Electrical Signaling in Neurons (4C)
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MCAT Chemical and Physical Foundations of Biological Systems › Electrical Signaling in Neurons (4C)
A study examines synaptic transmission at a single chemical synapse between one presynaptic neuron and one postsynaptic neuron. An action potential arriving at the presynaptic terminal opens voltage-gated Ca$^{2+}$ channels, allowing Ca$^{2+}$ influx that triggers vesicle fusion and neurotransmitter release. The neurotransmitter binds ligand-gated cation channels on the postsynaptic membrane, producing an excitatory postsynaptic potential (EPSP). In one condition, extracellular Ca$^{2+}$ is reduced from 2.0 mM to 0.2 mM while all other ions are unchanged, and the presynaptic action potential waveform is assumed unchanged.
Based on the scenario, which outcome is most consistent with synaptic transmission when extracellular Ca$^{2+}$ is lowered?
Smaller EPSPs occur because Ca$^{2+}$ normally leaves the presynaptic terminal to depolarize the postsynaptic neuron.
Smaller or absent EPSPs occur because reduced Ca$^{2+}$ decreases presynaptic vesicle fusion and neurotransmitter release.
Larger EPSPs occur because reduced Ca$^{2+}$ increases the opening probability of postsynaptic ligand-gated cation channels.
No change occurs because neurotransmitter release depends only on Na$^+$ influx through presynaptic ligand-gated channels.
Explanation
This question tests understanding of calcium's role in synaptic transmission. At chemical synapses, action potentials arriving at the presynaptic terminal open voltage-gated calcium channels, allowing calcium influx that triggers vesicle fusion and neurotransmitter release. When extracellular calcium is reduced from 2.0 mM to 0.2 mM, the driving force for calcium entry decreases dramatically, resulting in less calcium influx even when channels open. This reduction in calcium influx leads to decreased vesicle fusion and less neurotransmitter release, producing smaller or absent EPSPs in the postsynaptic cell. Option B correctly identifies this mechanism, while option A incorrectly suggests calcium affects postsynaptic receptors, option C misunderstands calcium's role, and option D incorrectly claims neurotransmitter release depends on sodium. A key principle to remember is that calcium concentration directly controls the probability of vesicle fusion at synapses.
A researcher investigates ion channel dynamics in a single neuron by selectively blocking voltage-gated K$^+$ channels along the axon while leaving voltage-gated Na$^+$ channels intact. The neuron is stimulated at the axon hillock to initiate an action potential. Resting potential is $-70\ \text{mV}$ and threshold is $-55\ \text{mV}$. Assume intracellular K$^+$ is high and extracellular K$^+$ is low, so K$^+$ tends to move out of the cell when K$^+$ channels open. Ignore changes in ion concentrations over the time course of a single action potential.
Which change would be expected if voltage-gated K$^+$ channels are blocked?
Action potentials propagate only toward the soma because K$^+$ channels determine the direction of saltatory conduction.
The rising phase is eliminated because K$^+$ influx is required to depolarize the membrane to threshold.
Synaptic neurotransmitter release increases because K$^+$ channel blockade directly opens postsynaptic ligand-gated receptors.
The falling phase is slowed and repolarization is delayed because reduced K$^+$ efflux opposes return toward resting potential.
Explanation
This question tests understanding of voltage-gated potassium channels' role in action potential repolarization. During an action potential, voltage-gated potassium channels open with a delay after sodium channels, allowing potassium efflux that repolarizes the membrane back toward resting potential. When these potassium channels are blocked, potassium cannot exit the cell efficiently during the falling phase of the action potential. This results in slower repolarization and a prolonged action potential duration, as the membrane takes longer to return to resting potential. Option A correctly identifies this effect, while option B incorrectly claims potassium influx is needed for depolarization, option C confuses axonal and synaptic mechanisms, and option D misunderstands saltatory conduction. The key insight is that potassium channels are responsible for repolarization, not depolarization, and blocking them extends the action potential duration.
A pharmacology group tests a drug that blocks voltage-gated Na$^+$ channels in a single peripheral neuron. The neuron normally initiates an action potential at the axon hillock when depolarization reaches threshold ($-55\ \text{mV}$). Resting membrane potential is $-70\ \text{mV}$. The drug binds preferentially to the open/inactivated states of Na$^+$ channels, reducing Na$^+$ conductance during depolarization. Assume K$^+$ channels are unaffected and ion gradients remain constant during the test.
Which change would be expected if voltage-gated Na$^+$ channels are blocked by the drug?
The neuron is more likely to reach threshold because blocking Na$^+$ channels decreases membrane resistance and amplifies current.
The action potential overshoot increases because reduced Na$^+$ influx allows more time for K$^+$ channels to depolarize the membrane.
