AP Biology Quiz: Facilitated Diffusion

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Facilitated DiffusionQuestion 1 of 20

Cells are exposed to a fluorescent dye that is polar and cannot cross lipid bilayers unaided. Fluorescence inside cells increases only when a specific membrane channel is present and open. The increase occurs without detectable ATP consumption and slows as internal dye concentration rises. Which explanation best accounts for dye entry into the cells?

The dye enters by facilitated diffusion through an open channel down its concentration gradient.

The dye enters by active transport because channels hydrolyze ATP to move solutes.

The dye enters by simple diffusion because polar molecules cross membranes rapidly.

The dye enters by phagocytosis because small solutes require vesicle uptake.

The dye enters by moving from low to high concentration due to channel selectivity.

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AP Biology Quiz

AP Biology Quiz: Facilitated Diffusion

Practice Facilitated Diffusion in AP 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 Facilitated Diffusion, giving you a quick way to practice the rules, question types, and explanations that matter most for AP 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

The dye enters by facilitated diffusion through an open channel down its concentration gradient. (correct answer)
The dye enters by active transport because channels hydrolyze ATP to move solutes.
The dye enters by simple diffusion because polar molecules cross membranes rapidly.
The dye enters by phagocytosis because small solutes require vesicle uptake.
The dye enters by moving from low to high concentration due to channel selectivity.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as the polar dye enters via facilitated diffusion down its gradient through the open channel without ATP, slowing as internal concentration rises toward equilibrium. The channel provides a selective pathway. No energy is consumed. A tempting distractor is B, which wrongly attributes ATP hydrolysis to channels, mixing with active transport. For polar solutes, verify gradient-driven entry via proteins without energy to identify facilitated diffusion.

Question 2

A researcher studies transport of polar solute R across a membrane. With a functional transporter present, R enters cells rapidly when outside concentration exceeds inside concentration. When the transporter gene is deleted, R entry becomes negligible despite the same gradient, and ATP inhibitors have no effect in either strain. Which explanation best accounts for these results?

R requires a membrane protein to cross; it moves down its gradient by facilitated diffusion. (correct answer)
R crosses by simple diffusion, and deleting the gene increases membrane rigidity.
R is imported by an ATP-driven pump, and inhibitors fail because cells store ATP.
R enters by endocytosis, and deleting the gene prevents vesicle fusion with the membrane.
R moves from low to high concentration, and deleting the gene removes the energy source.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as polar R requires the transporter for facilitated diffusion down its gradient, negligible without it, and ATP inhibitors have no effect, confirming passivity. Deletion removes the protein pathway. Gradients drive entry. A tempting distractor is C, which wrongly suggests ATP-driven pumping, misapplying energy needs. When gene deletion halts gradient-driven transport without ATP effects, recognize facilitated diffusion dependency on proteins.

Question 3

A membrane carrier transports solute Q. When extracellular Q is 60 mM and intracellular Q is 6 mM, net influx occurs. Adding a molecule structurally similar to Q reduces influx, but ATP levels remain constant and no ATP is consumed by the carrier. Which explanation best accounts for reduced influx after adding the similar molecule?

The similar molecule competitively inhibits binding to the carrier used for facilitated diffusion. (correct answer)
The similar molecule blocks ATP production, preventing active transport of Q into cells.
The similar molecule dissolves the membrane, eliminating the concentration gradient for Q.
The similar molecule reverses the gradient, forcing Q to move into cells by osmosis.
The similar molecule increases endocytosis rate, diluting Q influx through the carrier.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as the similar molecule competitively inhibits Q binding to the carrier, reducing facilitated diffusion down the gradient from 60 mM to 6 mM without ATP consumption. Inhibition blocks sites for passive transport. Constant ATP confirms no energy role. A tempting distractor is B, which incorrectly links to ATP blocking, based on active transport misconception. Use competitive inhibition and gradient persistence without energy to confirm facilitated diffusion.

Question 4

A cell expresses a carrier for solute W. Outside W is 18 mM and inside W is 2 mM. When temperature is lowered, W uptake rate decreases, but ATP inhibitors still do not affect uptake. Which explanation best accounts for the temperature dependence of W uptake?

