chapter 5

TUTORIAL 5.0: POPULATION GENETICS

5.2 Hardy-Weinberg Law
a) State the Hardy-Weinberg Law. (CLO 1)
b) Explain FIVE assumptions of the Hardy-Weinberg Law for genetic equilibrium. (CLO 2)
c) Calculate allele & genotype frequency. (CLO 3)

1. a) State the Hardy-Weinberg principle. [CLO 3, C1] [2 marks]

The Hardy-Weinberg principle states that the frequencies of allele and
genotypes remain constant from generation to generation

b) In a randomly breeding population of mice, black coat (H) is dominant to white coat (h). In the

population,36% have white coats. Calculate the genotypic frequency of black coated mice in this

population.[CLO 3, C3] [4 marks]

allele H encode for black coat

allele h encode for white coat

frequency for homozygous recessive genotype, q2 = 36/100

= 0.36

frequency for allele recessive, q = √0.36
= 0.6

frequency for allele dominant, p = 1- q
= 1- 0.6
= 0.4

Frequency genotype for black coated mice = p2 + 2pq

= (0.4) 2 + 2 (0.4) (0.6)
= 0.64

c) In a human population, the frequency of the recessive individuals for extra-long eyelashes is 90

per 1000. What is the individual percentage of this population that carries recessive alleles but

displays the phenotype of short eyelashes? [CLO 3, C1] [4

marks]

frequency for homozygous recessive genotype, q2 = 90/1000
= 0.09

Frequency for recessive allele , q = √0.09
= 0.3

Frequency for dominant allele, p = 1- q
= 1 - 0.3
= 0.7

Percentage of individual that carries recessive allele but displays the phenotype of short

eyelashes = 2pq x 100
= 2(0.7)(0.3) x 100
= 42%

2. A species of insects may exist as a dark variety or a fair variety. The dark trait is dominant. In a population of
5000 insects, there are 950 dark insects. By using the Hardy-Weinberg equation (p2 + 2pq + q2 = 1) calculate:

i. The frequency of the ‘fair’ allele. [CLO 3, C3] [3 marks]
Number of ‘fair’ insects (recessive phenotype) = 5000 – 950
= 4050
Frequency of homozygous recessive genotype, q2 = 4050/5000
= 0.81

Frequency of ‘fair’ allele, q =√0.81
= 0.9

ii. The frequency of the ‘dark’ allele. [CLO 3, C3] [2 marks]

Total frequency allele, p + q = 1
frequency of the ‘dark’ allele, p = 1- q

= 1- 0.9
= 0.1

iii. The number of heterozygous and homozygous dominant insects in the population. [CLO 3, C3] [3 marks]

Frequency of heterozygous insects = 2pq
= 2(0.1)(0.9)
= 0.18

Number of heterozygous insects = 0.18 x 5000
= 900

Frequency of homozygous dominant genotype insects = p2
= (0.1)2
= 0.01

Number of homozygous dominant insects = 0.01 x 5000
= 50

3. The allele for black hair (B) is dominant over that for grey hair (b). The alleles for B and b have frequencies p
and q respectively. In a randomly mating population of 600 hamsters, 216 have grey hair.

a) What is meant by a randomly mating population? [CLO 3, C1] [1 mark]

Each individual has an equal chance to mate freely or randomly with any other individual (from

the opposite sex) within the population.

b) Assuming that the Hardy-Weinberg principle applied, calculate the frequency of the dominant allele and

recessive allele in the population. [CLO 3, C3] [2 marks]

Frequency of the recessive genotype bb, q2 = 216/600
= 0.36

Frequency of recessive allele b, q2 = √0.36
= 0.6

Frequency of dominant allele B, p = 1- q
= 1- 0.6
= 0.4

c) If all the 216 hamsters with grey hair were killed and the rest were allowed to mate randomly,

i. Calculate the frequency of the dominant and recessive alleles in the new generation. [7 marks]
[CLO 3, C3]

Frequency of homozygous dominant, p2 = (0.4)2 = 0.16
Frequency of heterozygous, 2pq = 2 x 0.4 x 0.6

= 0.48

Number of dominant homozygous (BB) individuals = 0.16 x 600
= 96

Number of heterozygous (Bb) individuals= 0.48 x 600
= 288

If 216 grey hamster were killed from the population, number of hamsters left
= 600 – 216
= 384

Total allele left = 2 x 384 = 768 allele
Total of dominant allele , B = BB (homozygous) + B (heterozygous)

= 2(96) + 288
= 480
Frequency of dominant allele, p = Total number of dominant allele,B/ Total number of allele

= 480/768
= 0.625
Frequency of recessive allele, q = 1- 0.625

= 0.375

ii. Calculate the various genotypes of the F1 offspring and their frequencies. [CLO 3, C3] [3 marks]
Frequency of dominant homozygous genotype, p2 = (0.625)2

= 0.3906
Frequency of recessive homozygous genotype, q2 = (0.375) 2

= 0.1406
Frequency of heterozygous genotype, 2pq = 2 x 0.625 x 0.375

= 0.4688

iii. Calculate the percentage of the hamsters with black hair and grey hair in the F1 population.

