MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4c Electrical Signaling Neurons

What this quiz covers

This quiz focuses on 4c Electrical Signaling Neurons, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Propagation slows because increased $C_m$ requires more charge to change $V_m$, reducing the rate of depolarization across that segment (correct answer)
2. Propagation speeds up because increased $C_m$ stores more charge, allowing the next node to reach threshold sooner
3. Propagation reverses direction because delayed K$^+$ channel opening causes depolarization to travel back toward the soma
4. Propagation is unchanged because myelin affects only neurotransmitter release at axon terminals, not axonal conduction

Explanation: This question tests understanding of how membrane capacitance affects action potential propagation speed. In electrical signaling, neurons propagate action potentials through sequential depolarization of membrane segments, where the rate depends on how quickly each segment can reach threshold voltage. In the demyelinated segment, increased capacitance from 1.0 to 2.5 μF/cm² means more charge must accumulate to achieve the same voltage change (Q = CV). Since the current from the previous segment remains the same but more charge is needed, it takes longer to depolarize this segment to threshold, slowing propagation. Choice B incorrectly assumes that storing more charge speeds propagation, but capacitance actually acts as a barrier to rapid voltage changes. To approach similar problems, remember that higher capacitance requires more time to charge, analogous to filling a larger bucket with the same water flow rate.

Question 2

1. The resting membrane potential becomes less negative (depolarizes) because $E_K$ shifts to a higher value (correct answer)
2. The resting membrane potential becomes more negative (hyperpolarizes) because K$^+$ efflux increases
3. The resting membrane potential becomes more positive because K$^+$ flows into the cell until $V_m$ reaches $E_{\text{Na}}$
4. The resting membrane potential is unchanged because only Na$^+$ concentration determines $V_m$ at rest

Explanation: This question tests understanding of how ion concentrations determine membrane potential through the Nernst equation. In electrical signaling, the resting membrane potential of neurons primarily depends on K+ permeability and follows the K+ equilibrium potential (EK). When extracellular K+ increases from 4 mM to 12 mM while intracellular K+ remains at 140 mM, the concentration ratio [K+]out/[K+]in increases from 4/140 to 12/140. According to the Nernst equation, this makes EK less negative (depolarizes) because the natural log of a larger ratio yields a less negative value. Choice B incorrectly predicts hyperpolarization, but increasing external K+ always depolarizes by reducing the K+ concentration gradient. To solve Nernst equation problems, remember that increasing external cation concentration or decreasing internal cation concentration makes the equilibrium potential more positive.

Question 3

1. Net outward positive current (K$^+$ efflux) dominates at $-70\ \text{mV}$, hyperpolarizing the cell toward $-90\ \text{mV}$.
2. Net inward positive current dominates at $-70\ \text{mV}$, depolarizing the cell toward $0\ \text{mV}$ (an EPSP). (correct answer)
3. A presynaptic action potential is generated because ligand-gated channels on the postsynaptic cell trigger presynaptic Na$^+$ channel opening.
4. Net inward Cl$^-$ current dominates at $-70\ \text{mV}$ because mixed cation channels primarily conduct chloride at negative voltages.

Explanation: This question tests understanding of synaptic currents and excitatory postsynaptic potentials (EPSPs). When ligand-gated channels with a reversal potential of 0 mV open at a membrane potential of -70 mV, there is a 70 mV driving force for net inward current. Since these channels are permeable to both Na+ and K+, and the reversal potential (0 mV) is between ENa (+60 mV) and EK (-90 mV), the net effect is inward positive current that depolarizes the cell toward 0 mV. The correct answer (B) accurately describes this depolarization as an EPSP. Answer A incorrectly suggests net outward current would occur, which would require the membrane potential to be above the reversal potential. To analyze synaptic currents, always compare the membrane potential to the channel's reversal potential: current flows to drive the membrane toward the reversal potential.

