{ "format": "recallweave-deck/1", "title": "Enzymes: energy, speed and control", "attribution": "Original questions, fictional numerical profiles and distractors created for RecallWeave, October 2026. Scientific checks: [R1] OpenStax, Biology 2e, §6.2, https://openstax.org/books/biology-2e/pages/6-2-potential-kinetic-free-and-activation-energy ; [R2] OpenStax, Biology 2e, §6.5, https://openstax.org/books/biology-2e/pages/6-5-enzymes ; [R3] Alberts et al., Molecular Biology of the Cell, 4th ed., Catalysis and the Use of Energy by Cells, https://www.ncbi.nlm.nih.gov/books/NBK26838/ ; [R4] Cooper, The Cell: A Molecular Approach, 2nd ed., The Central Role of Enzymes as Biological Catalysts, https://www.ncbi.nlm.nih.gov/books/NBK9921/ . These are concept references, not copied questions or evidence of learning effectiveness.", "license": "Original wording and fictional scenarios supplied for Jacob's RecallWeave project. No textbook passages, figures or assessment items are reproduced. Reference works retain their own copyright and license terms; this metadata does not grant rights in those works.", "concepts": [ "Reaction energy", "Catalytic pathways", "Binding and saturation", "Enzyme regulation" ], "items": [ { "id": "enz-energy-1", "concept": "Reaction energy", "prerequisites": [], "prompt": "A fictional cellular reaction has a free-energy change ΔG = −12 kJ/mol under the stated conditions, yet product forms slowly. Which explanation fits both observations?", "options": [ "A negative ΔG means each molecule must convert immediately.", "Slow conversion proves that the reported ΔG must be positive.", "A favorable energy change can coexist with a substantial activation barrier.", "The slow rate shows that no enzyme could accelerate conversion." ], "answer": 2, "explanation": "ΔG describes the free-energy difference between the reaction's products and reactants under the stated conditions. It does not specify how quickly the transition occurs. A barrier along the reaction pathway can make a thermodynamically favorable conversion slow. [R1; R3]", "transfer": "When a pathway is slow, distinguish a question about its energy balance from a question about its rate." }, { "id": "enz-energy-2", "concept": "Reaction energy", "prerequisites": [], "prompt": "In an invented energy profile, reactants are at 20, products at 8, and the transition state at 68 kJ/mol, on the same reference scale. What are ΔG and the forward activation barrier?", "options": [ "ΔG = −12 kJ/mol; forward barrier = 48 kJ/mol.", "ΔG = +12 kJ/mol; forward barrier = 48 kJ/mol.", "ΔG = −12 kJ/mol; forward barrier = 60 kJ/mol.", "ΔG = +48 kJ/mol; forward barrier = 12 kJ/mol." ], "answer": 0, "explanation": "The reaction difference is 8 − 20 = −12 kJ/mol. The forward barrier is measured from the reactants to the transition state: 68 − 20 = 48 kJ/mol. The 60 kJ/mol difference uses the products as its starting point and is the reverse barrier in this profile. [R1]", "transfer": "Sketch the three levels, then label the starting level for each subtraction before calculating." }, { "id": "enz-energy-3", "concept": "Reaction energy", "prerequisites": [], "prompt": "A reversible reaction A ⇌ B is away from equilibrium. A catalyst is added without changing temperature or the reaction's reactants and products; binding does not appreciably sequester either species. What can catalysis change?", "options": [ "The equilibrium ratio, so that only B remains at the end.", "The reaction's ΔG, by permanently lowering product energy.", "The direction of equilibrium, by making the reverse reaction impossible.", "The approach rate, while leaving the equilibrium ratio unchanged." ], "answer": 3, "explanation": "A catalyst provides a faster route for interconversion without changing the reaction's energy endpoints. It accelerates approach to the equilibrium set by those conditions rather than selecting a new equilibrium composition. [R3; R4]", "transfer": "Separate how soon a reversible system approaches equilibrium from the composition it approaches." }, { "id": "enz-catalysis-1", "concept": "Catalytic pathways", "prerequisites": [ "Reaction energy" ], "prompt": "For an invented reaction, reactants are at 5 and products at −3 kJ/mol. An enzyme lowers the transition-state level from 47 to 29 kJ/mol. Which pair describes the catalyzed forward pathway?", "options": [ "The forward barrier is 18 kJ/mol, and ΔG is −8 kJ/mol.", "The forward barrier is 24 kJ/mol, and ΔG is −8 kJ/mol.", "The forward barrier is 24 kJ/mol, and ΔG is −26 kJ/mol.", "The forward barrier is 32 kJ/mol, and ΔG is +8 kJ/mol." ], "answer": 1, "explanation": "The new forward barrier is 29 − 5 = 24 kJ/mol. ΔG remains −3 − 5 = −8 kJ/mol because the endpoints are unchanged. The enzyme reduced the barrier by 18 kJ/mol; that reduction is not the remaining barrier. [R1; R2]", "transfer": "Compare the barrier before and after catalysis without moving the reactant or product level." }, { "id": "enz-catalysis-2", "concept": "Catalytic pathways", "prerequisites": [ "Reaction energy" ], "prompt": "One enzyme molecule has completed many ordinary catalytic cycles and released each product. No inhibitor, damage or degradation is present. What permits another cycle?", "options": [ "A product must be converted into a new enzyme after every cycle.", "The enzyme must supply the net free energy consumed by each reaction.", "The enzyme becomes a permanent part of each released product molecule.", "The regenerated enzyme can bind another substrate and catalyze again." ], "answer": 3, "explanation": "A catalyst participates in reaction steps but is regenerated across a completed catalytic cycle. Product release allows another cycle under suitable conditions. This does not make an enzyme immortal: damage or degradation could end its activity, but those effects are excluded here. [R4]", "transfer": "Follow the enzyme separately from the substrate and product when accounting for repeated reaction cycles." }, { "id": "enz-catalysis-3", "concept": "Catalytic pathways", "prerequisites": [ "Reaction energy" ], "prompt": "A folded enzyme variant still binds its substrate tightly, but no longer helps the reacting atoms reach the transition state. Which conclusion is justified?", "options": [ "Tight substrate binding proves that its catalytic rate must stay unchanged.", "Substrate binding alone does not establish that the variant catalyzes efficiently.", "Any bound substrate must become product, even without a catalytic pathway.", "Losing transition-state assistance must change the reaction's equilibrium ratio." ], "answer": 1, "explanation": "Recognition and catalysis are related but distinct. Binding can position a substrate, yet efficient conversion also depends on interactions along the reaction pathway, including transition-state stabilization. The premise therefore does not support an unchanged catalytic rate. [R3; R4]", "transfer": "When testing an enzyme variant, distinguish evidence for binding from evidence for product formation." }, { "id": "enz-binding-1", "concept": "Binding and saturation", "prerequisites": [ "Catalytic pathways" ], "prompt": "An enzyme recognizes substrate S through complementary shape and charge. A mutation reverses a contact charge while the protein remains folded. What may follow?", "options": [ "Recognition of S and the observed reaction rate may change.", "All substrates must now bind equally because the protein is folded.", "The same reaction must acquire a different equilibrium constant.", "Every catalytic cycle must consume the mutated enzyme permanently." ], "answer": 0, "explanation": "A folded protein can still have altered active-site interactions. Changing a contact charge may affect recognition, orientation or subsequent catalysis. The observation does not establish the direction or size of the rate change; that would require further evidence. [R2; R4]", "transfer": "Avoid treating 'still folded' as proof that all molecular interactions remain unchanged." }, { "id": "enz-binding-2", "concept": "Binding and saturation", "prerequisites": [ "Catalytic pathways" ], "prompt": "In a fictional initial-rate experiment with fixed enzyme and suitable constant conditions, substrate concentrations of 1, 4 and 16 units give rates of 5.0, 8.0 and 9.4 product units per minute. Assume a simple saturating enzyme with no inhibition. Which reading fits?", "options": [ "The reaction's equilibrium ratio rises whenever substrate concentration rises.", "Each fourfold substrate increase must cause a fourfold rate increase.", "The rate approaches an enzyme-limited ceiling as substrate becomes abundant.", "The enzyme is necessarily being consumed because the rate does not scale." ], "answer": 