Anatomy Quiz: Membrane Structure And Transport
16 questions · exam conditions
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Membrane Structure And TransportQuestion 1 of 16

A cell biologist is studying endocytosis and notices that when she depletes cellular ATP, receptor-mediated endocytosis stops completely, but fluid-phase endocytosis continues at about 30% of normal rate. When she adds cytochalasin D (which disrupts actin filaments), both processes stop completely. When she adds colchicine (which disrupts microtubules), receptor-mediated endocytosis is reduced by 80% while fluid-phase endocytosis is reduced by 20%. What can be concluded about the energy and cytoskeletal requirements of these processes?

Both processes depend equally on actin filaments for membrane deformation, but receptor-mediated endocytosis requires microtubules for vesicle trafficking
Receptor-mediated endocytosis is entirely ATP-dependent, while fluid-phase endocytosis can partially proceed through passive membrane dynamics
Both processes require ATP for specific steps, but receptor-mediated endocytosis has additional energy-dependent requirements
Fluid-phase endocytosis relies primarily on membrane tension changes, while receptor-mediated endocytosis requires coordinated cytoskeletal rearrangements
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Anatomy Quiz

Anatomy Quiz: Membrane Structure And Transport

Practice Membrane Structure And Transport in Anatomy with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Membrane Structure And Transport, giving you a quick way to practice the rules, question types, and explanations that matter most for Anatomy.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A cell biologist is studying endocytosis and notices that when she depletes cellular ATP, receptor-mediated endocytosis stops completely, but fluid-phase endocytosis continues at about 30% of normal rate. When she adds cytochalasin D (which disrupts actin filaments), both processes stop completely. When she adds colchicine (which disrupts microtubules), receptor-mediated endocytosis is reduced by 80% while fluid-phase endocytosis is reduced by 20%. What can be concluded about the energy and cytoskeletal requirements of these processes?

  1. Both processes depend equally on actin filaments for membrane deformation, but receptor-mediated endocytosis requires microtubules for vesicle trafficking (correct answer)
  2. Receptor-mediated endocytosis is entirely ATP-dependent, while fluid-phase endocytosis can partially proceed through passive membrane dynamics
  3. Both processes require ATP for specific steps, but receptor-mediated endocytosis has additional energy-dependent requirements
  4. Fluid-phase endocytosis relies primarily on membrane tension changes, while receptor-mediated endocytosis requires coordinated cytoskeletal rearrangements
Explanation: When you encounter questions about cellular processes and experimental manipulations, focus on what each treatment specifically disrupts and how the processes respond differently. Let's analyze what each experimental condition reveals. ATP depletion completely stops receptor-mediated endocytosis but allows fluid-phase endocytosis to continue at 30% - this shows receptor-mediated endocytosis has absolute ATP requirements while fluid-phase endocytosis can partially proceed without energy. Cytochalasin D (actin disruptor) completely stops both processes, indicating both absolutely require actin filaments for the membrane deformation needed to form endocytic vesicles. Colchicine (microtubule disruptor) severely reduces receptor-mediated endocytosis (80% reduction) but barely affects fluid-phase endocytosis (20% reduction), suggesting receptor-mediated endocytosis depends heavily on microtubules for vesicle trafficking to specific cellular destinations. Answer A correctly identifies that both processes equally depend on actin for membrane deformation (both stopped completely with cytochalasin D), while receptor-mediated endocytosis specifically requires microtubules for vesicle trafficking (major reduction with colchicine). Answer B incorrectly suggests receptor-mediated endocytosis is "entirely" ATP-dependent when both processes show some ATP dependence. Answer C misses the key distinction about microtubule requirements. Answer D incorrectly downplays the role of coordinated cytoskeletal requirements in both processes. Remember: experimental questions test your ability to connect specific inhibitors with their molecular targets and interpret differential effects on related cellular processes.

Question 2

During facilitated diffusion of glucose through a GLUT transporter, which statement best describes the relationship between transport rate and glucose concentration?

