{ "format": "recallweave-deck/1", "title": "Membrane transport: direction, selectivity, and energy", "attribution": "Original RecallWeave questions, explanations, transfer prompts, and fictional scenarios (2026). Scientific reference: Mary Ann Clark, Matthew Douglas, and Jung Choi, OpenStax Biology 2e, sections 5.1–5.3, https://openstax.org/books/biology-2e/pages/5-1-components-and-structure, https://openstax.org/books/biology-2e/pages/5-2-passive-transport, and https://openstax.org/books/biology-2e/pages/5-3-active-transport. No textbook questions, passages, or figures are reproduced. See courses/membrane-transport.md for assumptions, worked transfer answers, and references.", "license": "This original course text is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International (CC BY-NC-SA 4.0): https://creativecommons.org/licenses/by-nc-sa/4.0/. OpenStax is a scientific reference and does not endorse this course. This course license does not change the application's code license.", "concepts": [ "Membrane selectivity", "Passive transport", "Osmosis", "Active transport" ], "items": [ { "id": "ms-1", "concept": "Membrane selectivity", "prerequisites": [], "prompt": "A phospholipid bilayer separates two watery solutions. Which arrangement is most stable for its phospholipids?", "options": [ "Hydrophobic tails face the water; hydrophilic heads face one another.", "All heads point toward one solution; all tails point toward the other.", "Hydrophilic heads face the water; hydrophobic tails face one another.", "Heads and tails alternate freely between water and the bilayer's center." ], "answer": 2, "explanation": "Each phospholipid has a water-interacting head and nonpolar tails. In a bilayer, heads face the watery solution on each surface while tails pack into the interior. Both surfaces therefore expose heads. The bilayer is not a single sheet with heads on one side and tails on the other.", "transfer": "Use the head-and-tail arrangement to explain why a cut edge of a bilayer is energetically unfavorable in water." }, { "id": "ms-2", "concept": "Membrane selectivity", "prerequisites": [], "prompt": "An artificial membrane contains phospholipids but no transport proteins. Oxygen and sodium ions are both more concentrated outside. Which comparison is best for direct passage through the bilayer?", "options": [ "Oxygen crosses readily; sodium ions cross very poorly.", "Oxygen crosses poorly; sodium ions cross readily.", "Both cross readily because both particles are small.", "Neither crosses unless the membrane hydrolyzes ATP." ], "answer": 0, "explanation": "Oxygen is small and nonpolar, so it can pass through the bilayer's nonpolar interior. Sodium is a charged ion and crosses that interior very poorly without a suitable protein route. Particle size alone does not predict membrane permeability. Oxygen's passive passage does not require ATP hydrolysis.", "transfer": "Would adding an open sodium-selective channel mainly change sodium's crossing route or its charge? Explain." }, { "id": "ms-3", "concept": "Membrane selectivity", "prerequisites": [], "prompt": "Aquaporins, which are water channels, are added to a membrane while its solute concentrations and pressures are held fixed. There is already a nonzero osmotic driving force. The solute cannot cross. What is expected?", "options": [ "Water flows in reverse; solute movement stays unchanged.", "Solute crosses faster; the net water movement stops.", "Water and solute both cross faster because the pores admit both.", "Water flows faster in the same direction; solute stays blocked." ], "answer": 3, "explanation": "Aquaporins increase the membrane's permeability to water. With the driving force held fixed, this permits faster net water movement in its existing direction. A water-selective route does not make the specified solute permeable, reverse the driving force, or act as an ATP-powered pump.", "transfer": "Explain why removing aquaporins can slow osmosis without making a lipid membrane absolutely impermeable to water." }, { "id": "pt-1", "concept": "Passive transport", "prerequisites": [ "Membrane selectivity" ], "prompt": "A membrane is permeable to neutral solute X in dilute, ideal solutions. Initially, X is 8 mmol/L on side A and 2 mmol/L on side B. Temperature and pressure match, and no pump acts on X. Which describes its initial diffusion?", "options": [ "X crosses only from A to B; reverse crossings are impossible.", "More X crosses from A to B than from B to A.", "More X crosses from B to A than from A to B.", "X stays on each side until ATP is supplied to the membrane." ], "answer": 1, "explanation": "Individual X molecules move randomly and can cross in either direction. Initially, the larger concentration on side A produces more A-to-B crossings on average than B-to-A crossings. This net movement down the concentration gradient is passive; it does not require a pump or ATP.", "transfer": "If X later reaches 5 mmol/L on both sides, what