Carrier-mediated facilitated diffusion slows at low temperature because protein conformational changes occur more slowly. (correct answer)
Facilitated diffusion requires ATP, and low temperature prevents ATP synthesis needed for transport.
Simple diffusion through lipids slows because carriers block the bilayer at low temperature.
Endocytosis slows at low temperature, eliminating the main pathway for W entry.
Transport slows because W must move from low to high concentration when temperature decreases.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as carrier-mediated facilitated diffusion of W down its gradient from 18 mM to 2 mM slows at low temperature due to reduced protein conformational flexibility, unaffected by ATP inhibitors. Temperature affects protein dynamics in passive transport. Gradients remain the driver. A tempting distractor is B, which incorrectly ties to ATP synthesis, confusing with active processes. When temperature impacts rate without energy effects, consider protein involvement in facilitated diffusion.

Question 5

Two cell types have different numbers of the same glucose transporter in their plasma membranes. Both are placed in 10 mM extracellular glucose with 1 mM intracellular glucose. Cell type 1 shows a faster initial glucose uptake rate than cell type 2, and neither shows a measurable change in ATP concentration during uptake. Which explanation best accounts for the difference in initial uptake rates?

Cell type 1 has more transport proteins, increasing facilitated diffusion capacity down the glucose gradient. (correct answer)
Cell type 1 has more mitochondria, providing more ATP to power active glucose pumping.
Cell type 1 has a lower extracellular glucose concentration, increasing diffusion into the cell.
Cell type 1 has a higher intracellular glucose concentration, increasing net influx by diffusion.
Cell type 1 has more vesicles, increasing glucose uptake by endocytosis of extracellular fluid.

Explanation: This question illustrates how transporter protein density affects facilitated diffusion rates. Both cell types have the same concentration gradient (10 mM outside to 1 mM inside), but cell type 1 has more glucose transporters in its membrane, allowing faster facilitated diffusion down the gradient without ATP consumption. The lack of ATP change confirms this is passive transport, not active pumping. Choice D incorrectly suggests higher intracellular glucose would increase influx, but this would actually decrease the gradient and slow transport. When comparing facilitated diffusion rates, consider both the concentration gradient magnitude and the number of available transport proteins.

Question 6

A cell is placed in a solution with 9 mM molecule U outside and 3 mM inside. The membrane has U carriers that bind and release U without ATP hydrolysis. If the number of carriers is reduced by half, the initial rate of U uptake decreases. Which explanation best accounts for the reduced uptake rate?

Fewer carriers lower the capacity for facilitated diffusion of U down its concentration gradient without ATP. (correct answer)
Fewer carriers lower ATP production, preventing ATP-dependent pumps from importing U into the cell.
Fewer carriers increase membrane thickness, stopping simple diffusion of U through the bilayer.
Fewer carriers cause U to move from low to high concentration, reducing net influx at the start.
Fewer carriers increase exocytosis of U, which directly blocks U entry through any remaining carriers.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as facilitated diffusion relies on carrier proteins to transport molecule U down its concentration gradient from 9 mM outside to 3 mM inside without requiring ATP. Reducing the number of carriers by half decreases the overall transport capacity, leading to a slower initial uptake rate because fewer proteins are available to bind and shuttle U across the membrane. The process is passive and driven solely by the concentration gradient, with carriers facilitating the movement without energy input. A tempting distractor is B, which incorrectly assumes ATP is involved, reflecting the misconception that all protein-mediated transport requires energy like active transport. To distinguish facilitated diffusion from other transport types, always check if movement is down the gradient and ATP-independent.

Question 7

A neuron's membrane contains ligand-gated ion channels permeable to Na+. Extracellular Na+ is 145 mM and intracellular Na+ is 12 mM. When a neurotransmitter binds, Na+ influx increases rapidly even when ATP synthesis is blocked, and the influx stops if the channel protein is chemically cross-linked closed. Which explanation best accounts for the Na+ influx observed after neurotransmitter binding?

