CAIE A-Level Biology 17.2.5 Hardy–Weinberg Principle

Practise deriving p and q from phenotype data, calculating 2pq or genotype numbers and deciding whether population conditions permit Hardy–Weinberg equilibrium.

[Maximum number: 3]

There are more than 600 plant species in the genus Ipomoea. Many species are grown for their attractive flowers, and some species are used as crop plants.

Fig. 3.1 shows Ipomoea purpurea, the common morning glory.

Fig. 3.1

The gene that determines flower colour in I. purpurea has two alleles:
- a dominant allele that results in purple flowers
- a recessive allele that results in red flowers.

A student recorded the flower colour of all the I. purpurea plants in a field.
The student recorded 660 plants with purple flowers and 440 plants with red flowers.
Assuming the Hardy-Weinberg principle applies to this population, calculate the number of plants in the field that are heterozygous.

Use the equations:
p+q=1
$p^{2}+2 p q+q^{2}=1$
Show your working and give your answer to the nearest whole number.

Correct answer for question 3(a):

Total number of plants = 660 + 440 = 1100.

The red-flowered plants are homozygous recessive, so:

$q^2 = \frac{440}{1100} = 0.4$ $q = \sqrt{0.4} = 0.632455532$ p = 1 - q = 1 - 0.632455532 = 0.367544468

The heterozygote frequency is:

$2pq = 2 \times 0.632455532 \times 0.367544468 = 0.4649110641$

Number of heterozygous plants:

$2pq \times 1100 = 0.4649110641 \times 1100 = 511.4$

Nearest whole number: 511 heterozygous plants.

Marking guidance: award 3 marks for: calculating q^2 = 440/1100 = 0.4 and q = sqrt(0.4) = 0.632455532; calculating p = 1 - q = 0.367544468 and 2pq = 0.4649110641; multiplying 2pq by 1100 and giving 511 or 512 as the final whole-number answer, as accepted by the official markscheme.

All question bank results loaded

17.2.5—Hardy-Weinberg principle question 1