For the following cross AaBBCcDd x AaBbCcdd what is the prob

For the following cross: AaBBCcDd x AaBbCcdd,

what is the probability that any individual offspring will have the dominant phenotype?

I did 4 separate punnet squares and then multiplied the ratios of pairs with a dominant allele for each respective cross which equalled,

A= 3/4 B=1 C=3/4 D=1/2, multiplying this equalled 9/32, correct?

Please explain reasoning, thanks in advance.

Solution

Answer:

Yes you are right.

Probability of AA or Aa = ¼ + ½ =¾
Probability of BB or Bb = 1/2 + ½ =1
Probability of CC or Cc = ¼ + ½ =¾
Probability of DD or Dd = 0 + ½ =½

Total probability = ¾ x 1 x ¾ x ½ =9/32