- In a study of synaptic integration, a neuron receives a train of identical excitatory presynaptic spikes at 50 Hz. Under control conditions, the postsynaptic potentials show temporal summation and eventually reach threshold. A drug is applied that increases postsynaptic membrane leak conductance (decreases input resistance) without changing synaptic receptor conductance or presynaptic release. What outcome would be expected?
Assume: membrane time constant τ=RmCm; higher leak conductance lowers Rm and reduces voltage change for a given synaptic current.
Answer: Temporal summation decreases because lowering Rm reduces EPSP amplitude and shortens τ, making threshold less likely to be reached
This question tests understanding of passive membrane properties and temporal summation. Temporal summation occurs when successive EPSPs overlap in time before decaying, allowing their amplitudes to add. The membrane time constant (τ = RmCm) determines how quickly potentials decay; decreasing Rm by increasing leak conductance reduces τ, causing faster decay. Additionally, lower Rm means the same synaptic current produces smaller voltage changes (V = I × R). The correct answer (B) recognizes both effects: smaller, faster-decaying EPSPs are less likely to summate to threshold. Answer A incorrectly claims lower Rm increases τ, when it actually decreases it according to the time constant equation. For integration problems, remember that increased leak conductance makes neurons less responsive by reducing both EPSP amplitude and duration. - At a glutamatergic synapse, miniature EPSCs (mEPSCs) are recorded in the postsynaptic neuron in the presence of tetrodotoxin (TTX) to block action potentials. A presynaptic-targeted drug is then applied that decreases the probability of vesicle fusion but does not change vesicle neurotransmitter content. Compared with baseline, mEPSC frequency decreases while mEPSC amplitude is unchanged. Which mechanism best explains this pattern?
Answer: Reduced presynaptic release probability lowers event frequency without altering quantal size, leaving individual mEPSC amplitudes unchanged
This question tests understanding of quantal analysis and the distinction between presynaptic and postsynaptic mechanisms. Miniature EPSCs represent the postsynaptic response to single vesicle release events, with frequency reflecting release probability and amplitude reflecting quantal size (vesicle content and postsynaptic receptor response). A drug that decreases vesicle fusion probability reduces how often vesicles spontaneously release, decreasing mEPSC frequency without affecting the neurotransmitter content per vesicle or postsynaptic response, leaving amplitude unchanged. The correct answer (D) accurately describes this presynaptic mechanism. Option B incorrectly suggests presynaptic changes affect postsynaptic receptor conductance, conflating pre- and postsynaptic mechanisms. For quantal analysis questions, remember that frequency changes typically indicate presynaptic effects while amplitude changes can be either pre- or postsynaptic.
- Cultured hippocampal neurons are voltage-clamped at −70 mV. A presynaptic terminal is stimulated once, evoking neurotransmitter release onto a postsynaptic neuron expressing ligand-gated receptors permeable to Na+ and K+. In control solution, the postsynaptic current is an inward (negative) deflection. The experiment is repeated after changing the extracellular ionic composition so that extracellular Na+ is markedly reduced while extracellular K+ is unchanged; membrane potential is still clamped at −70 mV. Which mechanism best explains the expected change in the postsynaptic current?
Assume: the receptor channel is nonselective for Na+ vs K+; current direction reflects net cation flux at the clamped voltage.
Answer: The current decreases in magnitude (less inward) because reduced extracellular Na+ reduces Na+ influx through the open receptor channels at −70 mV
This question tests understanding of synaptic current direction and ionic driving forces. The postsynaptic current through ligand-gated channels depends on the electrochemical gradient for each permeant ion. At -70 mV, Na+ has a large inward driving force (ENa ≈ +60 mV) while K+ has a small outward driving force (EK ≈ -90 mV), resulting in net inward current. When extracellular Na+ is reduced, the Na+ equilibrium potential becomes less positive, reducing the inward driving force for Na+ and thus reducing the magnitude of inward current. The correct answer (B) properly identifies that reduced extracellular Na+ decreases Na+ influx through open channels. Answer C incorrectly suggests the current would reverse to outward, but K+ efflux alone at -70 mV is insufficient to overcome residual Na+ influx. For similar problems, calculate how changing ion concentrations affects equilibrium potentials and driving forces. - A researcher compares conduction in two axons of equal diameter: Axon 1 is unmyelinated; Axon 2 is myelinated with regularly spaced nodes of Ranvier. A toxin is applied that increases the capacitance of the axonal membrane without changing ion channel densities. The toxin slows conduction much more in Axon 1 than in Axon 2. Which mechanism best explains why the myelinated axon is less affected?
