IB Biology · Theme C Interaction and interdependence · Molecules
C1.1 Enzymes and metabolism
Enzymes are globular proteins whose active sites catalyse one reaction each by lowering activation energy. Rate depends on collisions, so temperature, pH and substrate concentration all change it. Metabolism is a network of enzyme pathways, and controlling enzymes controls the network.
Compiled from the IB Biology guide (first assessment 2025, updated May 2026 for 2028) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Biology guide (first assessment 2025, updated May 2026 for 2028).
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C1.1.1 Enzymes speed up reactions without being used up
An enzyme is a globular protein that acts as a biological catalyst.
It raises the rate of one reaction by lowering its activation energy.
The enzyme leaves the reaction unchanged, so one molecule works many times.
Without enzymes, reactions at cell temperatures would be too slow for life.
Students often think enzymes are used up like reactants. In fact each enzyme is released unchanged and binds another substrate.
Students often think an enzyme supplies energy to the substrate. In fact it lowers activation energy so more substrate molecules already have enough.
C1.1.2 Metabolism is a network of reactions, each with its own enzyme
Metabolism is the whole network of interdependent chemical reactions in an organism.
A metabolic pathway is a chain or cycle of reactions: each product feeds the next.
Because of specificity, almost every reaction needs its own enzyme.
Control the amount or activity of one enzyme and you control one pathway.
Students often think metabolism just means releasing energy from food. In fact it includes every reaction, catabolic and anabolic alike.
Students often think one enzyme can catalyse many reactions. In fact each catalyses one reaction or type, so organisms need many enzymes.
C1.1.3 Anabolism builds and uses energy; catabolism breaks down and releases it
Anabolism builds large molecules from small ones and needs energy, usually from ATP.
Protein synthesis, glycogen formation and photosynthesis are anabolic.
Catabolism breaks large molecules into small ones and releases energy.
Hydrolysis in digestion and oxidation of glucose in respiration are catabolic.
Monomers join by condensation, which releases water; hydrolysis breaks the bond by adding water.
Students often think respiration is anabolic because it makes ATP. In fact glucose is broken down and oxidised, so it is catabolic.
Students often think water is released when a bond is broken. In fact hydrolysis uses water; condensation releases it.
C1.1.4 The active site is a few amino acids held in place by the whole fold
Enzymes are globular proteins, folded into compact shapes held by bonds between amino acids.
The active site is where substrate binds and catalysis happens.
It is made of only a few amino acids.
Interactions across the whole three-dimensional structure hold those few in position.
Students often think only the active-site amino acids matter. In fact the rest of the fold positions them, so a change elsewhere can alter the site.
C1.1.5 Enzyme and substrate both change shape as they bind
The active site is not a rigid, exact match for the substrate.
As substrate binds, both enzyme and substrate change shape: this is induced fit.
The closer fit strains bonds in the substrate, so the reaction is catalysed.
The enzyme-substrate complex converts substrate to product, then releases it.
Students often think the active site is a rigid lock for one key. In fact binding itself reshapes the site into a closer fit.
Students often think only the enzyme changes shape. In fact both enzyme and substrate change shape on binding.
C1.1.6 Random motion brings substrate and active site together
Substrate and enzyme move by random molecular motion in the surrounding fluid.
Binding follows a collision with the substrate in the right orientation.
Rate depends on how often such collisions happen.
An immobilised enzyme, fixed in a membrane, still works: substrate comes to it.
Students often think enzymes seek out or grab their substrate. In fact random motion brings them together; nothing is sought.
Students often think an enzyme that cannot move cannot work. In fact moving substrate molecules collide with its fixed active site.
C1.1.7 Active site shape explains specificity and why denaturation stops activity
Specificity: the active site's shape, charge and polarity are complementary to one substrate.
Other molecules do not bind in a way that allows catalysis.
Denaturation by heat or extreme pH breaks the weak bonds holding the fold.
The active site loses its shape, substrate cannot bind, and activity is usually lost for good.
Students often think denaturation breaks peptide bonds. In fact peptide bonds stay intact; the weaker bonds holding the shape break.
Students often think specificity is just about size. In fact chemical properties of the site must match the substrate too.
C1.1.8 Temperature, pH and substrate concentration change the rate
Warmer molecules collide more often and with more energy, so rate rises to an optimum.
Above it, rate falls steeply because the enzyme is denatured.
Away from the optimum pH, changed charges on amino acids alter the active site.
More substrate means more collisions, until active sites are saturated and rate plateaus.
A sketch graph of these relationships is a model; real experimental results can be used to evaluate it.
Students often think cold denatures enzymes. In fact cold only slows collisions; structure is unchanged and activity returns on warming.
Students often think the plateau means substrate has run out. In fact every active site is occupied; the enzyme is at maximum rate.
C1.1.9 Rate is change in product or substrate per unit time
Rate is the change in product or substrate amount per unit time.
Find it from the gradient of a product-time graph.
Or as 1/time taken for a fixed change to occur.
Initial rate is the steepest early gradient, before substrate falls or product builds up.
Students often think rate is the total product formed. In fact rate is change per unit time, taken from the gradient.
Students often think a longer time means a higher rate. In fact rate is proportional to 1/time, so the shorter time wins.
C1.1.10 Enzymes lower activation energy but not the energy released
Energy is needed to break bonds in the substrate; this is the activation energy.
Energy is released when the bonds of the products form.
The enzyme strains substrate bonds, so activation energy is lower.
