Question

In a dihybrid cross (\(AaBb \times AaBb\)), what is the probability of obtaining an individual heterozygous for both traits (\(AaBb\))?

• A. \(\dfrac{1}{16}\)
• B. \(\dfrac{1}{8}\)
• C. \(\dfrac{1}{4}\)
• D. \(\dfrac{1}{2}\)

Hint:

For independent genes, \[ \boxed{ P(AaBb)=P(Aa)\times P(Bb). } \] Since \[ Aa\times Aa \Rightarrow P(Aa)=\frac12, \] and \[ Bb\times Bb \Rightarrow P(Bb)=\frac12, \] therefore, \[ \boxed{ P(AaBb)=\frac14. } \]

Answer 1

The Correct Option is C

Step 1: Consider each gene separately. For the cross \[ Aa \times Aa, \] the probability of obtaining a heterozygous genotype is \[ P(Aa)=\frac{1}{2}. \] Similarly, for \[ Bb \times Bb, \] the probability of obtaining \[ Bb \] is \[ P(Bb)=\frac{1}{2}. \]

Step 2: Apply the multiplication rule. Since the two genes assort independently, \[ P(AaBb) = P(Aa)\times P(Bb) = \frac12\times\frac12 = \frac14. \] Hence, \[ \boxed{\frac14} \] is the required probability. Therefore, \[ \boxed{(C)} \] is the correct answer.