Biology HL

1. 1.

   A hypothetical axon trace rises from -70 mV to +30 mV, falls to -80 mV, then returns to rest. Explain the channel events during depolarization and repolarization and explain why the sodium-potassium pump is not the immediate cause of the rapid falling phase.

   [4 marks] · no calculator

   Worked explanation

   1. Use ion direction and channel state rather than saying positive charge disappears.
   2. Delayed potassium-channel closure can account for the undershoot to -80 mV.
   3. Distinguish the rapid electrical waveform from long-term maintenance of ion gradients by ATP-driven pumping.

   Common mistake: The falling voltage is not caused by the pump instantly transporting all the sodium that entered back out.

   Marking points

   • At threshold voltage-gated sodium channels open and sodium enters down its electrochemical gradient, depolarizing the membrane.
   • Sodium channels then inactivate while voltage-gated potassium channels open and potassium exits.
   • Potassium efflux repolarizes and can transiently hyperpolarize the membrane.
   • The pump maintains/restores concentration gradients over time but rapid repolarization primarily results from channel-mediated potassium efflux.

   Examiner tip: The falling voltage is not caused by the pump instantly transporting all the sodium that entered back out.

2. 2.

   Compare propagation in a myelinated axon with propagation in an unmyelinated axon of similar diameter. Explain the roles of myelin, local currents and nodes of Ranvier, and predict a consequence of loss of myelin.

   [4 marks] · no calculator

   Worked explanation

   1. The signal still depends on ion currents; it does not teleport through empty space between nodes.
   2. Regeneration at separated nodes speeds propagation compared with continuous regeneration along unmyelinated membrane.
   3. Without insulation, current leakage can prevent sufficient downstream depolarization.

   Common mistake: Saltatory conduction does not mean a molecule physically jumps from one node to another carrying an entire impulse.

   Marking points

   • Myelin insulates internodal membrane and reduces current leakage.
   • Local currents spread to the next node, where concentrated voltage-gated channels regenerate an action potential.
   • Unmyelinated axons regenerate action potentials along successive nearby membrane regions; saltatory conduction is generally faster at similar diameter.
   • Demyelination can slow or block propagation because current may fail to bring the next excitable region to threshold.

   Examiner tip: Saltatory conduction does not mean a molecule physically jumps from one node to another carrying an entire impulse.

3. 3.

   In a simplified hypothetical integration model, a neurone rests at -70 mV and fires at -55 mV. At the integration site two simultaneous excitatory inputs each depolarize it by 9 mV, while one inhibitory input hyperpolarizes it by 7 mV. Assume linear summation for this task. Calculate the resulting potential with all three inputs, decide whether it fires, and explain what happens if the inhibitory input is removed. If excitation is increased further once threshold is exceeded, explain whether the amplitude of each action potential increases proportionally.

   [4 marks] · no calculator

   Worked explanation

   1. Combine the signed voltage changes at the stated integration site; the simplified linear rule is supplied, not a universal quantitative neuron law.
   2. Compare -59 and -52 with -55, remembering that -52 is more positive.
   3. Postsynaptic integration controls the all-or-nothing firing decision, not a graded amplitude for each action potential.

   Common mistake: A more negative number such as -59 mV is farther below threshold than -55 mV, not above it.

   Marking points

   • With all inputs: -70 + 9 + 9 - 7 = -59 mV.
   • -59 mV is below the -55 mV threshold, so no action potential occurs in this model.
   • Without inhibition: -70 + 9 + 9 = -52 mV, reaching/exceeding threshold, so an action potential occurs.
   • Inputs affect whether threshold is reached; they do not produce a proportionally larger action-potential amplitude once it fires.

   Examiner tip: A more negative number such as -59 mV is farther below threshold than -55 mV, not above it.

4. 4.

   During the absolute refractory period after a neuronal action potential, many voltage-gated sodium channels are inactivated. Explain why a second action potential cannot immediately be triggered at that membrane region and how this contributes to forward propagation along an axon.

   [3 marks] · no calculator

   Worked explanation

   1. Distinguish an inactivated channel from a closed channel ready to open.
   2. An additional stimulus cannot instantly restore the unavailable channels during the absolute refractory period.
   3. Local current can spread in both directions, but a regenerative spike is normally supported ahead because the recently active region is refractory.

   Common mistake: Do not say the sodium-potassium pump instantly causes the refractory period or that electrical current can physically flow only forward.

   Marking points

   • Inactivated sodium channels cannot immediately reopen in response to another depolarisation.
   • Channels must recover before that region can initiate another action potential.
   • Membrane just behind a propagating action potential is refractory, favouring excitation of the still-excitable region ahead under normal propagation conditions.

   Examiner tip: Do not say the sodium-potassium pump instantly causes the refractory period or that electrical current can physically flow only forward.

5. 5.

   A hypothetical loss of myelin leaves most axonal ion-channel distribution initially unchanged. Explain why conduction may slow or fail and why saltatory conduction does not mean sodium ions physically jump from one node of Ranvier to the next.

   [3 marks] · no calculator

   Worked explanation

   1. Follow the electrical effect along the axon rather than following one ion through the entire length.
   2. Insulation helps a local depolarisation affect more distant excitable membrane; removing it can reduce the safety of that transmission.
   3. New transmembrane ion movements at successive nodes regenerate the signal, so the same sodium ions need not travel node to node.

   Common mistake: Saltatory conduction is not movement of a packet of sodium ions through the axon from one node to the next.

   Marking points

   • Myelin normally reduces current leakage across internodal membrane and supports effective passive depolarisation towards the next node.
   • Myelin loss can reduce depolarisation reaching the next node, slowing conduction or preventing threshold from being reached.
   • Action potentials are regenerated at nodes; saltatory conduction describes successive electrical activation rather than individual sodium ions jumping between nodes.

   Examiner tip: Saltatory conduction is not movement of a packet of sodium ions through the axon from one node to the next.

Original, unofficial practice questions from SubjectScout — not official past-paper questions or complete syllabus coverage. Added English teaching explanations are AI-authored; independent teacher review is pending.