Question

There are 23,832 people total in a population:

5,504 people have blood type A

5,072 people have blood type B

647 people have blood type AB

12,609 people have blood type O

Determine whether or not this population is in Hardy-Weinberg equilibrium. Please SHOW ALL of the calculations.

Answer 1

Firstly in this population I am assuuming Hardy-Weinberg equilibrium is maintaining.

Human blood group is controlled by three alleles A, B and O. A is dominant over O, B is dominant over O, A and B show codominance and O is completely recessive.

So, A blood group can have genotype AA or AO, B blood group can have genotype BB or BO, AB blood group and O blood group will have genotype AB and OO.

Now, lets take allele frequency of A, B and O are p, q and r

So, the genotypic frequency will be,

P² + q² +r² + 2pq + 2pr + 2qr = 1.

Now,

p² + 2pr = A genotype frequency,

q² + 2qr = B genotype frequency

2pq = AB genotype frequency, and

r² = O genotype frequency.

Given,

23,832 people total in a population:

5,504 people have blood type A

5,072 people have blood type B

647 people have blood type AB

12,609 people have blood type O

Now,

i) r² = 12609/23832 = 0.529

or, r = 0.73

ii) p² + 2pr = 5504/ 23832 = 0.23

or, p² + 1.46p - 0.23 = 0.

Solving the equation, p = 0.143 ,

iii) 2pq = 647/ 23832 = 0.027,

or, 0.286q = 0.027,

or, q = 0.094.

So, the value of p + q + r = 0.143 + 0.094 + 0.73 = 0.967.

Expected value of p + q + r = 1. As the observed value is close to 1, it is in this population is in Hardy-Weinberg equilibrium.