This is a real-world medical entrance exam question, please give the true answer based on the question and selection. Topic name: Nervous system, Subject name: Physiology
[{'content': 'In a myelinated nerve fiber, refractive period is 1/2500 seconds. What is the impulse rate? \nSelect and explain. \nA. 40 per sec, B. 250 per sec, C. 400 per sec, D. 2500 per sec', 'role': 'user'}, {'content': 'Answer: D \nExplanation: A new action potential cannot occur in an excitable fiber as long as the membrane is still depolarized from the preceding action potential. The only condition that will re-open them is for the membrane potential to return either to or almost to the original resting membrane potential level. That period is called refractory period.Refractory period for large myelinated nerve fibers is 1/2500 second; hence such a fiber can carry a maximum of about 2500 impulses a second.Ref: Ganong&;s review of medical physiology, 23rd edition, Page no:87', 'role': 'assistant'}]