Answer: Myelin reduces the membrane area that must be charged between nodes, limiting the impact of increased capacitance on propagation speed
This question tests understanding of how myelination affects action potential propagation and the role of membrane capacitance. In unmyelinated axons, the entire membrane must be charged to threshold for continuous propagation, making conduction speed highly sensitive to membrane capacitance. In myelinated axons, myelin acts as an insulator that dramatically reduces the membrane area exposed at internodal regions, so only the small nodal regions need to be charged to threshold during saltatory conduction. When capacitance increases, unmyelinated axons must charge much more membrane area, significantly slowing conduction, while myelinated axons only need to charge the small nodal areas, minimizing the impact. The correct answer (A) accurately explains how myelin reduces the membrane area requiring charging, limiting capacitance effects. Answer B incorrectly states myelin increases leak conductance, C incorrectly involves neurotransmitter diffusion along axons, and D contradicts the fundamental role of myelin in speeding conduction. When comparing myelinated and unmyelinated axons, remember that myelin reduces the effective membrane area participating in propagation.
- In a sensory pathway experiment, a brief touch stimulus to the fingertip activates a primary afferent neuron that synapses onto a second-order neuron in the dorsal horn. The second-order neuron then projects to the thalamus. A pharmacologic agent is applied locally at the dorsal horn synapse that increases presynaptic Ca2+ influx during an arriving action potential, without changing action potential shape or postsynaptic receptor properties. What outcome would be expected at the dorsal horn synapse?
Assume: transmitter release probability increases with presynaptic Ca2+ entry.
Answer: A larger postsynaptic potential because increased presynaptic Ca2+ entry increases neurotransmitter release per presynaptic spike
This question tests understanding of calcium's role in neurotransmitter release at synapses. At chemical synapses, the amount of neurotransmitter released is directly proportional to presynaptic Ca2+ influx during the action potential. When a drug increases Ca2+ channel opening or conductance, more Ca2+ enters the terminal, triggering fusion of more synaptic vesicles and greater transmitter release. The correct answer (B) properly identifies that increased presynaptic Ca2+ entry enhances neurotransmitter release, producing a larger postsynaptic potential. Answer A incorrectly suggests Ca2+ would shorten the action potential and reduce release, when the drug specifically increases Ca2+ influx without changing spike shape. For synaptic modulation questions, remember that presynaptic Ca2+ concentration is the key determinant of release probability and quantal content. - In a patch-clamp experiment on a neuron, an investigator injects a constant depolarizing current step. Under control conditions, the neuron fires an action potential. After applying a drug that selectively prolongs the open time of voltage-gated K+ channels during the falling phase of the action potential, the neuron still fires but the action potential is narrower and the after-hyperpolarization is larger.
Which mechanism best explains these changes in action potential shape?
Answer: Enhanced K+ efflux accelerates repolarization and deepens after-hyperpolarization by driving Vm toward EK
This question tests understanding of action potential repolarization in the nervous system. Voltage-gated potassium channels facilitate repolarization by allowing potassium efflux, restoring membrane potential to rest. Prolonging their open time enhances efflux, speeding repolarization and deepening after-hyperpolarization. The correct answer aligns because increased potassium conductance narrows the spike and amplifies hyperpolarization toward EK. A common distractor like option B fails by misunderstanding that enhanced potassium efflux accelerates, not slows, repolarization. For similar questions, analyze ion currents during AP phases. Verify that potassium modulation affects falling phase duration and refractory properties. - A researcher studies synaptic transmission at a cholinergic neuromuscular junction in an isolated muscle preparation. A motor neuron is stimulated at 1 Hz, and end-plate potentials (EPPs) are recorded in the muscle fiber. In control conditions, each stimulus evokes an EPP that reliably reaches threshold and triggers a muscle action potential. The researcher then adds a toxin that cleaves SNARE proteins in the presynaptic terminal, without affecting postsynaptic receptors or voltage-gated channels. Which outcome would be expected?
Assume: SNARE-mediated vesicle fusion is required for neurotransmitter exocytosis.
