301. Assertion: A fibre remains afferent even if a local membrane defect prevents its impulse from reaching the central neural system.
Reason: Afferent classification depends on the normal direction of information flow, whereas successful conduction depends on excitability and propagation along the axonal membrane.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion
Explanation: The assertion is true. Afferent describes the normal direction in which a fibre carries information—from peripheral tissues or organs towards the central neural system. A local conduction block can prevent an impulse from completing that route, but it does not reverse the fibre’s anatomical and functional orientation. The reason is also true and explains the distinction. Direction-based classification and moment-to-moment conduction success are separate biological properties. Propagation requires stimulus-linked sodium permeability, local current and regeneration of an action potential in adjacent membrane regions. If one of these processes fails, the signal may stop before reaching the CNS even though the pathway remains afferent. The same logic prevents a damaged afferent fibre from being reclassified as efferent merely because no central recording is obtained. Classification follows intended direction; conduction depends on membrane function.
302. Two axonal sites are depolarised by the same first stimulus. Site P undergoes normal \(\mathrm{K^+}\) efflux and regains resting polarity. At Site Q, potassium efflux is blocked and the membrane remains depolarised. When an identical second effective stimulus is applied later, which prediction follows most directly from the described membrane mechanism?
ⓐ. P responds again more readily because repolarisation restored its resting state.
ⓑ. Q is better prepared because prolonged depolarisation restores responsiveness.
ⓒ. Both sites respond similarly because sodium influx alone determines recovery.
ⓓ. Repolarisation prevents a second response by stabilising the resting membrane.
Correct Answer: P responds again more readily because repolarisation restored its resting state.
Explanation: After depolarisation, increased potassium permeability permits \(\mathrm{K^+}\) to leave the axoplasm. This outward movement restores the normal polarity of the excited membrane region. The biological significance of repolarisation is not merely a change in charge; restoration of the resting state allows the region to respond again. Site P completes this recovery and is better prepared for the later effective stimulus. At Site Q, the blocked potassium efflux prevents normal restoration of resting polarity, so the membrane remains in the altered depolarised condition. The conclusion does not require importing numerical voltage values or advanced refractory-period terminology. It follows directly from comparing whether the membrane has returned to the resting polarised state. Potassium movement occurs after sodium-dependent depolarisation, and successful recovery preserves repeated neuronal responsiveness rather than permanently removing it. Recovery is therefore a prerequisite for renewed responsiveness at the repeatedly stimulated site.
303. At one instant, four successive axonal regions show the following states: P is repolarising through \(\mathrm{K^+}\) efflux, Q is fully depolarised, R is beginning \(\mathrm{Na^+}\) influx, and S remains at resting polarity. Which direction of impulse propagation is consistent with this pattern?
ⓐ. S to R to Q to P
ⓑ. Q to P to S to R
ⓒ. P to Q to R to S
ⓓ. R to S to P to Q
Correct Answer: P to Q to R to S
Explanation: Impulse conduction is a moving wave of membrane-state changes. A region behind the active front has already been excited and is repolarising through potassium efflux. The currently active region is depolarised, while the immediately adjacent region ahead begins sodium influx as local current reverses its polarity. Regions farther ahead remain resting until the wave reaches them. In the stated snapshot, P is the recovering region, Q is the fully depolarised region, R is the next region beginning excitation, and S is still resting. The propagation sequence is from P through Q and R towards S. This does not mean that one unchanged action potential is physically carried as an object from P to S. Instead, a new action potential is generated successively at adjacent regions, while previously excited regions undergo repolarisation behind the advancing front.
304. An action potential reaches a chemical-synapse terminal. Synaptic vesicles move towards the pre-synaptic membrane but cannot merge with it. Which event is blocked most directly, and what follows?
ⓐ. Vesicle movement is blocked, so no vesicle reaches the pre-synaptic membrane.
ⓑ. Vesicle fusion is blocked, so normal neurotransmitter release into the cleft fails.
ⓒ. Transmitter synthesis is blocked, so empty vesicles fuse with the membrane.
ⓓ. Receptor activation is blocked, so released transmitter cannot act postsynaptically.
Correct Answer: Vesicle fusion is blocked, so normal neurotransmitter release into the cleft fails.
