Answer:
The correct answer is - zero.
Explanation:
In the given heterozygous mouse with the genotype of AaBbCcDd, there are 4 different genes are present with there recessive forms A, B, C, and D and a, b, c, and d. In the heterozygous condition, there is always only one copy of both variants of a gene present.
By independent assortment and law of segregation, there are multiple gametes can be formed but as there is only one copy of each gene form or allele, the chances of the having two copies any allele of particular gene such as AA, BB, CC, DD, or aa, bb, cc, dd together is not possible.
Thus, the correct answer is - zero.