​In a population with the ABO blood group system, if the frequency of the allele Iᴬ is 0.3, the frequency of the allele Iᴮ is 0.2, and the frequency of the allele i is 0.5, what would be the expected percentage of population with blood group A, considering that the population is under Hardy-Weinberg equilibrium?​

Correct option is C

To calculate the expected percentage of individuals with blood group A, we need to apply the Hardy-Weinberg equilibrium principle.

• The genotypes for blood group A are I^AI​^A and I^Ai

• The frequency of I^A is given as 0.3, I^B is 0.2, and iii is 0.5.

Using Hardy-Weinberg, the frequency of the genotypes is calculated as follows:

• The frequency of I^AI^A (homozygous A) is $(0.3)$^$2$$=0.09$.

• The frequency of I^A​i (heterozygous A) is $2\times (0.3)\times (0.5)=0.30$.

Thus, the total frequency of individuals with blood group A is the sum of these two genotypes:

$$\text{Frequency of blood group A}=0.09+0.30=\mathrm{0.39.}$$

Therefore, the expected percentage of the population with blood group A is 39%.