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Based on theHardy-Weinberg equilibrium, the frequencies of individuals in the population are0.81, 0.01 and 0.18.The Hardy-Weinberg equilibrium equation is used to determine the frequencies of individuals in a population at equilibrium.The hardy-weinberg equation is given asp² + 2pq + q² = 1where,p² is the frequency of the homozygousdominant individualsq² is the frequency of the homozygousrecessive individuals.Given:c = 0.1 and C = 0.9Then;p² = 0.9²p² = 0.81Also;q² = 0.1²q² = 0.01Then, substituting the values in the Hardy-Weinberg equation:0.81 + 2pq - 0.01 = 12pq = 0.18Therefore, thefrequencies of individualsin the population are 0.81, 0.01 and 0.18.Learn more about theHardy-Weinberg equationhere:brainly.com/question/5028378#SPJ4...