4, we can calculate the expected genotype frequencies for the second generation
AA: p^2 ≈ 0.6^2 ≈ 0.36
Aa: 2pq ≈ 2 * 0.6 * 0.4 ≈ 0.48
aa: q^2 ≈ 0.4^2 ≈ 0.16
These expected genotype frequencies for the second generation match the frequencies observed in the first generation, confirming that the population is in Hardy-Weinberg equilibrium.
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Answer:
Random mating between the individuals of population.
Explanation:
Random mating refers to a condition in which all the members of a population have equal chances of mating with the opposite sex and is not prevented by any type of physical, psychological, biological, genetical and other barriers and the gene flows in the population.
In the given question, since the genotype of the population has not changed in the next generation, therefore, the population is following the assumptions of the Hardy Weinberg equilibrium in which the most important is that the individual is randomly mating with each other or show equal potency to mate with the opposite sex.
Thus, random mating is the correct answer.
