College Biology Quiz: Membrane Transport In Physiology
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Membrane Transport In PhysiologyQuestion 1 of 8

A neuron's resting potential is maintained at -70 mV primarily through the action of the sodium-potassium pump, which moves 3 Na⁺ out and 2 K⁺ in per ATP molecule. If this pump suddenly stopped working but ion channels remained intact, what would happen to the membrane potential over time?

The potential would immediately jump to 0 mV because the pump directly creates the voltage difference
The potential would gradually depolarize toward 0 mV as ion gradients dissipate through leak channels
The potential would become more negative as potassium continues to leave through channels unopposed
The potential would remain at -70 mV because leak channels are independent of pump activity
The potential would oscillate between positive and negative values due to uncontrolled ion movements
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College Biology Quiz

College Biology Quiz: Membrane Transport In Physiology

Practice Membrane Transport In Physiology in College Biology 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 Transport In Physiology, giving you a quick way to practice the rules, question types, and explanations that matter most for College Biology.

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.

All questions

Question 1

A neuron's resting potential is maintained at -70 mV primarily through the action of the sodium-potassium pump, which moves 3 Na⁺ out and 2 K⁺ in per ATP molecule. If this pump suddenly stopped working but ion channels remained intact, what would happen to the membrane potential over time?

  1. The potential would immediately jump to 0 mV because the pump directly creates the voltage difference
  2. The potential would gradually depolarize toward 0 mV as ion gradients dissipate through leak channels (correct answer)
  3. The potential would become more negative as potassium continues to leave through channels unopposed
  4. The potential would remain at -70 mV because leak channels are independent of pump activity
  5. The potential would oscillate between positive and negative values due to uncontrolled ion movements
Explanation: When approaching membrane potential questions, focus on the relationship between the sodium-potassium pump and passive ion movement through leak channels. The pump creates concentration gradients, while leak channels allow ions to flow down those gradients. The sodium-potassium pump establishes steep concentration gradients by moving Na⁺ out and K⁺ in against their concentration gradients. This creates high K⁺ inside and high Na⁺ outside the cell. The -70 mV resting potential exists because K⁺ leak channels allow some potassium to flow out, leaving negative charges behind. However, the pump continuously restores the gradients that drive this process. If the pump stops while leak channels remain open, the concentration gradients would gradually dissipate. Sodium would leak in down its concentration gradient while potassium continues leaking out, but without the pump replenishing these gradients, the driving forces weaken. Eventually, Na⁺ and K⁺ concentrations would equalize across the membrane, eliminating the basis for the membrane potential. The potential would gradually depolarize toward 0 mV, making answer B correct. Answer A is wrong because the pump doesn't directly create voltage—it creates gradients that enable voltage. Answer C misses that sodium influx would counteract potassium efflux as gradients dissipate. Answer D incorrectly suggests leak channels work independently of concentration gradients, when they actually depend on the gradients the pump maintains. Remember: the pump creates the "battery" (concentration gradients) that powers the membrane potential. Without it, the battery slowly drains through leak channels until it's dead (0 mV).

Question 2

Red blood cells are placed in three different solutions: Solution A (0.9% NaCl), Solution B (0.3% NaCl), and Solution C (1.8% NaCl). After 30 minutes, which sequence correctly describes the relative cell volumes from largest to smallest?

  1. Solution A > Solution B > Solution C, because higher solute concentrations always increase cell volume
  2. Solution B > Solution A > Solution C, because cells swell in hypotonic solutions and shrink in hypertonic solutions (correct answer)
  3. Solution C > Solution A > Solution B, because hypertonic solutions cause maximum water uptake by cells
  4. Solution A > Solution C > Solution B, because isotonic solutions provide optimal conditions for cell expansion
  5. Solution B > Solution C > Solution A, because lower salt concentrations always correlate with smaller cell sizes
Explanation: When you encounter questions about cells in different salt solutions, you're dealing with osmosis and tonicity. The key is understanding how water moves across cell membranes based on solute concentration differences. Red blood cells have an internal salt concentration equivalent to about 0.9% NaCl. This makes Solution A (0.9% NaCl) isotonic—equal solute concentration inside and outside the cell, so no net water movement occurs. Solution B (0.3% NaCl) is hypotonic—lower solute concentration outside than inside, causing water to enter the cells and make them swell. Solution C (1.8% NaCl) is hypertonic—higher solute concentration outside than inside, causing water to leave the cells and make them shrink. Therefore, cell volume order is: Solution B (largest, swollen) > Solution A (normal size) > Solution C (smallest, shrunken). This confirms answer B is correct. Answer A incorrectly states that higher solute concentrations increase cell volume, when actually they decrease it by drawing water out. Answer C has the relationship completely backwards—hypertonic solutions cause water loss, not uptake, making cells smaller. Answer D wrongly suggests isotonic solutions cause expansion, when they actually maintain normal cell size. For osmosis questions, always compare the solution's concentration to the cell's internal concentration (0.9% NaCl for red blood cells). Remember: hypotonic solutions cause swelling, isotonic maintains size, and hypertonic causes shrinkage. Water always moves toward higher solute concentration.