Explanation: The terminal action potential has arrived and the vesicles have already moved towards the pre-synaptic membrane. The failed stage is not action-potential conduction or vesicle movement. The vesicles must next fuse with the pre-synaptic membrane before their neurotransmitter can be released into the fluid-filled synaptic cleft. Preventing membrane fusion interrupts this release step directly. Without normal transmitter in the cleft, specific receptors on the post-synaptic membrane are not activated normally, ion channels do not produce the expected response, and the post-synaptic potential is reduced or absent. The junction remains a chemical synapse despite the failure; blockage of one stage does not convert its structure into an electrical junction. The experiment distinguishes two closely related pre-synaptic events, vesicle approach and vesicle fusion, by showing that successful movement towards the membrane is insufficient unless fusion and release also occur.
305. Transmitters X and Y are separately applied to a post-synaptic membrane. X binds and opens ion channels, whereas Y produces no response. A second membrane that normally responds to neither begins responding to X after receiving the same receptor present on the first membrane. The strongest inference is:
ⓐ. post-synaptic response requires the matching receptor for the transmitter
ⓑ. every neurotransmitter opens every post-synaptic ion channel directly
ⓒ. transmitter Y must be an electrical current rather than a chemical messenger
ⓓ. receptor placement is unnecessary if synaptic vesicles release sufficient transmitter
Correct Answer: post-synaptic response requires the matching receptor for the transmitter
Explanation: The first membrane responds to X but not Y, even though both chemical substances are applied under comparable conditions. The decisive additional result comes from the second membrane. It initially fails to respond but gains responsiveness to X after receiving the same receptor found on the first membrane. This change links the effect of X to receptor presence rather than to nonspecific contact with the membrane. Chemical synaptic transmission requires neurotransmitter binding to specific post-synaptic receptors, followed by opening of ion channels and generation of a post-synaptic potential. A large amount of transmitter cannot replace the appropriate receptor if recognition is absent. The experiment does not identify a named receptor subtype and does not justify importing receptor pharmacology. It establishes only the supported principle of specificity: a transmitter produces its normal receiving effect where a compatible post-synaptic receptor and linked channel response are present.
306. At one chemical junction, Neuron P has transmitter-filled vesicles in its terminal facing receptors on Neuron Q. Stimulation of P produces a response in Q. Depolarising Q at the same contact does not produce equivalent reverse transmission to P. The directional behaviour is explained by:
ⓐ. equal vesicle and receptor distribution on both sides of the junction
ⓑ. direct electrical current flowing symmetrically between the membranes
ⓒ. myelin preventing sodium movement in Neuron Q
ⓓ. pre-synaptic release machinery on P and post-synaptic receptors on Q
Correct Answer: pre-synaptic release machinery on P and post-synaptic receptors on Q
Explanation: Chemical transmission has a structural orientation. Neuron P possesses the terminal vesicles and pre-synaptic machinery needed to release neurotransmitter into the cleft. Neuron Q possesses the specific post-synaptic receptors and linked ion channels needed to convert that chemical message into a membrane response. Stimulation of P follows the complete supported sequence from release to reception. Depolarising Q at the same membrane contact does not automatically reverse the junction, since the receptor-bearing side is not stated to contain the corresponding transmitter-filled terminal machinery facing P. The asymmetry establishes functional directionality at that chemical synapse. An electrical synapse would be identified through closely apposed membranes and direct current flow rather than vesicle-receptor organisation. Myelin affects conduction along an axon and does not explain why one side releases transmitter while the other side receives it at the junction.
307. Consider the following statements.
I. The brain supports emotions as part of its higher mental functions.
II. Limbic structures acting with the hypothalamus contribute to emotional expression and motivation.
III. Any emotional disturbance proves that one isolated limbic structure alone is damaged.
IV. Emotional regulation may involve interaction among several central regions rather than one exclusive centre.
ⓐ. I and III only
ⓑ. II and III only
ⓒ. I, II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: Statement I is valid because the brain supports higher mental functions that include emotions along with memory, intelligence and thought. Statement II is also supported: limbic structures interact with the hypothalamus in regulating emotional expression, motivation and sexual behaviour. Statement III converts an integrated functional relation into an unjustified one-structure diagnosis. An emotional disturbance by itself does not prove that one isolated limbic component is solely responsible, especially because regulation depends on interaction among limbic structures, the hypothalamus and broader brain activity. Statement IV preserves this systems-level interpretation. Different regions may contribute distinct but connected parts of emotional regulation. The valid combination distinguishes a supported association from an exclusive localisation claim. Hard regional diagnosis requires additional anatomical or functional evidence rather than using the word emotion as a unique label for one structure.
308. Arrange the following functional stages in a supported progression from incoming sensory information to an organised voluntary response.
