Using Hardy Weinberg formulas, calculate the value of A. AA = 0.6, B. Aa = 0.24, C. aa = 0.16

The frequency of the dominant allele ($p$) is calculated as the frequency of the homozygous dominant genotype plus half the frequency of the heterozygous genotype.

$$p = \text{freq(AA)} + \frac{1}{2} \text{freq(Aa)}$$ $$p = 0.6 + \frac{1}{2}(0.24)$$ $$p = 0.6 + 0.12$$ $$p = 0.72$$

The frequency of the recessive allele ($q$) is calculated as the frequency of the homozygous recessive genotype plus half the frequency of the heterozygous genotype.

$$q = \text{freq(aa)} + \frac{1}{2} \text{freq(Aa)}$$ $$q = 0.16 + \frac{1}{2}(0.24)$$ $$q = 0.16 + 0.12$$ $$q = 0.28$$

We can verify that $p + q = 1$:

$$0.72 + 0.28 = 1.00$$

According to the Hardy-Weinberg principle:
The frequency of the homozygous dominant genotype (AA) is $p^2$:

$$AA = p^2 = (0.72)^2$$ $$AA = 0.5184$$

The frequency of the heterozygous genotype (Aa) is $2pq$:

$$Aa = 2pq = 2 \times 0.72 \times 0.28$$ $$Aa = 0.4032$$

The frequency of the homozygous recessive genotype (aa) is $q^2$:

$$aa = q^2 = (0.28)^2$$ $$aa = 0.0784$$

The values of the genotypes calculated using the Hardy-Weinberg formula are:

• AA = 0.5184
• Aa = 0.4032
• aa = 0.0784