How many types of gametes would be formed by the F $_{1}$ generation obtained by the cross between AABBCC and aabbcc?
[0.7 mark]

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Solution

The correct option is B 8
The cross can be represented as follows:

Total number of heterozygous in F $_{1}$ individuals (n) = 3.
Therefore, types of gametes produced by $F_{1} individuals = 2^{n} = 2^{3} = 8$ .

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