MarksRiser
MarksRiser
Back to Practice
Zoology-Hardy-Weinberg Principle

NEET Zoology Hardy-Weinberg Principle MCQ Question

Type: MCQ-numerical-Hard-Class 12

In a population at Hardy-Weinberg equilibrium, the frequency of allele A is 0.6. What is the expected frequency of the heterozygous genotype Aa?

A

0.24

B

0.36

C

0.48

D

0.64

Correct Answer

Option C

Detailed Explanation

Using the Hardy-Weinberg equation p^2 + 2pq + q^2 = 1, where p is the frequency of allele A, q is the frequency of allele a, and 2pq is the frequency of the heterozygous genotype Aa. Here, p = 0.6, hence q = 1 - 0.6 = 0.4. Therefore, 2pq = 2 * 0.6 * 0.4 = 0.48.

Found an issue with this question?

Related Questions

Explore More Questions

All Zoology QuestionsEvolution Questions