How many different types of gametes can be formed by $F_{1}$ progeny, resulting from the following cross?
AABBCC x aabbcc

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Solution

The correct option is A 8
In a cross between AABBCC and aabbcc, the progeny obtained in $F_{1}$ generation is AaBbCc only. This produce eight different types of gametes.
For different type of gametes in $F_{1}$ generation = $2^{n}$ (n = number of traits),
In AaBbCc, n = 3
So, $2^{3}$ = 8
Hence, the correct answer is option A, i.e., 8.

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