Question 3

During intense exercise, muscle cells experience a dramatic increase in carbon dioxide production. How does this affect the transport of CO₂ across the muscle cell membrane, and what is the primary driving force?

  1. CO₂ transport decreases because high intracellular concentrations create a reverse gradient opposing efflux
  2. CO₂ transport increases through facilitated diffusion via specific carbonic anhydrase channels in the membrane
  3. CO₂ transport increases due to the steeper concentration gradient driving faster simple diffusion across the lipid bilayer (correct answer)
  4. CO₂ transport remains constant because membrane transporters become saturated at high CO₂ concentrations
  5. CO₂ transport requires active transport mechanisms that consume ATP to move CO₂ against its concentration gradient
Explanation: When you encounter questions about gas transport across cell membranes, focus on the fundamental principles of diffusion and the unique properties of different gases. Carbon dioxide is a small, lipophilic molecule that crosses cell membranes through simple diffusion directly across the phospholipid bilayer. During intense exercise, muscle cells dramatically increase cellular respiration to meet energy demands, producing much more CO₂ as a metabolic waste product. This creates a steep concentration gradient with high CO₂ inside the cell and lower concentrations in the blood. According to Fick's law of diffusion, the rate of diffusion is directly proportional to the concentration gradient - the steeper the gradient, the faster the diffusion rate. Choice A is incorrect because a higher intracellular CO₂ concentration actually enhances the outward gradient, promoting efflux rather than opposing it. Choice B contains a fundamental error: carbonic anhydrase is an enzyme that catalyzes the conversion between CO₂ and bicarbonate, not a membrane channel. CO₂ doesn't require facilitated diffusion. Choice D is wrong because simple diffusion doesn't involve saturable transporters - the process depends only on the concentration gradient and membrane permeability. The correct answer is C because the increased CO₂ production creates a steeper concentration gradient that drives faster simple diffusion across the lipid bilayer. Study tip: Remember that CO₂, O₂, and other small lipophilic molecules use simple diffusion, while polar molecules typically require transporters that can become saturated. The diffusion rate always increases with steeper gradients.

Question 4

In the human lung, oxygen must cross the respiratory membrane to reach the bloodstream. If the partial pressure of oxygen in alveolar air is 100 mmHg and in venous blood is 40 mmHg, but a patient has pulmonary edema (fluid in alveoli), how would this condition primarily affect oxygen transport?

  1. Transport rate increases because the fluid provides additional dissolved oxygen for diffusion across the membrane
  2. Transport rate decreases because the increased diffusion distance through fluid reduces the efficiency of gas exchange (correct answer)
  3. Transport rate remains unchanged because the partial pressure gradient is the only factor determining diffusion rate
  4. Transport rate increases because fluid creates a more favorable concentration gradient for oxygen movement
  5. Transport rate decreases because fluid prevents the establishment of partial pressure gradients necessary for diffusion
Explanation: When you encounter questions about gas exchange in the lungs, focus on Fick's law of diffusion, which states that the rate of gas transfer depends on surface area, the concentration gradient, and inversely on the membrane thickness or diffusion distance. In healthy lungs, oxygen moves efficiently from alveolar air (100 mmHg) to venous blood (40 mmHg) across the thin respiratory membrane. This 60 mmHg gradient drives diffusion. However, pulmonary edema introduces fluid into the alveolar spaces, creating a significant barrier that oxygen must traverse before reaching the respiratory membrane. This additional fluid layer dramatically increases the diffusion distance, making gas exchange much less efficient despite the unchanged pressure gradient. Answer B correctly identifies that the increased diffusion distance through fluid reduces gas exchange efficiency. The fluid acts as an additional barrier that slows oxygen movement. Answer A incorrectly suggests fluid helps by providing dissolved oxygen. While fluid can contain dissolved gases, the primary effect is creating a diffusion barrier that impedes transport. Answer C wrongly claims only the pressure gradient matters for diffusion rate. This ignores diffusion distance, a critical component of Fick's law. Answer D incorrectly states that fluid improves the concentration gradient. The pressure gradient between alveolar air and blood remains the same; the problem is the physical barrier the fluid creates. Remember: In respiratory physiology questions, always consider all factors affecting gas exchange—not just concentration gradients, but also surface area and diffusion distance. Pathological conditions typically impair one or more of these factors.

