Find out the number of plants produced with genotype AabbCc out of 256 seeds collected from `F_(2)` progenines of a trihybrid cross -

Options

A. 32
B. 16
C. 12
D. 8

Answer

To solve the problem of finding the number of plants with the genotype AabbCc from 256 seeds collected from the F2 progeny of a trihybrid cross, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Genotype**: The genotype AabbCc consists of three traits:
- A: heterozygous (A/a)
- B: homozygous recessive (b/b)
- C: heterozygous (C/c)

2. **Determine the Probability for Each Trait**:
- For the trait A (A/a):
- The probability of getting A (homozygous dominant or heterozygous) is 1/2 (A/a).
- For the trait B (b/b):
- The probability of getting b/b (homozygous recessive) is 1/4.
- For the trait C (C/c):
- The probability of getting C (heterozygous) is 1/2.

3. **Calculate the Combined Probability**:
- Since these traits assort independently, we multiply the probabilities for each trait:
\[ P(AabbCc) = P(A) \times P(bb) \times P(C) = \left(\frac{1}{2}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{2}\right) \]
- This gives:
\[ P(AabbCc) = \frac{1}{2} \times \frac{1}{4} \times \frac{1}{2} = \frac{1}{16} \]

4. **Calculate the Number of Plants**:
- Out of 256 seeds, we can find the expected number of plants with the genotype AabbCc by multiplying the total number of seeds by the probability:
\[ \text{Number of plants} = 256 \times \frac{1}{16} = 16 \]

5. **Conclusion**:
- Therefore, the number of plants produced with the genotype AabbCc out of 256 seeds is **16**.

### Final Answer:
The number of plants with the genotype AabbCc is **16**.

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