IB Biology C2.2: Neural signalling worked solutions

The axon is the long fibre that carries impulses away from the cell body (B). Dendrites bring signals towards the cell body, the cell body holds the nucleus, and a Schwann cell makes the myelin sheath.

The inside is negative relative to the outside, about -70 mV (A). The pump moves 3\,\mathrm{Na^+} out and 2\,\mathrm{K^+} in, not the reverse (C); sodium is higher outside (B); and the pump keeps working at rest, so ATP is used (D).

Depolarization opens voltage-gated calcium channels and Ca^{2+} entering the terminal triggers exocytosis (B). Sodium entry is the postsynaptic event (A), binding to receptors is on the postsynaptic membrane (C), and breakdown ends the signal (D).

Binding opens the receptor channel, Na^+ diffuses in and the membrane depolarizes (B). Calcium entry is presynaptic (A), the pump is not triggered directly (C), and acetylcholinesterase is already in the cleft (D).

Speed rises with axon width and with myelination, and D has both advantages (D). B is wider than A but has no sheath, and C has a sheath but is narrower than D.

During repolarization the voltage-gated Na^+ channels have closed and the voltage-gated K^+ channels are open, so K^+ diffuses out (B).

The channels and pumps are clustered at the nodes of Ranvier, so depolarization happens only there (B). The sheath insulates the axon and has no channels beneath it.

Cocaine binds the dopamine transporter in the presynaptic membrane and blocks reuptake, so dopamine stays in the cleft (C). It does not stimulate release (A), act as a receptor ligand (B) or break dopamine down (D).

Inhibitory input hyperpolarizes the membrane: the inside becomes more negative, which moves it away from threshold (B).

Frequency =\dfrac{6}{0.120\ \mathrm{s}}=50\ \mathrm{s^{-1}} (C). Convert milliseconds to seconds first: 0.05, 20 and 720 come from dividing or multiplying the wrong way.

(a) P is a dendrite (1); Q is the cell body (1); S is a node of Ranvier, the gap between two sheath segments (1). (b) R is a segment of the myelin sheath, which insulates the axon (1).

(a) Potassium has the higher concentration inside, 140 against 4 (1). (b) The sodium-potassium pump maintains the gradients by active transport (1). (c) The pump moves three Na^+ out for two K^+ in, so a net positive charge leaves (1); K^+ also leaks out through channels faster than Na^+ leaks in, so the inside ends up negative (1). (d) Without ATP the pump stops and the gradients slowly run down, so the potential moves towards 0 mV and is less negative (1).

(a) Speed increases as width increases (1); each fourfold increase in width doubles the speed, 1.1\to2.2\to4.4 (1). (b) \dfrac{24.2}{2.2}=11 times (1). (c) Axon D has a myelin sheath and B does not, and they have the same width (1); the sheath insulates the axon, so conduction is much faster (1); HL: the impulse jumps from node to node. (d) A fit of R^2=0.96 means 96\% of the variation in speed is explained by axon width (1).

(a) The action potential depolarizes the presynaptic membrane, which opens voltage-gated calcium channels (1); Ca^{2+} diffuses into the terminal (1); vesicles fuse with the membrane and acetylcholine leaves by exocytosis (1). (b) Acetylcholine crosses the synaptic cleft by diffusion (1). (c) Acetylcholinesterase breaks the acetylcholine down (1), so the receptor channels close and the muscle fibre is not stimulated continuously (1).

(a) At R the potential is +40 mV (1). (b) Between R and S the voltage-gated Na^+ channels have closed (1) and the voltage-gated K^+ channels are open (1), so K^+ diffuses out and the potential falls: repolarization (1). (c) At T the potential is below the resting value of -70 mV, which is hyperpolarization (1), because the K^+ channels close slowly and extra K^+ leaves the axon (1).

(a) The sheath is divided into segments (1) separated by nodes of Ranvier, where the ion channels and pumps are clustered (1). (b) The impulse can no longer jump from node to node (1); every section of membrane has to depolarize in turn (1); so conduction is slower, or fails (1).

(a) The control response stays at about 100\% (1), while the response with the neonicotinoid falls steadily at each time (1). (b) \dfrac{100-12}{100}\times100=88\% (1). (c) The neonicotinoid binds to the acetylcholine receptors and is not broken down by acetylcholinesterase, so it stays bound (1); acetylcholine can no longer act on them, so the response falls (1).

(a) +7+6=+13 mV, so the potential is -70+13=-57 mV (1). (b) -57 mV is below the threshold of -55 mV, so no action potential is generated (1). (c) +7+6+5=+18 mV gives -52 mV (1), which is above threshold, so an action potential fires (1). (d) +18-6=+12 mV gives -58 mV (1), below threshold, so the inhibitory input stops the neuron firing (1).

