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MBBS QUESTION #10221
Question 1
Using the Nernst equation for potassium, $E_K = 61 \log_{10}\frac{[K^+]_o}{[K^+]_i}$ (mV, approx. at body temperature), a patient develops severe hyperkalemia. How does the resting membrane potential of excitable cells change, and what is the clinical consequence?
  • $E_K$ becomes less negative (membrane depolarizes), initially increasing excitability but ultimately causing dangerous conduction abnormalities✔️
  • $E_K$ becomes more negative (membrane hyperpolarizes), reducing excitability with no cardiac risk
  • Resting membrane potential is unaffected by extracellular K+ changes
  • $E_K$ reaches exactly 0 mV, abolishing all electrical activity instantly
Correct Answer Explanation
Raising extracellular K+ reduces the outward K+ concentration gradient, shifting $E_K$ toward less negative values and depolarizing the resting membrane potential. Initially this increases excitability, but sustained depolarization inactivates voltage-gated Na+ channels, ultimately causing dangerous cardiac conduction defects and arrhythmias — the basis for ECG changes (peaked T waves, widened QRS) in hyperkalemia.