Question

In a population of 1000 people in Canada, the frequencies of MN blood types are 100 MM, 740 MN, and 160 NN.

a. What is the frequency of each allele? Show your calculations.

a. Is this population in Hardy-Weinberg equilibrium? Explain your answer and show your calculations.

In a population of 1000 people in Canada, the frequencies of MN blood types are 100 MM, 740 MN, and 160 NN. a. What is the frequency of each allele? Show your calculations. a. Is this population in Hardy-Weinberg equilibrium? Explain your answer and show your calculations.

Answer 1

a. Frequency of M allele = f(MM) + 1/2 f(MN)

In MM genotype individuals, there are two M alleles present so it is = 100

In MN genotype individuals, there is one M allele present so it is = 1/2 * 740 = 370

So, it is 100 + 370 = 470.

Similarly, for N allele, Frequency = f(NN) + 1/2 f(MN)

In NN genotype individuals, there are two N alleles present so it is = 160

In MN genotype individuals, there is one N allele present so it is = 1/2 * 740 = 370

So, it is 160 + 370 = 530.

b. For following the hardy-weinberg principle, it should be able to prove the equation p² + q² + 2pq = 1

where, p² = frequency of MM genotype.

q² = frequency of NN genotype.

2pq = frequency of MN genotype.

So, 100 + 740 + 160 = 1000 (no. of individuals present)

or it can be stated as 0.1 + 0.74 + 0.16 = 1

This indicates that the population is in hardy-weinberg equilibrium. According to this, both the genotype and phenotypic frequencies in a population remain in equilibrium unless they are changed by external forces. It defines a state of non-evolution of a population.