Answer: EPP amplitude decreases or disappears because vesicle fusion is impaired, reducing acetylcholine release into the synaptic cleft
This question tests understanding of synaptic vesicle fusion and neurotransmitter release mechanisms. SNARE proteins are essential for vesicle fusion with the presynaptic membrane, allowing neurotransmitter exocytosis into the synaptic cleft. When SNARE proteins are cleaved by toxin, vesicles cannot fuse even when Ca2+ enters during an action potential, preventing acetylcholine release. The correct answer (A) recognizes that without vesicle fusion, neurotransmitter cannot be released, leading to decreased or absent EPPs and failure of muscle contraction. Answer C incorrectly suggests direct cytosolic diffusion could substitute for vesicular release, which doesn't occur at chemical synapses. When analyzing synaptic transmission problems, trace the sequence from action potential to Ca2+ entry to vesicle fusion to transmitter release.
- A researcher records from a postsynaptic neuron receiving excitatory input. Two conditions are compared: (i) a single presynaptic stimulus producing one EPSP; (ii) five presynaptic stimuli at 100 Hz producing a larger, temporally summated depolarization. In condition (ii), the neuron reaches threshold and fires an action potential, whereas in condition (i) it does not. What outcome would be expected if the postsynaptic membrane time constant (τ=RmCm) is decreased by 50% (with resting potential unchanged) while stimulation protocols remain the same?
Answer: Temporal summation becomes less effective because the membrane potential decays faster between inputs, reducing the likelihood of reaching threshold in condition (ii)
This question tests understanding of temporal summation and membrane time constants. The membrane time constant (τ = RmCm) determines how quickly voltage changes decay back to baseline - a smaller τ means faster decay. During temporal summation, successive EPSPs build upon residual depolarization from previous inputs, but with a reduced time constant, each EPSP decays more rapidly, leaving less residual depolarization for subsequent inputs to sum with. The correct answer (D) correctly identifies that faster decay between inputs reduces the effectiveness of temporal summation, making it less likely that the summed depolarization will reach threshold. Option B incorrectly suggests smaller τ increases charge storage, when actually it decreases the duration of stored charge. For integration questions, remember that temporal summation depends critically on the membrane time constant matching the inter-stimulus interval.
- In a cortical circuit, an excitatory neuron synapses onto a target neuron. A drug is applied that selectively blocks GABAA receptors but does not affect glutamate receptors, voltage-gated channels, or neurotransmitter release probability. In response to the same excitatory presynaptic firing pattern, the target neuron shows a higher probability of firing action potentials. Which mechanism best explains this change in target neuron output?
Answer: Blocking GABAA receptors reduces inhibitory Cl− conductance, increasing membrane resistance and allowing excitatory inputs to produce larger depolarizations
This question tests understanding of synaptic integration and the role of GABAergic inhibition. GABA_A receptors are ligand-gated chloride channels that typically produce inhibitory postsynaptic potentials by increasing membrane conductance and shunting excitatory currents. When GABA_A receptors are blocked, the inhibitory conductance is removed, increasing the neuron's input resistance and allowing excitatory postsynaptic potentials to produce larger depolarizations that more readily reach action potential threshold. The correct answer (A) accurately describes this disinhibition mechanism. Option C incorrectly states that blocking receptors increases Cl- influx, when blocking actually prevents chloride conductance. Understanding synaptic integration requires recognizing that inhibition works both by hyperpolarization and by conductance changes that shunt excitatory currents. - A patient has a focal lesion that selectively disrupts decussation (crossing) of second-order somatosensory neurons in the spinal cord, while leaving peripheral sensory neurons and the thalamus intact. A light touch stimulus is applied to the right hand. Based on the scenario, what outcome would be expected for cortical perception of that touch?
Assume: in the intact pathway, second-order touch neurons cross to the contralateral side before ascending to the thalamus and then to somatosensory cortex.
Answer: Touch from the right hand is perceived in the right somatosensory cortex because the pathway fails to cross and remains ipsilateral
This question tests understanding of sensory pathway organization and decussation in the nervous system. In the normal dorsal column-medial lemniscal pathway for touch, second-order neurons cross the midline (decussate) in the medulla or spinal cord before ascending to the contralateral thalamus and cortex. When decussation is disrupted, these neurons cannot cross and instead project ipsilaterally, meaning sensory information from the right hand reaches the right thalamus and right cortex instead of the left. The correct answer (A) recognizes that blocking decussation causes the pathway to remain on the same side, resulting in ipsilateral rather than contralateral perception. Answer B incorrectly suggests enhanced crossing when the mechanism for crossing is damaged. When analyzing sensory pathways, trace the anatomical route including where crossing occurs to predict effects of lesions.
