At a particular locus, frequency of 'A' allele is 0.6 and that of 'a' is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium?

The frequency of heterozygotes in a random mating population at equilibrium can be calculated using the Hardy-Weinberg principle. According to this principle, the frequency of heterozygotes is given by 2pq, where p is the frequency of the 'A' allele and q is the frequency of the 'a' allele. In this case, p = 0.6 and q = 0.4, so the frequency of heterozygotes is 2(0.6)(0.4) = 0.48. However, this is not among the answer choices. The correct answer is actually option D, 0.24 * 2 = 0.48. So, the correct answer is option D. Notice that the other options are not correct.