Discussion
Match Group-I (selection agent) and Group-II (gene) in CORRECT combination.
| Group-I | Group-II |
|---|---|
| P. Kanamycin | I. pmi |
| Q. Hygromycin | II. bar |
| R. Phosphinothricin | III. nptII |
| S. Mannose | IV. ptxD |
| V. dhfr | |
| VI. hpt |
Answer: A
The correct combination is P-III, Q-VI, R-II, S-I.
- Kanamycin is detoxified by neomycin phosphotransferase II, encoded by nptII (III).
- Hygromycin B is inactivated by hygromycin phosphotransferase, hpt (VI).
- Phosphinothricin (glufosinate) is acetylated and inactivated by phosphinothricin acetyltransferase, encoded by bar (II).
- Mannose is used for positive selection: cells carrying phosphomannose isomerase, pmi (I), can convert mannose-6-phosphate to fructose-6-phosphate and grow on it.
The extra genes are distractors: ptxD (phosphite dehydrogenase) is used with phosphite, and dhfr (dihydrofolate reductase) with methotrexate. Options B, C and D each pair at least one agent with one of these or with the wrong gene.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development