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This is what a neuron sounds like: every click is one action potential.
Ion gradients set the voltage. K+ leaking out holds the cell near −70 mV; Na+ rushing in drives the peak to about +30 mV. A stronger stimulus means more clicks per second, not bigger ones.
In a neuron at 37 °C, [K+] is 140 mM inside and 4 mM outside, so its K+ equilibrium potential is E_K = 61.5 mV × log10(4/140) ≈ −95 mV. If extracellular [K+] rises to 8 mM while intracellular [K+] stays at 140 mM, E_K becomes about:
Pick one. The red pen does the rest.
The deciding lineThe Nernst equation for K+ at 37 °C is E_K = 61.5 mV × log10([K+]out/[K+]in); E_K depends on the logarithm of the concentration ratio, so doubling [K+]out shifts it by 61.5 × log10 2 ≈ 18.5 mV toward zero.
Why C: 61.5 × log10(8/140) = 61.5 × (−1.243) = −76 mV. Equivalently, doubling [K+]out adds 61.5 × log10 2 = 18.5 mV to −95 mV: the smaller K+ gradient makes E_K less negative.
Why each choice fails
AScale slip: doubling [K+]out does not double E_K. The potential tracks the logarithm of the ratio, and a smaller K+ gradient makes it less negative: −95 + 18.5 ≈ −76 mV, not 2 × (−95) = −190 mV.
BDirection reversed: the step size, 18.5 mV, is right, but raising [K+]out shrinks the K+ gradient, so E_K moves toward zero (−76 mV), not away from it (−113 mV).
CHolds: 61.5 × log10(8/140) = −76 mV, 18.5 mV less negative than −95 mV.
DScale slip: the K+ gradient does halve (140/4 = 35 to 140/8 = 17.5), but E_K depends on the logarithm of that ratio, not the ratio itself, so it shifts by only 61.5 × log10 2 = 18.5 mV, to −76 mV, not to −95/2 ≈ −47 mV.
The plasma membrane of a resting neuron is 7.0 nm thick, and the cytosol is at −70 mV relative to the extracellular fluid. Treating the electric field within the membrane as uniform, what is the field's magnitude, and in which direction does it push a Cl− ion crossing the membrane?
Pick one. The red pen does the rest.
The deciding lineFor a uniform field between two parallel surfaces, E = ΔV/d, and the field points from higher to lower potential.
Why C: E = 0.070 V / (7.0 × 10−9 m) = 1.0 × 107 V/m. The field points from the extracellular side (0 mV) toward the cytosol (−70 mV), that is, into the cell, so the force on the negative Cl− ion points the other way, toward the outside of the cell.
Why each choice fails
ADirection reversed: the field does point into the cell, toward the lower-potential cytosol, but Cl− is negative, so the force on it, F = qE, points the opposite way, toward the outside.
BScale slip: 70/7.0 = 10 is in millivolts per nanometer, and 1 mV/nm = 106 V/m, so the field is 1.0 × 107 V/m, not 10 V/m.
CHolds: 0.070 V / 7.0 × 10−9 m = 1.0 × 107 V/m, pointing into the cell toward the lower-potential cytosol, so a negative ion is pushed toward the outside.
DScale slip, with the direction reversed as well: 1 mV/nm is 106 V/m, so the field is 1.0 × 107 V/m, and the force on a negative ion opposes the inward-pointing field, pushing it outward.
After an action potential, a neuron cannot fire again during an absolute refractory period of 2.0 ms. What is the highest possible firing rate, in action potentials per second?
Pick one. The red pen does the rest.
The deciding lineThe absolute refractory period is the interval after an action potential during which voltage-gated Na+ channels are inactivated and no new action potential can start, which sets an upper limit on firing rate.
Why C: Maximum rate = 1 / refractory period = 1 / 0.0020 s = 500 per second.
Why each choice fails
AScale slip: 50 per second would correspond to a refractory period of 20 ms; the period is 2.0 ms.
BWrong comparison: 200 per second corresponds to 5 ms between spikes, not 2.0 ms.
CHolds: 1 s = 1,000 ms; 1,000 ÷ 2.0 ms = 500 action potentials per second at most.
DDirection reversed: 2,000 per second would need a period of 0.5 ms; the ratio was inverted (2.0 ÷ 1,000 read as 2 per 1,000).
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