Final answer:
A Hardy-Weinberg equilibrium population is one that does not evolve, with constant allele frequencies; it requires no mutations, random mating, no gene flow, large population size, and no selection.
Step-by-step explanation:
A Hardy-Weinberg equilibrium population is one that is not evolving and the allele frequencies remain constant over time. For a population to be in Hardy-Weinberg equilibrium, it must meet five conditions: no mutation, random mating, no gene flow, infinite population size, and no selection. When given a population that is in Hardy-Weinberg equilibrium for alleles A and a, distinguished by frequencies p and q respectively, and where p + q = 1, calculating the genotype frequencies involves squaring p for the AA genotype (p²), squaring q for the aa genotype (q²), and multiplying p and q by 2 for the Aa genotype (2pq).
For instance, if the frequency of the homozygous recessive individuals (aa) is 0.04 (which represents q²), to derive q we take the square root of 0.04, resulting in q as 0.2. Subsequently, p can be found by subtracting q from 1, which would give p an equivalent of 0.8. Then, the frequency of the heterozygote genotype (Aa) can be determined by calculating 2pq, which would be 2 x 0.8 x 0.2, yielding a frequency of 0.32 for the heterozygote genotype.