2, "explanation": "The increases become progressively smaller. In the stated saturating model, the fixed enzyme population spends more time occupied and has a limited turnover capacity. More substrate therefore gives diminishing increases in initial rate. These fictional values do not identify an exact mechanism outside the stated model. [R3; R4]", "transfer": "Look for a plateau in initial-rate data, then ask which capacity is held fixed." }, { "id": "enz-binding-3", "concept": "Binding and saturation", "prerequisites": [ "Catalytic pathways" ], "prompt": "In a simple initial-rate assay with maintained saturating substrate, 0.10 μmol/L of active enzyme produces 10 μmol/L of product per second. Only active enzyme concentration doubles; turnover and all other conditions stay the same. What rate does this model predict?", "options": [ "5 μmol/L/s", "20 μmol/L/s", "10 μmol/L/s", "40 μmol/L/s" ], "answer": 1, "explanation": "At saturating substrate, the stated model's rate scales with the amount of active enzyme when turnover per enzyme is unchanged. Doubling 0.10 to 0.20 μmol/L doubles 10 to 20 μmol/L/s. This is a model prediction, not a measured response for an arbitrary biological system. [R3; R4]", "transfer": "State what remains constant before using a proportional prediction." }, { "id": "enz-regulation-1", "concept": "Enzyme regulation", "prerequisites": [ "Binding and saturation" ], "prompt": "In a simple reversible competitive-inhibition model, substrate S and inhibitor I cannot occupy the enzyme together. Which change can reduce I's fractional effect on the initial rate?", "options": [ "Lower substrate while keeping inhibitor and active enzyme fixed.", "Add enough inhibitor to occupy a larger fraction of the active sites.", "Remove active enzyme while keeping substrate and inhibitor fixed.", "Raise substrate while keeping inhibitor and active enzyme fixed." ], "answer": 3, "explanation": "In this specified competition, more substrate increases the opportunity for substrate rather than inhibitor to occupy the enzyme. That can reduce inhibition relative to the uninhibited rate at the same substrate concentration. It is not a claim that every inhibitor can be overcome this way. [R2; R4]", "transfer": "Check the stated binding mechanism before predicting whether additional substrate can offset inhibition." }, { "id": "enz-regulation-2", "concept": "Enzyme regulation", "prerequisites": [ "Binding and saturation" ], "prompt": "A reversible regulator binds away from an enzyme's active site. Substrate can still bind, but catalytic turnover decreases. What does this observation support?", "options": [ "The regulator must compete with substrate for the same binding position.", "The regulator has changed the free-energy difference of the reaction endpoints.", "Binding at a separate site can change the enzyme's catalytic behavior.", "The regulator must have irreversibly destroyed every enzyme it contacts." ], "answer": 2, "explanation": "Binding at a regulatory site can influence enzyme activity without occupying the substrate's site. The observation supports allosteric regulation. It does not by itself determine a complete kinetic inhibition class or imply irreversible damage. [R2; R4]", "transfer": "Describe what the evidence shows before attaching a more specific kinetic label." }, { "id": "enz-regulation-3", "concept": "Enzyme regulation", "prerequisites": [ "Binding and saturation" ], "prompt": "In the fictional pathway A → B → C → D, product D reversibly inhibits the enzyme for A → B. A is available and other conditions remain fixed. If free D is removed, what immediate regulatory response is expected?", "options": [ "Inhibition can ease, allowing the first enzyme's activity to increase.", "The first enzyme must stop because D was its required substrate.", "The first reaction's equilibrium constant must reverse its value.", "Every downstream reaction must instantly run at the same maximum rate." ], "answer": 0, "explanation": "The pathway's own product suppresses an earlier step: this is feedback inhibition. Lowering free D can relieve reversible inhibition. That predicts a regulatory response of the first enzyme, not an instant final pathway flux or guaranteed downstream production. [R2; R4]", "transfer": "Identify the regulated step, the feedback molecule and the time scale of the prediction." } ] }