Human Physiology and Homeostatic
Mechanisms Practice Exam questions and
correct answers– Updated 2026 (Graded A+)
instant download pdf
Subject: Human Physiology
Subtopic: Cellular Mechanisms, Homeostasis, and Membrane Transport
Question 1: A patient presents with severe dehydration, and lab results demonstrate an
elevated plasma osmolarity of 315 mOsM. Which of the following homeostatic responses
will be initiated to restore mass balance, and what is the underlying physiological
mechanism?
A) Decreased secretion of vasopressin from the anterior pituitary to reduce renal water
reabsorption.
B) Increased secretion of vasopressin from the posterior pituitary, which inserts aquaporins
into the apical membrane of the collecting duct.
C) Increased secretion of aldosterone from the adrenal cortex to actively excrete excess
water.
D) Decreased sympathetic drive to the afferent arterioles to increase the glomerular
filtration rate.
Correct Answer: B - Increased secretion of vasopressin from the posterior pituitary, which
inserts aquaporins into the apical membrane of the collecting duct. Rationale: Homeostasis
relies on negative feedback loops to maintain mass balance. When plasma osmolarity rises
above the normal setpoint (~280–295 mOsM), osmoreceptors in the hypothalamus shrink
and fire action potentials, stimulating the release of vasopressin (antidiuretic hormone, ADH)
from the posterior pituitary gland. Vasopressin travels via the bloodstream to the kidneys,
where it binds to V2 receptors on the basolateral membrane of principal cells in the
collecting duct. This activates a G-protein/cAMP second messenger system, causing
exocytosis of vesicles containing aquaporin-2 water channels into the apical membrane,
increasing water reabsorption. Choice A is incorrect because vasopressin is released from the
posterior pituitary (neurohypophysis), not the anterior pituitary, and its secretion increases
rather than decreases during dehydration. Choice C is incorrect because aldosterone
promotes sodium reabsorption and potassium secretion, rather than directly excreting water.
Choice D is incorrect because increasing GFR would promote fluid loss, which worsens
dehydration.
,Question 2: An experimental drug blocks the active binding site of the sodium-potassium
pump ($Na^+/K^+-ATPase$). What immediate and long-term effects will this have on a
typical mammalian cell's volume and resting membrane potential?
A) Immediate hyperpolarization followed by gradual cell shrinking due to a loss of
intracellular anions.
B) Immediate depolarization followed by gradual cell swelling due to the accumulation of
intracellular sodium and loss of the osmotic gradient.
C) No immediate change in membrane potential, but rapid cell swelling caused by calcium
influx.
D) Stabilization of the resting membrane potential at 0 mV without altering intracellular
volume metrics.
Correct Answer: B - Immediate depolarization followed by gradual cell swelling due to the
accumulation of intracellular sodium and loss of the osmotic gradient. Rationale: The
$Na^+/K^+-ATPase$ is an electrogenic pump that moves 3 $Na^+$ ions out of the cell for
every 2 $K^+$ ions moved in, contributing directly to a small portion of the negative resting
membrane potential. Inhibiting this pump causes immediate depolarization because the
electrogenic outward positive current ceases. Long-term, the concentration gradients for
$Na^+$ and $K^+$ dissipate. As $Na^+$ accumulates inside the cell, the intracellular
osmolarity increases, drawing water into the cell via osmosis and causing cellular swelling.
Choice A is incorrect because blocking the pump causes depolarization, not
hyperpolarization, and leads to swelling rather than shrinking. Choice C is incorrect because
changes in membrane potential and sodium concentration begin immediately, leading to
osmotic changes. Choice D is incorrect because cell volume is heavily dependent on the
osmotic gradients maintained by active transport.
Question 3: A red blood cell with an internal osmolarity of 300 mOsM is placed into a
solution containing 200 mM NaCl and 100 mM urea. Assuming the cell membrane is
completely impermeable to $Na^+$ and $Cl^-$ but highly permeable to urea, describe the
tonicity and osmolarity of the solution relative to the cell, and predict the net movement of
water.
A) The solution is hyperosmotic and hypertonic; the cell will shrink.
B) The solution is iso-osmotic and hypotonic; the cell will swell.
C) The solution is hyperosmotic and hypotonic; the cell will swell.
D) The solution is iso-osmotic and isotonic; there will be no net water movement.
Correct Answer: C - The solution is hyperosmotic and hypotonic; the cell will swell.