  1. Transport rate increases linearly with glucose concentration because diffusion follows Fick's law directly
  2. Transport rate increases with glucose concentration but plateaus at high concentrations due to transporter saturation (correct answer)
  3. Transport rate remains constant regardless of glucose concentration because transporters work at fixed rates
  4. Transport rate decreases with increasing glucose concentration due to competitive inhibition by excess substrate
  5. Transport rate oscillates with glucose concentration due to conformational changes in the transporter protein
Explanation: When you encounter questions about facilitated diffusion through membrane transporters, think about how these proteins behave differently from simple diffusion. Unlike passive diffusion through the lipid bilayer, transporter proteins have specific binding sites and can become saturated. GLUT transporters work by binding glucose molecules, undergoing a conformational change, and releasing glucose on the other side of the membrane. As glucose concentration increases, more transporters become occupied and transport rate increases. However, once all available transporters are working at maximum capacity, adding more glucose won't increase the transport rate further – this creates the characteristic saturation curve. This kinetic behavior follows Michaelis-Menten kinetics, similar to enzyme reactions, making option B correct. Option A is wrong because simple Fick's law applies to passive diffusion through membranes, not protein-mediated transport. While transport does increase with concentration initially, it's not a linear relationship throughout. Option C incorrectly suggests transporters work at fixed rates regardless of substrate concentration – this would mean glucose concentration has no effect on transport, which contradicts how facilitated diffusion works. Option D describes competitive inhibition, but glucose isn't inhibiting its own transporter; excess glucose simply can't be transported once all binding sites are occupied. Remember that facilitated diffusion questions often test whether you understand saturation kinetics versus simple diffusion. Look for keywords like "transporter" or "carrier protein" – these signal that you should expect a saturation curve, not linear kinetics.

Question 3

A researcher measures the initial rate of Na+ transport across a cell membrane under different conditions. In the presence of ATP and normal Na+/K+-ATPase function, 100 Na+ ions are transported out per second. When ATP is completely depleted but the Na+ gradient remains, 20 Na+ ions still move out per second. What best explains this observation?

  1. All Na+ transport requires ATP, so the 20 ions/second represents measurement error in the experimental setup
  2. The remaining transport occurs through passive Na+ channels driven by the electrochemical gradient established by previous ATP-dependent pumping (correct answer)
  3. Na+/K+-ATPase can function briefly without ATP by using stored conformational energy from previous ATP binding events
  4. The cell switches to anaerobic metabolism to produce alternative energy sources that can drive continued active transport
  5. Na+ transport reverses direction and the 20 ions/second actually represents inward movement through the pump running backward
Explanation: When you encounter questions about membrane transport, focus on distinguishing between active transport (requires energy) and passive transport (driven by gradients). This question tests whether you understand that cells use multiple transport mechanisms simultaneously. The key insight is that Na⁺/K⁺-ATPase creates concentration and electrical gradients by actively pumping Na⁺ out and K⁺ in. When ATP is present, this pump works at full capacity. However, even when ATP is depleted, the gradients previously established don't instantly disappear. Na⁺ can still move out through passive channels, driven by these existing electrochemical gradients - just at a much slower rate. Answer B correctly identifies this passive transport mechanism. The dramatic reduction from 100 to 20 ions/second shows most transport was ATP-dependent, but the remaining 20% occurs through gradient-driven passive channels. Answer A incorrectly assumes all Na⁺ transport requires ATP. While the pump does, passive channels don't. Answer C suggests the pump stores energy from previous ATP binding, but proteins don't work this way - they need continuous ATP for active transport. Answer D proposes alternative energy production, but anaerobic metabolism still produces ATP (just less efficiently), and the question states ATP is "completely depleted." Remember this principle: cells typically use both active pumps and passive channels for the same ion. When you see transport continuing after energy depletion, think about whether existing gradients could drive passive movement through channels or transporters.

Question 4

A lipid bilayer membrane separates two compartments. Compartment A contains 0.1 M glucose and 0.2 M sucrose. Compartment B contains 0.3 M glucose and 0.1 M sucrose. If both solutes can freely cross the membrane, what will be the net direction of solute movement when the system first begins to equilibrate?