Question 4

1. Conduction velocity decreases because increased internodal capacitance and decreased resistance increase current leak and slow depolarization of the next node. (correct answer)
2. Conduction velocity increases because current spreads more easily across the membrane when resistance is reduced.
3. Propagation reverses direction because myelin determines whether Na$^+$ influx occurs at the hillock or at the terminal.
4. Synaptic transmission is enhanced because demyelination increases Ca$^{2+}$ influx at the postsynaptic membrane.

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.

Question 5

1. Outward Na$^+$ current because $V_m$ is negative, pushing Na$^+$ out of the cell
2. Inward Na$^+$ current because $V_m<E_{\text{Na}}$, driving Na$^+$ into the cell (correct answer)
3. No Na$^+$ current because Na$^+$ moves only when voltage-gated (not ligand-gated) channels open
4. Inward K$^+$ current because opening Na$^+$ channels causes K$^+$ to enter to maintain electroneutrality

Explanation: This question tests understanding of driving force and current direction in electrical signaling. The driving force for any ion equals the difference between membrane potential and that ion's equilibrium potential (Vm - Eion). For Na+ at Vm = -40 mV with ENa = +60 mV, the driving force is -40 - (+60) = -100 mV. This negative driving force means Na+ flows inward (into the cell) when channels open, as the electrochemical gradient favors Na+ entry. The current is inward because both the electrical gradient (negative inside attracts positive Na+) and concentration gradient (higher Na+ outside) drive Na+ into the cell. Choice A incorrectly assumes negative Vm pushes Na+ out, but driving force direction depends on the difference between Vm and ENa, not Vm alone. To determine current direction, calculate driving force: negative values produce inward current for cations, positive values produce outward current.

Question 6

1. The postsynaptic membrane hyperpolarizes because Na$^+$ exits the cell through ligand-gated channels at rest
2. The postsynaptic membrane depolarizes because net cation current drives $V_m$ toward a value between $E_{\text{Na}}$ and $E_{\text{K}}$ (correct answer)
3. Neurotransmitter release directly opens postsynaptic voltage-gated Ca$^{2+}$ channels, producing an action potential without a graded potential
4. The postsynaptic response occurs only if neurotransmitter binds presynaptic receptors to open presynaptic Na$^+$ channels

Explanation: This question tests understanding of excitatory synaptic transmission and equilibrium potentials. In electrical signaling between neurons, neurotransmitter binding opens ligand-gated channels that allow specific ions to flow according to their electrochemical gradients. When channels permeable to both Na+ and K+ open, the membrane potential moves toward a weighted average of their equilibrium potentials (ENa = +60 mV and EK = -90 mV), which lies between these values. Since the resting potential is -70 mV, opening these channels causes net positive charge influx (more Na+ enters than K+ exits), resulting in depolarization. Choice A incorrectly states Na+ exits at rest, but the negative membrane potential and positive ENa drive Na+ inward. To solve similar problems, identify which ions can flow through opened channels and compare the membrane potential to each ion's equilibrium potential to determine current direction.

Question 7

1. The neuron fires an action potential with a larger overshoot because K$^+$ channels now dominate the rising phase
2. The neuron fails to generate a normal action potential because the rapid inward Na$^+$ current needed for the upstroke is reduced (correct answer)
3. The neuron hyperpolarizes immediately to $E_{\text{Na}}$ because Na$^+$ can no longer enter the cell
4. The neuron still fires normally because synaptic neurotransmitter release compensates for the missing Na$^+$ current

Explanation: This question tests understanding of voltage-gated ion channel roles in action potential generation. Electrical signaling in neurons requires rapid Na+ influx through voltage-gated channels to generate the characteristic fast upstroke of an action potential. When these Na+ channels are blocked, the depolarizing current injection can still move the membrane potential positive, but without the positive feedback of Na+ channel opening, the membrane cannot generate the explosive depolarization that defines an action potential. The neuron fails to fire because the rapid inward Na+ current, which normally drives the membrane from threshold (-55 mV) to peak (+30-40 mV), is absent. Choice A incorrectly suggests K+ channels could substitute for Na+ in the rising phase, but K+ channels cause repolarization, not depolarization. To analyze ion channel contributions, remember that Na+ channels are essential for the rising phase while K+ channels drive the falling phase of action potentials.