Na+ moves down its concentration gradient through an opened channel protein without ATP use. (correct answer)
Na+ is transported by a pump that requires ATP hydrolysis to open the channel.
Na+ crosses the membrane by dissolving in phospholipids and diffusing directly.
Na+ moves from low to high concentration because channels reverse diffusion direction.
Na+ enters by cotransport with glucose, requiring a proton gradient as energy input.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as Na+ moves down its concentration gradient from 145 mM outside to 12 mM inside through an opened ligand-gated channel without ATP, even when ATP synthesis is blocked. The neurotransmitter binding opens the channel, allowing passive influx, and chemical cross-linking closes it, stopping movement. This highlights how channels provide a pathway for ions to diffuse based on gradients without energy. A tempting distractor is B, which incorrectly implies ATP is needed for channel opening, confusing facilitated diffusion with active transport. To identify facilitated diffusion, confirm if transport is gradient-driven and unaffected by ATP inhibition.

Question 8

A cell is placed in a solution containing 20 mM glucose while the cytosol contains 2 mM glucose. The plasma membrane includes a specific glucose transporter protein. When ATP production is inhibited, glucose still enters the cell until intracellular and extracellular concentrations become similar, and uptake stops when the transporter is blocked by a competitive inhibitor. Which explanation best accounts for glucose movement into the cell under these conditions?

Glucose diffuses through the lipid bilayer because it is nonpolar and small.
A glucose pump hydrolyzes ATP to move glucose into the cell against its gradient.
Glucose moves down its concentration gradient through a membrane transporter without ATP input. (correct answer)
Glucose exits the cell through the transporter because transporters only export solutes.
Glucose enters by endocytosis, forming vesicles that bypass the membrane proteins.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is C, as glucose moves down its concentration gradient from 20 mM outside to 2 mM inside through a specific transporter protein without ATP, continuing until equilibrium even when ATP is inhibited. The transporter facilitates passive diffusion, and blocking it with a competitive inhibitor stops uptake, confirming protein involvement. This process equalizes concentrations without energy input, as gradients drive the net movement. A tempting distractor is B, which wrongly suggests ATP-dependent pumping against the gradient, stemming from the misconception that all solute transport is active. When evaluating transport mechanisms, assess if net movement aligns with the concentration gradient and persists without ATP.

Question 9

Red blood cells contain aquaporin proteins in the plasma membrane. When cells are placed in a hypotonic solution, water enters quickly; when aquaporins are blocked, water entry is much slower. No ATP is consumed during either condition. Which explanation best accounts for the faster water movement when aquaporins are present?

Aquaporins hydrolyze ATP to pump water into the cell against its gradient.
Aquaporins provide a hydrophilic pathway for water to move down its gradient. (correct answer)
Aquaporins convert water into ions that diffuse through the membrane.
Aquaporins increase membrane surface area by forming vesicles for water import.
Aquaporins bind water and carry it from low to high concentration by diffusion reversal.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is B, as aquaporins form hydrophilic channels that allow water to move down its osmotic gradient into the cell in hypotonic solutions without ATP consumption. Blocking aquaporins slows water entry, showing their role in facilitating faster passive movement across the otherwise impermeable bilayer. The absence of ATP use confirms the process is passive and gradient-driven. A tempting distractor is A, which falsely claims aquaporins use ATP for pumping, based on the misconception that all membrane proteins require energy. For water transport questions, evaluate if movement follows osmotic gradients without energy input to recognize facilitated diffusion.

Question 10

A bacterial cell membrane contains a channel protein selective for glycerol. When extracellular glycerol is 50 mM and intracellular glycerol is 5 mM, glycerol enters rapidly. When extracellular glycerol is reduced to 5 mM, net glycerol movement stops even though the channel remains present. ATP levels are unchanged across treatments. Which explanation best accounts for the change in net glycerol movement?

The channel requires ATP, so glycerol entry stops when ATP becomes limiting at 5 mM.
Net glycerol movement depends on the concentration gradient driving diffusion through the channel. (correct answer)
Glycerol must be endocytosed, and vesicle formation stops at lower solute concentrations.
The channel changes glycerol into lipids, preventing movement when concentrations equalize.
The channel pumps glycerol out of the cell whenever extracellular glycerol decreases.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is B, as net glycerol movement through the channel is driven by the concentration gradient, entering rapidly from 50 mM to 5 mM but stopping when concentrations equalize at 5 mM without ATP changes. The channel allows passive diffusion down the gradient, and equilibrium halts net flux. Unchanged ATP levels emphasize the energy-independent nature. A tempting distractor is A, which wrongly attributes stopping to ATP limitation, confusing it with active transport. To analyze transport cessation, check if equilibrium eliminates the driving gradient in passive processes.

