Questions on the action potential and its propagation

Question 1

Explain how the sodium-potassium pump and the permeability of the axon membrane together establish a resting potential of about −70 mV.

Accepted Answer

B1 the pump hydrolyses ATP to move three sodium ions out for every two potassium ions in, both against their concentration gradients B1 three positive charges leave for every two that enter, so each cycle of the pump makes the inside slightly more negative B1 the membrane has many potassium channels open at rest and very few sodium channels, so it is far more permeable to potassium than to sodium A1 potassium therefore leaks back out down the steep gradient the pump has built, carrying positive charge with it, while large negatively charged proteins are trapped inside and cannot leave

Question 2

Describe the changes in membrane potential during an action potential, from threshold to the recovery of the resting value, in terms of which voltage-gated channels are open at each point.

Accepted Answer

B1 at about −55 mV voltage-gated sodium channels open and sodium ions diffuse in, so the inside becomes less negative B1 the depolarisation opens more sodium channels, which is positive feedback, and the potential overshoots zero to reach about +40 mV B1 at the peak the sodium channels inactivate and the slower voltage-gated potassium channels are now open, so potassium leaves and the potential falls back: repolarisation A1 the potassium channels are slow to shut, so potassium keeps leaving past the resting value to about −80 mV, and the resting potential returns as they close, set by the resting potassium permeability

Question 3

Explain what is meant by the all-or-nothing principle, and explain how the nervous system distinguishes a light touch from a heavy blow if every action potential is the same size.

Accepted Answer

B1 below threshold no action potential is produced at all, and the small depolarisation simply fades away B1 at or above threshold a full action potential occurs and reaches the same peak of about +40 mV whatever the size of the stimulus, because the positive feedback runs to completion on its own B1 a stronger stimulus increases the frequency of impulses along a neurone A1 a stronger stimulus also recruits more neurones, as receptors with higher thresholds are brought in, and nothing about the amplitude of any impulse changes

Question 4

Explain what causes the refractory period after an action potential, and explain two consequences it has for the impulses a neurone can carry.

Accepted Answer

B1 during the absolute refractory period the voltage-gated sodium channels are inactivated, so no stimulus of any size can produce another action potential B1 during the relative refractory period that follows the membrane is hyperpolarised, so an unusually strong stimulus is needed B1 the patch of membrane behind the impulse has just fired and is refractory, so only the membrane ahead can be depolarised and the impulse travels in one direction A1 the refractory period sets a minimum interval between impulses, so it keeps them as separate countable events and puts an upper limit on the frequency a single neurone can reach

Question 5

An action potential is recorded 8.4 cm along an axon from the point of stimulation, 1.2 ms after the stimulus was applied. Calculate the conduction velocity of this axon in metres per second, and calculate how long an impulse would take to travel 1.4 m along an unmyelinated axon conducting at 1.0 m s⁻¹.

Accepted Answer

M1 velocity is distance divided by time, with both converted to metres and seconds: 8.4 cm is 0.084 m and 1.2 ms is 1.2 × 10⁻³ s A1 0.084 ÷ 0.0012 gives a conduction velocity of 70 m s⁻¹ M1 time is distance divided by velocity, so 1.4 ÷ 1.0 A1 1.4 s, which is seventy times the 20 ms the myelinated axon would take over the same distance

Question 6

Lidocaine is injected into the gum before a tooth is filled. Lidocaine blocks voltage-gated sodium ion channels. Explain why the patient feels no pain from the tooth during the treatment, although they remain fully conscious.

Accepted Answer

B1 lidocaine prevents the voltage-gated sodium channels in the membranes of the sensory neurones from opening, so sodium ions cannot diffuse in B1 the membrane cannot be depolarised to threshold, so no action potentials are produced B1 local currents cannot depolarise the next region of membrane either, so no impulses are propagated along the axons to the central nervous system and the pain is not perceived B1 the drug acts only on the neurones in the injected area, so neurones in the brain and elsewhere keep working and the patient stays conscious

Question 7

Describe how myelination increases the conduction velocity of an axon, using the term saltatory conduction in your answer.

Accepted Answer

B1 the layers of membrane wrapped round the axon by Schwann cells act as an electrical insulator, so ions cannot cross the membrane where the myelin is present B1 depolarisation can therefore only happen at the nodes of Ranvier, the gaps between adjacent Schwann cells B1 local currents flow from node to node, so the impulse appears to jump between them, which is called saltatory conduction B1 far fewer patches of membrane have to be depolarised than on a bare axon of the same length, which is why conduction is dramatically faster

Question 8

A patch of myelin around an axon in the spinal cord is destroyed by disease, but the axon itself is undamaged. Suggest what this does to the speed of conduction along the damaged stretch, and suggest why transmission can fail altogether if enough consecutive nodes of Ranvier are affected.

