1
PRINCIPLES OF NEUROSCIENCE | BSCI 353, EXAMS OF
BIOLOGY FULL PACKAGE QUESTIONS ANSWERS AND
RATIONALES 2026-27 LATEST UPDATED VERSION
INSTANT DOWNLOAD PDF..!!
Q1: An electrophysiologist patches an isolated mammalian cortical neuron at
room temperature. The internal concentration of K⁺ is set to 140 mM and the
external concentration is set to 5 mM. The resting membrane potential is
measured at -80 mV. If the temperature of the recording bath is elevated from
20°C to 37°C without changing ionic concentrations, how will the calculated
Nernst potential for K⁺ (\(E_{K}\)) shift, and what happens to the driving force
on K⁺ if the membrane potential (\(V_{m}\)) is held steady at -80 mV?
A) \(E_{K}\) becomes more positive; driving force decreases.
B) \(E_{K}\) becomes more negative; driving force increases.
C) \(E_{K}\) remains perfectly unchanged; driving force drops to zero.
D) \(E_{K}\) shifts to exactly 0 mV; driving force reverses direction.
Rationale: The correct answer is B because the Nernst equation states that
\(E_K = (RT/zF) \cdot \ln([K^+]_o/[K^+]_i)\). Since T (temperature in Kelvin) is in
the numerator, increasing the temperature increases the absolute magnitude of
the calculated equilibrium potential. Given that the concentration gradient
favors an efflux of K⁺, the value of \(\ln([K^+]_o/[K^+]_i)\) is negative, meaning
\(E_{K}\) becomes more negative (hyperpolarized). The driving force is
calculated as \((V_m - E_K)\). If \(V_{m}\) is held at -80 mV and \(E_{K}\)
becomes more hyperpolarized (e.g., shifts from -84 mV to -89 mV), the absolute
value of \((V_m - E_K)\) grows larger, resulting in an increased driving force.
Options A, C, and D mathematically misapply the relationship between
temperature, the equilibrium potential, and driving force calculation.
,2
Q2: A neuropharmacologist isolates a novel peptide toxin from a marine cone
snail. In a voltage-clamp recording of a spinal motor neuron, application of the
toxin prevents the rapid downstroke of the action potential and blocks the
macroscopic outward current during step-depolarizations, while leaving the
transient inward current unaffected. What is the molecular target of this toxin?
A) Voltage-gated sodium channels (\(Na_{V}\)).
B) Delayed rectifier voltage-gated potassium channels (\(K_{V}\)).
C) Inwardly rectifying potassium channels (Kir).
D) Low-threshold T-type calcium channels (\(Ca_{V}\)).
Rationale: The correct answer is B. The downstroke of a classic action potential
is driven by the inactivation of voltage-gated Na⁺ channels and the
simultaneous delayed opening of voltage-gated K⁺ channels, which generate an
outward hyperpolarizing current. Because the toxin blocks this macroscopic
outward current under voltage-clamp step-depolarizations while leaving the
fast inward current (Na⁺ influx) intact, it selectively blocks the delayed rectifier
K⁺ channels. Option A is wrong because blocking \(Na_{V}\) would eliminate
the inward current and upstroke. Option C is wrong because Kir channels
primarily pass inward currents at hyperpolarized potentials. Option D is wrong
because T-type calcium channels modulate low-threshold firing bursts, not the
standard macroscopic outward repolarization current.
Q3: A researcher calculates quantal parameters at a central excitatory synapse.
Under normal control conditions, the mean amplitude of an evoked excitatory
postsynaptic potential (EPSP) is 4.0 mV, and the mean amplitude of
spontaneous miniature EPSPs (mEPSPs) is 0.5 mV. Following the application of a
drug, the evoked EPSP amplitude drops to 1.5 mV, but the mEPSP amplitude
remains exactly 0.5 mV. What structural or functional change does this drug
induce?
,3
A) It down-regulates postsynaptic AMPA receptor density.
B) It decreases the presynaptic quantal content (m) by lowering the
probability of vesicle release or the number of release sites.
C) It non-competitively blocks postsynaptic NMDA receptors.
D) It increases presynaptic quantal size (q) by packing more neurotransmitter
into each vesicle.
Rationale: The correct answer is B. According to quantal analysis models, the
mean amplitude of an evoked potential is defined as M = m ⋅ q, where m is the
quantal content (the number of vesicles released) and q is the quantal size (the
postsynaptic response to a single vesicle). Miniature EPSPs (mEPSPs) provide a
direct measure of quantal size (q). Because the drug decreased the total evoked
EPSP but left the mEPSP amplitude completely unaltered (q = 0.5 mV), the
postsynaptic sensitivity remained steady. Thus, the change must be entirely
presynaptic, reducing the quantal content (m = M / q dropped from 8 to 3).
