AP BIOLOGY

Detailed explanation-1: -Given that the 1% error from this is negligible, the population supports this and therefore the remaining 102 individuals, 51% of the population, would be heterozygotes expressing the dominant phenotype.

Detailed explanation-2: -Answer and Explanation: If 72 out of 200 individuals in the animal population are homozygous recessive, 48% of the population would be heterozygotes, and 16% would be homozygous dominant.

Detailed explanation-3: -The Hardy-Weinberg Equation q = the frequency of the recessive allele in a population. 2pq = the frequency of the heterozygous dominant genotype. p2 = the frequency of homozygous dominant genotype. q2 = the frequency of homozygous recessive genotype.

Detailed explanation-4: -In a population of 200 mice, 98 are homozygous dominant for brown coat color (BB), 84 are heterozygous (Bb) and 18 are homozygous recessive (bb).

Detailed explanation-5: -Answer: The frequency of heterozygous individuals is equal to 2pq. In this case, 2pq equals 0.32, which means that the frequency of individuals heterozygous for this gene is equal to 32% (i.e. 2 (0.8)(0.2) = 0.32).