2. Write a paragraph of 3-5 sentences answering the following questions. Your re
ID: 148251 • Letter: 2
Question
2. Write a paragraph of 3-5 sentences answering the following questions. Your response should first answer the question and then go onto state physiological interpretation's behind your answer. Paragraphs that demonstrate complete and thorough reasoning will receive full points. You discover a drug that prevents vaitage-gated Na+ channel inactivation gates from opening when they are closed but has no effect on nactivation gates when they are open. What would happen to the action potential in a neuron exposed to this drug? Explain your answer. Assume the neuron is at resting membrane potential (e.g., -70 mV) at the time it is exposed to the drugExplanation / Answer
Neuronal membrane will have no effects at the resting membrane potential or at -70mV and will be much more depolarised during action potential.
Explanation: when the cells are at resting membrane potential it is at -70mV which is determined by concentration and charges of ions present at both sides. When a stimuli comes, an action potential generates, causing the cell membrane to depolarise. The voltage gated ions channels are usually shut during the resting phase but during membrane depolarisation, voltage gated sodium channels open and Na+ ions start to come inside to make membrane more depolarised. After continuous ion flow, Na+ channels are inactivated and K+ channels open to flow the K+ ions outward and make membrane as the previous resting state. When a drug is exposed which can prevent opening of voltage gated Na+ channel inactivation gates from opening, will have no effects at -70mV. At resting phase, the channels are closed. It is mentioned that if the inactivation gate is closed the drug will prevent it from opening. So the drug will bind to the gates. But when action potential generates Na+ channels will open continually without normal inactivation step as the drug will prevent the inactivation gates from opening. Hence membrane will become further and further depolarised or positively charged.
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