Human blood type is determined by three alleles IA, IB, and IO. The alleles IA and IB are co-dominant to each other, and both are dominant to IO. Within a large, randomly mating population (540,000 individuals), the frequencies for the blood type alleles are 0.3 for the IA allele, 0.6 for IO, and 0.1 for the IB allele.
a. Calculate the expected numbers of people in the population having each blood type A, B, AB, and O.
b. Determine the percentage of type B people that are heterozygotes (IBIO).

Explanation:

According to p²+q²+2pq=1 (Hardy-Weinberg equilibrium)

heterozygote allele will be always 2pq

1. Population

1. A blood group has two genotypes- IAIA and IAIO

The genotype frequency will be-

IAIA= 0.3 X 0.3 = 0.9

IAIO= (0.3 x 0.6) x 2= 0.36

Total genotype frequency = 0.9+ 0.36 =0.45 or 45%

Now 45% of 540,000= 2,43000 (A blood group)

2. B blood group has two genotypes- IAIB and IBIO

The genotype frequency will be-

IBIB= 0.1 x 0.1 = 0.1

IBIO= (0.1 x 0.6)-=0.12

Total genotype frequncy= 0.13 or 13%

popluation with B blood group=  13% of 540,000 will be 70,200

3. AB blood group is IAIB

2 x IAIB= (0.3 x 0.1) x 2= 0.6

popluation with B blood group= 6% of 540,000= 32,400

4. O blood group

2. Heterozygote

IBIO blood group

= (0.1 x 0.6) x 2  

= 0.12 or 12%

= 12%

answer: dna

because all cells have genes