​In a population, the frequency of allele 'a' is 0.2 and that of allele 'b' is 0.1. Consider that there are two alleles for each of the genes. What would be the expected percentage of population with genotype AaBb, considering that the population is under Hardy-Weinberg equilibrium?​

Correct option is D

The correct answer is (d)

Under Hardy-Weinberg equilibrium, the genotype frequencies can be calculated based on allele frequencies for each gene independently.

Frequency of allele a = 0.2
So, frequency of allele A = 1 - 0.2 = 0.8

Frequency of allele b = 0.1
So, frequency of allele B = 1 - 0.1 = 0.9

For a gene with two alleles (A and a), the frequency of heterozygous genotype Aa is:

$$2pq = 2 \times (frequency\ of\ A) \times (frequency\ of\ a) = 2 \times 0.8 \times 0.2 = 0.32$$

Similarly, for the other gene (B and b), the frequency of heterozygous genotype Bb is:

$$2pq=2×(frequency of B)×(frequency of b)=2×0.9×0.1=0.18$$

Since the two genes assort independently, the frequency of the combined genotype AaBb is the product of the frequencies of Aa and Bb:​

$$Frequency of AaBb=0.32×0.18=0.0576$$

​​To express this as a percentage:

$$0.0576×100=5.76$$​​
However, since the options do not have 5.76%, but have 57.6%, we need to verify whether there is a decimal place confusion.
Looking at the problem carefully, the calculation is correct for 5.76%. The option 4 is 57.6%, which seems to be a decimal error.
Since none of the options matches exactly 5.76%, it is possible the option 4 intends 5.76% but is incorrectly written.
If option 4 is 5.76%, then option 4 is correct.
If not, we must look for a closest match.
Since 5.76% is close to 57.6% divided by 10

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