How many different types of gametes can be formed by F $_{1}$ progeny, resulting from the following cross : AABBCC $\times$ aabbcc

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Solution

The correct option is B 8
AABBCC x aabbcc has three forms of alleles.
n = 3. The number of gametes produced by this genotype is calculated by 2 $^{n}$ , i.e., 2 $^{3}$ = 8
Thus, maximum number of gametes = 8.

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