Evolution — practice questions

We use the Hardy-Weinberg equations: p + q = 1 and p^2 + 2pq + q^2 = 1. Given: frequency of homozygous recessive genotype = 1% = 0.01.

(a) Recessive allele (q): We know that q^2 = frequency of homozygous recessive genotype. So, q^2 = 0.01 q = sqrt(0.01) q = 0.1 The frequency of the recessive allele is 0.1.

(b) Dominant allele (p): We know that p + q = 1. So, p + 0.1 = 1 p = 1 - 0.1 p = 0.9 The frequency of the dominant allele is 0.9.

(c) Heterozygous individuals (2pq): We know that the frequency of heterozygous individuals is 2pq. 2pq = 2 * 0.9 * 0.1 2pq = 0.18 The frequency of heterozygous individuals is 0.18 (or 18%).

We use the Hardy-Weinberg equations. The total population size is 500.

(a) Frequency of homozygous recessive genotype (q²): The number of susceptible (homozygous recessive) insects is 20. The frequency is the number of individuals with the genotype divided by the total population. q² = 20 / 500 q² = 0.04

(b) Frequency of the recessive allele (q): The frequency of the recessive allele 'q' is the square root of the homozygous recessive frequency 'q²'. q = √q² q = √0.04 q = 0.2

(c) Frequency of the dominant allele (p): The sum of allele frequencies is 1. p + q = 1 p = 1 - q p = 1 - 0.2 p = 0.8

(d) Number of heterozygous insects (2pq): First, calculate the frequency of the heterozygous genotype (2pq). Frequency of heterozygotes = 2 * p * q = 2 * 0.8 * 0.2 = 0.32 To find the number of heterozygous insects, multiply this frequency by the total population size. Number of heterozygotes = 0.32 * 500 = 160 insects.