Membrane Dynamics and Endocrine Signalling
Practice Exam questions and correct answers–
Updated 2026 (Graded A+) instant download pdf
Subject: Human Physiology
Subtopic: Cellular Membrane Dynamics and Transport
Question 1: A researcher is studying a novel pharmacological agent that selectively inhibits
the activity of the $Na^+$-$K^+$-ATPase pump in skeletal muscle cells. Following
administration of this drug, which of the following immediate intracellular alterations would
most likely occur?
A) An increase in intracellular sodium concentration and a decrease in intracellular
potassium concentration.
B) A decrease in intracellular sodium concentration and an increase in intracellular
potassium concentration.
C) Hyperpolarization of the resting membrane potential due to rapid potassium efflux.
D) Accelerated secondary active transport of glucose via SGLT transporters due to an
optimized sodium gradient.
Correct Answer: A - An increase in intracellular sodium concentration and a decrease in
intracellular potassium concentration. Rationale: The $Na^+$-$K^+$-ATPase primary active
transport pump structurally moves three $Na^+$ ions out of the cell and two $K^+$ ions into
the cell per ATP molecule hydrolyzed, maintaining high extracellular $Na^+$ and high
intracellular $K^+$ concentrations. Inhibiting this pump prevents this active exchange,
causing intracellular sodium to accumulate because passive leak channels allow sodium to
enter along its electrochemical gradient, while intracellular potassium decreases as it leaks
out without being replenished. Option B describes the exact opposite effect. Option C is
incorrect because stopping the electrogenic pump (which normally pumps out more positive
charges than it brings in) leads to depolarization, not hyperpolarization. Option D is incorrect
because SGLT relies on a steep extracellular-to-intracellular sodium gradient; flattening this
gradient impairs secondary active transport.
Question 2: A 42-year-old patient presents with severe dehydration due to cholera. The
cholera toxin constitutively activates Gs proteins, leading to high intracellular cAMP levels in
intestinal epithelial cells, opening apical $Cl^-$ channels (CFTR). Water follows osmotic
gradients into the intestinal lumen. To treat this effectively oral rehydration therapy
containing glucose and sodium is administered. What mechanism explains the efficacy of
this formulation?
,A) Glucose and sodium utilize simple diffusion to bypass the disrupted epithelial tight
junctions.
B) Sodium and glucose enter the enterocytes via the SGLT symporter, creating an osmotic
gradient that drives water reabsorption via aquaporins.
C) Glucose directly blocks the CFTR channels, stopping chloride secretion into the intestinal
lumen.
D) Sodium stimulates primary active transport via the sodium-potassium pump on the apical
membrane to directly pump water molecules.
Correct Answer: B - Sodium and glucose enter the enterocytes via the SGLT symporter,
creating an osmotic gradient that drives water reabsorption via aquaporins. Rationale:
Oral rehydration therapy leverages the SGLT (sodium-glucose cotransporter) protein located
on the apical membrane of intestinal epithelial cells. This is a secondary active transport
system that functions independently of the cAMP pathway disrupted by cholera toxin. The
simultaneous transport of sodium and glucose into the cell increases intracellular osmolarity,
which draws water from the lumen back into the body through aquaporins and paracellular
pathways. Option A is incorrect because glucose and sodium are polar/charged and cannot
efficiently cross through simple diffusion or tight junctions under these conditions. Option C is
incorrect because glucose does not act as a direct blocker of CFTR. Option D is incorrect
because the $Na^+$-$K^+$-ATPase pump is located on the basolateral membrane, not the
apical membrane, and it does not pump water directly.
Question 3: In an experimental biophysics laboratory, the resting membrane potential of an
artificial neuron is measured at -70 mV. If the extracellular potassium concentration is
experimentally doubled while keeping all other ion concentrations constant, what will
happen to the resting membrane potential, and which equation best predicts this change?
A) The membrane will hyperpolarize; predicted by the Fick's Law equation.