  1. Glucose moves A→B and sucrose moves A→B, both following their individual concentration gradients
  2. Glucose moves A→B and sucrose moves B→A, each following their individual concentration gradients (correct answer)
  3. Glucose moves B→A and sucrose moves A→B, both moving to equalize total solute concentration
  4. Glucose moves B→A and sucrose moves B→A, both following the steeper glucose gradient
  5. No net movement occurs because the total solute concentrations are equal in both compartments
Explanation: When you encounter membrane transport questions, focus on the fundamental principle that each solute moves independently down its own concentration gradient, from high to low concentration, until equilibrium is reached. Let's analyze each solute separately. For glucose, compartment A has 0.1 M while compartment B has 0.3 M. Since concentration is higher in B, glucose will move from B→A. For sucrose, compartment A contains 0.2 M and compartment B contains 0.1 M. Since concentration is higher in A, sucrose will move from A→B. Each molecule follows its individual gradient independently. Choice A incorrectly suggests glucose moves A→B, but glucose concentration is actually lower in A (0.1 M) than B (0.3 M), so it must move the opposite direction. Choice C makes the same error about glucose direction and incorrectly implies that molecules move to balance total solute concentrations rather than following individual gradients. Choice D correctly identifies glucose moving B→A but wrongly states sucrose also moves B→A - sucrose has a higher concentration in A, so it moves toward B, not away from it. The correct answer is B: glucose moves B→A following its gradient (0.3 M → 0.1 M), while sucrose moves A→B following its gradient (0.2 M → 0.1 M). Remember this key principle: in passive transport, each solute species moves independently according to its own concentration gradient, regardless of what other solutes are doing. Don't get distracted by total concentrations or assume one gradient influences another.

Question 5

Two solutions are separated by a membrane permeable only to water. Solution A contains 0.2 M NaCl and 0.1 M glucose. Solution B contains 0.15 M NaCl and 0.25 M glucose. What will be the direction of net water movement and the driving force?

  1. Water moves A→B because solution A has higher NaCl concentration creating greater osmotic pressure
  2. Water moves B→A because solution B has higher glucose concentration creating greater osmotic pressure
  3. Water moves B→A because solution B has higher total solute concentration creating greater osmotic pressure (correct answer)
  4. Water moves A→B because solution A has lower glucose concentration creating lower osmotic pressure
  5. No net water movement occurs because both solutions contain the same types of solutes
Explanation: When you encounter osmosis problems, the key is understanding that water moves from areas of lower solute concentration to areas of higher solute concentration. The driving force is always the total osmotic pressure created by all dissolved particles. To determine water movement direction, you need to calculate the total solute concentration in each solution. For Solution A: 0.2 M NaCl + 0.1 M glucose = 0.3 M total solutes. For Solution B: 0.15 M NaCl + 0.25 M glucose = 0.4 M total solutes. Since Solution B has the higher total solute concentration (0.4 M vs 0.3 M), it creates greater osmotic pressure and will draw water from Solution A. Answer C correctly identifies that water moves from B←A (meaning toward A, or A←B in the direction shown) because Solution B's higher total solute concentration creates greater osmotic pressure. Answer A focuses only on NaCl concentration while ignoring glucose, missing that osmotic pressure depends on all dissolved particles. Answer B makes the same mistake in reverse, considering only glucose concentration and incorrectly predicting water movement toward the lower total concentration. Answer D incorrectly suggests that lower glucose concentration in Solution A would cause water to move away from A, when actually water moves toward higher total solute concentrations. Study tip: In osmosis problems, always calculate the total molar concentration of all solutes combined. Individual solute concentrations don't matter—only the sum determines osmotic pressure and water movement direction. Water always flows toward the solution with more total dissolved particles.

Question 6

A researcher adds ouabain, a specific inhibitor of Na+/K+-ATPase, to cells in culture. After 30 minutes, which change would be most expected?

