Alright, let's solve the problem of finding the fraction of offspring that will be homozygous recessive when two heterozygous parents are crossed in a dihybrid cross.
Here's a step-by-step explanation:
1. Understand the Genotypes:
- In a dihybrid cross, we are dealing with two traits. Let's denote them as A/a and B/b.
- Each parent is heterozygous for both traits, meaning their genotype is AaBb.
2. Determine the Probability for Each Trait:
- The probability of an offspring being homozygous recessive for the first trait (aa) can be found using a Punnett square. Each heterozygous parent (Aa) can produce gametes A and a.
- The possible combinations of those gametes for one trait (A and a) are: AA, Aa, Aa, and aa.
- Therefore, the probability of getting aa is 1 out of 4 gametes (1/4).
3. Repeat the Process for the Second Trait:
- Similarly, for the second trait (B and b), where each parent is Bb, the probability of getting the combination bb is also 1 out of 4 (1/4).
4. Combining the Probabilities:
- Because the traits are independent (one trait does not affect the other), we multiply the probabilities of each independent event.
- The combined probability of an offspring being homozygous recessive for both traits (aabb) is calculated by multiplying the probabilities:
[tex]\( \text{Probability of aabb} = \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) = \frac{1}{16} \)[/tex].
5. Conclusion:
- The fraction of offspring that will be homozygous recessive (aabb) when two heterozygous (AaBb) parents are crossed is:
[tex]\( \frac{1}{16} \)[/tex].
Therefore, the correct answer is 1/16.
