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

148

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.

white blood cells or wbc s are mainly 2 types-

1.agranulocytes

2.granulocytes

the monocytes or macrophages of agranulocytes , dendritic cells which are accessory leucocytes, and b cells which is one type of lymphocytes   identify the antigens present on the pathogen's surface. these three cells are called antigen presenting cells(apc) which are phagocytes and destroy the antigens by secreting antitoxins. besides this leukocyte is always plays vital role in the immunity of the body.