A test cross of `F_(1)` flies +a/+b produced the following offspring
`++/ab=9`
`ab//ab=9`
`+b//ab=41`
`a+//ab=41`
What will be distance between linked gene :-

Options

A. 82 cM
B. 18 cM (cis)
C. 20 cM
D. 18 cM (trans)

Answer

To solve the problem of determining the distance between linked genes based on the offspring produced from a test cross of F1 flies, we can follow these steps:

### Step-by-step Solution:

1. **Identify Offspring Types and Counts**:
- From the test cross, we have the following offspring:
- `++/ab = 9` (Parental type)
- `ab//ab = 9` (Parental type)
- `+b//ab = 41` (Recombinant type)
- `a+//ab = 41` (Recombinant type)

2. **Calculate Total Offspring**:
- Total offspring = 9 (++) + 9 (ab) + 41 (+b) + 41 (a+) = 100

3. **Identify Recombinant Offspring**:
- The recombinant offspring are `+b//ab` and `a+//ab`, which together total:
- Recombinant offspring = 41 + 41 = 82

4. **Identify Non-Recombinant Offspring**:
- The non-recombinant offspring are `++/ab` and `ab//ab`, which total:
- Non-recombinant offspring = 9 + 9 = 18

5. **Calculate Recombination Frequency**:
- Recombination frequency (RF) is calculated as:
\[ RF = \frac{\text{Number of Recombinant Offspring}}{\text{Total Offspring}} \times 100 \]
- Here, the number of recombinant offspring is 82, and the total offspring is 100:
\[ RF = \frac{82}{100} \times 100 = 82\% \]

6. **Calculate Distance Between Linked Genes**:
- The distance between linked genes is expressed in centimorgans (cM), where 1% recombination frequency equals 1 cM. Therefore, the distance is:
\[ \text{Distance} = 82 \text{ cM} \]

7. **Conclusion**:
- The distance between the linked genes is 18 centimorgans.

### Final Answer:
The distance between the linked genes is **18 centimorgans**.

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