Question

3. You are helping two of your friends with their homework for another biology class. They are also learning about population genetics and Hardy-Weinberg equilibrium. This is the question they are working on: ‘‘There are 200 individuals in the population: 20 are homozygous dominant (A/A), 20 are heterozygous (A/a), and 160 are homozygous recessive (a/a). Is this population in Hardy-Weinberg equilibrium with respect to gene A?’’ Each of your friends has tried to solve the problem in a different way: Friend 1's Method: # A/A individuals = 20 # A/a individuals = 20 # a/a individuals = 160 Total # of individuals = 200 Observed allele frequencies Observed freq(A) = p = (2×20 + 20) ÷ (2×200) = 60 ÷ 400 = 0.150 Observed freq(a) = q = (2×160 + 20) ÷ (2×200) = 340 ÷ 400 = 0.850 Predicted genotype freq based on observed allele freq If pop' is in H-W eq: Predicted freq(A/A) = p² = 0.150² = 0.02 Predicted freq(A/a) = 2pq = 2×0.150×0.850 = 0.26 Predicted freq(a/a) = q² = 0.850² = 0.72 Observed genotype frequencies Based on data provided in the question: Observed freq(A/A) = 20/200 = 0.10 Observed freq(A/a) = 20/200 = 0.10 Observed freq(a/a) = 160/200 = 0.80 Observed genotype frequencies do not match the genotype frequencies that are predicted under H-W equilibrium; therefore, the population is not in Hardy-Weinberg equilibrium. Friend 2's Method: # A/A individuals = 20 # A/a individuals = 20 # a/a individuals = 160 Total # of individuals = 200 If pop' is in H-W eq: Freq(A/A) = p² Therefore, assume observed freq(A/A) = p² = 20/200 = 0.100 Predicted allele frequencies If pop' is in H-W eq, p² = 0.100 Predicted freq(A) = p = ?(p²) = ?(0.10) = 0.316 Predicted freq(a) = q = 1 - p = 1 - 0.32 = 0.684 Predicted genotype freq based on predicted allele freq If pop' is in H-W eq: Predicted freq(A/A) = p² = 0.316² = 0.10 Predicted freq(A/a) = 2pq = 2×0.316×0.684 = 0.43 Predicted freq(a/a) = q² = 0.684² = 0.47 Observed genotype frequencies Based on data provided in the question: Observed freq(A/A) = 20/200 = 0.10 Observed freq(A/a) = 20/200 = 0.10 Observed freq(a/a) = 160/200 = 0.80 Observed genotype frequencies do not match the genotype frequencies that are predicted under H-W equilibrium; therefore, the population is not in Hardy-Weinberg equilibrium. Which of your friends has solved the question using the correct method (or are both methods equally correct)? Please explain briefly. (3 marks)

          3. You are helping two of your friends with their homework for another biology class. They are also learning about population genetics and Hardy-Weinberg equilibrium. This is the question they are working on:
‘‘There are 200 individuals in the population: 20 are homozygous dominant (A/A), 20 are heterozygous (A/a), and 160 are homozygous recessive (a/a). Is this population in Hardy-Weinberg equilibrium with respect to gene A?’’

Each of your friends has tried to solve the problem in a different way:

Friend 1's Method:
# A/A individuals = 20
# A/a individuals = 20
# a/a individuals = 160
Total # of individuals = 200

Observed allele frequencies
Observed freq(A) = p = (2×20 + 20) ÷ (2×200) = 60 ÷ 400 = 0.150
Observed freq(a) = q = (2×160 + 20) ÷ (2×200) = 340 ÷ 400 = 0.850

Predicted genotype freq based on observed allele freq
If pop' is in H-W eq:
Predicted freq(A/A) = p² = 0.150² = 0.02
Predicted freq(A/a) = 2pq = 2×0.150×0.850 = 0.26
Predicted freq(a/a) = q² = 0.850² = 0.72

Observed genotype frequencies
Based on data provided in the question:
Observed freq(A/A) = 20/200 = 0.10
Observed freq(A/a) = 20/200 = 0.10
Observed freq(a/a) = 160/200 = 0.80

Observed genotype frequencies do not match the genotype frequencies that are predicted under H-W equilibrium; therefore, the population is not in Hardy-Weinberg equilibrium.

