Answer: E. none of the above
Explanation:
given data:
AabbDd x AaBbDd
For gene A, there will be cross of Aa X Aa. The probability of having a phenotype, that is either AA or aa is 2/4 = 1/2
For gene B, there will be cross of Bb X BB. The probability of having a phenotype, that is BB is 1/4
For gene D, there will be cross of Dd X Dd The probability of having a phenotype, that is either DD or dd is 2/4 = 1/2.
Hence, the probability of the off spring having all the genes.
= 1/2 * 1/4 * 1/2
= 1/16