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 finding the probability of obtaining the genotype AaBB in the F2 generation from the cross AABB x aabb, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Parent Genotypes**:
- The first parent is AABB (homozygous dominant for both traits).
- The second parent is aabb (homozygous recessive for both traits).

2. **Determine Gametes from Parents**:
- The gametes produced by AABB will only be AB (since both alleles are dominant).
- The gametes produced by aabb will only be ab (since both alleles are recessive).

3. **F1 Generation**:
- When we cross AABB with aabb, all offspring in the F1 generation will be AaBb (heterozygous for both traits).
- The genotype of the F1 generation is AaBb.

4. **F2 Generation**:
- To find the F2 generation, we need to self-cross the F1 generation (AaBb x AaBb).
- The gametes produced by AaBb will be: AB, Ab, aB, ab.

5. **Punnett Square for F2 Generation**:
- Set up a 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 Genotypes**:
- From the Punnett square, we can count the occurrences of each genotype:
- AABB: 1
- AABb: 2
- AaBB: 2
- AaBb: 4
- AAbb: 1
- Aabb: 2
- aaBB: 1
- aaBb: 2
- aabb: 1

7. **Total Combinations**:
- There are a total of 16 combinations in the F2 generation.

8. **Calculate Probability of AaBB**:
- The genotype AaBB appears 2 times in the 16 combinations.
- Therefore, the probability of obtaining AaBB is \( \frac{2}{16} \) or \( \frac{1}{8} \).

### Final Answer:
The probability of obtaining the genotype AaBB in the F2 generation is \( \frac{2}{16} \) or \( \frac{1}{8} \).