Question 5

During hemodialysis, a patient's blood is passed through a semipermeable membrane to remove waste products. If the dialysate (cleaning solution) has a urea concentration of 0 mg/dL and the patient's blood has a urea concentration of 80 mg/dL, which factor would most effectively increase the rate of urea removal?

  1. Increasing the urea concentration in the dialysate to 40 mg/dL to optimize the concentration gradient
  2. Decreasing the surface area of the dialysis membrane to concentrate the diffusion process
  3. Increasing the flow rate of dialysate to maintain the concentration gradient more effectively (correct answer)
  4. Adding ATP to the dialysate to provide energy for active transport of urea across the membrane
  5. Cooling the dialysate to reduce molecular motion and improve selective urea transport
Explanation: When you encounter hemodialysis questions, focus on the principles of passive diffusion across membranes. Urea removal depends on maintaining an optimal concentration gradient and maximizing the conditions that favor diffusion according to Fick's law. The correct answer is C because increasing dialysate flow rate continuously removes urea that has crossed into the dialysate, preventing buildup and maintaining the maximum concentration gradient (80 mg/dL difference). Fresh dialysate constantly replaces the used solution, keeping the "driving force" for diffusion at its peak throughout the treatment. Choice A is incorrect because adding urea to the dialysate (40 mg/dL) would reduce the concentration gradient from 80 mg/dL to only 40 mg/dL, cutting the driving force for diffusion in half and slowing urea removal. Choice B is wrong because decreasing membrane surface area reduces the total area available for diffusion, directly limiting the rate of urea transfer regardless of concentration gradients. Choice D misunderstands the transport mechanism—urea removal in dialysis relies on passive diffusion down its concentration gradient, not active transport, so ATP would provide no benefit. Remember that hemodialysis efficiency depends on three key factors: concentration gradient, membrane surface area, and membrane permeability. For college biology exams, always identify whether transport is passive or active first, then apply the appropriate principles. Passive processes like diffusion are enhanced by maintaining gradients and maximizing surface area, never by reducing the driving force.

Question 6

A researcher studying intestinal absorption finds that glucose uptake by epithelial cells is completely blocked when sodium ions are removed from the external solution, even though the glucose concentration gradient favors uptake. Which transport mechanism explains this observation?

  1. Glucose moves via simple diffusion, and sodium ions are required to maintain membrane fluidity for optimal transport
  2. Glucose uses facilitated diffusion through GLUT transporters that require sodium ions as allosteric activators
  3. Glucose uptake occurs through secondary active transport coupled to sodium movement down its electrochemical gradient (correct answer)
  4. Glucose requires primary active transport by ATP-powered pumps that also transport sodium ions simultaneously
  5. Glucose transport involves endocytosis mechanisms that depend on sodium-induced changes in membrane curvature
Explanation: When you encounter a transport question where removing one substance completely blocks another's movement despite a favorable concentration gradient, think about coupled transport mechanisms. The key clue here is that glucose uptake stops entirely when sodium is removed, even though glucose could theoretically move down its concentration gradient. This observation points to secondary active transport, where glucose uptake is directly coupled to sodium movement. In intestinal epithelial cells, the sodium-glucose cotransporter (SGLT) uses the energy from sodium moving down its electrochemical gradient to drive glucose uptake against its concentration gradient. When sodium is removed from the external solution, this driving force disappears, completely blocking glucose transport regardless of the glucose gradient. This explains why answer C is correct. Answer A is wrong because simple diffusion wouldn't require sodium ions for the transport process itself, only potentially for membrane properties. Answer B incorrectly describes the mechanism - while GLUT transporters do facilitate glucose movement, they don't require sodium and operate independently of sodium gradients. Answer D describes primary active transport, which would use ATP directly rather than coupling to sodium movement, and wouldn't explain the complete dependence on external sodium. Remember this pattern: when one molecule's transport is completely dependent on another ion's presence, look for secondary active transport (cotransport). The blocking effect when the driving ion is removed is a classic experimental result that distinguishes coupled transport from independent facilitated diffusion.