- A neuron is tested for excitability while extracellular Na+ is reduced, with osmolarity maintained by replacing Na+ with an impermeant cation. The threshold for firing increases, and action potentials that do occur have reduced peak amplitude.
Which mechanism best explains these observations?
Answer: Lower extracellular Na+ reduces the Na+ electrochemical driving force, decreasing inward current during the upstroke and reducing spike amplitude
This question tests understanding of ion gradients in action potential generation. Sodium driving force determines inward current magnitude during the upstroke. Reducing extracellular sodium weakens this force, lowering peak amplitude. The correct answer aligns because diminished gradient reduces current and spike height. A common distractor like option B fails by misunderstanding that lower external sodium decreases, not increases, influx. For similar questions, apply Goldman equation principles. Verify that gradient manipulations affect amplitude proportionally. - In an isolated frog sciatic nerve preparation, a single suprathreshold stimulus is delivered at one end of a myelinated axon. The extracellular solution initially contains (in mM): Na+ 145, K+ 4, Cl− 110, Ca2+ 2. The axon is then perfused with a solution in which extracellular Na+ is reduced to 20 mM and replaced with an impermeant cation to maintain osmolarity; all other ions are unchanged. Resting membrane potential remains near −70 mV, but the recorded action potential peak amplitude decreases substantially and conduction fails in some trials. Which mechanism best explains the reduced action potential amplitude and intermittent propagation failure under low extracellular Na+ conditions?
Answer: Decreased Na+ electrochemical driving force reduces inward Na+ current through voltage-gated Na+ channels, limiting depolarization and impairing regenerative propagation
This question tests understanding of action potential generation and the role of sodium ions in neural conduction. Action potentials depend on the electrochemical gradient for sodium ions, which drives Na+ influx through voltage-gated sodium channels during depolarization. When extracellular Na+ is reduced from 145 mM to 20 mM, the driving force for sodium (Vm - ENa) decreases substantially because ENa becomes less positive. The correct answer (A) explains that this reduced driving force limits the inward Na+ current during depolarization, resulting in smaller action potentials that may fail to reach threshold at subsequent nodes. Option C incorrectly suggests increased K+ efflux would cause a larger overshoot, when actually the overshoot depends on Na+ influx. A key principle for similar questions is that action potential amplitude and propagation depend critically on the sodium electrochemical gradient, not just the presence of functional channels. - A sensory afferent neuron synapses onto a spinal interneuron. In a slice preparation, the afferent is stimulated twice with an inter-stimulus interval of 40 ms. The second excitatory postsynaptic potential (EPSP) recorded in the interneuron is larger than the first, despite identical presynaptic stimuli. Postsynaptic membrane resistance and resting potential are unchanged between pulses. Which mechanism best explains this short-term increase in EPSP amplitude?
Answer: Residual Ca2+ in the presynaptic terminal after the first stimulus increases neurotransmitter release during the second stimulus
This question tests understanding of short-term synaptic plasticity, specifically paired-pulse facilitation. When two stimuli are delivered in rapid succession (40 ms interval), residual calcium from the first stimulus remains in the presynaptic terminal when the second stimulus arrives. This residual Ca2+ adds to the Ca2+ influx from the second stimulus, resulting in higher total calcium concentration and increased neurotransmitter release probability during the second pulse. The correct answer (A) identifies this calcium accumulation mechanism, which is a fundamental form of short-term enhancement at many synapses. Option D incorrectly invokes endocrine signaling, which operates on much slower timescales than the millisecond dynamics of synaptic transmission. When analyzing synaptic plasticity, consider whether the timescale (milliseconds to seconds) is consistent with local calcium dynamics versus slower processes like gene expression or hormonal signaling. - In a spinal cord slice, a sensory afferent makes a monosynaptic excitatory connection onto a motor neuron. Investigators record excitatory postsynaptic potentials (EPSPs) in the motor neuron while stimulating the afferent. When two identical stimuli are delivered 20 ms apart, the second EPSP is larger than the first. No changes are detected in postsynaptic input resistance or resting membrane potential.
