A dihybrid plant with genotype PpNn. It produces four types of gametes in following number PN = 200, pn = 200, Pn = 800, pN = 800 then what is the distance between linked genes?

Options

A. 30 cM
B. 10 cM
C. 25 cM
D. 20 cM

Answer

To solve the problem of determining the distance between linked genes in a dihybrid plant with the genotype PpNn, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Gametes and Their Frequencies:**
- The gametes produced by the plant are:
- PN = 200
- pn = 200
- Pn = 800
- pN = 800

2. **Determine the Total Number of Gametes:**
- Calculate the total number of gametes produced:
\[ \text{Total} = PN + pn + Pn + pN = 200 + 200 + 800 + 800 = 2000 \]

3. **Identify the Recombinant Gametes:**
- The recombinant gametes are those that show a combination of alleles different from the parental types. In this case, the recombinant gametes are Pn and pN:
- Pn = 800
- pN = 800
- Total recombinant gametes:
\[ \text{Total Recombinant} = Pn + pN = 800 + 800 = 1600 \]

4. **Calculate the Distance Between the Genes:**
- The distance between linked genes can be calculated using the formula:
\[ \text{Distance} = \left( \frac{\text{Total Recombinant}}{\text{Total Gametes}} \right) \times 100 \]
- Plugging in the values:
\[ \text{Distance} = \left( \frac{1600}{2000} \right) \times 100 = 80 \]

5. **Final Calculation:**
- Since we are looking for the distance in centimorgans (cM), we can conclude that the distance between the linked genes is 80 cM.

### Conclusion:
The distance between the linked genes in the dihybrid plant with genotype PpNn is **80 cM**.