Question

in the cross aabbcc × aabbcc, what is the probability of producing the genotype aabbcc? 1/4 1/8 1/16 1/64

Explanation:

Step1: Calculate probability for AA

For the \(Aa\times Aa\) cross, the probability of getting \(AA\) is \(\frac{1}{4}\) (using Punnett square: \(AA:Aa:Aa:aa = 1:2:1\)).

Step2: Calculate probability for BB

For the \(Bb\times Bb\) cross, the probability of getting \(BB\) is \(\frac{1}{4}\) (Punnett square: \(BB:Bb:Bb:bb=1:2:1\)).

Step3: Calculate probability for CC

For the \(Cc\times Cc\) cross, the probability of getting \(CC\) is \(\frac{1}{4}\) (Punnett square: \(CC:Cc:Cc:cc = 1:2:1\)).

Step4: Use multiplication rule

Since these are independent events (Mendel's law of independent assortment), the probability of \(AABBCC\) is \(P(AA)\times P(BB)\times P(CC)\). So, \(\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}=\frac{1}{64}\).

Answer:

\(\frac{1}{64}\) (the fourth option)