Question 8

1. EPSP amplitude decreases because less Ca$^{2+}$ entry reduces vesicle fusion and neurotransmitter release (correct answer)
2. EPSP amplitude increases because reduced Ca$^{2+}$ prevents K$^+$ efflux from the postsynaptic neuron
3. Action potential propagation reverses direction in the presynaptic axon because Ca$^{2+}$ channels determine spike polarity
4. Neurotransmitter release is unchanged because ligand-gated channels on the postsynaptic cell open presynaptic vesicles directly

Explanation: This question tests understanding of calcium's role in neurotransmitter release. In electrical signaling at synapses, voltage-gated Ca2+ channels at presynaptic terminals open during action potentials, allowing Ca2+ influx that triggers vesicle fusion and neurotransmitter release. When the drug blocks 80% of Ca2+ entry, fewer vesicles fuse with the membrane, releasing less neurotransmitter into the synaptic cleft. This reduced neurotransmitter concentration activates fewer postsynaptic receptors, generating a smaller excitatory postsynaptic potential (EPSP). Choice B incorrectly suggests Ca2+ affects postsynaptic K+ channels, but the drug specifically targets presynaptic Ca2+ channels. To approach synaptic transmission problems, trace the sequence: action potential → Ca2+ entry → vesicle fusion → neurotransmitter release → postsynaptic receptor activation → postsynaptic response.

Question 9

1. A second action potential occurs with reduced amplitude because fewer vesicles are available for neurotransmitter release
2. A second action potential cannot be elicited because a critical fraction of Na$^+$ channels remain inactivated (correct answer)
3. A second action potential occurs but propagates backward because K$^+$ channels open first on the second pulse
4. A second action potential is larger because Na$^+$ channels have increased conductance after the first spike

Explanation: This question tests understanding of the absolute refractory period in neuronal electrical signaling. During an action potential, voltage-gated Na+ channels undergo a cycle of closed → open → inactivated states, and inactivated channels cannot reopen until the membrane repolarizes sufficiently. The absolute refractory period occurs when most Na+ channels remain inactivated, preventing action potential generation regardless of stimulus strength. A second suprathreshold stimulus during this period fails to elicit an action potential because insufficient Na+ channels are available to generate the rapid depolarization required. Choice D incorrectly suggests Na+ conductance increases after a spike, but channel inactivation actually reduces available conductance. To understand refractory periods, remember that Na+ channel inactivation creates an absolute refractory period (no action potential possible) followed by a relative refractory period (higher threshold required).

Question 10

1. The action potential upstroke is abolished because Na$^+$ channels must inactivate to open.
2. The membrane remains depolarized longer because persistent Na$^+$ influx opposes repolarizing K$^+$ efflux. (correct answer)
3. The action potential becomes smaller because Na$^+$ influx reverses and becomes outward at positive voltages.
4. Repolarization occurs faster because Na$^+$ inactivation normally delays K$^+$ channel opening.

Explanation: The skill being tested is the effect of sodium channel inactivation on action potential duration. In neurons, electrical signaling during action potentials involves sodium channels opening for depolarization and then inactivating, while potassium channels open for repolarization. In this scenario, preventing sodium channel inactivation allows persistent sodium influx, prolonging depolarization as it opposes the repolarizing potassium efflux. This results in a longer-lasting positive membrane potential, as in choice B. Choice A is incorrect because the upstroke requires sodium channel opening, not inactivation, which normally limits duration. For similar toxin effects, compare the altered waveform to normal phases and ion currents. A useful strategy is to recall that channel states (open, inactivated) dictate spike shape, so disruptions predict specific changes like extended plateaus.

Question 11

1. The resting membrane potential becomes less negative (depolarizes) because $E_K$ shifts upward. (correct answer)
2. The resting membrane potential becomes more negative (hyperpolarizes) because the K$^+$ gradient increases.
3. The resting membrane potential is unchanged because only Na$^+$ determines resting potential.
4. The resting membrane potential depolarizes because K$^+$ flows into the cell down its concentration gradient.