P. Association areas relate information to memory and communication.
Q. A motor area participates in control of the intended movement.
R. A sensory area receives and processes incoming information.
S. The thalamus coordinates major sensory and motor signalling.
T. Regulatory output leaves the CNS towards skeletal muscle.
ⓐ. R, S, Q, P, T
ⓑ. S, R, P, Q, T
ⓒ. S, P, R, T, Q
ⓓ. P, R, S, Q, T
Correct Answer: S, R, P, Q, T
Explanation: The thalamus is a deep forebrain structure that coordinates major sensory and motor signalling, so S represents an early central coordinating stage in the stated progression. The information is then received and processed in an appropriate cortical sensory area at R. Association areas at P contribute higher integration by relating information across senses and with memory or communication. Once an intentional response is organised, a cortical motor area at Q participates in controlling the movement. The command finally leaves the CNS through an efferent somatic pathway and reaches skeletal muscle at T. This sequence is a functional progression rather than a detailed named tract. It distinguishes signalling coordination, sensory processing, association and motor control as related but non-identical operations. Moving motor output before sensory and association processing would not fit the supplied evidence for an organised response to incoming information.
309. In a midbrain specimen, the dorsal surface containing four rounded swellings is removed, but the canal through the region and its position between the forebrain and pons remain identifiable. Which conclusion is correct?
ⓐ. The cerebral aqueduct is lost, so the region can no longer be recognised as midbrain.
ⓑ. The hypothalamus is removed, while the corpora quadrigemina remain intact.
ⓒ. The cerebellar convolutions are removed from an otherwise intact hindbrain.
ⓓ. The corpora quadrigemina are removed, but other midbrain features remain.
Correct Answer: The corpora quadrigemina are removed, but other midbrain features remain.
Explanation: The corpora quadrigemina are the four rounded swellings located on the dorsal surface of the midbrain. Removal of that dorsal landmark removes the corpora quadrigemina. The specimen still contains a canal passing through the region, which is the cerebral aqueduct, and it remains positioned between the forebrain and pons. These independent features continue to identify the remaining region as midbrain. The hypothalamus lies at the base of the thalamus within the forebrain and is not the dorsal four-swellings structure. Cerebellar convolutions belong to the hindbrain and form an extensively folded surface rather than four named midbrain elevations. The conclusion uses both missing and preserved evidence. Loss of one landmark does not erase the anatomical identity of the entire division when its position and internal canal remain demonstrable. The remaining midbrain pathway can persist even though the specialised dorsal sensory-integration surface has been removed. Position and residual function together support the inference.
310. Information from the semicircular canals is experimentally interrupted, while auditory input to the cerebellum, skeletal-muscle responsiveness and medullary control of breathing remain intact. Which outcome is most justified?
ⓐ. Balance and voluntary movement both fail because all cerebellar inputs are lost.
ⓑ. Breathing becomes unstable because medullary control depends on semicircular input.
ⓒ. Balance may be impaired because one relevant cerebellar input has been lost.
ⓓ. Auditory input fully compensates for the missing semicircular information.
Correct Answer: Balance may be impaired because one relevant cerebellar input has been lost.
Explanation: The cerebellum integrates information related to balance, including input from the semicircular canals and the auditory system. The experiment removes one relevant source while preserving another. This changed condition is expected to reduce the completeness of information available for cerebellar integration and may impair coordinated balance. The conclusion should not be expanded into total loss of movement. Skeletal muscles remain responsive, and the cerebellum coordinates movement and balance rather than directly serving as the contractile tissue. Medullary respiratory control is explicitly intact, so breathing need not fail merely because a cerebellar input is interrupted. The auditory pathway also retains its sensory role and does not become an autonomic efferent pathway. The strongest prediction respects partial loss: removal of one balance-related input can weaken coordination even though other neural inputs and motor structures remain functional.
311. Two selective medullary interventions are compared.
| Intervention | Observation |
|---|
| P | Longitudinal communication through the medulla is interrupted, but respiratory, cardiovascular and gastric centres remain structurally intact |
| Q | The respiratory centre is suppressed, but conduction between higher brain regions and the spinal cord remains detectable |
The comparison most strongly supports which conclusion?
ⓐ. The medulla performs only one indivisible function.
ⓑ. Brain-spinal continuity and medullary control are distinct functions.
ⓒ. Respiratory control is performed exclusively by cerebral white matter.
ⓓ. Any medullary lesion must abolish every medullary activity simultaneously.