[CLO 3, C3] [2 marks]

Percentage of the hamsters with black hair in F1 population,

= (p2 + 2pq) x 100

= (0.3906 + 0.4688) x 100

= 85.94%

Percentage of the hamsters with grey hair in F1 population,

= 100 – 85.94

= 14.06%

4. In the large population of butterflies, 557 are white while 396 are brown. Assume that white color is dominant,
determine the following.

i. Allelic frequencies of each allele [CLO 3, C3] [4 marks]
Number of butterflies in the population = 557 + 396 = 953
Frequency of homozygous recessive genotype, q2 = 396/ 953
= 0.416

Frequency of recessive allele, q = √0.416
= 0.645

Frequency of dominant allele, p= 1- 0.645
= 0.355

ii. Expected genotypic frequencies [CLO 3, C3] [3 marks]

Frequency of homozygous dominant genotype, p2 = (0.355)2
= 0.126

Frequency of heterozygous genotype, 2pq = 2(0.355)(0.645)
= 0.458

Frequency of homozygous recessive genotype, q2= (0.645)2
= 0.416

iii. Number of heterozygous butterflies [CLO 3, C3] [2 marks]
Number of heterozygous butterflies = 2pq x 953
= 0.458 x 953
= 436

iv. Expected phenotypic frequencies [CLO 3, C3] [2 marks]
Frequency of white phenotype = p2 + 2pq
= 0.126 + 0.458
= 0.584

Frequency of brown phenotype, q2 = 0.416

v. Assuming that Hardy-Weinberg equilibrium is met, how many white butterflies would you expect to find

among the 1234 butterflies of the next generation? [CLO 3, C3] [2 marks]

Frequency of white phenotype = 0.584
Butterflies with white phenotype = 0.584 x 1234

= 721

5. Hardy and Weinberg developed a mathematical relationship between the frequencies of the alleles and
genotypes in populations.

a) State three conditions that must be fulfilled when using the Hardy –Weinberg equation

[CLO 3, C1] [3 marks]

• Large population size

• Random mating

• No migration

• No mutation

• No natural selection

b) In sheep, the characteristic of normal fur is controlled by a dominant allele R, while non-uniform fur
controlled by a recessive allele, r. In a population of 1000 sheep, it is found that 750 sheep have normal
fur. By using the Hardy -Weinberg equation, calculate: [CLO 3, C3]

i. The frequency of the recessive allele.
Number of individual non-uniform fur sheep is = 1000 – 750
= 250
Frequency of homozygous recessive genotype, q2 = 250/ 1000
= 0.25

Frequency of recessive allele, q = √0.25 [2 marks]
= 0.5

ii. Calculate the percentage of sheep in the population which have the heterozygous genotype for the

normal fur characteristic. [CLO 3, C3] [5 marks]

Total number of allele ,p + q = 1

Frequency heterozygous genotype, 2pq = 2 x 0.5 x 0.5
= 0.5

Frequency of recessive allele, q = 1- 0.5
= 0.5

Number of heterozygous sheep= 0.5 x 1000 = 500
Percentage of heterozygous sheep = (500/1000) x 100 = 50%

6. The groundsel plant can produce three types of flowers. The type of flowers produced is controlled by a pair of

alleles, both have no dominance. The type of flowers and their genotypes are as shown below.

phenotype Genotype

Floret with long spikes :RR

Floret with short spikes :Rr

Floret without spikes :rr

In a population of 500 plants, it is found that:
142 plants have Florets with long spikes
248 plants have Florets with short spikes
110 plants have Floret without spikes

a) Calculate the frequencies of allele R and r. [CLO 3, C3] [2 marks]
frequencies of allele R, p = Total number of allele R
Total number of allele
= (142x2) +248
500 X 2
= 0.532

frequencies of allele r, q = (110x2) + 248
500x 2

= 0.468

b) Calculate the observed frequencies of different genotype in the population and by using the equation
p+q =1, calculate the frequencies expected for the different genotypes in the population.[CLO 3, C3]
[5 marks]

Frequency of homozygous dominant genotype observed = 142/ 500
= 0.284

Frequency of homozygous dominant genotype expected = p2 = (0.532)2
= 0.283

Frequency of heterozygous genotype observed = 284 / 500
= 0.496

Frequency of heterozygous genotype expected = 2pq
= 2 x 0.532 x 0.468

= 0.498
Frequency of homozygous recessive genotype observed = 110/ 500

= 0.220

Frequency of homozygous recessive genotype expected = q2 = (0.468)2 = 0.219

c) Is the population obeying the principles of the hardy-Weinberg Law? [CLO 3, C1] [2 marks]

Yes, this is because the frequency of genotype observed is equal to the frequency of genotype

expected

7. Phenylketonuria (PKU) is a genetic disease controlled by the recessive allele. It is also an inborn error of
phenylalanine metabolism and leads to severe mental retardation. Table 1 shows the number of the normal and
the abnormal individuals in a population.

YEAR 1995 2005
The number of the normal individuals 4996 4992
The number of the abnormal individuals (having (PKU)
Total 48
5000 5000

Table 1 [4 marks]

a) Calculate the frequency of the recessive allele in 1995 and 2005. [CLO 3, C3] [1 mark]
Frequency of recessive homozygous genotype in 1995, q² = 4/5000 = 0.0008 [3 marks]
Frequency of recessive allele in 1995, q= 0.028

Frequency of recessive homozygous genotype in 2005, q² = 8/5000 = 0.0016
Frequency of recessive allele in 2005, q= 0.04

b) Give the conclusion of the allele frequency at (a) (i). [CLO 3, C1]
The allele frequency changes / increases/ is not in equilibrium.

c) Give three assumptions to calculate the allele frequency. [CLO 3, C1]
∙ The size of population is large
∙ Random fertilization or random mating
∙ No mutation
∙ No migration
∙ No natural selection