A2 Biology 9700: Hardy-Weinberg Allele and Genotype Frequency Calculations

In a large, randomly mating population of a wildflower species, petal colour is controlled by a single gene with two alleles: the dominant allele R produces red petals, and the recessive allele r produces white petals, but only in plants with the homozygous recessive genotype rr. A survey of a large sample from this population finds that 1% of the plants have white petals.

Assume that this population satisfies all the conditions required for the Hardy-Weinberg principle to apply. You may use the Hardy-Weinberg equations, written here in plain text: "p + q = 1" and "p squared + 2pq + q squared = 1", where p is the frequency of the dominant allele R and q is the frequency of the recessive allele r.

(a) Calculate the frequency of the recessive allele (q) and the frequency of the dominant allele (p) in this population. Show your working. [3]

(b) Calculate the percentage of the population that is heterozygous (genotype Rr) and the percentage that is homozygous dominant (genotype RR). Show your working, and check that your three genotype frequencies add up to 1. [3]

(c) State two conditions (assumptions) that must apply to a population for the Hardy-Weinberg principle to give an accurate prediction of its allele and genotype frequencies. [2]

Part (a): Finding q and p

The recessive phenotype (white petals) only occurs in the homozygous recessive genotype, rr. In the Hardy-Weinberg principle, the frequency of this genotype is given by “q squared” (the q squared term in “p squared + 2pq + q squared = 1”).

We are told that 1% of the plants are white-petalled, so:

q squared = 0.01

Taking the square root of both sides:

q = square root of 0.01 = 0.1

Using “p + q = 1”:

p = 1 minus q = 1 minus 0.1 = 0.9

So the frequency of the recessive allele r is q = 0.1 (10%), and the frequency of the dominant allele R is p = 0.9 (90%).

Part (b): Finding the genotype frequencies

The heterozygous genotype (Rr) has frequency “2pq”:

2pq = 2 times 0.9 times 0.1 = 0.18

So 18% of the population is heterozygous (Rr). These plants have red petals but carry the recessive allele.

The homozygous dominant genotype (RR) has frequency “p squared”:

p squared = 0.9 times 0.9 = 0.81

So 81% of the population is homozygous dominant (RR).

Check: the three genotype frequencies should sum to 1:

p squared + 2pq + q squared = 0.81 + 0.18 + 0.01 = 1.00

This equals 1, as required, and the homozygous recessive term (0.01, i.e. 1%) matches the phenotype frequency given in the question, so the working is consistent.

Part (c): Assumptions of the Hardy-Weinberg principle

The Hardy-Weinberg principle only gives accurate predictions if a population meets certain conditions. Any two of the following are acceptable:

The population is very large, so allele frequencies are not changed by chance (no genetic drift).
Mating is random with respect to this gene. Individuals do not preferentially choose mates with a particular genotype.
There is no selective advantage or disadvantage for any of the three genotypes (no natural selection acting on this gene).
There is no mutation occurring at this gene locus (no new alleles being created or lost by mutation).
There is no migration of individuals into or out of the population (no gene flow altering allele frequencies).

Final answers

(a) q (frequency of r) = 0.1 (10%); p (frequency of R) = 0.9 (90%).
(b) Heterozygous (Rr) = 18%; homozygous dominant (RR) = 81%; check: 0.81 + 0.18 + 0.01 = 1.00. ✓
(c) Any two of: large population (no genetic drift); random mating; no selection; no mutation; no migration.