Correct option is C
Camptothecin is a specific inhibitor of Topoisomerase I, not Topoisomerase II. Therefore, in the presence of Camptothecin, Topoisomerase I is inhibited, and if Topoisomerase II is still functional, it will carry out its activity.
Topoisomerase II, which cuts both strands of DNA simultaneously, changes the linking number by ±2 per catalytic cycle. So, when Topoisomerase I is inhibited by Camptothecin, the only enzyme able to act on DNA topology is Topoisomerase II, which will lead to changes in linking number by ±2.
Hence, option (c) correctly reflects the expected outcome: Topoisomerase II will remain active and change linking number by ±2 in the presence of Camptothecin.
Information Booster:
- Topoisomerase I creates transient single-strand breaks and changes the linking number by ±1.
- Topoisomerase II introduces double-strand breaks and changes linking number by ±2.
- Camptothecin specifically inhibits Topoisomerase I, not Topoisomerase II.
- Etoposide specifically inhibits Topoisomerase II, not Topoisomerase I.
- Linking number (Lk) refers to the number of times one strand of DNA wraps around the other in a closed circular DNA.
- Topoisomerase activity is crucial during DNA replication and transcription to relieve torsional stress.
- Selective inhibition of one topoisomerase allows the activity of the other to be measured based on changes in linking number.
Additional Information:
- (a) In the presence of Camptothecin, Topoisomerase I will lead to change in the linking number by ±2: Incorrect — Topoisomerase I only changes Lk by ±1, and it's inhibited in this case.
- (b) In the presence of Etoposide, Topoisomerase I will lead to change in the linking number by ±2: Incorrect — Topoisomerase I changes Lk by ±1, not ±2.
- (d) In the presence of Etoposide, Topoisomerase II will lead to change in the linking number by ±2: Incorrect — Etoposide inhibits Topoisomerase II, so no change in Lk by ±2 will occur; only Topoisomerase I will be active, changing Lk by ±1.
