Problems: Module 2
Write your answer first, then open the solution. Assignment key items show the points the grader awards; chart problems use the course figures; variants change the numbers or flip the premise; check-yourself items answer the end-of-module review questions.
Assignment key¶
Solution to Exercise 1
Non-excitable membranes simply have a membrane potential; it exists but does not drive their function. Excitable membranes are designed to let that potential fluctuate deliberately, and the fluctuation is how the cell does its job. One point per side, so describing only one type scores half. (slide 3)
Solution to Exercise 2
Pressure sensors: anatomically visceral receptors, functionally baroreceptors. (slides 7–8)
Heart rate command: motor, autonomic, sympathetic.
Hot metal: anatomically somatic receptors, functionally thermoreceptors.
Route into the CNS: sensory, or afferent.
Leg commands: motor, somatic.
Solution to Exercise 3
Electrochemical equilibrium is the membrane potential at which the chemical driving force and the electrical driving force offset each other in opposite directions. Neither gradient is zero, and the concentrations are not equal; the two forces simply cancel, so there is no net movement. (slides 12–14)
Solution to Exercise 4
One point per blank. The diagram is not reproduced here, so learn what each feature looks like on a trace instead.
Term | What it looks like on the trace |
|---|---|
Individual EPSP | A small upward bump from one excitatory input that stays below threshold |
Individual IPSP | A small downward dip from one inhibitory input |
Temporal summation | Bumps stacking from the same input firing in quick succession |
Spatial summation | A larger deflection from different inputs firing at the same moment |
Graded potential depolarization / repolarization / hyperpolarization | The rising edge of a bump / its return to baseline / the falling edge of a dip, all subthreshold |
Action potential depolarization | The steep upstroke after threshold, up to about +30 mV |
Action potential hyperpolarization | The undershoot below resting after repolarization |
Absolute refractory phase | From threshold through the spike and most of repolarization |
Relative refractory phase | The hyperpolarized undershoot that follows |
slides 29–39
Solution to Exercise 5
Solution to Exercise 6
Hypokalemia: lower outside K+ enlarges the concentration gradient, so more K+ leaves, leaving the inside more negative. Vm becomes more negative.
Hyperkalemia: higher outside K+ shrinks the gradient, so less K+ leaves and more positive charge stays inside. Vm becomes less negative. Each prediction is 2 points and each rationale 2 points, and rationale scores even if the prediction is wrong.
Why it’s dangerous: a smaller gap means threshold is easier to reach, so the membrane is more excitable. Too small and it becomes irritable, firing at inappropriate times, giving hyper-reactive nerve signaling and cardiac arrhythmias. A larger gap means less excitable.
Solution to Exercise 7
Ouabain (3): Vm becomes less negative, but only gradually, because K+ leak channels carry most of the resting conductance and the pump contributes modestly. (slide 22)
Electrogenic (2): it moves Na+ out and K+ in, but lopsidedly, 3 out for 2 in, so the inside loses one net positive charge per cycle.
No equilibrium (1): the pumps keep carrying K+ back in, restoring the gradient the leak channels dissipate.
Solution to Exercise 8
Amplitude: decays with distance, because nothing boosts or refreshes it once initiated. (slide 31)
Refractory: none at all, because graded potentials do not use channels with inactivation gates, which are what make refractory periods. (slides 38–40)
Threshold: none required to start one. (slides 29–30)
X three times: temporal summation, 3 × 4 = +12 mV.
X and Y together: spatial summation, +4 − 5 = −1 mV. (slide 32)
Solution to Exercise 9
Dendrite (4): λ is small, because graded potentials decay quickly and are subthreshold by the axon hillock. With ri roughly fixed, a small λ requires a small rm, meaning a leaky membrane that loses current as the signal travels. (slide 31)
Myelin (2): raises rm, since the wrapping stops ions leaking across the membrane. (slide 41)
Distance (3): unmyelinated axons leak current quickly, so it falls below threshold over a short distance and the action potential must regenerate at every adjacent patch. In myelinated axons the high rm raises λ, so current travels much farther above threshold and regenerates only at the nodes of Ranvier. (slides 40–41)
Solution to Exercise 10
Absolute (2): activation gate open, inactivation gate closed. The channel cannot open again until it resets, no matter how strong the stimulus. (slides 38–39)
Relative (1): the membrane is hyperpolarized, so the gap to threshold is larger and the membrane less excitable.
Mid-axon (2): propagates in both directions, because there is no trailing patch of refractory membrane to block one side. (slide 40)
Coding (2): refractory periods cap the firing rate so action potentials stay distinct. Without them, they would merge like summing graded potentials and the brain could not read their frequency.