  1. Immediate cell death due to complete loss of membrane potential within minutes of ouabain addition
  2. Decreased activity of Na+-glucose cotransporter due to dissipation of the Na+ gradient (correct answer)
  3. Increased K+ efflux through leak channels due to enhanced driving force from pump inhibition
  4. Hyperpolarization of the membrane potential due to accumulation of negative charges inside the cell
  5. Immediate reversal of all secondary active transport processes due to instant gradient collapse
Explanation: When you encounter questions about Na+/K+-ATPase inhibition, focus on the pump's role in maintaining ion gradients that drive secondary transport processes. The Na+/K+-ATPase creates a steep Na+ gradient (low inside, high outside) that powers many cotransporters. Ouabain blocks the Na+/K+-ATPase, preventing it from pumping Na+ out and K+ in. Over 30 minutes, the Na+ gradient gradually dissipates as Na+ accumulates inside the cell. Since the Na+-glucose cotransporter depends on this gradient to drive glucose uptake against its concentration gradient, the transporter's activity decreases significantly. This makes option B correct. Option A is wrong because cells don't die immediately from pump inhibition. While the situation is serious, cells can survive for hours before critical damage occurs. The membrane potential changes gradually, not within minutes. Option C incorrectly suggests increased K+ efflux. Actually, as the pump stops removing intracellular Na+ and stops bringing in K+, the driving force for K+ to leave through leak channels decreases rather than increases, since the membrane potential becomes less negative. Option D describes hyperpolarization, but the opposite occurs. As Na+ accumulates inside and K+ levels drop, the membrane actually depolarizes (becomes less negative), not hyperpolarizes. Remember: Na+/K+-ATPase questions often test your understanding of how primary active transport enables secondary transport. Always consider the time course—gradual changes over minutes to hours, not immediate effects.

Question 7

A cell membrane contains equal numbers of K+ leak channels and Na+ leak channels, but the K+ permeability is 10 times greater than Na+ permeability. If [K+]in = 140 mM, [K+]out = 5 mM, [Na+]in = 10 mM, and [Na+]out = 145 mM, which statement best predicts the membrane potential?

  1. The membrane potential will be approximately 0 mV because equal numbers of K+ and Na+ channels balance each other
  2. The membrane potential will be closer to the Na+ equilibrium potential because of the larger Na+ concentration gradient
  3. The membrane potential will be closer to the K+ equilibrium potential because K+ permeability dominates membrane conductance (correct answer)
  4. The membrane potential will fluctuate rapidly between K+ and Na+ equilibrium potentials due to competing ion flows
  5. The membrane potential cannot be determined without knowing the exact channel densities and open probabilities
Explanation: When analyzing membrane potential, remember that permeability, not just concentration gradients, determines which ions have the greatest influence. The membrane potential will be closest to the equilibrium potential of the most permeable ion. First, let's calculate the equilibrium potentials using the Nernst equation. For K+: EK=−61log⁡(140/5)=−90 mVE_K = -61 \log(140/5) = -90 \text{ mV}. For Na+: ENa=−61log⁡(10/145)=+71 mVE_{Na} = -61 \log(10/145) = +71 \text{ mV}. Even though there are equal numbers of K+ and Na+ channels, K+ permeability is 10 times greater than Na+ permeability. This means K+ ions flow much more readily across the membrane, making K+ the dominant influence on membrane potential. The membrane potential will therefore be much closer to the K+ equilibrium potential (-90 mV) than the Na+ equilibrium potential (+71 mV). Option A incorrectly assumes equal channel numbers means equal influence, ignoring the critical permeability difference. Option B focuses only on concentration gradients while missing that Na+ has low permeability despite its large gradient. Option D suggests unrealistic rapid fluctuations—membrane potential reaches a stable value determined by the relative permeabilities and driving forces of all ions. The correct answer is C because K+ permeability dominates membrane conductance, making the membrane potential closer to the K+ equilibrium potential. Study tip: Always remember that permeability trumps concentration gradient when predicting membrane potential. The most permeable ion has the strongest "voice" in determining the final voltage.

Question 8

An epithelial cell maintains an internal K+ concentration of 140 mM when the external K+ concentration is 5 mM. The membrane potential is -70 mV (inside negative). Given that the equilibrium potential for K+ is approximately -90 mV, what can be concluded about K+ transport in this cell?

