In the `F_(2)` generation a Mendelian dihybrid cross, the number of phenotypes and genotypes are

Options

A. phenotypes-2 , genotypes -16
B. phenotypes-9 , genotypes -4
C. phenotypes-4 , genotypes -8
D. phenotypes-4 , genotypes -9

Explanation

To solve the question regarding the number of phenotypes and genotypes in the F2 generation of a Mendelian dihybrid cross, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Dihybrid Cross**:
A dihybrid cross involves two traits, each controlled by different genes. For example, consider traits such as seed shape (round vs. wrinkled) and seed color (yellow vs. green).

2. **Identifying the Number of Traits (n)**:
In a dihybrid cross, we are studying two traits. Therefore, \( n = 2 \).

3. **Calculating the Number of Phenotypes**:
The formula to calculate the number of phenotypes is:
\[ \text{Number of phenotypes} = 2^n \]
Substituting \( n = 2 \):
\[ \text{Number of phenotypes} = 2^2 = 4 \]

4. **Calculating the Number of Genotypes**:
The formula to calculate the number of genotypes is:
\[ \text{Number of genotypes} = 3^n \]
Substituting \( n = 2 \):
\[ \text{Number of genotypes} = 3^2 = 9 \]

5. **Conclusion**:
In the F2 generation of a Mendelian dihybrid cross, there are 4 phenotypes and 9 genotypes.

### Final Answer:
- **Number of Phenotypes**: 4
- **Number of Genotypes**: 9