B) The membrane will depolarize; predicted by the Nernst equation or Goldman-Hodgkin-
Katz (GHK) equation.
C) The membrane potential will remain unchanged because the cell is impermeable to
potassium at rest; predicted by Starling's equation.
D) The membrane will immediately drop to 0 mV due to complete loss of the sodium driving
force; predicted by the Michaelis-Menten equation.
Correct Answer: B - The membrane will depolarize; predicted by the Nernst equation or
Goldman-Hodgkin-Katz (GHK) equation. Rationale: The resting membrane potential of a cell
is highly sensitive to the concentration gradient of potassium because membranes at rest are
highly permeable to $K^+$ via leak channels. Increasing the extracellular potassium
concentration reduces the concentration gradient between the inside and outside of the cell,
,which reduces the chemical driving force for potassium efflux. As less positive charge leaves
the cell, the interior becomes more positive (depolarized). The Nernst and GHK equations
mathematically describe how ionic concentrations and membrane permeability determine
equilibrium and membrane potentials. Option A is incorrect because reducing the gradient
causes depolarization, not hyperpolarization. Option C is incorrect because resting
membranes are highly permeable to potassium. Option D is incorrect because the sodium
gradient is not directly modified here, and the potential changes progressively rather than
dropping immediately to exactly 0 mV.
Question 4: A cell is placed in a solution containing 150 mOsM of non-penetrating solute
(NaCl) and 150 mOsM of a highly penetrating solute (urea). How would you describe the
initial osmolarity of the solution relative to the cell (assume cell osmolarity is 300 mOsM),
and what will happen to the cell volume at equilibrium?
A) The solution is hyperosmotic, and the cell will swell.
B) The solution is isosmotic, and the cell volume will remain unchanged.
C) The solution is isosmotic, and the cell will shrink.
D) The solution is isosmotic, and the cell will swell.
Correct Answer: D - The solution is isosmotic, and the cell will swell. Rationale: Total
osmolarity counts all solutes regardless of whether they can penetrate the cell membrane or
not. Since $150 \text{ mOsM NaCl} + 150 \text{ mOsM urea} = 300 \text{ mOsM}$, the
solution is initially isosmotic to a standard 300 mOsM cell. Tonicity, however, depends solely
on the concentration of non-penetrating solutes. The cell contains 300 mOsM of non-
penetrating solutes, while the solution contains only 150 mOsM of non-penetrating solutes
(NaCl). Because urea is a penetrating solute, it will diffuse down its concentration gradient
into the cell until its concentration equilibrates. This leaves a net excess of non-penetrating
solutes inside the cell, creating an osmotic gradient that drives water into the cell, causing it
to swell. Therefore, the solution is hypotonic, despite being isosmotic initially. Options A, B,
and C fail to properly separate the definitions of osmolarity and tonicity.
Question 5: Consider a cell that expresses a specific carrier protein performing facilitated
diffusion for a particular amino acid. Which of the following features uniquely distinguishes
this facilitated diffusion from simple diffusion through the lipid bilayer?
A) The rate of transport is directly proportional to the concentration gradient without any
upper threshold limits.
B) Transport can move the amino acid against its electrochemical gradient by directly
utilizing metabolic energy from ATP.
C) The transport rate demonstrates saturation kinetics, reaching a maximum transport
maximum ($T_m$) when all carriers are occupied.
, D) Transport requires the presence of a hydrostatic pressure gradient across the plasma
membrane to drive the molecules.
Correct Answer: C - The transport rate demonstrates saturation kinetics, reaching a
maximum transport maximum ($T_m$) when all carriers are occupied. Rationale:
Facilitated diffusion relies on a finite number of specialized carrier proteins within the plasma
membrane. Because the number of transporters is fixed, the rate of transport shows
saturation kinetics; as the solute concentration increases, the transport rate increases until
all binding sites are continuously occupied, reaching a transport maximum ($T_m$). Simple
diffusion does not show saturation because it does not rely on protein binding sites. Option A
describes simple diffusion, which is linear and unsaturable. Option B describes primary active
transport, not facilitated diffusion, which is passive and cannot move solutes against a
gradient. Option D describes filtration, not facilitated diffusion.