Explanation: The skill being tested is how changes in extracellular ion concentrations affect resting membrane potential via the Nernst equation. In neurons, the resting potential is largely set by potassium permeability through leak channels, approximated by the potassium equilibrium potential EK, which depends on the potassium gradient. Increasing extracellular potassium from 4 mM to 12 mM reduces the gradient, shifting EK less negative and thus depolarizing the resting potential, as in choice A. This logically follows because Vm tracks EK when potassium dominates permeability. Choice B is incorrect as it misstates the gradient change; higher external potassium decreases, not increases, the gradient, leading to depolarization, not hyperpolarization. For similar problems, use the Nernst formula to predict EK shifts qualitatively. A key strategy is to remember that resting Vm moves toward the EK of the most permeable ion when concentrations change.

Question 12

1. The membrane hyperpolarizes toward $E_{Cl}$ because Cl$^-$ influx drives $V_m$ more negative. (correct answer)
2. The membrane depolarizes because Cl$^-$ efflux makes the inside less negative.
3. The presynaptic terminal releases less neurotransmitter because postsynaptic Cl$^-$ channels lower presynaptic Ca$^{2+}$ entry.
4. An action potential is triggered because Cl$^-$ channel opening increases membrane resistance and amplifies depolarization.

Explanation: The skill being tested is the role of chloride channels in inhibitory synaptic potentials. In neurons, electrical signaling at inhibitory synapses often involves opening chloride channels, driving Vm toward ECl, which is typically near or below resting potential. With Vm at -60 mV and ECl near -70 mV, opening channels causes chloride influx, hyperpolarizing the membrane as in choice A. This logically inhibits by moving Vm away from threshold. Choice B is incorrect because at -60 mV > ECl, net influx (not efflux) occurs, hyperpolarizing, not depolarizing. In similar scenarios, compare Vm to Eion to predict direction. A key check is recalling that inhibitory effects can be hyperpolarizing or shunting depending on ECl relative to Vm.

Question 13

1. The inhibitory input can reduce firing by shunting depolarizing current and keeping $V_m$ closer to $E_{Cl}$. (correct answer)
2. The inhibitory input increases firing because Cl$^-$ influx always depolarizes the membrane.
3. The excitatory input prevents firing because Na$^+$ channel opening hyperpolarizes toward $E_{Na}$.
4. Neither input affects firing because only presynaptic action potentials determine postsynaptic spikes.

Explanation: The skill being tested is synaptic integration of excitatory and inhibitory inputs. In neurons, graded potentials from synapses sum at the hillock, with excitatory inputs depolarizing toward threshold and inhibitory ones hyperpolarizing or shunting. Simultaneous excitatory and inhibitory inputs allow the Cl- conductance to shunt depolarization, reducing spike likelihood, as in choice A. This follows because inhibition clamps Vm near ECl, countering excitation. Choice B is incorrect as Cl- influx typically inhibits, not excites, unless ECl > Vm. For similar questions, compare reversal potentials to resting Vm. A key strategy is to consider shunting inhibition's role in modulating excitability without strong hyperpolarization.

Question 14

1. Postsynaptic potentials are greatly reduced because presynaptic Ca$^{2+}$ entry is required for neurotransmitter release. (correct answer)
2. Postsynaptic potentials increase because blocking Ca$^{2+}$ channels increases Na$^+$ influx into the postsynaptic cell.
3. Synaptic transmission is unchanged because Ca$^{2+}$ acts only to repolarize the presynaptic membrane.
4. Presynaptic action potentials fail because Ca$^{2+}$ channels generate the rising phase of the spike.

Explanation: The skill being tested is the necessity of presynaptic calcium for neurotransmitter release. In neurons, action potentials at terminals open calcium channels, with influx triggering vesicle fusion and release. Blocking these channels prevents calcium entry, greatly reducing release and postsynaptic potentials, as in choice A. This logically follows as fusion is calcium-dependent. Choice B is incorrect because blocking calcium doesn't increase sodium influx; it halts transmission. To check, recall calcium's role in exocytosis machinery. Remember, presynaptic blockers affect release, distinguishable from postsynaptic effects in experiments.

