Genetics Problems - Population Genetics - Calculating Gene Frequency

Calculating Gene Frequency

If in a given sample of N individuals of which D is dominant allele (A¹A²), H is heterozygous (A¹A²) and R is recessive allele (A² A²).

Then N=D+H+R

Each A¹A¹ genotype has two A alleles. Heterozygotes (A¹A²) have only one A¹ 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.

Population Genetics: Hardy Weinberg Law / How to solve Hardy Weinberg Problems

Problem: 1

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

Each individual will have two homologous chromosomes each carrying a particular allele, frequency of 'M' can be calculated by doubling the number of homologous 'M' blood type and adding to it the frequencies of heterozygous 'MN' blood type.
Then frequency of 'M' will be (50 x 2) + 20 = 120, in a similar manner the frequency 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