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

Options

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

Answer

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 1: Understand the Dihybrid Cross
A dihybrid cross involves two traits, each with two alleles. For example, let's consider two traits: seed shape (round vs. wrinkled) and seed color (yellow vs. green). The round seed shape (R) is dominant over wrinkled (r), and yellow seed color (Y) is dominant over green (y).

### Step 2: Determine the Parental Generation (P)
In the parental generation, we can have the following genotypes:
- Round Yellow (RRYY) x Wrinkled Green (rryy)

### Step 3: Determine the F1 Generation
The F1 generation will all be heterozygous for both traits:
- All F1 offspring will be RrYy (Round Yellow).

### Step 4: Perform the F2 Generation Cross
When we cross two F1 individuals (RrYy x RrYy), we can use a Punnett square or the forked-line method to find the genotypes and phenotypes of the F2 generation.

### Step 5: Count the Phenotypes
From the dihybrid cross, the phenotypes in the F2 generation will be:
1. Round Yellow (RY)
2. Round Green (Rg)
3. Wrinkled Yellow (rY)
4. Wrinkled Green (rg)

This results in a total of **4 distinct phenotypes**.

### Step 6: Count the Genotypes
The genotypes from the F2 generation can be summarized as follows:
- 1 RRYY (Round Yellow)
- 2 RRYy (Round Yellow)
- 2 RrYY (Round Yellow)
- 4 RrYy (Round Yellow)
- 1 RRyy (Round Green)
- 2 Rryy (Round Green)
- 1 rryy (Wrinkled Green)

When we count these, we find a total of **9 different genotypes**.

### Conclusion
Thus, in the F2 generation of a Mendelian dihybrid cross, the number of phenotypes is **4**, and the number of genotypes is **9**.

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

---