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

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The Hardy-Weinberg equation is p^2 + 2pq + q^2 = 1, where p is the frequency of the dominant allele and q is the frequency of the recessive allele. Given p = 0.8, q = 1 - p = 0.2. The frequency of heterozygotes is 2pq = 2(0.8)(0.2) = 0.32 or 32%.

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

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