Correct Answer: Brain-spinal continuity and medullary control are distinct functions.
Explanation: Intervention P selectively disrupts communication through the medullary region while leaving the listed involuntary-control centres structurally intact. Intervention Q produces the complementary result: respiratory regulation is suppressed, but detectable conduction between higher brain regions and the spinal cord remains. Together, the observations show that the medulla has more than one functional relation. It forms continuity between the brain and spinal cord and also contains centres involved in respiration, cardiovascular reflexes and gastric secretion. These roles occur in the same anatomical region but are not an indivisible single operation. A selective intervention may disturb one relation more strongly than another. The comparison does not place respiratory control in cerebral white matter and does not justify claiming that every medullary lesion abolishes all functions. Preserved functions act as controls that allow the affected role to be distinguished from neighbouring roles within the medulla.
312. A local stimulus is applied to three neural organisations. In P, activity spreads through a diffuse neuronal network without a brain. In Q, responses are organised through a brain, ganglia and connecting neural tissues. In R, a brain and spinal cord process input received through peripheral nerves. The order of increasing neural centralisation is:
ⓐ. Q, P, R
ⓑ. R, Q, P
ⓒ. P, Q, R
ⓓ. P, R, Q
Correct Answer: P, Q, R
Explanation: Organisation P represents the diffuse nerve-net condition exemplified by Hydra. Neurons communicate through a distributed network, but no highly centralised brain or ganglionic processing centre is described. Organisation Q represents the insect condition, in which a brain, ganglia and connecting neural tissues provide greater concentration and organisation of neural processing. Organisation R shows the vertebrate pattern, with a developed CNS consisting of brain and spinal cord and a PNS connecting central structures with peripheral tissues. The progression concerns increasing concentration and organisation of neural elements, not the first appearance of stimulus responsiveness. All three systems can coordinate responses, but Q and especially R provide more centralised processing and controlled information flow than the diffuse arrangement in P.
313. Two lesions are produced in the same sensory pathway. Lesion P severs a nerve outside the brain and spinal cord, so peripheral stimulation no longer reaches the CNS. Lesion Q damages the spinal cord, so an impulse delivered directly to the intact peripheral nerve reaches the lesion but is not processed or transmitted normally through central pathways. P and Q respectively affect:
ⓐ. a PNS communication route and a CNS control structure
ⓑ. a CNS afferent centre and a PNS somatic target
ⓒ. a spinal CNS structure and a cranial PNS nerve
ⓓ. an autonomic centre and an endocrine signalling route
Correct Answer: a PNS communication route and a CNS control structure
Explanation: Lesion P lies in a nerve outside the brain and spinal cord. Such nerves belong to the peripheral neural system and connect peripheral tissues with central structures. Since peripheral stimulation fails to reach the CNS, the severed route is functioning as an afferent communication pathway. Lesion Q lies in the spinal cord, which belongs to the CNS together with the brain. The CNS is responsible for information processing and control, while the PNS provides connecting nerves. Direct stimulation of the intact peripheral nerve allows the signal to reach the damaged central structure, separating peripheral conduction from central processing or transmission. The observations demonstrate that CNS and PNS are classified by anatomical boundaries as well as by complementary roles in one information pathway. Comparing the two lesions separates peripheral entry of information from central relay after the impulse has already reached the CNS. The two sites produce different consequences because they lie on opposite sides of central entry.
314. A retinal neuron has one dendrite and one axon arising from its cell body. The axon later divides into \(8\) terminal branches, each ending in synaptic knobs. The neuron remains:
ⓐ. multipolar, since the total number of distal branches exceeds \(2\)
ⓑ. unipolar, because all distal branches arise from one axon at the soma
ⓒ. non-myelinated, since bipolar neurons cannot possess terminal branches
ⓓ. bipolar, because class depends on axon-dendrite arrangement at the soma
Correct Answer: bipolar, because class depends on axon-dendrite arrangement at the soma
Explanation: A bipolar neuron is characterised by one axon and one dendrite arising from the cell body. Distal branching of the axon does not create additional dendrites and does not convert the neuron into a multipolar type. Axonal branches remain parts of the same axon and terminate in synaptic knobs containing neurotransmitter-filled vesicles. Multipolar classification requires one axon and two or more dendrites, not several terminal branches at the end of one axon. The retinal location is also consistent with the representative occurrence of bipolar neurons. Myelination is a separate property based on axonal covering and cannot be inferred from the process arrangement alone. The decisive evidence is the number and identity of processes associated with the soma, not the number of terminal contacts formed by the distal axon.
