Question

Consider a randomly breeding population of squirrels with two morphs – white-striped and brown-striped. In a population, 16\% are white-striped individuals, while the rest are all brown-striped. The trait for stripes is governed by one gene where the allele for brown stripes is dominant. Assuming Hardy–Weinberg equilibrium, the frequency of the allele for white stripes would be \_\_\_\_\_\_\_\_\_\_\_. (Round off to two decimal places)

Hint:

To solve Hardy–Weinberg equilibrium problems: 1. Use \(q^2\) for recessive phenotypes to find \(q\). 2. Calculate \(p\) as \(1 - q\). 3. Verify using \(p^2 + 2pq + q^2 = 1\).

Answer 1

Step 1: Understand the problem. The trait for white stripes is recessive, and 16\% of the population is white-striped (\(q^2 = 0.16\)). To find the frequency of the allele for white stripes (\(q\)), we take the square root of \(q^2\): \[ q = \sqrt{q^2} = \sqrt{0.16} = 0.4 \] Step 2: Verify the frequency of the dominant allele. The frequency of the dominant allele (\(p\)) is given by: \[ p = 1 - q = 1 - 0.4 = 0.6 \] Step 3: Confirm Hardy–Weinberg equilibrium. Under Hardy–Weinberg equilibrium, the allele frequencies should satisfy: \[ p^2 + 2pq + q^2 = 1 \] \[ (0.6)^2 + 2(0.6)(0.4) + (0.4)^2 = 0.36 + 0.48 + 0.16 = 1 \] The calculation confirms Hardy–Weinberg equilibrium.