Q4. In a population in equilibrium, 36 per cent of individuals show a recessive trait (aa). Find q, p and the frequencies of AA and Aa.
q = 0.36, so q = 0.6 and p = 0.4. AA = p = 0.16 and Aa = 2pq = 2 × 0.4 × 0.6 = 0.48 (check: 0.16 + 0.48 + 0.36 = 1).
Q5. Name the five factors that NCERT says affect Hardy–Weinberg equilibrium.
Gene migration or gene flow, genetic drift, mutation, genetic recombination and natural selection (NCERT page 121).
Q6. A sample of 400 individuals has 196 AA, 168 Aa and 36 aa. Find the frequency of allele A.
A alleles: 2 × 196 + 168 = 560 out of 2 × 400 = 800, so p = 560 ÷ 800 = 0.7 (and q = 0.3).
Q7. Why does NCERT call p + 2pq + q = 1 a binomial expansion?
Because it is (p + q) multiplied out; since p + q = 1, the three genotype frequencies add up to 1 (NCERT page 121).
Q8. In which kind of selection do more individuals acquire the mean character value, and how does the curve change?
Stabilising selection; the peak of the curve gets higher and narrower (NCERT page 121, Figure 6.8).