Crossing AABB & aabb, the probability of AaBB would be in `F_(2)` generation :-

Options

A. `1//16`
B. `2//16`
C. `8//16`
D. `4//16`

Answer

To solve the problem of determining the probability of obtaining the genotype AaBB in the F2 generation from a cross between AABB and aabb, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parent Genotypes**:
- The parent genotypes are AABB (homozygous dominant) and aabb (homozygous recessive).

2. **Determine the F1 Generation**:
- When we cross AABB with aabb, all offspring in the F1 generation will be heterozygous for both traits. Thus, the F1 generation will be AaBb.

3. **Determine the Gametes from F1 Generation**:
- The gametes produced by the F1 generation (AaBb) will be AB and ab.

4. **Cross the F1 Generation to Obtain F2 Generation**:
- To find the F2 generation, we will perform a dihybrid cross of AaBb x AaBb.
- The possible gametes from each parent are AB, Ab, aB, and ab.

5. **Set Up a Punnett Square**:
- Create a 4x4 Punnett square with the gametes from both parents:
```
AB Ab aB ab
---------------------
AB | AABB AABb AaBB AaBb
Ab | AABb AAbb AaBb Aabb
aB | AaBB AaBb aaBB aabB
ab | AaBb Aabb aabB aabb
```

6. **Count the Desired Genotype (AaBB)**:
- From the Punnett square, we can see that the genotype AaBB appears twice:
- In the first row: AaBB
- In the third row: AaBB
- Therefore, there are 2 occurrences of AaBB out of a total of 16 possible genotypes.

7. **Calculate the Probability**:
- The probability of obtaining the genotype AaBB in the F2 generation is:
\[ \text{Probability} = \frac{\text{Number of AaBB}}{\text{Total Offspring}} = \frac{2}{16} = \frac{1}{8} \]

### Final Answer:
The probability of obtaining the genotype AaBB in the F2 generation is **2 out of 16** or simplified as **1 out of 8**.