Ouabain, Membrane Potential, and Cell Swelling
In a physiology lab, cultured neurons are bathed in normal saline and then exposed to 10 µM ouabain, a specific inhibitor of the Na+/K+-ATPase. Within 10 minutes, intracellular Na+ rises from 12 mM to 38 mM, intracellular K+ falls from 140 mM to 105 mM, and the resting membrane potential changes from −70 mV to −58 mV. After 30 minutes, the cells swell and some lyse. A student concludes that ouabain directly opens a nonselective cation channel and that the membrane potential change is due to the K+ equilibrium potential alone. The lab manual states that ouabain blocks the pump, not a channel. Extracellular Na+ is 145 mM and extracellular K+ is 5 mM. A classmate argues that the pump is electrogenic.
1. Evaluate the student's two claims. Which parts are correct, which are wrong, and what is the mechanism?
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The claim that ouabain directly opens a nonselective cation channel is wrong. Ouabain is a specific inhibitor of the Na+/K+-ATPase; it binds the pump's extracellular K+ site and stops active transport. The observed Na+ rise and K+ fall are exactly what pump inhibition predicts: Na+ leaks in and is no longer pumped out, while K+ leaks out and is no longer pumped in. The claim that the membrane-potential change is due to the K+ equilibrium potential alone is also incomplete. The pump is electrogenic, moving three Na+ out per two K+ in, so blocking it removes a small hyperpolarizing contribution. More importantly, the K+ gradient falls, shifting E_K toward a less negative value, while the rising Na+ gradient and any increased Na+ permeability pull Vm toward a more positive E_Na. The resting potential is a weighted average described by the Goldman equation, not by E_K alone. The observed −58 mV is consistent with pump inhibition plus changed ion gradients, not with a direct channel-opening action.
2. Calculate the Nernst equilibrium potential for K+ before and after ouabain using [K+]out = 5 mM. Does the change explain the observed −58 mV? What else must be considered?
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Using E_K = 61.5 log10([K+]out/[K+]in) at body temperature, before ouabain E_K = 61.5 log10(5/140) = 61.5 × (−1.447) ≈ −89 mV. After ouabain, E_K = 61.5 log10(5/105) = 61.5 × (−1.322) ≈ −81 mV. The K+ equilibrium potential therefore depolarizes by only about 8 mV, from −89 to −81 mV. That is not enough to explain a change from −70 to −58 mV, which is 12 mV, and it does not explain why Vm was −70 rather than −89 before treatment. At rest, the membrane is permeable to Na+ as well, so Vm is positive to E_K; the Goldman equation weights E_K and E_Na by their permeabilities. Ouabain also stops the electrogenic pump, removing a few millivolts of hyperpolarization, and the rising intracellular Na+ shifts E_Na. Thus E_K alone is insufficient.
3. Predict what would happen if extracellular Na+ were replaced by an impermeant cation at the same time as ouabain. Explain your reasoning.
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If extracellular Na+ is replaced by an impermeant cation, Na+ entry is greatly reduced. Ouabain still inhibits the pump, so K+ continues to leak out and intracellular K+ falls, but the cell cannot accumulate Na+ from outside. The depolarization should be smaller because the inward Na+ leak is removed; Vm may stay closer to E_K, although the falling K+ gradient and loss of the electrogenic pump still shift it. Cell swelling should also be reduced or delayed. Swelling normally occurs because accumulated Na+ and the Cl− that follows it raise intracellular osmolarity, drawing water in; without Na+ entry, the osmotic load does not build as much. The cells may still lose K+ and shrink transiently or fail to regulate volume, but widespread lysis at 30 min should be less. A control with normal Na+ and ouabain is needed to separate the Na+ substitution from the pump block; replacing Na+ also changes the electrochemical gradient, so the experiment tests the role of Na+ influx, not ouabain specificity.