315. A neuron receives a natural stimulus through its dendrites, and its cell body develops normal electrical activity. No impulse reaches the terminal knobs after the axon is severed midway. Direct stimulation of the isolated distal axon produces transmitter release and a normal response in the next neuron. The findings indicate:
ⓐ. dendritic reception and axonal conduction are both interrupted
ⓑ. axonal continuity is interrupted, but distal terminal function is preserved
ⓒ. post-synaptic receptors are blocked, while axonal continuity is preserved
ⓓ. the cell body loses its nucleus, but distal vesicle release is preserved
Correct Answer: axonal continuity is interrupted, but distal terminal function is preserved
Explanation: The natural stimulus reaches the dendrites and produces activity in the cell body, so the receiving processes and soma remain functional. The impulse fails only after encountering the severed axon, localising the primary defect to continuity of the outward-conducting process. Direct stimulation of the isolated distal portion bypasses the cut and causes transmitter release followed by a normal response in the next neuron. This control shows that the terminal synaptic knobs, vesicle-fusion mechanism, neurotransmitter and post-synaptic receptors are still capable of functioning. The absent response after natural stimulation is thus not a primary synaptic failure. Neural communication requires an unbroken directional sequence from dendrites and soma through the axon to the terminal. Preserved function at both ends does not restore communication when the conducting link between them has been physically interrupted.
316. A non-myelinated axon is enclosed by a Schwann cell but has no myelin sheath or nodes of Ranvier. A short membrane region in its middle is made incapable of depolarisation, while regions on both sides remain structurally intact. Stimulation at one end is expected to:
ⓐ. propagation stops at the non-depolarisable segment
ⓑ. saltatory bypass through newly formed nodes
ⓒ. accelerated continuous conduction through the blocked segment
ⓓ. electrical bypass through the surrounding Schwann cell
Correct Answer: propagation stops at the non-depolarisable segment
Explanation: In a non-myelinated fibre, the Schwann cell may enclose the axon without forming a myelin sheath. The exposed axonal membrane is depolarised successively along its length, producing continuous conduction. A region that cannot undergo depolarisation breaks this sequence. Local current may reach the blocked region, but the action potential cannot be regenerated there normally, so activation of more distal membrane becomes ineffective. Saltatory conduction cannot appear merely because one segment is non-excitable; it requires myelin segments separated by exposed nodes of Ranvier. The fibre also does not become a synapse, since the blocked site remains part of one axon and lacks pre-synaptic and post-synaptic membranes. The result distinguishes ordinary Schwann-cell enclosure from actual myelin-dependent conduction and shows why membrane excitability is required throughout a continuously conducting fibre.
317. A resting axon initially has high internal \(\mathrm{K^+}\), high external \(\mathrm{Na^+}\), retained negatively charged proteins and normal ion-selective permeability. A treatment makes the membrane freely permeable to the intracellular proteins but initially leaves sodium and potassium permeability unchanged. The most direct prediction is:
ⓐ. the negative character of the axoplasm increases as negative proteins leave
ⓑ. sodium-potassium pumping immediately changes to \(2\) \(\mathrm{Na^+}\) out for \(3\) \(\mathrm{K^+}\) in
ⓒ. loss of retained negative proteins weakens one contributor to the negative inner surface
ⓓ. the treated region necessarily produces an action potential without any change in sodium permeability
Correct Answer: loss of retained negative proteins weakens one contributor to the negative inner surface
Explanation: The resting membrane normally retains negatively charged proteins within the axoplasm and does not allow them to cross freely. Their presence contributes to the negative charge of the inner membrane surface. If the membrane suddenly becomes permeable to these proteins, their loss reduces that intracellular negative contribution. The treatment does not directly alter the initial sodium and potassium concentration gradients or the membrane’s ion-selective permeability, so a complete numerical prediction of membrane potential is not supported. It also does not reverse the sodium-potassium pump, whose standard exchange remains \(3\) sodium ions outward for every \(2\) potassium ions inward. An action potential specifically requires a brief sharp increase in sodium permeability and inward sodium movement. Protein escape may destabilise resting polarisation, but it is not equivalent to the normal sodium-dependent depolarising event.
318. A sodium-potassium pump completes \(4\) standard cycles before an effective stimulus produces an action potential. Which report is fully justified by the pump data alone?