Postsynaptic potentials increase because Na$^+$ channel blockade in the presynaptic neuron directly activates postsynaptic receptors.
Action potential initiation fails or requires a larger stimulus because reduced Na$^+$ influx prevents rapid depolarization.
Explanation
This question tests understanding of sodium channels' critical role in action potential initiation. Voltage-gated sodium channels are essential for the rapid depolarization phase of action potentials, as they allow massive sodium influx when the membrane reaches threshold. When these channels are blocked by a drug, the neuron cannot generate the rapid depolarization needed to initiate an action potential, or requires a much larger stimulus to achieve the same effect through remaining unblocked channels. Option C correctly identifies that action potential initiation fails or requires larger stimuli, while option A incorrectly suggests potassium channels depolarize the membrane, option B misunderstands how channel blockade affects membrane properties, and option D confuses presynaptic and postsynaptic mechanisms. The principle to remember is that sodium channel availability directly determines the threshold for action potential initiation.
In a single neuron, the resting membrane potential is primarily determined by K$^+$ leak channels. A researcher alters extracellular K$^+$ concentration to study effects on membrane potential. Assume temperature $T=310\ \text{K}$ and use the Nernst equation for K$^+$: $$E_K=\frac{RT}{zF}\ln\left(\frac{K^+{out}}{K^+{in}}\right)$$ with $z=+1$, $R=8.314\ \text{J mol}^{-1}\text{K}^{-1}$, and $F=96485\ \text{C mol}^{-1}$. Intracellular $K^+{in}$ remains $140\ \text{mM}$. Extracellular $K^+{out}$ is increased from $4\ \text{mM}$ to $12\ \text{mM}$.
Which outcome is most consistent with this manipulation?
No change occurs because resting potential depends only on Na$^+$ concentration gradients, not K$^+$.
The membrane hyperpolarizes because the increased extracellular K$^+$ strengthens the outward driving force for K$^+$ efflux.
The membrane depolarizes because K$^+$ flows into the cell through voltage-gated Na$^+$ channels at rest.
The membrane depolarizes (resting potential becomes less negative) because $E_K$ becomes less negative as $[K^+]_{out}$ increases.
Explanation
This question tests understanding of how ion concentration gradients affect membrane potential through the Nernst equation. The resting membrane potential is primarily determined by potassium permeability and the potassium equilibrium potential (EK). According to the Nernst equation, EK depends on the ratio of extracellular to intracellular potassium concentrations. When extracellular potassium increases from 4 mM to 12 mM while intracellular remains at 140 mM, the concentration ratio decreases, making EK less negative (closer to zero). Since the membrane potential follows EK due to high potassium permeability at rest, the membrane depolarizes. Option B correctly identifies this depolarization, while option A incorrectly predicts hyperpolarization, option C ignores potassium's dominant role at rest, and option D proposes an impossible mechanism. The key principle is that increasing extracellular potassium always depolarizes neurons.
A lab studies saltatory conduction in a single myelinated axon. Myelin increases membrane resistance and decreases membrane capacitance in internodal regions, whereas nodes of Ranvier have high densities of voltage-gated Na$^+$ channels. The axon is stimulated at the hillock to initiate an action potential. In one condition, a demyelinating insult reduces myelin integrity along the internodes without directly affecting Na$^+$ channel function at the nodes. Assume the axon diameter and ion gradients are unchanged.
Which statement is most consistent with the expected effect on action potential propagation?
Conduction velocity increases because current spreads more easily across the membrane when resistance is reduced.
Propagation reverses direction because myelin determines whether Na$^+$ influx occurs at the hillock or at the terminal.
Synaptic transmission is enhanced because demyelination increases Ca$^{2+}$ influx at the postsynaptic membrane.
Conduction velocity decreases because increased internodal capacitance and decreased resistance increase current leak and slow depolarization of the next node.
Explanation
This question tests understanding of saltatory conduction in myelinated axons. Myelin acts as an insulator that increases membrane resistance and decreases capacitance in internodal regions, forcing current to jump between nodes of Ranvier where sodium channels are concentrated. When myelin is damaged, the internodal membrane resistance decreases and capacitance increases, allowing more current to leak across the membrane between nodes. This current leak means less depolarizing current reaches the next node, slowing the rate of depolarization and reducing conduction velocity. Option A correctly identifies these effects, while option B incorrectly suggests reduced resistance speeds conduction, option C misunderstands action potential directionality, and option D confuses axonal and synaptic mechanisms. The principle is that myelin integrity is crucial for maintaining fast conduction velocity through efficient current flow between nodes.
