**N**

**individuals of which**

**D**is dominant allele (A

^{1}A

^{2}),

**H**is heterozygous

**(A**and

^{1}A^{2})**R**is recessive allele (A

^{2}A

^{2}).

Then

**N=D+H+R****Each A**

^{1}A^{1}

**genotype has two A alleles.**

**Heterozygotes (A**

^{1}A^{2}) have only one A^{1}

**allele.**

Let

**‘p’**

**represents the**

**frequency of the A allele**

**and**

**‘q’**

**represents**

**the frequency of the a allele**

**Then**

**p= [**2D+H]

**/**2N

**= [D+1/2H]/N**

**q= [**2R+H]

**/**2N

**= [R+1/2H]/N**

**Frequency of gene = Frequency of homozygote for that gene + 1/2 frequency of heterozygotes.**

**In human MN blood groups, we sample a population of 100 individuals with 50MM, 20MN, and 30NN, frequencies of 'M' and 'N' can be calculated by using above equation**

**Frequency of gene = Frequency of homozygote for that gene + 1/2 frequency of heterozygotes**

**Frequency of M = Frequency of homozygote for that gene( MM) + 1/2 frequency of heterozygotes (MN)**

Frequency of M = 50/100 MM + 1/2 (20/100)MN

Frequency of M = 0.5 MM + 1/2 (0.2) MN

Frequency of M = 0.5+0.1

**Frequency of M**

**= 0.6**

Then frequency of N = 1- M

frequency of N = 1 - 0.6

**Frequency of**

**N=0.4**

**or**

**otherwise**

Frequency of N = 30/100 NN + 1/2 (20/100)MN

Frequency of N = 0.3 NN + 1/2 (0.2) MN

Frequency of N = 0.3 + 0.1

**Frequency of N**

**= 0.4**

**or**

**other wise**

Then

**frequency of 'M**' will be (50 x 2) + 20 =

**120**, in a similar manner the f

**requency of N**will be (30 x 2 ) + 20 =

**80**.

**The relative frequencies of M allele**can be worked as

**M / (M+N)**

**and that of**

**N allele**can be worked out as

**N / (M+N).**

**M =**

**M / (M+N)**

M = 120 / (120 +80)

M = 120 / 200

**M = 0.6**

**N =N / (M+N)**

N= 80 / 120 + 80

N= 80 / 200

**N = 0.4**

**Learn more : MCQ on Population Genetics - Hardy Weinberg Equilibrium**