Question 6: Red blood cells (RBCs) are isolated and suspended in a test tube containing a
solution of 0.9% sodium chloride (NaCl). A researcher then adds a small amount of
concentrated solution containing a substance that blocks aquaporin channels completely.
What is the predicted consequence of this modification on the cell volume when the cells
remain in the 0.9% NaCl solution?
A) The cells will immediately burst due to the uncontrolled influx of sodium ions.
B) The cells will shrink significantly because water will be forced out through the lipid bilayer
by active transport.
C) The cell volume will remain largely unchanged because 0.9% NaCl is isotonic, meaning
there is no net osmotic driving force for water movement.
D) The cells will swell slowly because blocking aquaporins reverses the structural orientation
of the sodium-potassium pump.
Correct Answer: C - The cell volume will remain largely unchanged because 0.9% NaCl is
isotonic, meaning there is no net osmotic driving force for water movement. Rationale: A
0.9% NaCl solution is clinically isotonic to human red blood cells, meaning the concentration
of non-penetrating solutes outside matches the concentration inside. Because there is no
osmotic gradient, there is no net movement of water across the membrane. Blocking
aquaporins reduces the permeability of the membrane to water, but since there is no driving
force for water movement in an isotonic solution, the volume remains stable. Option A is
incorrect because sodium transport is unaffected and does not cause bursting. Option B is
incorrect because water is not actively transported by pumps. Option D is incorrect because
aquaporin inhibition does not structurally alter or reverse the $Na^+$-$K^+$-ATPase pump.
Question 7: A multi-drug resistant tumor cell line overexpresses ABC (ATP-Binding Cassette)
transporters, specifically P-glycoprotein. These transporters actively pump
Practice Exam questions and correct answers–
Updated 2026 (Graded A+) instant download pdf
Subject: Human Physiology
Subtopic: Cellular Membrane Dynamics and Transport
Question 1: A researcher is studying a novel pharmacological agent that selectively inhibits
the activity of the $Na^+$-$K^+$-ATPase pump in skeletal muscle cells. Following
administration of this drug, which of the following immediate intracellular alterations would
most likely occur?
A) An increase in intracellular sodium concentration and a decrease in intracellular
potassium concentration.
B) A decrease in intracellular sodium concentration and an increase in intracellular
potassium concentration.
C) Hyperpolarization of the resting membrane potential due to rapid potassium efflux.
D) Accelerated secondary active transport of glucose via SGLT transporters due to an
optimized sodium gradient.
Correct Answer: A - An increase in intracellular sodium concentration and a decrease in
intracellular potassium concentration. Rationale: The $Na^+$-$K^+$-ATPase primary active
transport pump structurally moves three $Na^+$ ions out of the cell and two $K^+$ ions into
the cell per ATP molecule hydrolyzed, maintaining high extracellular $Na^+$ and high
intracellular $K^+$ concentrations. Inhibiting this pump prevents this active exchange,
causing intracellular sodium to accumulate because passive leak channels allow sodium to
enter along its electrochemical gradient, while intracellular potassium decreases as it leaks
out without being replenished. Option B describes the exact opposite effect. Option C is
incorrect because stopping the electrogenic pump (which normally pumps out more positive
charges than it brings in) leads to depolarization, not hyperpolarization. Option D is incorrect
because SGLT relies on a steep extracellular-to-intracellular sodium gradient; flattening this
gradient impairs secondary active transport.