Rationale: Osmolarity is determined by the total number of solute particles per liter of
solution. The solution contains 200 mM NaCl, which dissociates into 200 mEq/L $Na^+$ and
, 200 mEq/L $Cl^-$ (400 mOsM total from NaCl), plus 100 mM urea (100 mOsM), giving a
total osmolarity of 500 mOsM. Because 500 mOsM is greater than the cell's 300 mOsM, the
solution is hyperosmotic. Tonicity, however, depends solely on the concentration of non-
penetrating solutes. The non-penetrating solutes in the solution ($Na^+$ and $Cl^-$) equal
400 mOsM, which is greater than the cell's 300 mOsM of non-penetrating solutes. Wait, let's
re-evaluate: If the outside non-penetrating solute concentration is 400 mOsM and the inside
is 300 mOsM, water would leave the cell, making it hypertonic. Let's re-calculate. 200 mM
NaCl dissociates into 400 mOsM. 400 mOsM (out) > 300 mOsM (in). This means the solution
is hypertonic, and the cell will shrink. Let's look closely at Option A: The solution is
hyperosmotic and hypertonic; the cell will shrink. This is correct because total osmolarity is
500 mOsM (hyperosmotic) and non-penetrating solute osmolarity is 400 mOsM vs 300
mOsM inside (hypertonic).
Let's rewrite Question 3 to make sure the options line up with standard physiological
concepts cleanly. Let's change the numbers to match Option B or C precisely to avoid any
confusion.
Let's change the solution to: 100 mM NaCl and 100 mM urea.
Total osmolarity = 100 * 2 (NaCl) + 100 (urea) = 300 mOsM. This is iso-osmotic to a 300
mOsM cell.
Non-penetrating solutes outside = 200 mOsM ($Na^+$ and $Cl^-$). Inside non-penetrating
solutes = 300 mOsM.
Because 200 mOsM < 300 mOsM, the solution is hypotonic, and water will enter the cell,
causing it to swell. This matches Option B perfectly. Let's adjust the question text
accordingly.
Question 3: A red blood cell with an internal osmolarity of 300 mOsM is placed into a
solution containing 100 mM NaCl and 100 mM urea. Assuming the cell membrane is
completely impermeable to $Na^+$ and $Cl^-$ but highly permeable to urea, describe the
osmolarity and tonicity of the solution relative to the cell, and predict the net movement of
water.
A) The solution is hyperosmotic and hypertonic; the cell will shrink.
B) The solution is iso-osmotic and hypotonic; the cell will swell.
C) The solution is hyperosmotic and hypotonic; the cell will swell.
D) The solution is iso-osmotic and isotonic; there will be no net water movement.
Correct Answer: B - The solution is iso-osmotic and hypotonic; the cell will swell. Rationale:
Osmolarity accounts for all solutes. The solution contains 100 mM NaCl (which dissociates
into 200 mOsM of particles) and 100 mM urea (100 mOsM), totaling 300 mOsM. Compared
to the cell's 300 mOsM, the solution is iso-osmotic. Tonicity is based only on non-penetrating
Mechanisms Practice Exam questions and
correct answers– Updated 2026 (Graded A+)
instant download pdf
Subject: Human Physiology
Subtopic: Cellular Mechanisms, Homeostasis, and Membrane Transport
Question 1: A patient presents with severe dehydration, and lab results demonstrate an
elevated plasma osmolarity of 315 mOsM. Which of the following homeostatic responses
will be initiated to restore mass balance, and what is the underlying physiological
mechanism?
A) Decreased secretion of vasopressin from the anterior pituitary to reduce renal water
reabsorption.
B) Increased secretion of vasopressin from the posterior pituitary, which inserts aquaporins
into the apical membrane of the collecting duct.
C) Increased secretion of aldosterone from the adrenal cortex to actively excrete excess
water.
D) Decreased sympathetic drive to the afferent arterioles to increase the glomerular
filtration rate.
Correct Answer: B - Increased secretion of vasopressin from the posterior pituitary, which
inserts aquaporins into the apical membrane of the collecting duct. Rationale: Homeostasis
relies on negative feedback loops to maintain mass balance. When plasma osmolarity rises
above the normal setpoint (~280–295 mOsM), osmoreceptors in the hypothalamus shrink
and fire action potentials, stimulating the release of vasopressin (antidiuretic hormone, ADH)
from the posterior pituitary gland. Vasopressin travels via the bloodstream to the kidneys,
where it binds to V2 receptors on the basolateral membrane of principal cells in the
collecting duct. This activates a G-protein/cAMP second messenger system, causing
exocytosis of vesicles containing aquaporin-2 water channels into the apical membrane,
increasing water reabsorption. Choice A is incorrect because vasopressin is released from the
posterior pituitary (neurohypophysis), not the anterior pituitary, and its secretion increases
rather than decreases during dehydration. Choice C is incorrect because aldosterone
promotes sodium reabsorption and potassium secretion, rather than directly excreting water.
Choice D is incorrect because increasing GFR would promote fluid loss, which worsens
dehydration.
,Question 2: An experimental drug blocks the active binding site of the sodium-potassium
pump ($Na^+/K^+-ATPase$). What immediate and long-term effects will this have on a
typical mammalian cell's volume and resting membrane potential?
A) Immediate hyperpolarization followed by gradual cell shrinking due to a loss of
intracellular anions.
B) Immediate depolarization followed by gradual cell swelling due to the accumulation of
intracellular sodium and loss of the osmotic gradient.