At a single excitatory synapse, neurotransmitter binds postsynaptic ligand-gated cation channels permeable to both Na$^+$ and K$^+$. The postsynaptic neuron has resting potential $V_m=-70\ \text{mV}$. Assume $E_{Na}=+60\ \text{mV}$ and $E_K=-90\ \text{mV}$. The ligand-gated channel has a reversal potential near $0\ \text{mV}$ due to mixed Na$^+$/K$^+$ permeability. A brief presynaptic action potential triggers neurotransmitter release.
Which postsynaptic response is most consistent with these parameters immediately after receptor activation?
No postsynaptic potential occurs because ligand-gated channels require presynaptic Ca$^{2+}$ influx into the postsynaptic neuron to open.
An EPSP occurs because net inward current drives $V_m$ toward the channel reversal potential near $0\ \text{mV}$.
An IPSP occurs because opening cation channels drives $V_m$ toward $E_K=-90\ \text{mV}$.
An EPSP occurs because neurotransmitter directly opens presynaptic voltage-gated Na$^+$ channels to depolarize the postsynaptic cell.
Explanation
This question tests understanding of ligand-gated channels and their reversal potentials. When neurotransmitter binds to ligand-gated cation channels permeable to both sodium and potassium, the channels open and allow current flow based on the electrochemical gradients for both ions. The reversal potential of these mixed cation channels (near 0 mV) represents the membrane potential at which no net current flows through the channels. Since the resting potential (-70 mV) is more negative than the reversal potential (0 mV), opening these channels causes net inward current (mainly sodium influx exceeding potassium efflux), depolarizing the membrane toward the reversal potential and creating an EPSP. Option B correctly identifies this mechanism, while option A incorrectly focuses only on potassium, option C misunderstands receptor activation, and option D confuses pre- and postsynaptic mechanisms. The key principle is that current flow through ligand-gated channels always drives the membrane potential toward the channel's reversal potential.
A researcher examines refractory periods in a single unmyelinated axon. During an action potential, voltage-gated Na$^+$ channels transition from closed to open and then to an inactivated state; they must return to the closed state before they can reopen. Voltage-gated K$^+$ channels open with a delay and contribute to repolarization and after-hyperpolarization. A second depolarizing stimulus is delivered shortly after the peak of the first action potential.
Which outcome is most consistent with delivering the second stimulus during the relative refractory period (not the absolute refractory period)?
A second action potential can occur but typically requires a larger-than-normal stimulus because K$^+$ conductance remains elevated and the membrane is hyperpolarized.
A second action potential cannot occur regardless of stimulus strength because all Na$^+$ channels are permanently inactivated after one spike.
A second action potential is guaranteed with the same stimulus because neurotransmitter binding at the synapse resets Na$^+$ channels.
A second action potential occurs but propagates only backward toward the soma because K$^+$ channels enforce reverse conduction.
Explanation
This question tests understanding of absolute versus relative refractory periods. During the relative refractory period, some sodium channels have recovered from inactivation and returned to the closed state, making them available for activation, but potassium conductance remains elevated from the previous action potential. This elevated potassium conductance causes membrane hyperpolarization below resting potential, requiring a larger-than-normal stimulus to reach threshold for a second action potential. Option A correctly describes this requirement for a stronger stimulus during the relative refractory period, while option B incorrectly describes the absolute refractory period, option C ignores refractory mechanisms, and option D misunderstands action potential propagation. The key distinction is that during the relative refractory period, action potentials are possible but require stronger stimuli due to residual potassium conductance and membrane hyperpolarization.
A pharmacological agent is applied to a single inhibitory synapse to test its impact on synaptic transmission. Normally, presynaptic action potentials trigger Ca$^{2+}$ influx and release of an inhibitory neurotransmitter that opens postsynaptic ligand-gated Cl$^-$ channels. Resting membrane potential is $-70\ \text{mV}$ and the Cl$^-$ equilibrium potential is approximately $E_{Cl}=-75\ \text{mV}$. The drug selectively blocks the postsynaptic Cl$^-$ channels but does not affect presynaptic action potentials or neurotransmitter release.
Which postsynaptic change is most consistent with applying this drug?
Inhibitory postsynaptic potentials increase because blocking Cl$^-$ channels forces more Cl$^-$ influx through presynaptic voltage-gated Ca$^{2+}$ channels.
No change occurs because inhibition is determined only by presynaptic Ca$^{2+}$ entry, not postsynaptic receptor conductance.
Inhibitory postsynaptic potentials decrease because the postsynaptic membrane can no longer increase Cl$^-$ conductance in response to neurotransmitter.
Excitatory postsynaptic potentials increase because the inhibitory neurotransmitter now binds and opens postsynaptic voltage-gated Na$^+$ channels.