Which mechanism best explains the increased amplitude of the second EPSP?
(Assume the synapse uses a typical fast excitatory neurotransmitter and that postsynaptic receptor number is unchanged.)
Answer: Residual presynaptic Ca2+ from the first spike increases vesicle release probability for the second stimulus
This question tests understanding of short-term synaptic plasticity in the nervous system. Paired-pulse facilitation occurs when residual calcium from the first stimulus elevates release probability for the second. In this excitatory synapse, the brief interval allows calcium accumulation to enhance vesicle fusion for the subsequent EPSP. The correct answer aligns because elevated presynaptic calcium boosts neurotransmitter release, increasing the second EPSP amplitude. A common distractor like option B fails by misunderstanding that facilitation is presynaptic, not due to postsynaptic sodium channel inactivation. For similar questions, assess if timing suggests calcium-mediated plasticity. Verify that unchanged postsynaptic properties point to presynaptic mechanisms in facilitation. - A cultured neuron receives a brief excitatory synaptic input at a dendritic spine. Under control conditions, the postsynaptic potential (PSP) rises to a peak of +6 mV at the soma and decays back to baseline within 30 ms. A researcher then applies a compound that increases membrane leak conductance uniformly across the soma and dendrites without changing the synaptic receptor current at the spine. The same synaptic input now produces a +2 mV peak at the soma. Which outcome would be expected if leak conductance increases in this neuron?
Answer: The membrane time constant decreases, causing greater attenuation of the PSP as it spreads to the soma
This question tests understanding of passive membrane properties and how changes in membrane conductance affect synaptic potential propagation. The membrane time constant (τ = RC) determines how quickly voltage changes decay over time, where R is membrane resistance and C is capacitance. When leak conductance increases, membrane resistance decreases, which reduces the time constant and causes faster decay of the postsynaptic potential as it spreads from the dendritic spine to the soma. The correct answer (A) properly identifies that a decreased time constant leads to greater attenuation of the PSP amplitude by the time it reaches the soma. Answer B incorrectly states the time constant increases when conductance increases, C incorrectly involves presynaptic mechanisms when the change is postsynaptic, and D misunderstands how leak conductance affects resting potential. When analyzing passive spread of synaptic potentials, remember that increased conductance means decreased resistance, faster decay, and more attenuation over distance.
- A neuron receives two simultaneous synaptic inputs on its dendrites: one excitatory input that opens cation channels with a reversal potential near 0 mV, and one inhibitory input that opens Cl− channels with a reversal potential of −70 mV. The neuron's resting membrane potential is −70 mV. When both inputs are activated together, the depolarization produced by the excitatory input alone is reduced.
Which mechanism best explains the reduced depolarization during co-activation?
Answer: Inhibitory input increases membrane conductance, reducing the voltage change produced by the excitatory synaptic current (shunting inhibition)
This question tests understanding of synaptic integration in the nervous system. Shunting inhibition increases conductance, reducing voltage changes from excitatory currents without strong hyperpolarization. Co-activation allows inhibitory conductance to dampen excitatory depolarization at rest. The correct answer aligns because elevated conductance shunts the excitatory input, limiting net depolarization. A common distractor like option B fails by misunderstanding that inhibition can shunt without depolarizing the resting potential. For similar questions, consider reversal potentials and conductance changes. Verify that simultaneous inputs integrate via summation and shunting mechanisms.
- A neuron is held at a resting membrane potential of approximately −70 mV. A brief inhibitory synaptic input opens postsynaptic Cl− channels. The measured reversal potential for Cl− in this neuron is −75 mV. During the inhibitory input, the membrane potential shifts slightly negative and the neuron becomes less likely to fire in response to an excitatory input.
Which mechanism best explains how opening Cl− channels reduces firing probability in this neuron?