  1. K+ is in electrochemical equilibrium since the concentration gradient matches the electrical gradient perfectly
  2. Active transport must be moving K+ into the cell because the membrane potential favors K+ efflux
  3. K+ tends to leak out through channels, but active transport pumps it back in to maintain the gradient (correct answer)
  4. The cell uses secondary active transport to accumulate K+ using the energy from Na+ gradients
  5. K+ movement is blocked because the membrane potential is more positive than the equilibrium potential
Explanation: When you encounter questions about ion transport and membrane potentials, you need to compare the actual membrane potential with the equilibrium potential to determine whether ions are in balance or require active transport. Here's the key analysis: The equilibrium potential for K+ is -90 mV, which represents where K+ would be in electrochemical equilibrium given the concentration gradient (140 mM inside vs 5 mM outside). However, the actual membrane potential is -70 mV. Since -70 mV is less negative than -90 mV, the membrane potential favors K+ efflux - K+ ions will naturally leak out of the cell down both their concentration gradient and the electrical gradient. To maintain the high intracellular K+ concentration (140 mM) against this tendency to leak out, the cell must actively transport K+ back in. This is exactly what the Na+/K+-ATPase pump does - it uses ATP to pump K+ into the cell while pumping Na+ out, working against the natural leak of these ions. Answer C correctly describes this process: K+ leaks out through channels but is actively pumped back in. Answer A is wrong because the ion isn't in equilibrium - the -70 mV membrane potential doesn't match the -90 mV equilibrium potential. Answer B incorrectly states that active transport moves K+ into the cell "because" the membrane potential favors efflux, when actually it's "despite" this tendency. Answer D mentions secondary active transport, but K+ accumulation primarily relies on the primary active transport of the Na+/K+-ATPase pump. Remember: Compare actual membrane potential to equilibrium potential to determine if active transport is needed.

Question 9

A cell is placed in a solution with a solute concentration of 0.9% NaCl. The cell's internal solute concentration is 0.3% NaCl. Assuming the membrane is permeable to water but not to NaCl, what will happen to the cell over time, and what is the primary driving force?

  1. The cell will shrink because water moves out due to the higher external osmotic pressure (correct answer)
  2. The cell will swell because water moves in due to the higher internal hydrostatic pressure
  3. The cell will shrink because NaCl moves out due to the concentration gradient favoring efflux
  4. The cell will remain unchanged because the membrane is selectively permeable to water only
  5. The cell will swell because water moves in due to the lower internal osmotic pressure
Explanation: When you encounter osmosis problems, focus on water movement across selectively permeable membranes. Water always moves from areas of lower solute concentration (higher water concentration) to areas of higher solute concentration (lower water concentration) to establish equilibrium. In this scenario, the external solution has 0.9% NaCl while the cell's internal concentration is only 0.3% NaCl. Since the membrane is permeable to water but not NaCl, water will move from inside the cell (lower solute concentration) to outside the cell (higher solute concentration). This water loss causes the cell to shrink, and the driving force is osmotic pressure - specifically, the higher osmotic pressure of the external solution pulling water outward. Answer A correctly identifies both the outcome (cell shrinkage) and the mechanism (higher external osmotic pressure driving water movement). Answer B is wrong because it confuses the direction of water movement and incorrectly mentions hydrostatic pressure instead of osmotic pressure as the driving force. Answer C is incorrect because NaCl cannot move across the membrane - only water can move, and the question states the membrane is impermeable to NaCl. Answer D is wrong because selective permeability to water is exactly what allows osmosis to occur; the cell will definitely change as water redistributes. Remember: in osmosis questions, always identify which side has higher solute concentration, then predict water will move toward that side. The solution with more solutes has greater osmotic pressure and "pulls" water molecules across the membrane.

Question 10

A membrane protein undergoes a conformational change that alternately exposes a binding site to either side of the membrane. The protein can bind substrate when the site faces outward but releases it when the site faces inward, regardless of substrate concentration. This mechanism best describes:

  1. A passive ion channel that opens and closes in response to voltage changes across the membrane
  2. Primary active transport that uses ATP to drive substrate movement against its concentration gradient (correct answer)
  3. A facilitated diffusion carrier that allows bidirectional substrate movement down concentration gradients
  4. Secondary active transport that couples substrate movement to an existing ion gradient
  5. Simple diffusion enhancement through temporary membrane pores created by protein conformational changes
Explanation: When you encounter questions about membrane transport mechanisms, focus on the key clues about energy requirements and directional movement. This question describes a protein that forces substrate release regardless of concentration—a hallmark of active transport. The correct answer is B because this mechanism describes primary active transport. The protein's conformational change that forces substrate release "regardless of substrate concentration" is the critical phrase. This means the protein is working against the natural tendency of the substrate, which requires energy input (ATP). The alternating exposure of binding sites with mandatory release creates unidirectional transport against concentration gradients—exactly what primary active transport accomplishes. Let's examine why the other options don't fit: A is incorrect because ion channels allow passive flow through an open pore, not the alternating conformational changes with forced substrate binding and release described here. C describes facilitated diffusion, but this process is bidirectional and follows concentration gradients—it wouldn't force substrate release regardless of concentration. D represents secondary active transport, which couples to existing ion gradients rather than directly using ATP, and the question doesn't mention any coupling to ion movement. The key strategy for transport questions is identifying whether the process requires energy and moves substances against their gradients. When you see "regardless of concentration," think active transport. The conformational change mechanism with forced release specifically points to primary active transport using ATP, distinguishing it from secondary active transport that relies on pre-existing gradients.