Question 2: A 42-year-old patient presents with severe dehydration due to cholera. The
cholera toxin constitutively activates Gs proteins, leading to high intracellular cAMP levels in
intestinal epithelial cells, opening apical $Cl^-$ channels (CFTR). Water follows osmotic
gradients into the intestinal lumen. To treat this effectively oral rehydration therapy
containing glucose and sodium is administered. What mechanism explains the efficacy of
this formulation?
,A) Glucose and sodium utilize simple diffusion to bypass the disrupted epithelial tight
junctions.
B) Sodium and glucose enter the enterocytes via the SGLT symporter, creating an osmotic
gradient that drives water reabsorption via aquaporins.
C) Glucose directly blocks the CFTR channels, stopping chloride secretion into the intestinal
lumen.
D) Sodium stimulates primary active transport via the sodium-potassium pump on the apical
membrane to directly pump water molecules.
Correct Answer: B - Sodium and glucose enter the enterocytes via the SGLT symporter,
creating an osmotic gradient that drives water reabsorption via aquaporins. Rationale:
Oral rehydration therapy leverages the SGLT (sodium-glucose cotransporter) protein located
on the apical membrane of intestinal epithelial cells. This is a secondary active transport
system that functions independently of the cAMP pathway disrupted by cholera toxin. The
simultaneous transport of sodium and glucose into the cell increases intracellular osmolarity,
which draws water from the lumen back into the body through aquaporins and paracellular
pathways. Option A is incorrect because glucose and sodium are polar/charged and cannot
efficiently cross through simple diffusion or tight junctions under these conditions. Option C is
incorrect because glucose does not act as a direct blocker of CFTR. Option D is incorrect
because the $Na^+$-$K^+$-ATPase pump is located on the basolateral membrane, not the
apical membrane, and it does not pump water directly.
Question 3: In an experimental biophysics laboratory, the resting membrane potential of an
artificial neuron is measured at -70 mV. If the extracellular potassium concentration is
experimentally doubled while keeping all other ion concentrations constant, what will
happen to the resting membrane potential, and which equation best predicts this change?
A) The membrane will hyperpolarize; predicted by the Fick's Law equation.
B) The membrane will depolarize; predicted by the Nernst equation or Goldman-Hodgkin-
Katz (GHK) equation.
C) The membrane potential will remain unchanged because the cell is impermeable to
potassium at rest; predicted by Starling's equation.
D) The membrane will immediately drop to 0 mV due to complete loss of the sodium driving
force; predicted by the Michaelis-Menten equation.
Correct Answer: B - The membrane will depolarize; predicted by the Nernst equation or
Goldman-Hodgkin-Katz (GHK) equation. Rationale: The resting membrane potential of a cell
is highly sensitive to the concentration gradient of potassium because membranes at rest are
highly permeable to $K^+$ via leak channels. Increasing the extracellular potassium
concentration reduces the concentration gradient between the inside and outside of the cell,
,which reduces the chemical driving force for potassium efflux. As less positive charge leaves
the cell, the interior becomes more positive (depolarized). The Nernst and GHK equations
mathematically describe how ionic concentrations and membrane permeability determine
equilibrium and membrane potentials. Option A is incorrect because reducing the gradient
causes depolarization, not hyperpolarization. Option C is incorrect because resting
membranes are highly permeable to potassium. Option D is incorrect because the sodium
gradient is not directly modified here, and the potential changes progressively rather than
dropping immediately to exactly 0 mV.
Question 4: A cell is placed in a solution containing 150 mOsM of non-penetrating solute
(NaCl) and 150 mOsM of a highly penetrating solute (urea). How would you describe the
initial osmolarity of the solution relative to the cell (assume cell osmolarity is 300 mOsM),
and what will happen to the cell volume at equilibrium?
A) The solution is hyperosmotic, and the cell will swell.
B) The solution is isosmotic, and the cell volume will remain unchanged.
C) The solution is isosmotic, and the cell will shrink.
D) The solution is isosmotic, and the cell will swell.