Options A and C would reduce both evoked EPSPs and mEPSPs by blunting
postsynaptic reception. Option D contradicts the steady mEPSP data.
Q4: A mutation in the gene encoding the calcium sensor synaptotagmin-1
alters its binding affinity so that it requires threefold higher intracellular
calcium levels to trigger conformational shifts. How will this mutation affect
transmission at a fast chemical synapse during a single presynaptic action
potential?
A) It will increase the frequency of spontaneous miniature end-plate potentials
(MEPPs).
B) It will prevent the propagation of the action potential down the axon shaft.
C) It will severely blunt or abolish synchronous neurotransmitter release
while leaving asynchronous release largely intact.
D) It will double the velocity of synaptic vesicle endocytosis.
, 4
Rationale: The correct answer is C. Synaptotagmin-1 is the primary low-affinity
calcium sensor anchored in the synaptic vesicle membrane that mediates fast,
synchronous exocytosis upon binding calcium entering via voltage-gated
calcium channels (\(Ca_{V}\)). A mutation reducing its calcium affinity means
the localized calcium microdomain created during an action potential will fail to
fully activate it, severely reducing or abolishing fast synchronous release.
Asynchronous release, which relies on high-affinity calcium sensors like double
C2 domain proteins (doc2) or stromal interaction molecules, will be less
affected or remain intact. Option A is wrong because spontaneous release
(MEPPs) is mostly independent of high-calcium synaptotagmin activation.
Option B is incorrect because synaptotagmin does not affect axonal action
potential conduction. Option D is unsupported by vesicle recycling physics.
Q5: In an experimental setup, you express wild-type Drosophila Shaker
potassium channels in Xenopus oocytes. You apply a patch-clamp pipette and
record macroscopic currents using a step protocol from -80 mV to +40 mV. You
then introduce an intracellular protease that cleaves the first 20 amino acids of
the channel’s N-terminus. What operational change do you observe in the
recorded potassium currents?
A) The channels fail to open at any voltage step, acting as a complete loss-of-
function block.
B) The channels open normally upon depolarization but completely lose their
rapid inactivation property, remaining open for the duration of the voltage
step.
C) The activation threshold of the channel shifts from -40 mV to +60 mV.
D) The channel reverses its selectivity filter, passing sodium ions instead of
potassium ions.
Rationale: The correct answer is B. The Shaker potassium channel utilizes an N-
PRINCIPLES OF NEUROSCIENCE | BSCI 353, EXAMS OF
BIOLOGY FULL PACKAGE QUESTIONS ANSWERS AND
RATIONALES 2026-27 LATEST UPDATED VERSION
INSTANT DOWNLOAD PDF..!!
Q1: An electrophysiologist patches an isolated mammalian cortical neuron at
room temperature. The internal concentration of K⁺ is set to 140 mM and the
external concentration is set to 5 mM. The resting membrane potential is
measured at -80 mV. If the temperature of the recording bath is elevated from
20°C to 37°C without changing ionic concentrations, how will the calculated
Nernst potential for K⁺ (\(E_{K}\)) shift, and what happens to the driving force
on K⁺ if the membrane potential (\(V_{m}\)) is held steady at -80 mV?
A) \(E_{K}\) becomes more positive; driving force decreases.
B) \(E_{K}\) becomes more negative; driving force increases.
C) \(E_{K}\) remains perfectly unchanged; driving force drops to zero.
D) \(E_{K}\) shifts to exactly 0 mV; driving force reverses direction.
Rationale: The correct answer is B because the Nernst equation states that
\(E_K = (RT/zF) \cdot \ln([K^+]_o/[K^+]_i)\). Since T (temperature in Kelvin) is in
the numerator, increasing the temperature increases the absolute magnitude of
the calculated equilibrium potential. Given that the concentration gradient
favors an efflux of K⁺, the value of \(\ln([K^+]_o/[K^+]_i)\) is negative, meaning
\(E_{K}\) becomes more negative (hyperpolarized). The driving force is
calculated as \((V_m - E_K)\). If \(V_{m}\) is held at -80 mV and \(E_{K}\)
becomes more hyperpolarized (e.g., shifts from -84 mV to -89 mV), the absolute
value of \((V_m - E_K)\) grows larger, resulting in an increased driving force.
Options A, C, and D mathematically misapply the relationship between
temperature, the equilibrium potential, and driving force calculation.
,2
Q2: A neuropharmacologist isolates a novel peptide toxin from a marine cone
snail. In a voltage-clamp recording of a spinal motor neuron, application of the
toxin prevents the rapid downstroke of the action potential and blocks the
macroscopic outward current during step-depolarizations, while leaving the
transient inward current unaffected. What is the molecular target of this toxin?
A) Voltage-gated sodium channels (\(Na_{V}\)).
