A neurotoxin blocks all voltage-gated ion channels in an axon membrane, leaving only the passive membrane resistance and capacitance intact. When current is injected at one point, what happens?
AAction potentials propagate normally because passive spread is sufficient to carry the signal the full length of the axon
BVoltage decays exponentially with distance from the injection site, described by V(x) = V₀ exp(−x/λ), with no regeneration at distant sites
CThe membrane potential does not change at all because all ion movement is blocked
DThe signal propagates faster than normal because the slow sodium channel activation step is bypassed
Without voltage-gated channels, only passive cable spread operates. The voltage change decreases exponentially with distance, decaying to 37% of its original value at x = λ (the length constant). No new action potentials can be initiated at distant sites because there are no channels to regenerate the signal. This is the 'leaky hose' model: current enters at one point, leaks out through passive membrane conductance along the way, and the pressure (voltage) drops steadily with distance. Action potentials require active channel gating — they are not passively conducted.
Question 2 Multiple Choice
Myelination increases the length constant (λ) of an axon, which speeds action potential conduction. The mechanism is:
AMyelin adds voltage-gated sodium channels at regular intervals, reducing the distance current must travel passively
BMyelin increases intracellular axial resistance, forcing current to flow more rapidly down the axon interior
CMyelin increases membrane resistance (reducing current leak) and decreases membrane capacitance (reducing the charge needed to change membrane voltage), both of which increase λ
DMyelin decreases axon diameter, which concentrates current flow and increases signal amplitude
λ = √(rₘ/rᵢ). Myelin wraps around the axon, making the membrane much thicker — this dramatically increases rₘ (less current leaks out per unit length, like a better-insulated hose) and decreases cₘ (thicker insulation reduces capacitance, so less charge is needed to change the membrane voltage). Both effects increase λ. The result is that passive depolarization can spread much farther before decaying below threshold, allowing the next node of Ranvier to be triggered from much greater distance. Fewer nodes = fewer regeneration steps = faster conduction.
Question 3 True / False
The length constant (λ) of a given axon is a fixed physical property that cannot be altered by changes in the axon's structural or molecular properties.
TTrue
FFalse
Answer: False
λ = √(rₘ/rᵢ) depends on membrane resistance per unit length (rₘ) and intracellular axial resistance per unit length (rᵢ). Both are variable: rₘ changes with myelination and with the number and type of open ion channels in the membrane; rᵢ changes with axon diameter (wider axons have lower axial resistance, increasing λ). Myelination is the clearest biological example — it dramatically increases rₘ and decreases cₘ, increasing λ by orders of magnitude compared to unmyelinated axons of similar diameter.
Question 4 True / False
Passive current spread governed by cable theory is essential for action potential propagation even though action potentials are active, regenerative events.
TTrue
FFalse
Answer: True
Between active channel clusters (between nodes of Ranvier in myelinated axons, or between channel-dense patches in unmyelinated ones), current must travel passively. The depolarization from one action potential must spread passively far enough to bring the next cluster of voltage-gated sodium channels to threshold. The length constant determines how far this passive spread reaches before decaying. If λ is too small (as in demyelination), the passive current does not reach the next node at sufficient amplitude, and propagation fails. Cable theory describes the passive infrastructure on which active propagation depends.
Question 5 Short Answer
Explain why a larger length constant (λ) speeds action potential conduction in a myelinated axon, even though action potentials themselves are not passive signals.
Think about your answer, then reveal below.
Model answer: Action potentials require active regeneration at each node of Ranvier — voltage-gated sodium channels must be triggered to sustain the signal. Between nodes, current spreads passively. A larger λ means passive depolarization travels farther before decaying below the threshold for triggering the next node. In myelinated axons, myelin increases rₘ and decreases cₘ, both increasing λ, so the passive current can jump across longer internodal distances and still trigger the next node. Fewer nodes are needed to cover the same axon length, and each passive 'jump' covers more ground, making conduction faster overall.
This is the mechanistic explanation for saltatory conduction (the action potential 'jumping' from node to node). Cable theory explains *why* the jump works: sufficient passive depolarization reaches each node because λ is large enough. Without this understanding, myelin's effect on speed is mysterious; with it, the physics is transparent.