Correct Answer: D - The solution is isosmotic, and the cell will swell. Rationale: Total
osmolarity counts all solutes regardless of whether they can penetrate the cell membrane or
not. Since $150 \text{ mOsM NaCl} + 150 \text{ mOsM urea} = 300 \text{ mOsM}$, the
solution is initially isosmotic to a standard 300 mOsM cell. Tonicity, however, depends solely
on the concentration of non-penetrating solutes. The cell contains 300 mOsM of non-
penetrating solutes, while the solution contains only 150 mOsM of non-penetrating solutes
(NaCl). Because urea is a penetrating solute, it will diffuse down its concentration gradient
into the cell until its concentration equilibrates. This leaves a net excess of non-penetrating
solutes inside the cell, creating an osmotic gradient that drives water into the cell, causing it
to swell. Therefore, the solution is hypotonic, despite being isosmotic initially. Options A, B,
and C fail to properly separate the definitions of osmolarity and tonicity.
Question 5: Consider a cell that expresses a specific carrier protein performing facilitated
diffusion for a particular amino acid. Which of the following features uniquely distinguishes
this facilitated diffusion from simple diffusion through the lipid bilayer?
A) The rate of transport is directly proportional to the concentration gradient without any
upper threshold limits.
B) Transport can move the amino acid against its electrochemical gradient by directly
utilizing metabolic energy from ATP.
C) The transport rate demonstrates saturation kinetics, reaching a maximum transport
maximum ($T_m$) when all carriers are occupied.
, D) Transport requires the presence of a hydrostatic pressure gradient across the plasma
membrane to drive the molecules.
Correct Answer: C - The transport rate demonstrates saturation kinetics, reaching a
maximum transport maximum ($T_m$) when all carriers are occupied. Rationale:
Facilitated diffusion relies on a finite number of specialized carrier proteins within the plasma
membrane. Because the number of transporters is fixed, the rate of transport shows
saturation kinetics; as the solute concentration increases, the transport rate increases until
all binding sites are continuously occupied, reaching a transport maximum ($T_m$). Simple
diffusion does not show saturation because it does not rely on protein binding sites. Option A
describes simple diffusion, which is linear and unsaturable. Option B describes primary active
transport, not facilitated diffusion, which is passive and cannot move solutes against a
gradient. Option D describes filtration, not facilitated diffusion.
Question 6: Red blood cells (RBCs) are isolated and suspended in a test tube containing a
solution of 0.9% sodium chloride (NaCl). A researcher then adds a small amount of
concentrated solution containing a substance that blocks aquaporin channels completely.
What is the predicted consequence of this modification on the cell volume when the cells
remain in the 0.9% NaCl solution?
A) The cells will immediately burst due to the uncontrolled influx of sodium ions.
B) The cells will shrink significantly because water will be forced out through the lipid bilayer
by active transport.
C) The cell volume will remain largely unchanged because 0.9% NaCl is isotonic, meaning
there is no net osmotic driving force for water movement.
D) The cells will swell slowly because blocking aquaporins reverses the structural orientation
of the sodium-potassium pump.
Correct Answer: C - The cell volume will remain largely unchanged because 0.9% NaCl is
isotonic, meaning there is no net osmotic driving force for water movement. Rationale: A
0.9% NaCl solution is clinically isotonic to human red blood cells, meaning the concentration
of non-penetrating solutes outside matches the concentration inside. Because there is no
osmotic gradient, there is no net movement of water across the membrane. Blocking
aquaporins reduces the permeability of the membrane to water, but since there is no driving
force for water movement in an isotonic solution, the volume remains stable. Option A is
incorrect because sodium transport is unaffected and does not cause bursting. Option B is
incorrect because water is not actively transported by pumps. Option D is incorrect because
aquaporin inhibition does not structurally alter or reverse the $Na^+$-$K^+$-ATPase pump.
Question 7: A multi-drug resistant tumor cell line overexpresses ABC (ATP-Binding Cassette)
transporters, specifically P-glycoprotein. These transporters actively pump