Question 15

1. The resting membrane potential tends to depolarize as Na$^+$ accumulates inside and K$^+$ is lost, reducing the K$^+$ gradient. (correct answer)
2. The resting membrane potential hyperpolarizes because the pump normally brings Na$^+$ into the cell.
3. Action potentials become larger because reduced pumping increases the Na$^+$ gradient immediately.
4. Synaptic transmission increases because ATP inhibition directly opens postsynaptic ligand-gated channels.

Explanation: The skill being tested is the role of the sodium-potassium pump in maintaining resting potential. In neurons, the Na/K ATPase pumps sodium out and potassium in, sustaining gradients that set resting Vm via leak permeabilities. Inhibiting the pump allows gradients to rundown, with sodium accumulation and potassium loss depolarizing Vm, as in choice A. This occurs gradually as leaks dissipate ions. Choice B is incorrect as the pump expels sodium, so inhibition doesn't hyperpolarize. In similar scenarios, consider long-term gradient effects. Remember, while not directly electrogenic in steady state, pump failure leads to depolarization over time.

Question 16

1. Current leaks across the membrane through resting conductances and is limited by internal resistance, reducing voltage with distance. (correct answer)
2. The EPSP attenuates because neurotransmitter is degraded as it diffuses down the dendrite.
3. The EPSP attenuates because action potentials can only travel from axon terminal to soma.
4. The EPSP attenuates because voltage-gated Na$^+$ channels inactivate in the dendrite during subthreshold signals.

Explanation: The skill being tested is passive decrement of graded potentials in dendrites. In neurons, subthreshold signals like EPSPs spread passively, decaying due to current leak through membrane resistance and axial resistance limiting spread. Thus, the EPSP attenuates with distance as current leaks and voltage drops, as in choice A. This follows cable theory principles. Choice B is incorrect; EPSPs are electrical, not diffusing neurotransmitter. For similar questions, apply length constant lambda = sqrt(rm/ra). A key strategy is distinguishing passive dendritic spread from active axonal propagation.

Question 17

1. The membrane hyperpolarizes because K$^+$ efflux drives $V_m$ toward $E_K$. (correct answer)
2. The membrane depolarizes because K$^+$ influx drives $V_m$ toward $+60\ \text{mV}$.
3. Neurotransmitter release increases because postsynaptic K$^+$ channels directly activate presynaptic Ca$^{2+}$ channels.
4. An action potential is triggered because opening K$^+$ channels increases Na$^+$ permeability.

Explanation: This question tests understanding of electrical signaling in neurons, specifically the mechanisms underlying inhibitory postsynaptic potentials (IPSPs) through ligand-gated ion channels. Electrical signaling in neurons relies on changes in membrane potential driven by ion fluxes through channels, where the direction of ion movement depends on the electrochemical gradient. In this scenario, a neurotransmitter binds to postsynaptic ligand-gated K+ channels, opening them and permitting K+ ions to flow according to the gradient. The correct answer follows logically because the resting Vm (-65 mV) is more positive than EK (-90 mV), causing K+ efflux that hyperpolarizes the membrane toward EK, producing an IPSP as stated in choice A. Choice B is incorrect because it misstates that K+ influx depolarizes the membrane toward +60 mV, ignoring that +60 mV approximates the Na+ equilibrium potential, not K+, and influx would require Vm to be more negative than EK, which it is not here. A useful check for similar questions is to always compare Vm to the ion's equilibrium potential to determine the direction of flow and resulting potential change. Additionally, remember that IPSPs typically involve hyperpolarization or Cl- influx to inhibit action potential generation in the postsynaptic neuron.