Question 11

A cell has a membrane channel selective for a neutral sugar. Outside sugar is 12 mM, inside is 1 mM. When the channel is closed, sugar uptake is near zero; when opened, uptake increases without ATP use. Which explanation best accounts for the dependence of uptake on channel state?

Opening the channel provides a pathway for sugar to diffuse down its gradient without energy. (correct answer)
Opening the channel activates ATP hydrolysis, converting diffusion into active transport.
Closing the channel increases lipid solubility of sugar, preventing entry when open.
Opening the channel forces sugar to move from low to high concentration by pumping.
Closing the channel triggers endocytosis, which is the primary route for sugar entry.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as opening the channel allows sugar to diffuse down its gradient from 12 mM to 1 mM without energy, with closure preventing entry. The open state provides a passive pathway. Uptake depends on channel availability. A tempting distractor is B, which incorrectly adds ATP activation, mixing with active mechanisms. Evaluate state-dependent uptake aligned with gradients and no energy to confirm facilitated diffusion.

Question 12

A membrane protein transports solute Z only when Z binds to a specific site. With 40 mM Z outside and 4 mM inside, net Z influx occurs. When the same cells are treated with a reagent that locks the protein in one conformation, Z influx decreases even though the gradient remains. No ATP consumption is detected. Which explanation best accounts for reduced Z influx?

Z influx decreases because carrier conformational changes are required for facilitated diffusion. (correct answer)
Z influx decreases because ATP hydrolysis can no longer power transport after locking.
Z influx decreases because solute gradients stop affecting diffusion when proteins are present.
Z influx decreases because vesicle-mediated transport requires membrane proteins to bend.
Z influx decreases because Z must move from low to high concentration to enter cells.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as Z influx via facilitated diffusion requires carrier conformational changes to move Z down its gradient from 40 mM to 4 mM, reduced when locked without ATP use. Locking prevents necessary protein flexibility for transport. Gradients drive the passive process. A tempting distractor is B, which incorrectly invokes ATP powering, based on active transport misconception. Assess if conformational inhibition affects gradient-driven transport without energy to recognize facilitated diffusion.

Question 13

A cell membrane contains a channel that allows Ca2+ to pass when open. Extracellular Ca2+ is 2 mM and cytosolic Ca2+ is $10^{-4}$ mM. Opening the channel produces Ca2+ influx even when ATP is depleted. Which explanation best accounts for Ca2+ influx under ATP-depleted conditions?

Ca2+ influx occurs because channels use ATP to move ions from low to high concentration.
Ca2+ influx occurs because Ca2+ diffuses through the lipid bilayer due to its charge.
Ca2+ influx occurs because Ca2+ moves down its gradient through an open channel protein. (correct answer)
Ca2+ influx occurs because vesicles import extracellular Ca2+ during exocytosis.
Ca2+ influx occurs because the channel pumps Ca2+ out, lowering cytosolic calcium.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is C, as Ca2+ moves down its steep gradient from 2 mM to 10^{-4} mM through the open channel protein without ATP, even under depletion. The channel allows passive influx driven by concentration differences. Energy independence is evident. A tempting distractor is A, which falsely claims channels use ATP for uphill movement, confusing with pumps. In ion transport, check steep gradients and ATP depletion effects to identify facilitated diffusion.

Question 14

A scientist tracks movement of lactose into cells with a lactose-specific permease. When lactose is 15 mM outside and 1 mM inside, lactose enters rapidly in the dark and in the presence of ATP synthase inhibitors. When outside lactose is lowered to 1 mM, net transport becomes zero. Which explanation best accounts for these observations?

