Question

If we cross AaBBCcDdeeFf ×AABBccDdEeFf . Then what will be the chance of genotype AaBBccddEeFf ?

Answer 1

You can do punnet squares for each individual gene, the multiply the probabilities.

So for AA x Aa, you have 1/2 AA and 1/2 Aa. So probability of getting a child with Aa is 0.5.
For Bb x BB, similar Punnet square for AA x Aa, so probability for BB is 0.5.
cc x CC, all offspring will be Cc. Probability for Cc is 1.
Dd x Dd: 1/2 Dd, 1/4 DD, and 1/4 dd. Probability for dd is 0.25.
Ee x Ee: similar as Dd x Dd. Probablity for EE is 1/4 or 0.25.
FF x ff : similar to cc x CC. All offspring will be Ff so probability is 1.

Then multiply:
1/2* 1/2 * 1 * 1/4 * 1/4 *1 = 1/64

so 1/64