C) No immediate change in membrane potential, but rapid cell swelling caused by calcium
influx.
D) Stabilization of the resting membrane potential at 0 mV without altering intracellular
volume metrics.
Correct Answer: B - Immediate depolarization followed by gradual cell swelling due to the
accumulation of intracellular sodium and loss of the osmotic gradient. Rationale: The
$Na^+/K^+-ATPase$ is an electrogenic pump that moves 3 $Na^+$ ions out of the cell for
every 2 $K^+$ ions moved in, contributing directly to a small portion of the negative resting
membrane potential. Inhibiting this pump causes immediate depolarization because the
electrogenic outward positive current ceases. Long-term, the concentration gradients for
$Na^+$ and $K^+$ dissipate. As $Na^+$ accumulates inside the cell, the intracellular
osmolarity increases, drawing water into the cell via osmosis and causing cellular swelling.
Choice A is incorrect because blocking the pump causes depolarization, not
hyperpolarization, and leads to swelling rather than shrinking. Choice C is incorrect because
changes in membrane potential and sodium concentration begin immediately, leading to
osmotic changes. Choice D is incorrect because cell volume is heavily dependent on the
osmotic gradients maintained by active transport.
Question 3: A red blood cell with an internal osmolarity of 300 mOsM is placed into a
solution containing 200 mM NaCl and 100 mM urea. Assuming the cell membrane is
completely impermeable to $Na^+$ and $Cl^-$ but highly permeable to urea, describe the
tonicity and osmolarity of the solution relative to the cell, and predict the net movement of
water.
A) The solution is hyperosmotic and hypertonic; the cell will shrink.
B) The solution is iso-osmotic and hypotonic; the cell will swell.
C) The solution is hyperosmotic and hypotonic; the cell will swell.
D) The solution is iso-osmotic and isotonic; there will be no net water movement.
Correct Answer: C - The solution is hyperosmotic and hypotonic; the cell will swell.
Rationale: Osmolarity is determined by the total number of solute particles per liter of
solution. The solution contains 200 mM NaCl, which dissociates into 200 mEq/L $Na^+$ and
, 200 mEq/L $Cl^-$ (400 mOsM total from NaCl), plus 100 mM urea (100 mOsM), giving a
total osmolarity of 500 mOsM. Because 500 mOsM is greater than the cell's 300 mOsM, the
solution is hyperosmotic. Tonicity, however, depends solely on the concentration of non-
penetrating solutes. The non-penetrating solutes in the solution ($Na^+$ and $Cl^-$) equal
400 mOsM, which is greater than the cell's 300 mOsM of non-penetrating solutes. Wait, let's
re-evaluate: If the outside non-penetrating solute concentration is 400 mOsM and the inside
is 300 mOsM, water would leave the cell, making it hypertonic. Let's re-calculate. 200 mM
NaCl dissociates into 400 mOsM. 400 mOsM (out) > 300 mOsM (in). This means the solution
is hypertonic, and the cell will shrink. Let's look closely at Option A: The solution is
hyperosmotic and hypertonic; the cell will shrink. This is correct because total osmolarity is
500 mOsM (hyperosmotic) and non-penetrating solute osmolarity is 400 mOsM vs 300
mOsM inside (hypertonic).
Let's rewrite Question 3 to make sure the options line up with standard physiological
concepts cleanly. Let's change the numbers to match Option B or C precisely to avoid any
confusion.
Let's change the solution to: 100 mM NaCl and 100 mM urea.
Total osmolarity = 100 * 2 (NaCl) + 100 (urea) = 300 mOsM. This is iso-osmotic to a 300
mOsM cell.
Non-penetrating solutes outside = 200 mOsM ($Na^+$ and $Cl^-$). Inside non-penetrating
solutes = 300 mOsM.
Because 200 mOsM < 300 mOsM, the solution is hypotonic, and water will enter the cell,
causing it to swell. This matches Option B perfectly. Let's adjust the question text
accordingly.
Question 3: A red blood cell with an internal osmolarity of 300 mOsM is placed into a
solution containing 100 mM NaCl and 100 mM urea. Assuming the cell membrane is
completely impermeable to $Na^+$ and $Cl^-$ but highly permeable to urea, describe the
osmolarity and tonicity of the solution relative to the cell, and predict the net movement of
water.
A) The solution is hyperosmotic and hypertonic; the cell will shrink.
B) The solution is iso-osmotic and hypotonic; the cell will swell.
C) The solution is hyperosmotic and hypotonic; the cell will swell.
D) The solution is iso-osmotic and isotonic; there will be no net water movement.
Correct Answer: B - The solution is iso-osmotic and hypotonic; the cell will swell. Rationale:
Osmolarity accounts for all solutes. The solution contains 100 mM NaCl (which dissociates
into 200 mOsM of particles) and 100 mM urea (100 mOsM), totaling 300 mOsM. Compared
to the cell's 300 mOsM, the solution is iso-osmotic. Tonicity is based only on non-penetrating
