Consider a gene locus in a diploid population with two possible alleles, A and a. Let p and q represent the frequency of the A and a alleles in the population, respectively. As there are only two possible alleles, p and q, we know that

p + q = 1.

The three genotypes possible are AA, Aa, and aa. The Hardy-Weinberg Law states that after one generation of random mating, the frequencies of the three genotypes in the population are given by:

f_AA = p²,

f_Aa = 2pq,

f_aa = q².

An individual has genotype AA if both parents contribute an A allele. Likewise, if both parents contribute an a allele then the individual has genotype aa. This can happen one way, and the probability that an individual is homozygous for allele A is the frequency of the A allele in the population squared, and the probability that an individual is homozygous for allele a is the frequency of the a allele in the population squared.

An individual has genotype Aa if either an A allele is inherited maternally and an a allele is inherited paternally, or if an a allele is inherited maternally and an A allele is inherited paternally. The probability of having an A and a allele is pq (the product of their frequencies in the population), and since a heterozygous genotype can arise two different ways, the frequency of genotype Aa is 2pq. Notice that because p+q = 1, we also know that (p+q)² = p² +2pq +q² = 1.

Use the information above to answer the following questions:

Determine the frequencies of the aa genotype after one generation of random mating.

Determine the frequency of the Aa genotype after one generation of random mating.

Find the frequency of the A allele that will maximize the frequency of the Aa genotype.

Find the frequency of the A and a alleles in a given population.

Find the frequency of the A allele that will maximize the mean fitness of a population.

Find the frequency of the A allele that corresponds to a given mean fitness.

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