- 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?
Answer: K+ movement through the channel follows its gradient direction, consistent with facilitated diffusion without ATP.
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. - 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?
Answer: Net glycerol movement depends on the concentration gradient driving diffusion through the channel.
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.
- In a lab, two identical cells differ only in membrane proteins: Cell 1 expresses a chloride channel; Cell 2 lacks it. Both are placed in 100 mM Cl− solution while cytosolic Cl− is 10 mM. Cell 1 shows rapid Cl− influx; Cell 2 shows minimal influx. ATP inhibitors do not change Cell 1 influx. Which explanation best accounts for the difference between the cells?
Answer: Cell 1 has a channel enabling Cl− to diffuse down its gradient without ATP.
This question tests understanding of facilitated diffusion. The correct answer is B, as Cell 1's chloride channel enables Cl− to diffuse down its gradient from 100 mM to 10 mM without ATP, while Cell 2 lacks this pathway, leading to minimal influx. Channels provide a passive route for ions. ATP inhibitors not affecting influx confirms passivity. A tempting distractor is A, which wrongly contrasts with ATP pumps, misapplying energy requirements. Compare protein presence and gradient effects without ATP changes to pinpoint facilitated diffusion.
- A cell is placed in a solution with 5 mM amino acid X outside and 0.2 mM inside. Net influx of X occurs and stops when inside and outside concentrations become equal. A membrane protein that binds X is required for rapid influx; when it is blocked, influx slows and ATP levels remain stable. Which explanation best accounts for why net influx stops at equal concentrations?
Answer: At equilibrium, the concentration gradient is eliminated, so facilitated diffusion has no net direction.
This question demonstrates the equilibrium principle in facilitated diffusion of amino acid X. Net influx occurs initially because X moves down its concentration gradient (5 mM outside to 0.2 mM inside) through a membrane protein without ATP consumption, as ATP levels remain stable when the protein is blocked. Net influx stops when concentrations equalize because facilitated diffusion requires a concentration gradient as the driving force - at equilibrium, forward and reverse transport rates are equal. Choice B incorrectly attributes the stop to ATP depletion, but ATP levels were shown to be stable throughout. Remember that facilitated diffusion reaches equilibrium when the concentration gradient is eliminated.
- 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?
Answer: Fewer carriers lower the capacity for facilitated diffusion of U down its concentration gradient without ATP.
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.
- 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?
Answer: R requires a membrane protein to cross; it moves down its gradient by facilitated diffusion.
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.
- 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?
Answer: Z influx decreases because carrier conformational changes are required for facilitated diffusion.
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.
- 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?
Answer: Glucose moves down its concentration gradient through a membrane transporter without ATP input.
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.
- A cell expresses a transporter that allows fructose to cross the membrane. When extracellular fructose is 25 mM and intracellular fructose is 2 mM, fructose enters. When intracellular fructose is experimentally raised to 25 mM, net fructose movement becomes zero, although the transporter remains functional. Which explanation best accounts for the change in net movement?
Answer: Net fructose movement stops when concentrations equalize because diffusion through transporters is gradient-driven.
This question tests understanding of facilitated diffusion. The correct answer is A, as net fructose movement through the transporter is gradient-driven facilitated diffusion, stopping at equal concentrations of 25 mM despite functional transporters. Equilibrium eliminates the driving force for net flux. The process remains passive. A tempting distractor is B, which falsely blames ATP depletion, confusing with energy-dependent transport. When net movement halts at equilibrium, consider facilitated diffusion as the mechanism and check for persistent gradients.
- 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?
Answer: T enters by facilitated diffusion through the transporter, moving down its concentration gradient.
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.
- A researcher measures uptake of a polar solute X that cannot cross the lipid bilayer. Cells express a specific carrier protein for X. When extracellular X is increased, uptake rate rises and then reaches a maximum even though the gradient continues to increase; ATP inhibitors do not reduce uptake. Which explanation best accounts for the uptake pattern of solute X?
Answer: X enters by facilitated diffusion through carriers that become saturated at high X levels.
This question tests understanding of facilitated diffusion. The correct answer is B, as X enters via carrier-mediated facilitated diffusion down its gradient, with uptake saturating at high concentrations due to limited carrier sites, unaffected by ATP inhibitors. The polar nature prevents simple diffusion, and saturation indicates protein involvement. Gradients drive the process passively until carriers are fully occupied. A tempting distractor is C, which incorrectly suggests active transport and ATP depletion, reflecting confusion between passive and active mechanisms. When uptake plateaus without ATP effects, consider carrier saturation in facilitated diffusion as a key indicator.
- Cells are exposed to 100 mM fructose outside and 10 mM inside. A saturable uptake rate is observed: increasing extracellular fructose from 100 mM to 200 mM produces little additional increase in uptake. Adding an inhibitor that binds a membrane protein decreases uptake immediately, while ATP levels remain constant. Which explanation best accounts for the saturable fructose uptake under these conditions?
Answer: Fructose uptake is limited by the number of membrane transport proteins mediating diffusion down the gradient.
This question illustrates facilitated diffusion showing saturation kinetics due to limited transport proteins. Fructose moves down its concentration gradient (100 mM outside to 10 mM inside) through membrane transport proteins without ATP consumption, as ATP levels remain constant when the inhibitor is added. The saturable uptake rate occurs because all transport proteins become occupied at high substrate concentrations, limiting the maximum transport rate regardless of further concentration increases. Choice C incorrectly attributes saturation to simple diffusion through lipids, but the inhibitor's effect on a membrane protein proves protein-mediated transport. To recognize facilitated diffusion with saturation, look for plateau effects at high concentrations and protein involvement without energy use.
- 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?
Answer: The pore provides a selective route for V+ to diffuse down its concentration gradient without ATP.
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.
- Researchers measure water movement across a membrane containing aquaporins. The extracellular solution has lower solute concentration than the cytosol, and water enters the cell rapidly. When aquaporins are chemically blocked, the rate of water entry decreases, but the direction of net water movement remains the same. Cellular ATP concentration is unchanged during the assay. Which explanation best accounts for the reduced water entry rate when aquaporins are blocked?
Answer: Blocking aquaporins reduces facilitated diffusion of water down its gradient through membrane proteins.
This question tests understanding of facilitated diffusion of water through aquaporin channels. Water moves from the hypotonic solution (lower solute concentration outside) to the hypertonic cytosol (higher solute concentration inside) through aquaporins, following its concentration gradient without ATP input. When aquaporins are blocked, water entry slows but doesn't stop completely because some water can still cross the lipid bilayer directly, though much more slowly. Choice B incorrectly suggests ATP-driven pumping, contradicting the observation that ATP levels remain unchanged. Remember that facilitated diffusion increases the rate of movement down a gradient but doesn't change the direction of net transport.
- 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?
Answer: Cell type 1 has more transport proteins, increasing facilitated diffusion capacity down the glucose gradient.
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.
- 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?
Answer: Aquaporins provide a hydrophilic pathway for water to move down its gradient.
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.
- 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?
Answer: Ca2+ influx occurs because Ca2+ moves down its gradient through an open channel protein.
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. - 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?
Answer: With equal concentrations, facilitated diffusion through carriers produces no net flux despite bidirectional movement.
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.
- 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?
Answer: The dye enters by facilitated diffusion through an open channel down its concentration gradient.
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.
- 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?
Answer: Y moves down its gradient via carrier-mediated facilitated diffusion needing binding sites.
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.
