Question

In the F2 generation of a homozygous round (AA) × homozygous wrinkled (aa) cross in peas, two round seeds are chosen at random. What is the probability that one is AA and the other Aa?

(1/3)2

(2/3)2

(2/3)(1/3)

2(2/3)(1/3)

Answer 1

All will be Aa so the probability of it being Aa is 1 and the probability of it being AA is 0

Step-by-step explanation:

When crossing with a punnett square you can see that all the seeds are Aa, and probability is the required outcome divided by total outcome, when this is done, it wil be seen that the probability for Aa is 1 and that of AA is 0