1. **Problem statement:** Calculate allele frequencies and genotype/phenotype frequencies for cystic fibrosis and lizard populations using Hardy-Weinberg principles. 2. **Relevant formulas:** - Hardy-Weinberg equation: $$p^2 + 2pq + q^2 = 1$$ where $p$ is the frequency of the dominant allele and $q$ is the frequency of the recessive allele. - Allele sum: $$p + q = 1$$ 3. **Part 4a: Frequency of the recessive allele ($q$) in cystic fibrosis population** - Given: 1 in 2500 babies affected means $$q^2 = \frac{1}{2500} = 0.0004$$ - Calculate $q$: $$q = \sqrt{0.0004} = 0.02$$ 4. **Part 4b: Frequency of the dominant allele ($p$)** - Using $$p + q = 1$$ - Substitute $q=0.02$: $$p = 1 - 0.02 = 0.98$$ 5. **Part 4c: Percentage of carriers (heterozygous, $2pq$)** - Calculate $2pq$: $$2pq = 2 \times 0.98 \times 0.02 = 0.0392 = 3.92\%$$ 6. **Part 5: Allele frequencies in lizard population** - Total lizards: $$396 + 557 = 953$$ - Frequency of recessive genotype ($q^2$): $$q^2 = \frac{396}{953} \approx 0.4157$$ - Calculate $q$: $$q = \sqrt{0.4157} \approx 0.6449$$ - Calculate $p$: $$p = 1 - 0.6449 = 0.3551$$ 7. **Part 6: Expected genotype frequencies in lizard population** - Given counts: BBS = 30, BS = 96, SS = 120 - Total individuals: $$30 + 96 + 120 = 246$$ - Given frequencies: - $$p^2 = 0.12$$ - $$2pq = 0.46$$ - $$q^2 = 0.42$$ - These sum to 1: $$0.12 + 0.46 + 0.42 = 1$$ 8. **Part 6d: Expected phenotype frequencies and heterozygous individuals** - Total population: 438 - Number of heterozygous (BS) individuals predicted: $$2pq \times 438 = 0.46 \times 438 = 201.48 \approx 201$$ **Final answers:** - 4a: $$q = 0.02$$ - 4b: $$p = 0.98$$ - 4c: Carriers = $$3.92\%$$ - 5: $$p = 0.3551, q = 0.6449$$ - 6d: Heterozygous individuals = $$201$$