A2 Biology 9700: Hardy-Weinberg Carrier Frequency Calculation

A rare recessive metabolic condition is controlled by a single gene with two alleles: the recessive allele c causes the condition only in individuals with genotype cc, while the dominant allele C, present in genotypes CC or Cc, results in an unaffected phenotype (individuals with genotype Cc are unaffected but are carriers of allele c). In a large, randomly mating human population of 10,000 people that satisfies the conditions needed for the Hardy-Weinberg principle to apply, health records show that exactly 25 people have this condition.

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 C and q is the frequency of the recessive allele c.

(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 number of people, out of the 10,000, who are expected to be unaffected carriers (genotype Cc), and the number expected to be homozygous dominant (genotype CC). Show that your three genotype numbers add up to 10,000. [4]

(c) A genetic screening programme directly tests a large sample of this population and finds a carrier frequency noticeably different from your prediction in (b). Suggest one reason, other than sampling or measurement error, why the actual carrier frequency in a real population might differ from a Hardy-Weinberg prediction. [2]

Part (a): Finding q and p

The recessive phenotype (the metabolic condition) only occurs in the homozygous recessive genotype, cc, which has frequency $q^2$ in the Hardy-Weinberg principle.

We are told that 25 out of 10,000 people have the condition, so:

$q^2 = \frac{25}{10000} = 0.0025$

Taking the square root of both sides:

$q = \sqrt{0.0025} = 0.05$

Using $p + q = 1$ :

$p = 1 - q = 1 - 0.05 = 0.95$

So the frequency of the recessive allele c is $q = 0.05$ (5%), and the frequency of the dominant allele C is $p = 0.95$ (95%).

Part (b): Finding the number of carriers and homozygous dominant individuals

The heterozygous (carrier) genotype, Cc, has frequency $2pq$ :

$2pq = 2 \times 0.95 \times 0.05 = 0.095$

So the expected number of carriers in the population is:

$0.095 \times 10000 = 950 \text{ people}$

The homozygous dominant genotype, CC, has frequency $p^2$ :

$p^2 = 0.95 \times 0.95 = 0.9025$

So the expected number of homozygous dominant people is:

$0.9025 \times 10000 = 9025 \text{ people}$

Check: the three genotype counts should add up to the total population of 10,000:

$25 \text{ (cc)} + 950 \text{ (Cc)} + 9025 \text{ (CC)} = 10000$

This matches the given population size exactly, confirming the working is consistent.

Part (c): Why a real screening result might differ from the prediction

The Hardy-Weinberg prediction assumes the population meets all the model’s conditions exactly. In a real population, this is rarely perfectly true, so an observed carrier frequency could genuinely differ from the prediction (beyond simple sampling error) for reasons such as:

Non-random mating: for example, people aware of a family history of the condition, or of their own carrier status (through genetic counselling), may be more or less likely to have children with another carrier than random mating would predict.
Migration: people carrying different allele frequencies moving into or out of the population would change the local allele frequency away from the original prediction.
Selection: if carriers or affected individuals have, in reality, a slightly different survival or reproductive rate than non-carriers (contrary to the “no selection” assumption), allele frequencies would shift over generations.
Genetic drift: if the population is not effectively infinite, chance fluctuations in allele frequency between generations can cause the observed frequency to depart from the theoretical prediction.

Any one of these (or a similar valid reason) is an acceptable explanation.

Final answers

(a) q (frequency of c) = 0.05 (5%); p (frequency of C) = 0.95 (95%).
(b) Carriers (Cc) = 950 people; homozygous dominant (CC) = 9025 people; check: 25 + 950 + 9025 = 10,000. ✓
(c) Any one violation of a Hardy-Weinberg assumption, e.g. non-random mating, migration, selection, or genetic drift, could explain the discrepancy.