How much linkage strength is present between two genes A & B, which are 6 cM far from each other in a chromosome ?

Options

A. `6%`
B. `le50%`
C. `94%`
D. Data insufficient

Answer

To find the linkage strength between two genes A & B that are 6 centimorgans (cM) apart, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Linkage Map Distance**:
- The linkage map distance between two genes is expressed in centimorgans (cM). This distance is directly related to the recombination frequency (RF), which is the percentage of recombination events that occur between the two genes during meiosis.

2. **Identify the Given Distance**:
- In this case, the distance between genes A and B is given as 6 cM.

3. **Relate Linkage Map Distance to Recombination Frequency**:
- The recombination frequency (RF) is equal to the linkage map distance in centimorgans. Therefore, if the distance is 6 cM, the recombination frequency is also 6%.

4. **Calculate Linkage Strength**:
- Linkage strength can be calculated using the formula:
\[ \text{Linkage Strength} = 100\% - \text{Recombination Frequency} \]
- Substituting the recombination frequency:
\[ \text{Linkage Strength} = 100\% - 6\% = 94\% \]

5. **Conclusion**:
- The linkage strength between genes A and B, which are 6 cM apart, is 94%.

### Final Answer:
The linkage strength between genes A and B is **94%**.

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