One mark for each point, maximum 6: at rest the pump has made Na^+ high outside and K^+ high inside (1); a stimulus raises the membrane potential to threshold (1); voltage-gated Na^+ channels open and Na^+ diffuses in (1); the inside becomes positive: depolarization (1); Na^+ channels close and voltage-gated K^+ channels open, so K^+ diffuses out and the potential falls: repolarization (1); Na^+ spreads sideways inside the axon as a local current (1) and brings the next section of membrane to threshold, so the process repeats along the axon (1); the pump restores the gradients (1). Each ion movement must be tied to a channel; saying that sodium is pumped in earns no mark.

Signal, maximum 5: the action potential depolarizes the presynaptic membrane and opens voltage-gated Ca^{2+} channels (1); Ca^{2+} diffuses into the terminal (1); vesicles fuse with the membrane and release acetylcholine by exocytosis (1); acetylcholine diffuses across the cleft and binds to receptors on the postsynaptic membrane (1); the receptor channel opens, Na^+ enters and the membrane depolarizes, starting an action potential if threshold is reached (1); acetylcholinesterase breaks acetylcholine down, which ends the signal (1). Pesticide, maximum 3: the neonicotinoid binds to the acetylcholine receptors on the postsynaptic membrane (1); acetylcholinesterase cannot break it down, so it stays bound (1); acetylcholine cannot act, transmission is blocked and the insect is paralysed (1).

(a) Frequency =\dfrac{7}{0.100\ \mathrm{s}}=70\ \mathrm{s^{-1}} (1 for the method, 1 for the value). (b) Frequency increases as the stimulus strength increases, from 20 to 90\ \mathrm{s^{-1}} (1). (c) The height of every spike is 110 mV whatever the strength (1), so an action potential is all or nothing and a stronger stimulus changes only how often spikes occur (1). (d) \text{speed}=\dfrac{0.30\ \mathrm{m}}{6.0\times10^{-3}\ \mathrm{s}}=50\ \mathrm{m\,s^{-1}} (1 for the method, 1 for the value). (e) Speed depends on axon width and on myelination, and neither changes with stimulus strength (1).

(a) \dfrac{100}{1.0}=100 times (1). (b) Fibre B has a myelin sheath, so the impulse jumps from node to node (1) instead of depolarizing every patch of membrane (1); B is also wider, so ions flow along it with less resistance (1). (c) Equal steps on a logarithmic axis are tenfold changes, so the very wide range of widths and speeds fits on one graph (1). (d) The impulses from fast fibre B reach the brain before those from slow fibre C (1).

(a) 44.0-37.5=6.5\,{}^\circ\mathrm{C} lower with capsaicin (1). (b) Heat opens cation channels in the nerve ending (1); cations enter, the membrane depolarizes and reaches threshold (1); impulses travel along the sensory neuron to the brain, where the pain is perceived (1). (c) Capsaicin opens the same channels, so less heat is needed to reach threshold (1). (d) An anaesthetic prevents the receptors from becoming active, so no impulses are sent to the brain (1).

(a) The pump moves ions by active transport (1) and Na^+ crosses a channel by facilitated diffusion (1). (b) The pump moves Na^+ and K^+ against their concentration gradients, which needs the energy of ATP (1); Na^+ moves through the channel down its gradient, so no ATP is needed (1). (c) Without ATP from respiration the pump stops and the Na^+ and K^+ gradients slowly run down (1), so the resting potential fades and action potentials can no longer be generated (1).

(a) The substrate is acetylcholine (1) and the products are choline and acetate (1). (b) Acetylcholine is not broken down (1), so it stays bound to the receptors and their channels stay open (1); the muscle fibre is stimulated continuously and stays contracted (1). (c) A competitive inhibitor binds to the active site, which is why it blocks the substrate (1).

(a) The signal passes along a sensory neuron (1), a relay neuron in the spinal cord (1) and a motor neuron to the muscle (1). (b) Nerve impulses travel rapidly along the axons and cross only a few synapses (1), whereas a hormone must be released into the blood and carried to its target (1). (c) At each synapse the signal is passed by neurotransmitter in one direction only (1).

(a) Curare occupies the receptors, so acetylcholine cannot bind (1); the channels do not open, so no Na^+ enters and the muscle fibre is not depolarized (1); the fibre does not contract and the animal is paralysed (1). (b) Curare would be expected to leave the release of acetylcholine unchanged, since the nerve impulse still reaches the terminal and vesicles fuse by exocytosis (1). (c) Ca^{2+} released from the sarcoplasmic reticulum binds to troponin, which moves tropomyosin and lets myosin heads bind to actin (1).