Final answer:
The probability of obtaining an (Aa) zygote in a population at Hardy-Weinberg equilibrium with the given genotype frequencies is 48%, as that is the observed frequency of the heterozygous genotype.
Step-by-step explanation:
The probability of obtaining an (Aa) zygote, assuming Hardy-Weinberg equilibrium and given genotype frequencies of (AA) 36%, (Aa) 48%, and (aa) 16%, would be 48%. This is based on the actual given frequency of the heterozygous genotype (Aa) within the population. The Hardy-Weinberg principle, which includes equations such as p² + 2pq + q² = 1, is used to calculate and predict the genetic structure of a population that is not evolving. It requires random mating and assumes no factors that affect gene frequencies, like mutation, migration, genetic drift, or selection, are present.
According to the Hardy-Weinberg equilibrium equation, the frequency of the (Aa) genotype can be calculated as 2pq, where p = frequency of A and q = frequency of a. In this case, the given genotype frequencies are (AA) 36%, (Aa) 48%, and (aa) 16%. To find the frequency of A, we can use the equation p² + 2pq + q² = 1 and solve for p. Using this equation, we find that p = √(36/100) = 0.6.
Next, we can calculate the frequency of q (frequency of a) by subtracting p from 1: q = 1 - p = 1 - 0.6 = 0.4.
Finally, we can calculate the frequency of the (Aa) genotype using the equation 2pq: (Aa) = 2 × 0.6 × 0.4 = 0.48 or 48%.