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

Options

A. `1:2:1`
B. `1:1`
C. `9:3:3:1`
D. `27:9:9:9:3:3:3:1`

Answer

To determine the genotypic 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 F1 Generation**:
The genotype given is "AABbCC". This means:
- "AA" is homozygous dominant for gene A.
- "Bb" is heterozygous for gene B.
- "CC" is homozygous dominant for gene C.

2. **Determine Gamete Formation**:
Since the plant is selfing, we need to determine the types of gametes produced by this genotype:
- For "AA", the only gamete produced is "A".
- For "Bb", the gametes produced are "B" and "b".
- For "CC", the only gamete produced is "C".
Thus, the possible gametes from this genotype are:
- ABC (from AABbCC)
- ABc (from AABbCC)
- AbC (from AABbCC)
- Abc (from AABbCC)

3. **Set Up a Punnett Square**:
We will set up a Punnett square to visualize the cross between the gametes:
- The gametes from one parent (AABbCC) will be on one side, and the gametes from the other parent (AABbCC) will be on the other side.
- The gametes are: ABC, ABc, AbC, Abc.

4. **Fill in the Punnett Square**:
The Punnett square will have 4 rows and 4 columns, where we will fill in the combinations of the gametes:
```
ABC ABc AbC Abc
-------------------------------------
ABC | AABBCc | AABBCc | AABbCC | AABbCc |
ABc | AABBCc | AABBCc | AABbCc | AABbCc |
AbC | AABbCC | AABbCc | AABbCC | AABbCc |
Abc | AABbCc | AABbCc | AABbCc | AABbCc |
```

5. **Count the Genotypes**:
Now we will count the different genotypes produced:
- AABBCc: 1
- AABbCC: 1
- AABbCc: 2
- AABbCc: 2

6. **Determine the Genotypic Ratio**:
The genotypic ratio can be summarized as:
- 1 AABBCc : 2 AABbCc : 1 AABbCC

Therefore, the genotypic ratio in the F2 generation is:
**1 : 2 : 1**.

### Final Answer:
The genotypic ratio in the F2 generation will be **1 : 2 : 1**.

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