Explanation
This question tests understanding of inhibitory synaptic transmission mediated by chloride channels. At inhibitory synapses, neurotransmitter binding opens ligand-gated chloride channels, allowing chloride flux based on its electrochemical gradient. With ECl at -75 mV and resting potential at -70 mV, chloride channel opening causes slight hyperpolarization or stabilization near resting potential, producing an IPSP. When the drug blocks these postsynaptic chloride channels, the inhibitory neurotransmitter can still be released but cannot produce its normal effect because the channels cannot open. This results in decreased or absent IPSPs. Option A correctly identifies this reduction in inhibitory potentials, while option B proposes an impossible mechanism, option C incorrectly suggests the neurotransmitter changes its target, and option D ignores the role of postsynaptic receptors. The principle is that blocking postsynaptic receptors prevents the corresponding postsynaptic response regardless of presynaptic function.
In a simplified experiment on action potential initiation and propagation, a single myelinated motor neuron is stimulated at the axon hillock with a brief depolarizing current pulse. Voltage-gated Na$^+$ channels open rapidly when membrane potential reaches threshold (about $-55\ \text{mV}$), while voltage-gated K$^+$ channels open more slowly and contribute to repolarization. The neuron's resting membrane potential is $-70\ \text{mV}$. Myelin is present along the axon except at nodes of Ranvier, where Na$^+$ channel density is high. Assume extracellular fluid has high Na$^+$ and low K$^+$ relative to the cytosol, and that action potentials are all-or-none once threshold is reached. The investigator applies a second identical stimulus 1.5 ms after the first, at the same location.
Which statement best describes what is most consistent with action potential propagation under these conditions?
The second stimulus will propagate only backward toward the soma because saltatory conduction reverses direction at the nodes of Ranvier.
The second stimulus will trigger a larger-than-normal action potential because K$^+$ efflux increases the driving force for Na$^+$ influx above baseline.
The second stimulus will trigger a full action potential because neurotransmitter release from the axon terminal resets the membrane at the hillock within 1.5 ms.
The second stimulus is unlikely to trigger a full action potential because many Na$^+$ channels are still inactivated during the absolute refractory period.
Explanation
This question tests understanding of the refractory period following an action potential. During an action potential, voltage-gated sodium channels open rapidly, then inactivate, and must return to their closed state before they can open again. The absolute refractory period (typically 1-2 ms) occurs when most sodium channels are inactivated, making it impossible or very difficult to generate another action potential regardless of stimulus strength. Since the second stimulus arrives 1.5 ms after the first, many sodium channels are still in their inactivated state. Option B correctly identifies that the second stimulus is unlikely to trigger a full action potential due to sodium channel inactivation, while option A incorrectly suggests neurotransmitter release at the terminal affects the hillock, and options C and D propose physiologically impossible scenarios. To approach similar questions, remember that the absolute refractory period enforces the unidirectional propagation of action potentials and limits firing frequency.
A team evaluates signal integration at the axon hillock in a single postsynaptic neuron receiving one excitatory synapse and one inhibitory synapse. The excitatory synapse opens ligand-gated cation channels with a reversal potential near $0\ \text{mV}$. The inhibitory synapse opens ligand-gated Cl$^-$ channels with $E_{\text{Cl}}=-70\ \text{mV}$. The neuron rests at $-70\ \text{mV}$ and has threshold $-55\ \text{mV}$ at the hillock. The excitatory and inhibitory synapses are activated simultaneously with similar conductance changes. Which outcome is most consistent with these conditions?
An action potential is guaranteed because any excitatory neurotransmitter release forces $V_m$ to exceed threshold
An action potential is less likely because increased Cl$^-$ conductance counteracts depolarization and stabilizes $V_m$ near rest
An action potential is more likely because inhibitory Cl$^-$ channels depolarize the membrane toward $0\ \text{mV}$
Postsynaptic depolarization increases because the inhibitory synapse triggers presynaptic Ca$^{2+}$ entry at the excitatory terminal
Explanation
This question tests understanding of synaptic integration in neuronal electrical signaling. When excitatory and inhibitory synapses activate simultaneously, their effects compete at the axon hillock where action potentials initiate. The excitatory synapse attempts to depolarize toward 0 mV while the inhibitory Cl- channels, with ECl = -70 mV (same as resting potential), increase membrane conductance without changing voltage at rest. However, this increased Cl- conductance acts as a "shunt" that reduces the effectiveness of excitatory input by providing an alternative current path. The inhibitory conductance stabilizes the membrane near -70 mV, preventing the excitatory input from depolarizing the membrane to threshold (-55 mV). Choice C incorrectly suggests excitation guarantees threshold crossing, but inhibition can effectively cancel excitation. To analyze synaptic integration, consider both the driving forces and the relative conductances—inhibition near rest acts primarily through conductance increase rather than hyperpolarization.