Question 1 You collect 500 buccal swabs from a human population and use the tiss
ID: 220351 • Letter: Q
Question
Question 1
You collect 500 buccal swabs from a human population and use the tissue to isolate DNA from each individual you sampled. You then genotype each individual at a single nucleotide polymorphism (SNP) that you presume to be selectively neutral and obtain the following results.
A1A1 = 300
A1A2 = 50
A2A2 = 150
What is the estimated frequency of A1A2 heterozygotes in this population? Round your answer to two decimal places.
Question 2
Using the data from the previous question, calculate the frequency of A1. Round your answer to two decimal places.
Question 3
Using the data from the first question, calculate the frequency of A2A2 homozygotes expected at Hardy-Weinberg equilibrium. Round your answer to two decimal places.
Question 4
Using the data from the first question, calculate the Hardy-Weinberg expected heterozygosity. Round your answer to two decimal places.
Question 5
Using the data from the first question, calculate the inbreeding coefficient for the population you have sampled. Round your answer to two decimal places.
Explanation / Answer
A1A1 = 300; A1A2 = 50; A2A2 = 150. (500 Sample)
Allele: A1 = 600; A1 = 50, A2 = 50; A2 = 300. ===> Total Allele = 1000
A1 = 600 + 50 = 650,
A1 = 650/ 1000 = 0.65 (Frequency of A1 Allele = 0.65)
A2 = 300 + 50 = 350
A2 = 350/ 1000 = 0.35 (Frequency of A2 Allele= 0.35)
p + q = 1
0.65 + 0.35 = 1 (Hence proved)
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p = 0.65
p2 = 0.4225 (Frequency of A1A1 homozygotes= 0.4225)
q = 0.35
q2 = 0.1225 (Frequency of A2A2 homozygotes= 0.1225)
2pq = 2 * 0.65 * 0.35 = 0.455 (Frequency of A1A2 heterozygotes= 0.455)
P2 + 2PQ + Q2 = 1
0.4225 + 0.1225 + 0.455 = 1 (Hence proved)
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Expected Heterozygote:
2pq = 2 * 0.65 * 0.35 = 0.455
0.455 * 500 = 227.5
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Inbreeding coefficient = # Expected Heterozygotes - # Observed Heterozygotes
# Expected Heterozygotes
= 227.5 - 50 / 227.5
= 0.7802
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