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ABO blood type is examined in a Taiwanese population, nad the allele frequencies are determined. In the population, f(IA)=0.3, f(IB)=0.15 and f(i)=0.55.
Part A: Assuming Hardy-Weinberg conditions apply, what is the frequency of genotype IAIA? Enter your answer to two decimal places.
Part B: Assuming Hardy-Weinberg conditions apply, what is the frequency of genotype IBIB? Enter your answer to four decimal places.
Part C: Assuming Hardy-Weinberg conditions apply, what is the frequency of genotype IAIB? Enter your answer to two decimal places.
Part D: Assuming Hardy-Weinberg conditions apply, what is the frequency of genotype IAi? Enter your answer to two decimal places.
Part E: Assuming Hardy-Weinberg conditions apply, what is the frequency of genotype IBi? Enter your answer to three decimal places.
Part F: Assuming Hardy-Weinberg conditions apply, what is the frequency of genotype ii? Enter your answer to four decimal places.
Part G: What is the frequency of type A blood in this population? Enter your answer to two decimal places.
Part H: What is the frequency of type B blood in this population? Enter your answer to four decimal places.
Part I: What is the frequency of type AB blood in this population? Enter your answer to two decimal places.
Part J: What is the frequency of type O blood in this population? Enter your answer to four decimal places.
Answer: Part A: f(IAIA) = 0.09
Part B: f(IBIB) = 0.0225
Part C: f(IAIB) = 0.05
Part D: f(IAi) = 0.16
Part E: f(IBi) = 0.082
Part F: f(ii) = 0.3025
Part G: f(type A) = 0.25
Part H: f(type B) = 0.1050
Part I: f(type AB) = 0.05
Part J: f(type O) = 0.3025
Explanation: A population in Hardy-Weinberg equilibrium is a population with a constant value in the population's allele frequency.
In the population of Taiwan, frequencies are:
A: Frequency for genotype IAIA: Frequency of allele A is 0.3, then for genotype IAIA:
f(IAIA) = 0.3²
f(IAIA) = 0.09
B: Frequency of allele B is 0.15. Then, for genotype IBIB:
f(IBIB) = 0.15²
f(IBIB) = 0.0225
C: Frequency for genotype IAIB:
f(IAIB) = 0.3*0.15
f(IAIB) = 0.05
D: Frequency for genotype IAi:
f(IAi) = 0.3*0.55
f(IAi) = 0.16
E: frequency for genotype IBi:
f(IBi) = 0.15*0.55
f(IBi) = 0.082
F: frequency for allele i is 0.55. Then, for genotype ii:
f(ii) = 0.55²
f(ii) = 0.3025
Individuals with type A blood can be IAIA or IAi. Using probability's "OR" rule:
f(type A) = 0.09 + 0.165
f(type A) = 0.25
The same happens to individual with type B blood:
f(type B) = 0.0225 + 0.0825
f(type B) = 0.105
Since blood type is a codominance, i.e., allele A and allele B are expressed, frequency of type AB is equal to frequency of genotype AB:
f(type AB) = 0.3*0.15
f(type AB) = 0.05
Frequency for type O is:
f(type O) = 0.55²
f(type O) = 0.3025