Lactose transport is facilitated diffusion driven by the lactose concentration gradient. (correct answer)
Lactose transport is active transport driven directly by ATP hydrolysis at the permease.
Lactose transport is osmosis because lactose changes water potential across the membrane.
Lactose transport is exocytosis because lactose must be packaged into vesicles.
Lactose transport is simple diffusion through the bilayer because lactose is hydrophobic.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as lactose transport is facilitated diffusion down its gradient from 15 mM to 1 mM via permease, occurring in the dark and with ATP inhibitors, stopping at equilibrium. The permease enables passive entry driven by the gradient. No energy input is required. A tempting distractor is B, which claims direct ATP hydrolysis, confusing it with active transport. Test for gradient dependence and ATP independence to confirm facilitated diffusion in nutrient uptake.

Question 15

A cell has a membrane carrier that binds amino acid Y. Extracellular Y is 30 mM and intracellular Y is 3 mM. When a nonhydrolyzable ATP analog is added, Y uptake continues unchanged; when a mutation reduces Y binding affinity, uptake rate decreases at the same gradient. Which explanation best accounts for these results?

Y uptake requires ATP, but the analog provides extra energy to maintain transport.
Y crosses by simple diffusion, and binding affinity changes membrane fluidity.
Y moves down its gradient via carrier-mediated facilitated diffusion needing binding sites. (correct answer)
Y enters by endocytosis, and the mutation reduces vesicle formation frequency.
Y is pumped against its gradient, and reduced affinity increases pump turnover.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is C, as Y moves down its gradient from 30 mM to 3 mM via carrier-mediated facilitated diffusion, requiring binding sites, with unchanged uptake under ATP analog indicating no energy need, and reduced affinity lowering rate. Carriers facilitate passive transport via conformational changes. The mutation affects binding without energy involvement. A tempting distractor is E, which incorrectly assumes pumping against the gradient, stemming from active transport confusion. Examine if mutations affect binding and gradients drive movement to identify facilitated diffusion.

Question 16

A lab measures transport of solute U across a membrane using radiolabeled U. With U at 5 mM outside and 0.5 mM inside, net uptake occurs in cells expressing a U-specific carrier. When outside U is 0.5 mM, net uptake is zero, although labeled U still exchanges in both directions. ATP inhibitors do not change results. Which explanation best accounts for zero net uptake at 0.5 mM outside?

With equal concentrations, facilitated diffusion through carriers produces no net flux despite bidirectional movement. (correct answer)
With equal concentrations, carriers stop functioning entirely and cannot bind U molecules.
With equal concentrations, ATP-driven pumps reverse direction and cancel uptake of U.
With equal concentrations, U becomes hydrophobic and remains trapped outside the membrane.
With equal concentrations, endocytosis and exocytosis rates become identical for U vesicles.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as at equal concentrations of 0.5 mM, facilitated diffusion via carriers shows no net flux despite bidirectional exchange, driven by gradients that are absent at equilibrium. Carriers allow passive movement in both directions. ATP inhibitors not affecting confirms passivity. A tempting distractor is C, which wrongly invokes ATP pumps reversing, confusing with active transport. Use radiolabel exchange at equilibrium without net change to detect facilitated diffusion.

Question 17

A membrane contains a protein that forms a pore for a specific ion V+. Outside V+ is 30 mM and inside is 3 mM. When the pore is present, V+ enters rapidly; when the pore is absent, entry is minimal. Adding ATP does not change V+ entry when the pore is present. Which explanation best accounts for the role of the pore protein?

The pore provides a selective route for V+ to diffuse down its concentration gradient without ATP. (correct answer)
The pore uses ATP to change V+ into a neutral form that crosses the bilayer.
The pore pumps V+ out of the cell, lowering intracellular concentration below 3 mM.
The pore increases membrane thickness, causing ions to accumulate inside by osmosis.
The pore triggers vesicle formation that imports V+ independently of concentration gradients.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as the pore provides a selective pathway for V+ to diffuse down its gradient from 30 mM to 3 mM without ATP, minimal without it. The protein enables passive ion movement. ATP addition has no effect. A tempting distractor is B, which falsely claims ATP conversion, misapplying energy roles. For ions, confirm protein-dependent gradient movement without energy to recognize facilitated diffusion.