Question 18

1. Fewer available Na$^+$ channels can open on the next spike, reducing the peak inward current and lowering spike amplitude. (correct answer)
2. More available Na$^+$ channels can open on the next spike, increasing amplitude as frequency rises.
3. Spike amplitude decreases because neurotransmitter receptors on the postsynaptic cell are desensitized.
4. Spike amplitude decreases because K$^+$ channels become unable to open when Na$^+$ channels are inactivated.

Explanation: This question tests understanding of electrical signaling in neurons, particularly the role of voltage-gated Na+ channel inactivation in action potential amplitude during high-frequency firing. Electrical signaling in neurons involves action potentials generated by rapid Na+ influx through voltage-gated channels, followed by inactivation that requires time to recover at resting potential. In this scenario, repeated stimulation at high frequency shortens interspike intervals, limiting recovery time for inactivated Na+ channels. The correct answer follows logically because incomplete recovery means fewer Na+ channels are available to open, reducing peak inward current and thus lowering action potential amplitude, as described in choice A. Choice B is incorrect as it suggests more channels become available, which contradicts the mechanism of inactivation and would predict increased rather than decreased amplitude, highlighting a misconception about recovery dynamics. For similar questions, calculate or estimate recovery time relative to stimulation frequency to predict effects on channel availability. A transferable strategy is to recall that Na+ channel inactivation contributes to the refractory period, so high-frequency firing can accumulate inactivation, reducing excitability and spike amplitude.

Question 19

1. The action potential duration increases because inward Na$^+$ current persists longer before repolarization completes. (correct answer)
2. Neurotransmitter release decreases because Na$^+$ channels in the postsynaptic membrane are required for vesicle fusion in the presynaptic terminal.
3. The action potential amplitude decreases because prolonged Na$^+$ channel opening drives the membrane toward $E_{\text{K}}$ instead of $E_{\text{Na}}$.
4. Action potentials propagate only toward the soma because delayed Na$^+$ channel inactivation reverses the direction of current flow along the axon.

Explanation: This question tests understanding of action potential kinetics and the role of Na+ channel inactivation. During a normal action potential, Na+ channels rapidly inactivate after opening, allowing K+ channels to repolarize the membrane. When a toxin slows Na+ channel inactivation, Na+ channels remain open longer, continuing to allow Na+ influx even as K+ channels try to repolarize the membrane. This prolongs the depolarization phase and increases action potential duration. The correct answer (A) accurately describes this prolongation. Answer C incorrectly suggests the amplitude would decrease and that prolonged Na+ opening would drive the membrane toward EK rather than ENa. To analyze channel kinetics problems, consider how changes in channel opening/closing times affect the balance between depolarizing and repolarizing currents.

Question 20

1. The postsynaptic EPSP amplitude decreases because reduced presynaptic Ca$^{2+}$ influx lowers neurotransmitter release probability. (correct answer)
2. The postsynaptic EPSP amplitude increases because reduced presynaptic Ca$^{2+}$ influx directly opens postsynaptic ligand-gated channels.
3. The presynaptic action potential fails to propagate because Ca$^{2+}$ influx is required for voltage-gated Na$^+$ channel opening along the axon.
4. Neurotransmitter release increases because low extracellular Ca$^{2+}$ enhances vesicle fusion by reducing charge repulsion at the membrane.

Explanation: This question tests understanding of synaptic transmission and the role of calcium in neurotransmitter release. At chemical synapses, action potentials trigger opening of voltage-gated Ca2+ channels in the presynaptic terminal, and the resulting Ca2+ influx is essential for vesicle fusion and neurotransmitter release. When extracellular Ca2+ is reduced from 2.0 mM to 0.2 mM (a 10-fold reduction), less Ca2+ enters the presynaptic terminal, resulting in decreased neurotransmitter release and smaller postsynaptic EPSPs. The correct answer (A) accurately describes this relationship. Answer B incorrectly suggests that reduced presynaptic Ca2+ would directly affect postsynaptic channels, which is not how synaptic transmission works. When analyzing synaptic questions, always trace the sequence: presynaptic Ca2+ influx → vesicle fusion → neurotransmitter release → postsynaptic receptor activation.