Question

In humans, MN blood types are determined by an autosomal gene with two alleles: LM and LN. In one population of 592 people, there are 148 individuals with blood type N. Assuming the population is in Hardy-Weinberg equilibrium, how many individuals should have the blood type M

Answer 1

148

Step-by-step explanation:

According to Hardy-Weinberg equilibrium,

p + q = 1

p² + 2pq + q² = 1 where,

p = frequency of dominant allele

q = frequency of recessive allele

p² = frequency of homozygous dominant genotype

2pq = frequency of heterozygous genotype

q² = frequency of homozygous recessive genotype

Here,

Total population = 592

Number of NN people = 148

Frequency of N blood group or NN genotype (q²) = 148/592 = 0.25

q = √0.25 = 0.5

p = 1 - q

= 1 - 0.5 = 0.5

Hence, p = 0.5

Frequency of MM genotype = p² = 0.25

Number of people with MM genotype = 0.25*592 = 148

Hence, 148 people will have MM genotype or M blood group.