ⓐ. \(12\) \(\mathrm{Na^+}\) inward and \(8\) \(\mathrm{K^+}\) outward; later sodium influx equals \(4\)
ⓑ. \(12\) \(\mathrm{Na^+}\) outward and \(8\) \(\mathrm{K^+}\) inward; later sodium influx is undetermined
ⓒ. \(8\) \(\mathrm{Na^+}\) outward and \(12\) \(\mathrm{K^+}\) inward; later sodium influx equals \(20\)
ⓓ. \(20\) \(\mathrm{Na^+}\) move outward through the pump; later sodium influx equals zero
Correct Answer: \(12\) \(\mathrm{Na^+}\) outward and \(8\) \(\mathrm{K^+}\) inward; later sodium influx is undetermined
Explanation: Each pump cycle actively transports \(3\) \(\mathrm{Na^+}\) outward and \(2\) \(\mathrm{K^+}\) inward. Four cycles move \(4\times3=12\) sodium ions out and \(4\times2=8\) potassium ions in. The total number of transported ions is \(20\), while the net outward movement of positive charge is \(12-8=4\) equivalents. These pump calculations do not reveal how many sodium ions later enter during an action potential. Depolarisation occurs through a stimulus-induced increase in sodium permeability, and no numerical value for that channel-mediated influx is supplied. Pump stoichiometry describes active maintenance of ionic gradients, not the total ionic movement associated with one nerve impulse. A biologically valid calculation must preserve this distinction and avoid converting a known \(3:2\) exchange into an unsupported action-potential ion count. The pump relation determines only transport through the pump. Channel-mediated sodium influx during the later action potential depends on permeability, concentration gradient and duration, none of which is quantified under the stated conditions. Reporting an inferred influx would exceed the data.
319. Membrane potential is recorded at two sites, P and Q, on the same non-myelinated axon, with Q located farther from the stimulus. Normally, P shows a depolarising rise before Q, and both later repolarise. After local current flow between the sites is interrupted, P retains its normal curve but Q remains at resting baseline. This result shows that:
ⓐ. Q normally receives neurotransmitter released from P across a synaptic cleft
ⓑ. one unchanged action potential physically moves from P to Q
ⓒ. potassium efflux at P directly produces sodium influx at Q without current flow
ⓓ. local current must trigger a new action potential in the adjacent region
Correct Answer: local current must trigger a new action potential in the adjacent region
Explanation: The normal delay between P and Q indicates sequential activation along the axon. When current flow between them is interrupted, P still depolarises and repolarises, proving that stimulus detection and membrane events at P remain functional. Q, however, receives no effective influence and stays at resting polarity. This localises the missing mechanism to transmission of current from the depolarised site towards the adjacent resting site. Under normal conditions, internal and external local currents complete a circuit that reverses polarity at the next membrane region and generates a new action potential there. P and Q are regions of one axon, not two cells separated by a synapse. The experiment also rejects the idea that one fixed electrical event travels unchanged; propagation occurs through successive regeneration of depolarisation along the membrane. The unchanged response at P controls for stimulus effectiveness, while the absent response at Q localises the failure to propagation between the two sites.
320. Two pathways are identical in axonal length and number of junctions. The only difference is that Pathway P contains an electrical synapse, whereas Pathway Q contains a chemical synapse. Axonal conduction is normal in both. Which outcome is expected?
ⓐ. Q responds earlier because neurotransmitter crosses the cleft faster than direct current
ⓑ. Both pathways respond simultaneously because synapse type cannot affect transmission speed
ⓒ. P responds first because direct current bypasses vesicle release and receptor activation
ⓓ. P responds earlier because electrical synapses are the commonest synapses in humans
Correct Answer: P responds first because direct current bypasses vesicle release and receptor activation
Explanation: The pathways are controlled for axonal length and number of junctions, so the comparison isolates synapse type. At an electrical synapse, the membranes lie in close proximity and electrical current passes directly from one neuron to the next. A chemical synapse requires several sequential events: terminal arrival of the action potential, vesicle movement and fusion, neurotransmitter release, diffusion across the cleft, receptor binding and ion-channel activation. The additional stages make chemical transmission slower than electrical transmission under the stated comparison. Electrical synapses are rare in humans, but rarity does not explain the faster response and should not be confused with speed. The experiment supports a mechanistic inference rather than a frequency claim. Direct current transfer reduces the number of conversion stages between the pre-synaptic and post-synaptic electrical events. The difference in delay is therefore attributable to the mode of synaptic transfer rather than axonal length or junction number. All other stated variables are controlled by the design.
