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 find the distance between linked genes in the given dihybrid plant with genotype PpNn, we can follow these steps:

### Step-by-Step Solution:

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

2. **Determine the Total Number of Gametes:**
- Total gametes = PN + pn + Pn + pN
- Total gametes = 200 + 200 + 800 + 800 = 2000

3. **Identify Parental and Recombinant Gametes:**
- Parental gametes are the ones that are the same as the parent genotype:
- Parental gametes: Pn (800) and pN (800)
- Recombinant gametes are those that differ from the parental types:
- Recombinant gametes: PN (200) and pn (200)

4. **Calculate the Total Number of Recombinant Gametes:**
- Total recombinant = PN + pn
- Total recombinant = 200 + 200 = 400

5. **Calculate the Distance Between Linked Genes:**
- The formula to calculate the distance between linked genes (in centimorgans) is:
\[ \text{Distance} = \left( \frac{\text{Total Recombinant Gametes}}{\text{Total Gametes}} \right) \times 100 \]
- Plugging in the values:
\[ \text{Distance} = \left( \frac{400}{2000} \right) \times 100 = 20 \text{ centimorgans} \]

6. **Conclusion:**
- The distance between the linked genes is 20 centimorgans.

### Final Answer:
The distance between the linked genes is **20 centimeters**.

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