Cooling (3): slower gates lengthen the absolute refractory period, which lowers maximum firing frequency, so the brain perceives the cold as less intense. The beginning of numbing.
Solution to Exercise 11
Terminal (2): voltage-gated Ca2+ channels, Ca2+ flowing in. (slide 43)
Toxin (2): neurotransmission is blocked, so the skeletal muscle cell is never stimulated and cannot contract. (slide 44)
Receptors (4): ligand-gated ion channels on a neuron, G protein-coupled receptors on cardiac muscle. The ligand-gated channel is faster because far fewer steps must happen. (Module 1 slide 16)
Mixed channel (4): net EPSP. Ion flow is proportional to Vm − EX. At Vm ≈ −70 mV, ENa ≈ +60 mV sits 130 mV away while EK ≈ −90 mV sits only 20 mV away, so far more Na+ enters than K+ leaves. (slide 16)
Chart analysis¶
Solution to Exercise 12
Na+ permeability rises steeply right at threshold and peaks with the action potential, because voltage-gated Na+ channels respond almost instantly. It falls as the channels inactivate at the peak.
K+ permeability rises later and more gradually and stays elevated after the spike, because K+ channels are slow to open and slow to close.
Overshoot: K+ keeps leaving while its channels are still open after repolarization, so Vm dips below −70 mV until they close.
Relative refractory period: Na+ channels are resetting one by one and the membrane is hyperpolarized, so the gap between Vm and threshold (−55 mV on the figure) is larger. A larger-than-normal stimulus is needed to close it. (M2 slide 39)
Solution to Exercise 13
(a) The lines cross near 1 µm. Above that diameter myelinated axons conduct much faster, reaching about 50 m/s at 10 µm against under 10 m/s unmyelinated. Below it myelin gives no advantage, which is why the thinnest axons are left unmyelinated.
(b) The large-diameter axon has λ = 3 mm and a low internal resistance ra; the small axon has λ = 1 mm and a high ra. Lower internal resistance lets passive current travel farther before decaying to 37%, so thicker axons conduct faster.
(c) Velocity rises steeply with temperature, reaching about 5× the 20°C baseline near 40°C, because warmth speeds diffusion of ions. Past about 40–45°C the curves crash to zero: heat beyond physiological limits stops conduction. (M2 slide 42)
Solution to Exercise 14
1: A fires twice with a long gap. Each EPSP decays before the next arrives, so they do not add.
2: A fires twice in quick succession. Temporal summation stacks the second EPSP on the first.
3: A and B fire together. Spatial summation adds the two EPSPs.
4: A A B B in rapid succession combines both kinds of summation, the depolarization reaches threshold at the axon hillock, and an action potential fires.
5: C alone gives an IPSP (downward). A + C together cancel, so Vm barely changes. Summation applies to IPSPs too. (M2 slide 32)
Variant¶
Solution to Exercise 15
z = +2 halves the constant. Vm is more negative than ECa, so Ca2+ flows into the cell, with chemical and electrical gradients both pointing in.
Solution to Exercise 16
Solution to Exercise 17
Solution to Exercise 18
Solution to Exercise 19
(a) Temporal summation: 5 × 3 = +15 mV.
(b) Spatial summation: 3 + 3 − 4 = +2 mV.
(c) −70 + 15 = −55 mV. It reaches threshold, so yes. Threshold is “meets or exceeds.”
Solution to Exercise 20
The length constant doubles. Passive current travels twice as far before decaying to 37%.
Solution to Exercise 21
It has almost no affinity for resting, closed Na+ channels. It binds open (depolarizing) and inactivated (repolarizing) voltage-gated Na+ channels, blocking further Na+ influx. The pain neuron cannot depolarize, so pain signals stop reaching the brain. From the Assignment 2 key, Track 1.