Friend 2's Method:
# A/A individuals = 20
# A/a individuals = 20
# a/a individuals = 160
Total # of individuals = 200

If pop' is in H-W eq: Freq(A/A) = p²
Therefore, assume observed freq(A/A) = p² = 20/200 = 0.100

Predicted allele frequencies
If pop' is in H-W eq, p² = 0.100
Predicted freq(A) = p = ?(p²) = ?(0.10) = 0.316
Predicted freq(a) = q = 1 - p = 1 - 0.32 = 0.684

Predicted genotype freq based on predicted allele freq
If pop' is in H-W eq:
Predicted freq(A/A) = p² = 0.316² = 0.10
Predicted freq(A/a) = 2pq = 2×0.316×0.684 = 0.43
Predicted freq(a/a) = q² = 0.684² = 0.47

Observed genotype frequencies
Based on data provided in the question:
Observed freq(A/A) = 20/200 = 0.10
Observed freq(A/a) = 20/200 = 0.10
Observed freq(a/a) = 160/200 = 0.80

Observed genotype frequencies do not match the genotype frequencies that are predicted under H-W equilibrium; therefore, the population is not in Hardy-Weinberg equilibrium.

Which of your friends has solved the question using the correct method (or are both methods equally correct)? Please explain briefly. (3 marks)

3. You are helping two of your friends with their homework for another biology class. They are also learning about population genetics and Hardy-Weinberg equilibrium. This is the question they are working on:
‘‘There are 200 individuals in the population: 20 are homozygous dominant (A/A), 20 are heterozygous (A/a), and 160 are homozygous recessive (a/a). Is this population in Hardy-Weinberg equilibrium with respect to gene A?’’

Each of your friends has tried to solve the problem in a different way:

Friend 1's Method:
# A/A individuals = 20
# A/a individuals = 20
# a/a individuals = 160
Total # of individuals = 200

Observed allele frequencies
Observed freq(A) = p = (2×20 + 20) ÷ (2×200) = 60 ÷ 400 = 0.150
Observed freq(a) = q = (2×160 + 20) ÷ (2×200) = 340 ÷ 400 = 0.850

Predicted genotype freq based on observed allele freq
If pop' is in H-W eq:
Predicted freq(A/A) = p² = 0.150² = 0.02
Predicted freq(A/a) = 2pq = 2×0.150×0.850 = 0.26
Predicted freq(a/a) = q² = 0.850² = 0.72

Observed genotype frequencies
Based on data provided in the question:
Observed freq(A/A) = 20/200 = 0.10
Observed freq(A/a) = 20/200 = 0.10
Observed freq(a/a) = 160/200 = 0.80

Observed genotype frequencies do not match the genotype frequencies that are predicted under H-W equilibrium; therefore, the population is not in Hardy-Weinberg equilibrium.

Friend 2's Method:
# A/A individuals = 20
# A/a individuals = 20
# a/a individuals = 160
Total # of individuals = 200

If pop' is in H-W eq: Freq(A/A) = p²
Therefore, assume observed freq(A/A) = p² = 20/200 = 0.100

Predicted allele frequencies
If pop' is in H-W eq, p² = 0.100
Predicted freq(A) = p = ?(p²) = ?(0.10) = 0.316
Predicted freq(a) = q = 1 - p = 1 - 0.32 = 0.684

Predicted genotype freq based on predicted allele freq
If pop' is in H-W eq:
Predicted freq(A/A) = p² = 0.316² = 0.10
Predicted freq(A/a) = 2pq = 2×0.316×0.684 = 0.43
Predicted freq(a/a) = q² = 0.684² = 0.47

Observed genotype frequencies
Based on data provided in the question:
Observed freq(A/A) = 20/200 = 0.10
Observed freq(A/a) = 20/200 = 0.10
Observed freq(a/a) = 160/200 = 0.80

Observed genotype frequencies do not match the genotype frequencies that are predicted under H-W equilibrium; therefore, the population is not in Hardy-Weinberg equilibrium.

Which of your friends has solved the question using the correct method (or are both methods equally correct)? Please explain briefly. (3 marks)

Added by Mariano C.

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