Question

ConsiderapopulationinwhichthefrequencyofalleleAisp \(=0.7\) and thefrequencyofalleleais \(q=0.3\), andwheretheallelesarecodominant. What will be the allele frequencies after one generation if the following occurs? (a) \(w_{A A}=1, w_{A a}=0.9, w_{a a}=0.8\) (b) \(w_{A A}=1, w_{A a}=0.95, w_{a a}=0.9\) (c) \(w_{A A}=1, w_{A a}=0.99, w_{a a}=0.98\) (d) \(w_{A A}=0.8, w_{A a}=1, w_{a a}=0.8\)

Short answer

Question: Calculate the new allele frequencies after one generation for each given case. Answer: (a) p ≈ 0.692, q ≈ 0.308 (b) p ≈ 0.695, q ≈ 0.305 (c) p ≈ 0.698, q ≈ 0.302 (d) p ≈ 0.702, q ≈ 0.298

Step-by-step solution

Calculate the initial genotype frequencies

Using the given initial allele frequencies (\(p = 0.7\), \(q = 0.3\)), we can calculate the initial genotype frequencies: \(AA: p^2 = (0.7)^2 = 0.49\) \(Aa: 2pq = 2(0.7)(0.3) = 0.42\) \(aa: q^2 = (0.3)^2 = 0.09\)

Calculate the resultant genotype frequencies after selection

Now apply the selection coefficients given in each part of the problem to calculate the resultant genotype frequencies after selection: (a) \(AA: 1 \times 0.49 = 0.49\) \(Aa: 0.9 \times 0.42 = 0.378\) \(aa: 0.8 \times 0.09 = 0.072\) (b) \(AA: 1 \times 0.49 = 0.49\) \(Aa: 0.95 \times 0.42 = 0.399\) \(aa: 0.9 \times 0.09 = 0.081\) (c) \(AA: 1 \times 0.49 = 0.49\) \(Aa: 0.99 \times 0.42 = 0.4158\) \(aa: 0.98 \times 0.09 = 0.0882\) (d) \(AA: 0.8 \times 0.49 = 0.392\) \(Aa: 1 \times 0.42 = 0.42\) \(aa: 0.8 \times 0.09 = 0.072\)

Calculate the new allele frequencies after selection

To determine the new allele frequencies, normalize the genotype frequencies and then apply the Hardy-Weinberg equilibrium formulas: (a) Total frequency after selection: \(0.49 + 0.378 + 0.072 = 0.94\) New allele frequencies: \(p = \frac{0.49 + 0.5 \times 0.378}{0.94} \approx 0.692\), \(q = \frac{0.072 + 0.5 \times 0.378}{0.94} \approx 0.308\) (b) Total frequency after selection: \(0.49 + 0.399 + 0.081 = 0.97\) New allele frequencies: \(p = \frac{0.49 + 0.5 \times 0.399}{0.97} \approx 0.695\), \(q = \frac{0.081 + 0.5 \times 0.399}{0.97} \approx 0.305\) (c) Total frequency after selection: \(0.49 + 0.4158 + 0.0882 = 0.994\) New allele frequencies: \(p = \frac{0.49 + 0.5 \times 0.4158}{0.994} \approx 0.698\), \(q = \frac{0.0882 + 0.5 \times 0.4158}{0.994} \approx 0.302\) (d) Total frequency after selection: \(0.392 + 0.42 + 0.072 = 0.884\) New allele frequencies: \(p = \frac{0.392 + 0.5 \times 0.42}{0.884} \approx 0.702\), \(q = \frac{0.072 + 0.5 \times 0.42}{0.884} \approx 0.298\) Hence, we have found the new allele frequencies after one generation for each given case.