B) Delayed rectifier voltage-gated potassium channels (\(K_{V}\)).
C) Inwardly rectifying potassium channels (Kir).
D) Low-threshold T-type calcium channels (\(Ca_{V}\)).
Rationale: The correct answer is B. The downstroke of a classic action potential
is driven by the inactivation of voltage-gated Na⁺ channels and the
simultaneous delayed opening of voltage-gated K⁺ channels, which generate an
outward hyperpolarizing current. Because the toxin blocks this macroscopic
outward current under voltage-clamp step-depolarizations while leaving the
fast inward current (Na⁺ influx) intact, it selectively blocks the delayed rectifier
K⁺ channels. Option A is wrong because blocking \(Na_{V}\) would eliminate
the inward current and upstroke. Option C is wrong because Kir channels
primarily pass inward currents at hyperpolarized potentials. Option D is wrong
because T-type calcium channels modulate low-threshold firing bursts, not the
standard macroscopic outward repolarization current.
Q3: A researcher calculates quantal parameters at a central excitatory synapse.
Under normal control conditions, the mean amplitude of an evoked excitatory
postsynaptic potential (EPSP) is 4.0 mV, and the mean amplitude of
spontaneous miniature EPSPs (mEPSPs) is 0.5 mV. Following the application of a
drug, the evoked EPSP amplitude drops to 1.5 mV, but the mEPSP amplitude
remains exactly 0.5 mV. What structural or functional change does this drug
induce?
,3
A) It down-regulates postsynaptic AMPA receptor density.
B) It decreases the presynaptic quantal content (m) by lowering the
probability of vesicle release or the number of release sites.
C) It non-competitively blocks postsynaptic NMDA receptors.
D) It increases presynaptic quantal size (q) by packing more neurotransmitter
into each vesicle.
Rationale: The correct answer is B. According to quantal analysis models, the
mean amplitude of an evoked potential is defined as M = m ⋅ q, where m is the
quantal content (the number of vesicles released) and q is the quantal size (the
postsynaptic response to a single vesicle). Miniature EPSPs (mEPSPs) provide a
direct measure of quantal size (q). Because the drug decreased the total evoked
EPSP but left the mEPSP amplitude completely unaltered (q = 0.5 mV), the
postsynaptic sensitivity remained steady. Thus, the change must be entirely
presynaptic, reducing the quantal content (m = M / q dropped from 8 to 3).
Options A and C would reduce both evoked EPSPs and mEPSPs by blunting
postsynaptic reception. Option D contradicts the steady mEPSP data.
Q4: A mutation in the gene encoding the calcium sensor synaptotagmin-1
alters its binding affinity so that it requires threefold higher intracellular
calcium levels to trigger conformational shifts. How will this mutation affect
transmission at a fast chemical synapse during a single presynaptic action
potential?
A) It will increase the frequency of spontaneous miniature end-plate potentials
(MEPPs).
B) It will prevent the propagation of the action potential down the axon shaft.
C) It will severely blunt or abolish synchronous neurotransmitter release
while leaving asynchronous release largely intact.
D) It will double the velocity of synaptic vesicle endocytosis.
, 4
Rationale: The correct answer is C. Synaptotagmin-1 is the primary low-affinity
calcium sensor anchored in the synaptic vesicle membrane that mediates fast,
synchronous exocytosis upon binding calcium entering via voltage-gated
calcium channels (\(Ca_{V}\)). A mutation reducing its calcium affinity means
the localized calcium microdomain created during an action potential will fail to
fully activate it, severely reducing or abolishing fast synchronous release.
Asynchronous release, which relies on high-affinity calcium sensors like double
C2 domain proteins (doc2) or stromal interaction molecules, will be less
affected or remain intact. Option A is wrong because spontaneous release
(MEPPs) is mostly independent of high-calcium synaptotagmin activation.
Option B is incorrect because synaptotagmin does not affect axonal action
potential conduction. Option D is unsupported by vesicle recycling physics.
Q5: In an experimental setup, you express wild-type Drosophila Shaker
potassium channels in Xenopus oocytes. You apply a patch-clamp pipette and
record macroscopic currents using a step protocol from -80 mV to +40 mV. You
then introduce an intracellular protease that cleaves the first 20 amino acids of
the channel’s N-terminus. What operational change do you observe in the
recorded potassium currents?
A) The channels fail to open at any voltage step, acting as a complete loss-of-
function block.
B) The channels open normally upon depolarization but completely lose their
rapid inactivation property, remaining open for the duration of the voltage
step.
C) The activation threshold of the channel shifts from -40 mV to +60 mV.
D) The channel reverses its selectivity filter, passing sodium ions instead of
potassium ions.
Rationale: The correct answer is B. The Shaker potassium channel utilizes an N-