Question 18

A scientist places cells in a solution containing 100 mM solute T; cytosolic T is 10 mM. Cells express a transporter specific for T. When the transporter is inhibited, intracellular T remains low. When the inhibitor is removed, T increases until near 100 mM, with no detectable ATP consumption. Which explanation best accounts for the increase in intracellular T after inhibitor removal?

T enters by facilitated diffusion through the transporter, moving down its concentration gradient. (correct answer)
T enters by active transport, and removing inhibitor restores ATP hydrolysis by the pump.
T enters by simple diffusion, and inhibitor removal increases bilayer permeability to ions.
T enters by exocytosis, and inhibitor removal increases vesicle release into the cell.
T enters only when internal concentration exceeds external concentration, reversing diffusion.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as T enters via facilitated diffusion down its gradient from 100 mM to 10 mM through the transporter, increasing to equilibrium after inhibitor removal without ATP consumption. Inhibitor blocks the passive pathway. Gradients drive equalization. A tempting distractor is B, which claims active pumping restored, based on energy misconception. When inhibition reversal allows gradient equalization without energy, identify facilitated diffusion.

Question 19

A membrane contains a potassium channel that is open under the experimental conditions. Intracellular K $^+$ concentration is higher than extracellular K $^+$ . Net K $^+$ efflux is observed and decreases when the channel is blocked, while ATP concentration remains unchanged. When intracellular K $^+$ is experimentally lowered below extracellular K $^+$ , net K $^+$ movement becomes inward through the unblocked channel. Which explanation best accounts for the observed direction changes in K $^+$ movement?

K $^+$ movement through the channel follows its gradient direction, consistent with facilitated diffusion without ATP. (correct answer)
K $^+$ movement reverses because the channel switches into an ATP-powered pump when gradients change.
K $^+$ movement reverses because the lipid bilayer becomes permeable to ions when ATP is unchanged.
K $^+$ movement reverses because vesicles import or export K $^+$ depending on cytosolic concentration.
K $^+$ movement reverses because the channel forces ions to move from low to high concentration.

Explanation: This question demonstrates bidirectional facilitated diffusion through potassium channels. K+ initially moves out of the cell (efflux) because intracellular K+ concentration is higher than extracellular, following its concentration gradient through the channel without ATP consumption. When the gradient is experimentally reversed (intracellular K+ becomes lower), the direction reverses to influx, proving the channel facilitates passive movement in either direction based solely on the gradient. Choice E incorrectly claims channels force movement from low to high concentration, which would violate thermodynamics without energy input. To understand facilitated diffusion directionality, remember that channels allow passive movement in whichever direction the concentration gradient dictates.

Question 20

In a lab, cells are placed in a solution containing 20 mM glucose while cytosolic glucose is 2 mM. When a specific membrane protein is inhibited, glucose entry rapidly decreases, yet cellular ATP levels remain unchanged during the first minute. If the inhibitor is removed, glucose entry resumes immediately. No vesicle formation is observed, and the membrane remains intact. Which explanation best accounts for the glucose movement into the cells under normal conditions?

Glucose enters by diffusion through the phospholipid bilayer because it is nonpolar at physiological pH.
Glucose is transported by a carrier protein down its concentration gradient without direct ATP input. (correct answer)
Glucose is pumped into the cell against its gradient by an ATP-driven transporter.
Glucose enters only when endocytotic vesicles internalize extracellular fluid containing glucose.
Glucose moves from low to high concentration through a channel, powered by the membrane potential.

Explanation: This question tests understanding of facilitated diffusion, where molecules move down their concentration gradient through membrane proteins without direct ATP input. Glucose moves from high concentration (20 mM outside) to low concentration (2 mM inside) through a carrier protein, which explains why inhibiting this protein stops glucose entry while ATP levels remain unchanged. The immediate resumption of transport when the inhibitor is removed confirms the protein is not damaged but simply blocked. Choice A incorrectly claims glucose is nonpolar (it's actually polar due to hydroxyl groups), while choice C wrongly suggests ATP-driven pumping against the gradient. When identifying facilitated diffusion, look for movement down a concentration gradient through a protein without energy consumption.