Question

As per Hardy-Weinberg Law, in a random mating population what is the frequency of \(AA\) genotypes from a cross of \(Aa(2pq)\times Aa(2pq)\)?

• A. \(2p^2q^2\)
• B. \(2p^3q\)
• C. \(p^2q^2\)
• D. \(2pq^3\)

Hint:

For \(Aa\times Aa\), the fraction of \(AA\) progeny is always \(\frac{1}{4}\).

Answer 1

The Correct Option is C

Concept:
According to Hardy-Weinberg equilibrium: \[ AA=p^2,\quad Aa=2pq,\quad aa=q^2 \]

Step 1: Given cross. \[ Aa(2pq)\times Aa(2pq) \] The frequency of this mating is: \[ (2pq)(2pq)=4p^2q^2 \]

Step 2: Genotypic ratio of \(Aa\times Aa\). \[ Aa\times Aa \Rightarrow AA:Aa:aa=1:2:1 \] So, the fraction of \(AA\) progeny from this cross is: \[ \frac{1}{4} \]

Step 3: Frequency of \(AA\) from this mating. \[ \text{Frequency of }AA=4p^2q^2\times \frac{1}{4} \] \[ =p^2q^2 \] \[ \therefore \text{Correct Answer is (C)} \]