In guinea pig, black short hair (BBSS) is dominant over white long hair (bbss). In the dihybrid cross, `F_(2)` genotype BBSS, BbSS,BBSs and BbSs appears in the ratio of

Options

A. `9:3:3:1`
B. `4:2:1:2`
C. `1:2:1:4`
D. `1:2:2:4`

Answer

To solve the problem regarding the inheritance of traits in guinea pigs, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Traits and Genotypes**:
- Black short hair is represented by the genotype BBSS (dominant).
- White long hair is represented by the genotype bbss (recessive).

2. **Determine the Gametes of the Parents**:
- The parents are BBSS (black short hair) and bbss (white long hair).
- The gametes produced by the BBSS parent will be: BS (since both alleles are the same).
- The gametes produced by the bbss parent will be: bs (since both alleles are the same).

3. **F1 Generation**:
- The F1 generation results from the cross of the two parental genotypes:
- F1 genotype = BBSS × bbss = BbSs (all offspring will have this genotype).

4. **Self-Pollination of F1 Generation**:
- The F1 generation (BbSs) will be self-pollinated to produce the F2 generation.
- The gametes produced by BbSs will be: BS, Bs, bS, bs.

5. **Construct a Punnett Square**:
- Create a 4x4 Punnett square using the gametes from the F1 generation:
- Rows and columns will each have the gametes: BS, Bs, bS, bs.

6. **Fill in the Punnett Square**:
- Fill in the Punnett square to find the genotypes of the offspring:
- The combinations will yield:
- BBSS
- BBSs
- BbSS
- BbSs
- BbSs
- Bbss
- bbSS
- bbSs
- bbsS
- bbsS
- etc.

7. **Count the Genotypes**:
- Count the occurrences of each genotype:
- BBSS: 1
- BbSS: 2
- BBSs: 2
- BbSs: 4
- bbss: 1
- Thus, the ratio of the genotypes is:
- 1 (BBSS) : 2 (BbSS) : 2 (BBSs) : 4 (BbSs)

8. **Final Ratio**:
- The final ratio of the genotypes in the F2 generation is 1:2:2:4.

### Conclusion:
The correct answer to the question is the ratio of the genotypes in the F2 generation, which is **1:2:2:4**.