Question 7

An intestinal epithelial cell maintains a high intracellular potassium concentration (140 mM) compared to the extracellular concentration (5 mM). Despite this gradient, potassium continues to accumulate in the cell. Which combination of transport mechanisms best explains this observation?

  1. Passive potassium channels allow influx while sodium-potassium pumps provide additional potassium uptake against the gradient (correct answer)
  2. Facilitated diffusion through potassium transporters is enhanced by favorable membrane potential changes
  3. Potassium-chloride cotransporters use the chloride gradient to drive potassium accumulation beyond equilibrium
  4. Primary active transport exclusively maintains potassium levels through ATP-dependent pumps working against electrochemical gradients
  5. Secondary active transport couples potassium uptake to sodium efflux through specialized antiporter proteins
Explanation: When you encounter questions about ion transport across cell membranes, focus on distinguishing between passive transport (down gradients) and active transport (against gradients). The key here is understanding that multiple mechanisms can work simultaneously to maintain ion concentrations. The correct answer is A because intestinal epithelial cells use a combination of transport mechanisms. The sodium-potassium pump (Na⁺/K⁺-ATPase) actively transports potassium into the cell against its concentration gradient using ATP energy. This primary active transport can move K⁺ from 5 mM outside to 140 mM inside. Additionally, passive potassium channels allow some K⁺ influx when the membrane potential favors inward movement, supplementing the pump's action. Option B is incorrect because facilitated diffusion cannot move ions against their concentration gradient - it only accelerates movement down existing gradients. With K⁺ at 140 mM inside versus 5 mM outside, facilitated diffusion would cause K⁺ efflux, not accumulation. Option C misrepresents cotransporter function. While K⁺-Cl⁻ cotransporters exist, they typically move both ions in the same direction and cannot drive sustained accumulation against such a steep K⁺ gradient without additional energy input. Option D is partially correct about ATP-dependent pumps but incorrectly suggests this is the exclusive mechanism. The word "exclusively" makes this wrong - cells typically use multiple complementary transport systems. Remember: when you see extreme ion gradients being maintained, look for active transport mechanisms, and don't dismiss answers that combine multiple transport types - cells rarely rely on just one mechanism.

Question 8

A research team investigates water transport in desert plants by measuring water potential components in different plant tissues. They collect the following data from a drought-stressed cactus at midday:

Based on the data in the table, in which direction will water move between the root and stem tissues, and what is the driving force?

TissueSolute Potential (MPa)Pressure Potential (MPa)Water Potential (MPa)
Root-1.8+0.3-1.5
Stem-2.1+0.2-1.9
Leaf-2.4+0.1-2.3
  1. Water moves from stem to root because the stem has higher solute concentration creating osmotic drive
  2. Water moves from root to stem because the root has higher pressure potential providing force
  3. Water moves from root to stem because the root has a less negative water potential than the stem (correct answer)
  4. Water moves from stem to root because pressure potential difference overcomes solute potential difference
  5. No net water movement occurs because the pressure potentials are similar between tissues
Explanation: Water moves from areas of higher water potential to lower water potential. The root has a water potential of -1.5 MPa while the stem has -1.9 MPa. Since -1.5 > -1.9, water moves from root to stem. Choice A incorrectly focuses on solute concentration and gets the direction wrong. Choice B incorrectly identifies pressure potential as the driving force when water potential determines direction. Choice D incorrectly suggests stem-to-root movement. Choice E is wrong because there is a significant water potential difference (-0.4 MPa) driving movement.