Question 11

A cell's Na+/K+-ATPase pump is functioning normally, moving 3 Na+ out and 2 K+ in per ATP hydrolyzed. If the pump rate suddenly doubles while ATP availability remains constant, what is the most likely immediate consequence?

  1. The cell will quickly deplete its ATP stores and pump activity will cease within minutes
  2. The Na+ and K+ gradients will become exactly twice as steep as before the change occurred
  3. The membrane potential will become more negative due to increased electrogenic pump activity (correct answer)
  4. Ion gradients will remain unchanged because the stoichiometry of the pump has not been altered
  5. Secondary active transport processes will immediately double their rates to match the primary transport
Explanation: When analyzing Na+/K+-ATPase pump questions, focus on three key characteristics: stoichiometry (3 Na+ out, 2 K+ in), energy requirement (1 ATP per cycle), and electrogenicity (net positive charge removal creates negative membrane potential). If pump rate doubles while ATP remains constant, the pump will hydrolyze ATP twice as fast, moving ions at double the previous rate. Since each pump cycle removes one net positive charge from the cell (3+ out, 2+ in = -1 net), doubling the pump activity means twice as many positive charges are removed per unit time. This increased electrogenic activity will make the membrane potential more negative, making C correct. Let's examine why the other options fail: A assumes ATP depletion will immediately halt the pump, but cells maintain ATP through multiple metabolic pathways and have reserves. While ATP consumption increases, immediate depletion within minutes is unlikely under normal cellular conditions. B incorrectly suggests gradients become exactly twice as steep. Ion gradients depend on the balance between pump activity and passive leak - doubling pump rate doesn't simply double the gradient steepness. D misses the temporal aspect entirely. While the 3:2 stoichiometry hasn't changed, the rate of ion movement has doubled, which absolutely affects both gradients and membrane potential in the short term. Remember: Na+/K+-ATPase questions often test your understanding that this pump is electrogenic. Any change in pump activity directly affects membrane potential because of the unequal ion exchange ratio. Always consider both the stoichiometry and the rate of pump activity.

Question 12

A student measures the membrane potential of a cell using microelectrodes and finds it to be -60 mV. She then applies tetrodotoxin (TTX), which blocks voltage-gated Na⁺ channels, and observes no change in the resting potential. Next, she applies tetraethylammonium (TEA), which blocks voltage-gated K⁺ channels, and again sees no change. Finally, she applies ouabain to block the Na⁺/K⁺-ATPase, and the membrane potential gradually depolarizes to -20 mV over 10 minutes. What is the most likely explanation for these observations?

  1. The cell's resting potential is maintained entirely by the Na⁺/K⁺-ATPase, with no significant contribution from passive ion movements
  2. The cell has primarily leak channels for K⁺ and Na⁺, with the Na⁺/K⁺-ATPase maintaining the gradients that drive passive flux through these channels (correct answer)
  3. The cell's membrane contains only voltage-gated channels that are closed at rest, making the pump the sole determinant of membrane potential
  4. The cell has equal permeabilities to Na⁺ and K⁺, so the pump's electrogenic activity is the primary source of the membrane potential
Explanation: The key observations are: (1) blocking voltage-gated channels has no effect on resting potential, indicating they're not active at rest, (2) blocking the pump causes gradual depolarization to -20 mV, not 0 mV. This suggests the cell has leak channels that allow passive ion movement according to concentration gradients. The pump maintains these gradients, and when it's blocked, the potential moves toward the Goldman potential determined by the permeabilities and remaining gradients. The fact that it only goes to -20 mV (not 0 mV) indicates gradients haven't fully dissipated. Choice A can't explain why the potential stops at -20 mV. Choice C incorrectly assumes only voltage-gated channels exist. Choice D doesn't explain the specific final potential reached.

