In a Mendelian dihybrid cross kinds of recombinant phenotypes formed in `F_(2)` generation

Options

A. 2
B. 4
C. 6
D. 10

Answer

To solve the question regarding the kinds of recombinant phenotypes formed 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 represented by two alleles. For example, consider traits for seed shape (round vs. wrinkled) and seed color (yellow vs. green). The alleles can be represented as:
- Round (R) is dominant over wrinkled (r)
- Yellow (Y) is dominant over green (y)

### Step 2: Determine the Genotypes of the Parents
In a typical dihybrid cross, the parental generation (P) would have the following genotypes:
- One parent: RRYY (homozygous dominant for both traits)
- Other parent: rryy (homozygous recessive for both traits)

### Step 3: Create the F1 Generation
The F1 generation results from crossing the two parental genotypes:
- All offspring (F1) will be RrYy (heterozygous for both traits), exhibiting both dominant traits (round and yellow).

### Step 4: Self the F1 Generation to Obtain F2
To obtain the F2 generation, we self the F1 generation (RrYy x RrYy). This can be visualized using a Punnett square.

### Step 5: Analyze the Punnett Square
The Punnett square for the dihybrid cross will yield the following phenotypic ratio in the F2 generation:
- 9 Round Yellow (RRYY, RRYy, RrYY, RrYy)
- 3 Round Green (RRyy, Rryy)
- 3 Wrinkled Yellow (rrYY, rrYy)
- 1 Wrinkled Green (rryy)

### Step 6: Identify Recombinant Phenotypes
Recombinant phenotypes are those that show a combination of traits different from the parental generation. In this case:
- The recombinant phenotypes are:
- Round Green (Rryy) - 3
- Wrinkled Yellow (rrYY) - 3

Thus, the total number of recombinant phenotypes is 3 (Round Green) + 3 (Wrinkled Yellow) = 6.

### Conclusion
The kinds of recombinant phenotypes formed in the F2 generation of a Mendelian dihybrid cross is **6**.