On selfing a plant of `F_(1)`-generation with genotype "AaBbCC", the phenotypic ratio in `F_(2)`-generation will be

Options

A. `3:1`
B. `1:1`
C. `9:3:3:1`
D. `27:9:9:9:3:3:3:1`

Answer

To solve the problem of determining the phenotypic ratio in the F2 generation from a selfing of a plant with the genotype "AaBbCC", we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Genotype of the F1 Generation**:
The genotype given is AaBbCC. Here, A and B are heterozygous (Aa and Bb), while C is homozygous dominant (CC).

2. **Determine the Types of Gametes Produced**:
Since C is homozygous (CC), it will not contribute to variation in the gametes. Therefore, we only need to consider the alleles A and B.
- The possible gametes from Aa are: A and a.
- The possible gametes from Bb are: B and b.
- Since C is constant (CC), we can ignore it for the gamete combinations.

Thus, the gametes produced will be:
- AB
- Ab
- aB
- ab

3. **Perform a Dihybrid Cross**:
We can now perform a dihybrid cross using the gametes:
- The gametes from one parent: AB, Ab, aB, ab
- The gametes from the other parent (selfing the same plant): AB, Ab, aB, ab

This results in a 4x4 Punnett square, giving us 16 combinations.

4. **Count the Phenotypes**:
Since C is homozygous, it does not affect the phenotypic ratio. We can focus on the combinations of A and B:
- The phenotypic ratio for a typical dihybrid cross (AaBb x AaBb) is:
- 9 (A-B-): Dominant for both traits
- 3 (A-bb): Dominant for A and recessive for B
- 3 (aaB-): Recessive for A and dominant for B
- 1 (aabb): Recessive for both traits

5. **Final Phenotypic Ratio**:
Therefore, the phenotypic ratio in the F2 generation will be:
- **9 : 3 : 3 : 1**.