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(B) The cross between $AA BB CC$ and $aa bb cc$ produces an $F_1$ progeny with the genotype $Aa Bb Cc$.To find the number of different types of gamete

Vedclass · Accepted Answer

$8$

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(B) The cross between $AA BB CC$ and $aa bb cc$ produces an $F_1$ progeny with the genotype $Aa Bb Cc$.To find the number of different types of gametes produced by an individual with $n$ heterozygous gene pairs,we use the formula $2^n$.In the genotype $Aa Bb Cc$,there are $3$ heterozygous gene pairs $(n = 3)$.Therefore,the number of different types of gametes = $2^3 = 2 	imes 2 	imes 2 = 8$.The possible gametes are $ABC, ABc, AbC, Abc, aBC, aBc, abC, abc$.

How many different types of gametes can be formed by $F_1$ progeny,resulting from the following cross: $AA BB CC \times aa bb cc$?

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