Solution to Exercise 22
Solution to Exercise 23
False. Both gradients are still present, equal in strength and opposite in direction, so there is no net movement. (slides 12–14)
Check yourself¶
Solution to Exercise 24
Its job is to change voltage on purpose. Precise control over which ions cross, when, and how fast takes many channel types: leak channels to set rest, ligand-gated channels for graded potentials, voltage-gated channels for action potentials. A non-excitable membrane only needs a steady potential. (M2 slide 3)
Solution to Exercise 25
Depolarization (up): positive ions enter, as Na+ influx. Hyperpolarization (down below baseline): positive ions leave or negative ions enter, as K+ efflux or Cl− influx. Repolarization: return to baseline, typically K+ efflux after a depolarization. (M2 slide 3)
Solution to Exercise 26
Pain receptors in the fingertip (somatic receptors) → sensory, afferent neurons → CNS (spinal cord and brain) integrates → somatic motor, efferent neurons → skeletal muscle (effector) pulls the hand away. (M2 slide 7)
Solution to Exercise 27
Both are structures sensitive to their environment. Module 1: a membrane protein that binds a ligand, such as a G protein-coupled receptor binding norepinephrine. Module 2: a specialized cell that turns a stimulus into an electrical signal, such as a thermoreceptor or baroreceptor. (M2 slide 8)
Solution to Exercise 28
Na+: chemical in (more outside), electrical in (positive ion drawn to the negative inside), electrochemical strongly in. K+: chemical out (more inside), electrical in, electrochemical out because the chemical gradient is slightly stronger. (M2 slide 12)
Solution to Exercise 29
Solution to Exercise 30
It assumes the membrane is permeable to only one ion. Real membranes are permeable to several ions at once. Nernst potentials are still useful because they show which way each ion wants to move at a given Vm, and the chord conductance and Goldman equations are built from them. (M2 slide 15)
Solution to Exercise 31
Driving force is Vm − ECl = −70 − (−47) = −23 mV. The inside is more negative than chloride’s equilibrium, so electrical repulsion of the negative ion outweighs its inward chemical gradient and Cl− moves out. The lecture’s “more negative than EX means it flows in” rule is written for cations; flip it for an anion. (M2 slide 16)
Solution to Exercise 32
K+ leak channels: many, always open, steady K+ loss makes the inside negative. The main contributor. Na+ leak channels: few, a small trickle in that keeps Vm slightly above EK. Na+-K+ pump: restores the gradients and, being electrogenic (3 out, 2 in), adds a little negativity. (M2 slide 22)
Solution to Exercise 33
Each term is weighted by that ion’s share of total conductance, and at rest most open channels are K+ leak channels, so K+'s weight is about 0.70. If gNa matched gK, the two weights would be equal and Vm would land between EK and ENa: in the Assignment 2 neuron, about −21 mV, a large depolarization. (M2 slide 23)
Solution to Exercise 34
Spatial summation of an EPSP and an IPSP. If they are equal in size they cancel and Vm barely changes; otherwise the net change is their sum (for +4 and −5 mV, −1 mV). (M2 slides 30, 32)
Solution to Exercise 35
Graded potentials decay with distance because nothing refreshes them. A signal far out on a dendrite has more membrane to cross before reaching the axon hillock, so it arrives smaller and is less likely to meet threshold. (M2 slide 31)
Solution to Exercise 36
Spatial: several synapses fire at the same time. Temporal: one synapse fires repeatedly before earlier potentials decay. Both together: two excitatory terminals each firing in rapid bursts, as in panel 4 of slide 32 (A A B B), which reaches threshold. (M2 slide 32)
Solution to Exercise 37
Rest: both closed, Na+ channels resting and ready. Threshold: Na+ channels begin opening, K+ still closed. Depolarization: many Na+ open, K+ closed. Peak (+30 mV): Na+ inactivating, K+ finally opening. Repolarization and hyperpolarization: Na+ inactivated then resetting, K+ open then slowly closing. (M2 slide 38)
Solution to Exercise 38
Voltage-gated K+ channels are slow to close. K+ keeps leaving after Vm reaches resting, carrying it below −70 mV until the channels close. (M2 slides 38–39)
Solution to Exercise 39
At the peak, Na+ channels close with the activation gate open and the inactivation gate closed. In that state they cannot open again whatever the voltage or stimulus strength. Only when they reset to resting and ready (activation closed, inactivation open) can a new action potential start. (M2 slides 38–39)
Solution to Exercise 40
Na+ spreading from an active patch reaches both neighbors, but the patch behind is still refractory and cannot fire. Only the patch ahead can, so the signal moves away from the axon hillock. (M2 slide 40)
Solution to Exercise 41
Velocity falls sharply. Without myelin, membrane resistance drops, current leaks out, the length constant shrinks, and the action potential must regenerate at every patch of membrane. Voltage-gated channels are concentrated at the nodes, so a stripped internode may fail to conduct at all. (M2 slides 41–42)
Solution to Exercise 42
Thick and myelinated fastest, then thin and myelinated, then thick and unmyelinated, assuming the thin axon is above about 1 µm where myelin helps and temperature is equal. Myelin gives the largest gain (10–150 against 0.5–2 m/s); diameter adds to it. (M2 slides 41–42)
Solution to Exercise 43
The action potential only opens voltage-gated Ca2+ channels at the terminal. Ca2+ entry is the signal for vesicles to undergo exocytosis. Block the channels and the electrical signal never becomes a chemical one, so nothing reaches the next cell. (M2 slide 43)