AP BIOLOGY

Detailed explanation-1: -Calculate the percentage of individuals homozygous for the dominant allele. We know that the frequency of the recessive homozygote genotype is q2 and equal to 0.09. Thus, p = 1 – q ∴ p = 1 – 0.30 = 0.70 ∴The homozygote dominants are represented by p2 = (0.70)2 = 0.49 or 49% Page 2 4.

Detailed explanation-2: -A monohybrid cross is a cross between two individuals with different alleles at one genetic locus. If we cross two heterozygous individuals (Aa), then the percentage of offsprings with recessive phenotype will be: Aa x Aa = AA Aa Aa aa, so recessive phenotypes would be 1/4 = 25%.

Detailed explanation-3: -To determine q, which is the frequency of the recessive allele in the population, simply take the square root of q2 which works out to be 0.632 (i.e. 0.632 x 0.632 = 0.4). So, q = 0.63. Since p + q = 1, then p must be 1-0.63 = 0.37.

Detailed explanation-4: -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.