changes about net flow and what keeps happening at the molecular scale?" }, { "id": "pt-2", "concept": "Passive transport", "prerequisites": [ "Membrane selectivity" ], "prompt": "Two closed compartments have reached diffusion equilibrium for neutral solute X across a permeable membrane. The solutions are ideal, their temperatures match, and no pressure difference acts on X. What continues at equilibrium?", "options": [ "Molecules stop moving once the concentrations match.", "Only water can cross after solute concentrations match.", "Molecules cross both ways at equal average rates.", "ATP continuously forces matching numbers of X across." ], "answer": 2, "explanation": "Diffusion equilibrium is dynamic. X molecules still move and cross the membrane, but the average opposing crossing rates balance. The net flux is zero even though molecular motion continues. Neither a membrane closure nor an ATP-driven balancing mechanism is needed.", "transfer": "Would seeing one labeled X molecule cross the membrane prove that net diffusion is occurring? Explain what would need to be compared." }, { "id": "pt-3", "concept": "Passive transport", "prerequisites": [ "Membrane selectivity" ], "prompt": "A neutral nutrient is more concentrated outside a cell than inside. Its entry requires a specific carrier protein, but the carrier does not use ATP or couple entry to another solute. What best describes the uptake?", "options": [ "Primary active transport, because a protein is involved.", "Facilitated diffusion down the nutrient's concentration gradient.", "Secondary active transport, because every carrier uses an ion gradient.", "Simple diffusion through the lipid core, because ATP is not used." ], "answer": 1, "explanation": "A carrier can help a solute move down its own concentration gradient without supplying energy. That is facilitated diffusion. A protein route distinguishes it from simple diffusion through the lipid bilayer. Direct ATP use or coupling to another solute's gradient would describe different mechanisms, neither of which is present here.", "transfer": "If this carrier is the only route and external nutrient concentration rises while carrier number stays fixed, why might uptake eventually approach a maximum rate?" }, { "id": "os-1", "concept": "Osmosis", "prerequisites": [ "Membrane selectivity", "Passive transport" ], "prompt": "In an idealized model, a flexible artificial cell has 200 mmol/L of the same nonpenetrating, non-dissociating solute inside and 300 mmol/L outside. Only water can cross; hydrostatic pressure is initially equal on both sides. What happens first?", "options": [ "Net water enters the cell, so its volume initially increases.", "Solute leaves the cell until the outside is less concentrated.", "There is no net flow because the membrane blocks the solute.", "Net water leaves the cell, so its volume initially decreases." ], "answer": 3, "explanation": "The outside has the greater concentration of retained solute particles. With equal initial pressure and a water-permeable membrane, net water movement is initially outward toward that solution. The cell initially loses volume. Osmosis does not require the solute itself to cross; this prompt explicitly prevents it from doing so.", "transfer": "Repeat the initial prediction with the inside and outside concentrations exchanged. Which variable sets the direction when pressure remains equal?" }, { "id": "os-2", "concept": "Osmosis", "prerequisites": [ "Membrane selectivity", "Passive transport" ], "prompt": "A water-permeable membrane separates equal-volume ideal solutions. Side A has 100 mmol/L of nonpenetrating solute X plus 100 mmol/L of nonpenetrating solute Y. Side B has 200 mmol/L of nonpenetrating solute Z. None dissociates; pressure and temperature match. What is the initial net water flow?", "options": [ "No net water flow, although water crosses in both directions.", "Net flow to side A, because it contains more solute types.", "Net flow to side B, because 200 is greater than either 100.", "No water crosses, because the total particle concentrations match." ], "answer": 0, "explanation": "Count all retained particles per volume: side A has 100 + 100 = 200 mmol/L and side B has 200 mmol/L. Under these ideal, equal-pressure conditions there is no initial net water flow. Counting the number of solute names, or comparing Z with only one component of A, gives the wrong comparison. Water still crosses both ways.", "transfer": "If Z instead dissociated into two retained particles per formula unit while the stated formula-unit concentrations stayed the same, how would you revise the particle count and initial water direction?" }, { "id": "os-3", "concept": "Osmosis", "prerequisites": [ "Membrane selectivity", "Passive transport" ], "prompt": "A plant cell with an intact, rigid wall is placed in a solution with less nonpenetrating solute than its cytoplasm. Water can cross the cell membrane, but the solute cannot, and the wall limits