Answer :
Final answer:
The frequency of the homozygous recessive genotype should be q²; since p is given as 0.7, q would be 0.3, and q² would be 0.09. The population is not in Hardy-Weinberg equilibrium since the question states the recessive homozygote frequency is 0.20, contradicting the value calculated from p and q. Hence, the statement is false.
Explanation:
In a population that satisfies the conditions for Hardy-Weinberg equilibrium, the allele frequencies remain constant from one generation to the next, provided no evolutionary influences are acting upon the population. The equation representing Hardy-Weinberg equilibrium is p² + 2pq + q² = 1, where p is the frequency of the dominant allele and q is the frequency of the recessive allele. For a population to be in genetic equilibrium, it must experience random mating, no mutations, no gene flow, no genetic drift, and no selection.
In the scenario provided, if the frequency, p, of an allele is 0.7, we can deduce that q = 1 - p, giving us q = 0.3. According to the Hardy-Weinberg equation, the frequency of the homozygous recessive genotype (aa) should be q², which would be 0.3² or 0.09. However, the question states that the frequency of the homozygous recessive genotype is 0.20 in the next generation. This discrepancy indicates that the population is not in Hardy-Weinberg equilibrium, therefore the statement is B. False.