Accepted Answer

B1 without myelin over that stretch, saltatory conduction is lost there, so every patch of membrane along it has to be depolarised in turn rather than the impulse jumping between nodes B1 conduction along the damaged stretch slows towards the much lower speed typical of an unmyelinated axon B1 local currents decay with distance, and normally still reach the next node large enough to bring it to threshold B1 if several exposed nodes lie in a row, the current arriving at the next one may be too weak to reach threshold, so no action potential is generated there and transmission along that axon fails

Question 9

A sensory neurone has an absolute refractory period of 2.0 ms. Calculate the maximum frequency, in impulses per second, at which it could fire, and calculate how many impulses it could send in 5 seconds at that maximum frequency.

Accepted Answer

M1 maximum frequency is the reciprocal of the refractory period, converted to seconds: 1 ÷ 0.0020 A1 500 impulses per second M1 number of impulses in 5 s is frequency multiplied by time: 500 × 5 A1 2500 impulses in the 5 seconds

Question 10

A local anaesthetic injected at a tooth blocks the voltage-gated sodium channels in the sensory neurones nearby. Suggest why the patient feels no pain from the tooth, and suggest why the sense of touch in the lip on that side is lost as well.

Accepted Answer

B1 with the voltage-gated sodium channels blocked, sodium ions cannot rush into the axon, so the membrane cannot depolarise to +40 mV and no action potential is generated B1 no impulses therefore reach the brain from the receptors in the tooth, and pain is only felt when impulses arrive there B1 the drug cannot distinguish one neurone from another, so it blocks the channels in every axon it reaches, including those carrying impulses from touch receptors in the lip

Question 11

Give the resting potential and the peak potential from the trace, and calculate the size of the depolarisation in millivolts.

Accepted Answer

B1 the resting potential is −70 mV M1 the peak is +40 mV, and the depolarisation is the peak minus the resting potential A1 40 − (−70) = 110 mV

Question 12

The trace falls below the resting potential before returning to it. Explain what causes this undershoot, and explain what it means for a second stimulus arriving at that moment.

Accepted Answer

B1 the voltage-gated potassium channels are slow to close, so potassium ions keep leaving after the resting potential has been reached B1 the inside of the axon therefore becomes more negative than −70 mV, which is hyperpolarisation B1 this is the relative refractory period: a second stimulus has further to depolarise the membrane to reach threshold, so only a larger stimulus produces another action potential and the impulses stay discrete

Question 13

Explain why a wider axon conducts impulses faster than a narrower axon of the same type, even though the size of a single action potential is unaffected by the axon's diameter.

Accepted Answer

B1 a wider axon offers less resistance to the flow of ions along its length, and proportionally less charge leaks away across the membrane as it travels B1 local currents therefore spread further along the axon before they die away, and depolarise the membrane ahead to threshold more quickly B1 axon diameter affects how fast an existing action potential's local currents spread, not the size of the action potential itself, which is set by the same voltage-gated channels regardless of diameter

Question 14

Compare what happens to conduction velocity when an axon from a frog is warmed from 10 °C to 20 °C with what happens when it is warmed from its normal operating temperature to a temperature well above it.

Accepted Answer

B1 warming from 10 °C to 20 °C, within the physiological range, increases conduction velocity, because ions diffuse faster and the pump and channel proteins work faster, whereas warming well above the normal operating temperature does not continue that rise B1 at the higher extreme the proteins responsible for generating and propagating the impulse are disrupted, so conduction instead becomes unreliable and can fail B1 together the two results show that raising temperature raises conduction speed only up to a point set by the animal's normal operating range, beyond which the benefit is reversed

Question 15

Explain why the sodium-potassium pump could not, on its own, produce the rapid depolarisation seen during an action potential.

Accepted Answer

B1 the pump moves three sodium ions out for every two potassium ions in, a net loss of positive charge from the axon, so it makes the inside more negative and can never depolarise the membrane B1 the pump works continuously by active transport against the ion gradients, whereas depolarisation depends on sodium ions diffusing rapidly down a steep gradient once channels open, which is intrinsically far faster B1 depolarisation depends on large numbers of voltage-gated sodium channels opening together, which produces the positive feedback that drives the rapid rise; the pump has no such mechanism

Question 16

State the resting potential of a typical neurone in millivolts, and state the ratio in which the sodium-potassium pump moves the two ions across the membrane.

Accepted Answer

B1 the resting potential is about −70 mV, with the inside negative relative to the outside B1 the pump moves three sodium ions out for every two potassium ions in