Question 13

A student is examining red blood cells under different osmotic conditions. She prepares three solutions: Solution A (280 mOsm/L), Solution B (320 mOsm/L), and Solution C (240 mOsm/L). Normal red blood cells have an internal osmolarity of approximately 300 mOsm/L.

After 30 minutes in each solution, the student observes the cells and measures their volumes. In Solution A, cells appear slightly swollen; in Solution B, cells appear shrunken with a spiky appearance; in Solution C, some cells have lysed. If she then transfers cells from Solution B directly into Solution C, what is the most likely immediate outcome?

  1. The cells will gradually return to normal size as the osmotic gradient slowly equilibrates across the membrane
  2. The cells will remain shrunken because they have lost too much water to respond to the new osmotic gradient
  3. The cells will rapidly swell and many will likely undergo hemolysis due to the sudden large influx of water (correct answer)
  4. The cells will first swell slightly, then shrink again as the membrane adjusts its permeability to the new solution
Explanation: Red blood cells transferred from Solution B (320 mOsm/L, hypertonic) to Solution C (240 mOsm/L, hypotonic) will experience a dramatic osmotic gradient. The cells in Solution B were already dehydrated and shrunken. When suddenly placed in the hypotonic Solution C, water will rapidly rush into the cells down the steep concentration gradient. This rapid influx will cause the cells to swell quickly, and many will exceed their elastic limit and undergo hemolysis. Choice A incorrectly suggests a gradual process when the osmotic response is rapid. Choice B incorrectly assumes the cells cannot respond to osmotic gradients. Choice D describes an unrealistic membrane permeability adjustment that doesn't occur in red blood cells.

Question 14

A pharmaceutical researcher is testing a new drug that affects membrane transport. She finds that the drug reduces glucose uptake by 60% when added to cells in culture. However, when she adds the drug along with ouabain (a Na⁺/K⁺-ATPase inhibitor), glucose uptake is reduced by 95%. When ouabain alone is added, glucose uptake decreases by 40%. What is the most likely mechanism of action of the new drug?

  1. The drug directly blocks glucose transporters, and the additive effect with ouabain is due to independent mechanisms affecting different transport proteins
  2. The drug inhibits Na⁺/K⁺-ATPase activity partially, and the combination with ouabain produces complete inhibition of this pump
  3. The drug blocks ATP synthesis, reducing energy available for both glucose transport and Na⁺/K⁺-ATPase function, with ouabain providing additional pump inhibition
  4. The drug interferes with secondary active transport of glucose by disrupting ion gradients, and ouabain enhances this effect by further compromising the Na⁺ gradient (correct answer)
Explanation: When analyzing drug effects on cellular transport, you need to consider how different transport mechanisms interact and depend on each other. This question tests your understanding of primary versus secondary active transport and how they're interconnected. The key insight is recognizing that glucose uptake in most cells relies on secondary active transport, specifically the sodium-glucose cotransporter (SGLT). This transporter uses the sodium gradient established by the Na⁺/K⁺-ATPase pump to drive glucose into cells against its concentration gradient. When this sodium gradient is disrupted, glucose transport becomes severely impaired. Answer D is correct because it explains the experimental observations perfectly. The new drug disrupts ion gradients (likely the sodium gradient), reducing glucose uptake by 60%. When ouabain is added alongside the drug, it further compromises the Na⁺/K⁺-ATPase pump, nearly eliminating the sodium gradient and reducing glucose uptake to just 5% of normal levels. Answer A is wrong because truly independent mechanisms wouldn't show such dramatic synergistic effects - you'd expect simple addition, not the near-complete inhibition observed. Answer B incorrectly assumes the drug targets the Na⁺/K⁺-ATPase directly, but the drug alone causes more glucose uptake reduction than ouabain alone, which wouldn't make sense if they both targeted the same pump. Answer C suggests ATP depletion, but this would affect all cellular processes more uniformly and wouldn't explain the specific interaction pattern. Remember: when you see transport questions involving multiple drugs with synergistic effects, think about the interdependence between primary active transport (pumps) and secondary active transport (cotransporters).