Answer: Cl− conductance drives the membrane potential toward ECl and increases membrane conductance, shunting excitatory depolarization
This question tests understanding of inhibitory synaptic mechanisms in the nervous system. Inhibitory postsynaptic potentials often involve chloride channels, which increase conductance and drive membrane potential toward the chloride reversal potential. Here, opening chloride channels at rest hyperpolarizes slightly and shunts excitatory inputs by reducing membrane resistance. The correct answer aligns because the conductance increase clamps voltage near ECl, diminishing excitatory depolarization. A common distractor like option B fails by misunderstanding that chloride influx typically hyperpolarizes or shunts, not depolarizes to threshold. For similar questions, calculate if the reversal potential is below rest to predict hyperpolarization. Always check for shunting effects in inhibition even without net hyperpolarization. - A postsynaptic neuron expresses a metabotropic receptor that, when activated by neurotransmitter, increases intracellular cAMP and leads to phosphorylation that closes a subset of K+ leak channels. In current-clamp at rest, activation of this receptor causes the membrane potential to depolarize by 4 mV and increases the likelihood of firing in response to a subsequent excitatory input. Which mechanism best explains the increased excitability?
Answer: Closing K+ leak channels reduces outward K+ current at rest, depolarizing the membrane toward threshold
This question tests understanding of metabotropic receptor signaling and how modulation of leak channels affects neuronal excitability. K+ leak channels contribute to the resting membrane potential by allowing K+ efflux, keeping the membrane hyperpolarized. When the metabotropic receptor activates the cAMP pathway leading to phosphorylation that closes these K+ leak channels, the outward K+ current at rest decreases, causing the membrane to depolarize toward threshold and making the neuron more likely to fire in response to excitatory input. The correct answer (A) properly describes how closing K+ leak channels reduces outward current and depolarizes the membrane. Answer B incorrectly states that closing channels increases current, C incorrectly involves systemic hormone effects, and D nonsensically suggests cAMP creates Na+ ions. When analyzing metabotropic modulation of excitability, consider how changes in leak conductances shift the resting potential relative to threshold. - In a brainstem reflex circuit, an afferent sensory neuron makes an excitatory synapse onto an interneuron, which then makes an inhibitory synapse onto a motor neuron. In response to a brief sensory stimulus, the motor neuron normally shows a transient hyperpolarization (inhibitory postsynaptic potential, IPSP) starting 6 ms after the stimulus. A drug is applied that selectively blocks GABAA receptors on the motor neuron but does not affect GABA release or GABAB receptors. Which function is most consistent with the observed change in motor neuron response after drug application?
Answer: The motor neuron IPSP is reduced because fast inhibitory Cl− conductance is eliminated at the postsynaptic membrane
This question tests understanding of inhibitory synaptic transmission and the specific role of GABA_A receptors. GABA_A receptors are ligand-gated Cl- channels that, when activated by GABA, allow Cl- influx which typically hyperpolarizes the postsynaptic neuron, producing an inhibitory postsynaptic potential (IPSP). When these receptors are blocked by the drug, the inhibitory interneuron still releases GABA, but it cannot activate the blocked GABA_A receptors on the motor neuron, eliminating the fast Cl- conductance and thus the IPSP. The correct answer (A) properly identifies that blocking GABA_A receptors eliminates the postsynaptic Cl- conductance responsible for the IPSP. Answer B contradicts basic pharmacology by suggesting blocking increases function, C incorrectly describes GABA_A receptors as presynaptic Ca2+ channels, and D confuses ionotropic GABA_A receptors with metabotropic signaling. When analyzing inhibitory synaptic circuits, remember that GABA_A receptors are postsynaptic Cl- channels that produce fast inhibition. - A researcher records from a neuron that exhibits a refractory period after each action potential. Two identical suprathreshold current pulses are injected into the soma. When the second pulse is delivered 2 ms after the first, no second action potential occurs. When the second pulse is delivered 8 ms after the first, a second action potential occurs but requires a slightly larger pulse amplitude than baseline.
Which mechanism best explains the lack of a second action potential at 2 ms and the elevated threshold at 8 ms?
Answer: At 2 ms many voltage-gated Na+ channels remain inactivated (absolute refractory); at 8 ms increased K+ conductance persists (relative refractory)
This question tests understanding of refractory periods in neuronal excitability. Absolute refractoriness involves sodium channel inactivation, while relative refractoriness features lingering potassium conductance. At 2 ms, inactivated sodium channels prevent firing; at 8 ms, elevated potassium raises threshold. The correct answer aligns because these mechanisms explain the timing of excitability recovery. A common distractor like option B fails by misunderstanding channel states during recovery phases. For similar questions, recall absolute versus relative refractory timelines. Verify that sodium inactivation dominates early, with potassium effects persisting longer.
