In the Hardy-Weinberg equation,p² + 2pq + q² = 1.If the dominant allele frequency is 0.8,what percent of the population will be homozygous recessive?

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The frequency of the homozygous recessive genotype (q^2) can be calculated using the Hardy-Weinberg equation, where p + q = 1. If the dominant allele frequency (p) is 0.8, then the recessive allele frequency (q) is 1 - 0.8 = 0.2. Therefore, q^2 = (0.2)^2 = 0.04, or 4% of the population.

In the Hardy-Weinberg Equation,p2 + 2pq + Q2 = 1

In the Hardy-Weinberg Equation,p² + 2pq + Q² = 1