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Question #46929
The Question
In a modified Meselson-Stahl experiment, a culture of E. coli with fully 15N-labeled DNA is transferred to a 14N medium. Concurrently, a specific chemical inhibitor of DNA topoisomerase is introduced to the culture. After 20 minutes (one standard generation time), the DNA is extracted and analyzed via CsCl density gradient centrifugation. Which of the following represents the most likely outcome and its correct physiological basis?
Options
- A.
A single hybrid band (15N-14N); because helicase alone is sufficient to unwind the E. coli genome for one complete replication cycle.
- B.
A single heavy band (15N-15N); unresolved positive supercoiling ahead of the replication fork halts helicase activity, preventing duplication.
- C.
Two distinct bands (heavy and light); topological stress forces the DNA to replicate conservatively to bypass the replication fork stall.
- D.
A single light band (14N-14N); the inhibitor causes the parental heavy strands to be degraded and entirely replaced by light isotopes.
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Regulatory_Failure
Understand how NOT, EXCEPT, and logical constraints are used to manipulate exam results.
Concepts Behind This Question
Meselson-Stahl Density Gradient|Ori Sequence Stringency|Key Analytical Nature
Why This Question Is Tricky
Common Mistakes
Conceptual Pivot: Students often memorize the Meselson-Stahl results (heavy -> hybrid -> light/hybrid) without integrating the mechanical requirements of the replication machinery.
The trap is assuming replication will complete (Option A) or alter its mode (Option C), failing to realize that topological stress is an absolute physical barrier to replication fork progression.
Quick Facts About This Question
Question Information
| Subject | Biology |
| Chapter | The Living World - Class 11 |
| Concept | Key Analytical Nature |
| Difficulty | 8 |
| Cognitive Load | 5 |
| Target Solve Time | 120 sec |
| Access | Free |