Discussion
In a trihybrid cross (AaBbCc × AaBbCc). What is the probability of obtaining an offspring with all recessive genotypes (aabbcc) ?
Answer: A
The probability is 1/64.
The three genes assort independently, so the probability for each gene can be worked out separately and multiplied. In \(Aa \times Aa\) the chance of \(aa\) is 1/4, and likewise for \(bb\) and \(cc\):
\[ \begin{aligned} P(aabbcc) &= \tfrac{1}{4} \times \tfrac{1}{4} \times \tfrac{1}{4} \\ &= \tfrac{1}{64} \end{aligned} \]
Equivalently, each parent makes \(2^{3} = 8\) kinds of gamete, giving \(8 \times 8 = 64\) combinations, of which only one (abc from each parent) is fully recessive.
1/16 is the answer for a dihybrid cross, and 1/32 and 1/128 come from mixing up the powers of 2 and 4.
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