Question 15

A researcher is studying ion transport in neurons and observes that when extracellular Na⁺ concentration is reduced from 150 mM to 75 mM, the resting membrane potential changes from -70 mV to -85 mV. When extracellular K⁺ is reduced from 5 mM to 2.5 mM, the resting potential changes from -70 mV to -88 mV. What can be concluded about the relative membrane permeability to these ions?

  1. The membrane is equally permeable to Na⁺ and K⁺, as both concentration changes produce similar potential changes
  2. The membrane is more permeable to K⁺ than Na⁺, because the K⁺ change produced a larger hyperpolarization per unit concentration change (correct answer)
  3. The membrane is more permeable to Na⁺ than K⁺, because Na⁺ has a higher baseline concentration and contributes more to the resting potential
  4. The membrane permeability cannot be determined from this data because the concentration changes were not identical between the two experiments
Explanation: To determine relative permeability, we must consider both the magnitude of concentration change and the resulting potential change. Na⁺ was reduced by 75 mM (50% reduction) and caused a 15 mV hyperpolarization. K⁺ was reduced by 2.5 mM (50% reduction) and caused an 18 mV hyperpolarization. Despite the much smaller absolute concentration change for K⁺, it produced a larger effect on membrane potential, indicating higher membrane permeability to K⁺. This aligns with typical neuron physiology where K⁺ permeability dominates the resting potential. Choice A incorrectly focuses only on the magnitude of potential change. Choice C confuses concentration with permeability. Choice D is incorrect because relative effects can be compared even with different concentration changes.

Question 16

A researcher studying membrane lipid composition creates artificial vesicles with varying cholesterol content: 0%, 20%, and 40% cholesterol by mole fraction. She then measures the rate of passive glucose diffusion across each membrane at 25°C and 37°C. At 25°C, glucose flux rates are 2.0, 1.2, and 0.8 μmol/cm²/min for 0%, 20%, and 40% cholesterol, respectively. At 37°C, the rates are 4.1, 3.8, and 3.5 μmol/cm²/min. What can be concluded about cholesterol's effect on membrane properties?

  1. Cholesterol acts as a fluidity buffer, decreasing permeability more at low temperatures but maintaining functionality at high temperatures (correct answer)
  2. Cholesterol increases membrane stability at both temperatures, but has minimal effect on permeability at physiological temperature
  3. Cholesterol decreases membrane fluidity at both temperatures, with the effect being more pronounced at lower temperatures
  4. Cholesterol primarily affects membrane thickness rather than fluidity, as evidenced by the consistent reduction in glucose flux at both temperatures
Explanation: When you encounter questions about membrane composition and permeability, focus on how cholesterol acts as a fluidity buffer - it has opposite effects at different temperatures to maintain optimal membrane function. Looking at the data, cholesterol's impact varies dramatically with temperature. At 25°C, increasing cholesterol from 0% to 40% reduces glucose flux by 60% (from 2.0 to 0.8 μmol/cm²/min). However, at 37°C, the same cholesterol increase only reduces flux by 15% (from 4.1 to 3.5 μmol/cm²/min). This demonstrates cholesterol's buffering effect - it significantly restricts permeability when membranes would otherwise be too fluid at low temperatures, but has minimal impact at physiological temperatures where membrane function is already optimal. Option B incorrectly suggests cholesterol has minimal effect on permeability at both temperatures, but the 60% reduction at 25°C is substantial, not minimal. Option C focuses only on fluidity reduction without recognizing the buffering concept - cholesterol doesn't simply decrease fluidity uniformly. Option D misses the point entirely by attributing the effect to membrane thickness rather than temperature-dependent fluidity modulation. The correct answer is A because it captures cholesterol's dual role: dramatically reducing permeability at low temperatures (where membranes need stabilization) while maintaining near-normal function at physiological temperatures (where excessive restriction would impair cellular processes). Remember: cholesterol is nature's thermostat for membranes. When you see temperature-dependent permeability data, always consider how cholesterol maintains optimal membrane fluidity across different conditions rather than simply making membranes more or less permeable.