expansion. As water enters, what most directly opposes further net entry?", "options": [ "ATP-driven movement of water back through every aquaporin.", "The cell wall becoming a membrane that freely passes the solute.", "Rising hydrostatic pressure as the cell presses against its wall.", "Loss of random water motion as the cytoplasm becomes crowded." ], "answer": 2, "explanation": "The wall resists expansion, so water entry raises the cell's hydrostatic pressure, called turgor pressure in this setting. That pressure opposes further net entry and can balance the osmotic effect of the retained solute. Solute concentrations need not become equal for net water movement to stop. Water molecules continue moving.", "transfer": "Could net water entry stop while the solute concentration inside remains higher than outside? Explain what could balance that concentration effect." }, { "id": "at-1", "concept": "Active transport", "prerequisites": [ "Membrane selectivity", "Passive transport" ], "prompt": "An engineered membrane pump moves a neutral solute from 1 mmol/L on side A to 10 mmol/L on side B, with equal pressure and no other coupled solute. The pump hydrolyzes ATP during each transport cycle. How is this transport classified?", "options": [ "Facilitated diffusion, because a protein carries the solute.", "Simple diffusion, because individual molecules still move randomly.", "Secondary active transport, using a second solute's downhill gradient.", "Primary active transport, with ATP directly powering uphill movement." ], "answer": 3, "explanation": "The pump moves the neutral solute from lower to higher concentration and directly couples that movement to ATP hydrolysis. This is primary active transport. Facilitated and simple diffusion cannot sustain the stated uphill net flux without an energy source. Secondary active transport would couple transport to another solute's downhill movement, which the prompt excludes.", "transfer": "If ATP runs out but the 1-to-10 concentration difference remains, could this pump sustain the same uphill net flux without another energy source? Why?" }, { "id": "at-2", "concept": "Active transport", "prerequisites": [ "Membrane selectivity", "Passive transport" ], "prompt": "A cell has more potassium ions inside than outside, and its interior is electrically negative relative to the outside. A potassium channel opens. No magnitudes of the concentration difference or voltage are given. What can be concluded about the initial net potassium movement?", "options": [ "It must be outward, because ions always follow concentration alone.", "Its direction needs both the chemical and electrical effects to be compared.", "It must be inward, because electrical attraction always dominates.", "It must be zero, because two opposing effects always cancel exactly." ], "answer": 1, "explanation": "The potassium concentration difference favors outward movement, while the negative interior electrically favors inward movement of positive potassium ions. These contributions oppose each other. Their relative sizes determine the net electrochemical driving force; knowing their directions alone does not establish whether the net flux is inward, outward, or zero. The open channel itself is a passive route.", "transfer": "Predict the initial direction instead if positive sodium ions are more concentrated outside and the interior remains negative, assuming an open sodium channel. Which way do both contributions point?" }, { "id": "at-3", "concept": "Active transport", "prerequisites": [ "Membrane selectivity", "Passive transport" ], "prompt": "An operating cotransporter brings a neutral nutrient into a cell against its concentration gradient by coupling it to sodium moving inward down its electrochemical gradient. The cotransporter does not hydrolyze ATP. The only pump maintaining the sodium gradient stops, but the cotransporter stays intact and the gradient still exists initially. No other process restores it. What is the best initial prediction?", "options": [ "Uphill uptake can continue using the sodium gradient, then weaken as that gradient dissipates.", "Nutrient uptake must stop instantly because every transport protein needs its own supply of ATP.", "The nutrient must immediately diffuse outward even while inward cotransport remains coupled.", "Uphill nutrient uptake can continue indefinitely without any remaining source of usable energy." ], "answer": 0, "explanation": "The cotransporter's immediate energy source is sodium's electrochemical gradient. Stopping the pump does not erase energy already stored in that gradient, so secondary active uptake can continue while the gradient is sufficient. As the gradient dissipates without maintenance, it provides less driving force for uphill nutrient uptake. The pump and cotransporter have related but different energy roles.", "transfer": "Why can inhibiting an ATP pump eventually reduce the activity of a different transporter that never binds ATP?" } ] }