Questions 1 to 3 are based on the following scenario Question 1: The Isle of Soa
ID: 134819 • Letter: Q
Question
Questions 1 to 3 are based on the following scenario
Question 1:
The Isle of Soay (which means Sheep Island) is located in the North Atlantic Ocean off the coast of Scotland. This island is extremely isolated and is almost impossible to access. But despite this isolation, several populations of animals have survived here for thousands of years unmanaged and feral.
The texture and colour of the fleece of Soay Sheep can exhibit significant variation. Their fleece can be hairy or woolly, and their fleece can be blonde- or dark-coloured.
You are conducting a population genetic study of these sheep and you are interested in the distribution of these phenotypes and genotypes across this island.
Each sheep was assessed at both loci (i.e. fleece colour and fleece texture). After three years of data collection, you produced the following table from the 205 sheep you observed.
For simplicity, we will assume that blonde-coloured fleece (B) is dominant over dark-coloured fleece (b), and that hairy fleece (H) is dominant over woolly fleece (h).
(In reality, the inheritance of these traits is likely more complex and these sheep will exhibit a range of colours and textures in their fleeces in this population – but here we will keep it simple.)
Assuming the Soay population is in Hardy-Weinberg equilibrium for the two loci, what are the observed frequencies of each allele (B, b, H and h) in this sheep population?
Round your answer to three decimal places and submit your answer in the correct format (e.g. 0.062).
Locus 1: Fleece Colour
(a) Frequency of the b allele = (1 mark)
(b) Frequency of the B allele = (1 mark)
Locus 2: Fleece Texture
(c) Frequency of the h allele = (1 mark)
(d) Frequency of the H allele = (1 mark)
HINT: The assumption that this population is in Hardy-Weinberg equilibrium will allow you to begin this question.
4 points
QUESTION 2
Question 2:
Assuming the Soay population is in Hardy-Weinberg equilibrium for the two loci (fleece colour and fleece texture), and based on the allele frequencies you determined above (in Question 1), calculate the frequencies of each genotype in this sheep population.
Round your answer to three decimal places and submit your answer in the correct format (e.g. 0.062).
(a) Locus 1: Fleece colour
(i) The genotype frequency for blonde heterozygotes (i.e. Bb) is (1 mark)
(ii) The genotype frequency for blonde homozygotes (i.e. BB) is (1 mark)
(iii) The genotype frequency for dark homozygotes (i.e. bb) is (1 mark)
(b) Locus 2: Fleece texture
(i) The genotype frequency for hairy heterozygotes (i.e. Hh) is (1 mark)
(ii) The genotype frequency for hairy homozygotes (i.e. HH) is (1 mark)
(iii) The genotype frequency for woolly homozygotes (i.e. hh) is (1 mark)
6 points
QUESTION 3
Question 3:
Based on your calculations above in Questions 1 and 2, answer the following questions.
Round your answers to the nearest whole number (because you cannot have part of a sheep!).
(a) Locus 1: Fleece colour
(i) How many sheep would you expect to be heterozygous for blonde fleece colour (i.e. Bb)? (1 mark)
(ii) How many sheep would you expect to be homozygous dominant for blonde fleece colour (i.e. BB)? (1 mark)
(iii) How many sheep would you expect to be homozygous recessive for dark fleece colour (i.e. bb)? (1 mark)
(b) Locus 2: Fleece Texture
(i) How many sheep would you expect to be heterozygous for hairy fleece (i.e. Hh)? (1 mark)
(ii) How many sheep would you expect to be homozygous dominant for hairy fleece (i.e. HH)? (1 mark)
(iii) How many sheep would you expect to be homozygous recessive for woolly fleece (i.e. hh)? (1 mark)
6 points
QUESTION 4
Phenotype Number of Sheep Locus 1: Fleece Colour Blonde 117 Dark 88 Locus 2: Fleece Texture Hairy 154 Woolly 51Explanation / Answer
1. Frequency of the b allele
Total = 117 + 88 = 205
b2 = 88/205
b2 = 0.42
b = 0.64
(b) Frequency of the B allele
B+b = 1
B = 1-0.64
B = 0.36
(c) Frequency of the h allele
Total = 154 + 51 = 205
h2 = 51 / 205
h2 = 0.24
h = 0.48
(d) Frequency of the H allele = (1 mark)
H+h = 1
H = 1-0.48
H = 0.52
2.
i) The genotype frequency for blonde heterozygotes (i.e. Bb) is
2Bb = 2X0.36X0.64 = 0.46
(ii) The genotype frequency for blonde homozygotes (i.e. BB) is
B2 = B X B = 0.36 X 0.36 = 0.12
(iii) The genotype frequency for dark homozygotes (i.e. bb) is
b2 = 0.42
(i) The genotype frequency for hairy heterozygotes (i.e. Hh) is
2Hh = 2X0.52X0.48 = 0.49
(ii) The genotype frequency for hairy homozygotes (i.e. HH) is
H2 = HXH = 0.52 X 0.52 = 0.27
(iii) The genotype frequency for woolly homozygotes (i.e. hh) is (1 mark)
h2 = 0.24
3. (i) How many sheep would you expect to be heterozygous for blonde fleece colour (i.e. Bb)?
0.46 X 205 = 94
(ii) How many sheep would you expect to be homozygous dominant for blonde fleece colour (i.e. BB)?
0.12 X 205 = 25
(iii) How many sheep would you expect to be homozygous recessive for dark fleece colour (i.e. bb)?
0.42 X 205 = 86
(b) Locus 2: Fleece Texture
(i) How many sheep would you expect to be heterozygous for hairy fleece (i.e. Hh)?
0.49 X 205 = 100
(ii) How many sheep would you expect to be homozygous dominant for hairy fleece (i.e. HH)?
0.27 X 205 = 55
(iii) How many sheep would you expect to be homozygous recessive for woolly fleece (i.e. hh)?
0.24 X 205 = 50
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