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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

Step-by-step 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

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