The energy difference between substrate and products is unchanged.
Students often think an enzyme makes a reaction release more energy. In fact only the activation energy is lowered.
Students often think breaking bonds releases energy. In fact breaking bonds needs energy; forming bonds releases it.
C1.1.11 Enzymes act inside cells or are secreted to act outside HL
Intracellular enzymes act inside the cell that made them.
Glycolysis in the cytoplasm and the Krebs cycle in the matrix are intracellular.
Extracellular enzymes are made inside cells and secreted to act outside.
Digestive enzymes such as amylase, pepsin and lipase hydrolyse food in the gut lumen.
Students often think food is digested inside gut cells. In fact secreted enzymes digest it in the lumen, outside cells.
Students often think extracellular enzymes are made outside cells. In fact ribosomes make them inside; they are then secreted.
C1.1.12 Metabolism always makes heat, and some animals depend on it HL
No metabolic reaction transfers energy with 100% efficiency.
So some energy becomes heat in every reaction, in every living cell.
Mammals, birds and some other animals use this heat to keep a constant body temperature.
Students often think only shivering makes heat. In fact every reaction leaks heat, because none is perfectly efficient.
Students often think cold-blooded animals produce no heat. In fact all organisms do; mammals and birds differ in depending on it.
C1.1.13 Pathways run as a line or as a cycle HL
A linear pathway converts substrate through intermediates to product; nothing is regenerated.
Glycolysis, glucose to pyruvate, is linear.
In a cyclical pathway, the last product is the first substrate, so one intermediate is regenerated.
The Krebs cycle regenerates oxaloacetate; the Calvin cycle regenerates RuBP.
Students often think a cycle produces nothing new. In fact substrates enter and products leave each turn.
Students often think the starting compound of a cycle is used up. In fact the last reaction regenerates it.
C1.1.14 Binding at an allosteric site reshapes the active site HL
An allosteric site is away from the active site; only specific substances bind there.
Binding causes a conformational change that alters the active site and stops catalysis.
Binding is reversible: activity returns when the substance leaves.
A non-competitive inhibitor works this way, so extra substrate cannot overcome it.
Students often think non-competitive inhibition is permanent. In fact the inhibitor leaves and the enzyme regains its shape.
Students often think a non-competitive inhibitor denatures the enzyme. In fact the change is reversible; the enzyme is not denatured.
C1.1.15 Competitive inhibitors share the active site with the substrate HL
A competitive inhibitor resembles the substrate and binds reversibly to the active site.
Substrate and inhibitor compete; more substrate reduces the inhibitor's effect.
At high substrate concentration, rate approaches the uninhibited maximum.
Statins competitively inhibit HMG-CoA reductase, so liver cells make less cholesterol.
Students often think more substrate overcomes any inhibitor. In fact only competitive inhibition is overcome; non-competitive stays.
Students often think statins break down cholesterol. In fact they slow the enzyme that synthesises it.
C1.1.16 The end product switches off the first enzyme HL
In feedback inhibition, the pathway's end product inhibits its first enzyme.
It binds reversibly at an allosteric site; the pathway slows when product accumulates.
As product is used up, inhibition is released and the pathway resumes.
Isoleucine inhibits threonine deaminase, the first enzyme in its own synthesis.
Students often think the end product inhibits the last enzyme. In fact it inhibits the first, so no intermediates pile up.
Students often think feedback inhibition is permanent. In fact binding is reversible and the pathway restarts.
C1.1.17 Some inhibitors change the active site for good HL
In mechanism-based inhibition, the inhibitor binds and the enzyme's own action converts it.
The product bonds covalently to the active site; inhibition is irreversible.
Penicillin does this to bacterial transpeptidase, which cross-links peptidoglycan.
Without cross-links the wall weakens and the cell bursts.
Resistant bacteria have a mutated transpeptidase whose active site no longer binds penicillin well.
Students often think penicillin is a reversible competitive inhibitor. In fact it bonds covalently and the enzyme never recovers.
Students often think bacteria alter their own enzyme when exposed. In fact a mutation is selected; resistant cells survive and reproduce.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Which statement correctly describes how an enzyme acts as a catalyst in a cell?
Answer and reasoning
It increases the rate of a reaction that could occur without it, and is released unchanged. — Enzymes are catalysts: they increase the rate of reactions that would otherwise occur far too slowly at cell temperature, and the enzyme is not consumed, so each molecule catalyses the reaction repeatedly.
It is converted into product along with the substrate, so the cell must keep making it. — A student who treats the enzyme as a reactant picks this. In fact the enzyme is released unchanged when the product leaves the active site; it is not consumed.
It makes possible a reaction that could not take place at all in its absence in the cell. — A student who has only ever seen the catalysed reaction picks this. In fact the reaction can occur without the enzyme; the enzyme increases its rate, which at cell temperature is otherwise far too slow to sustain life.
It transfers energy to the substrate so that the reaction is driven forward faster. — A student who thinks enzymes 'help' by supplying energy picks this. In fact the enzyme lowers the activation energy; the energy for reaction comes from the substrate molecules' own kinetic energy.
2 Which statement best describes metabolism in a living organism?
Answer and reasoning
The set of reactions that release energy from food molecules to keep the organism alive and active. — A student who uses the everyday meaning of metabolism picks this. In fact metabolism includes anabolic reactions such as protein synthesis and photosynthesis as well as energy-releasing catabolic ones.
A series of reactions whose rate is set only by how much substrate the organism takes in from outside. — A student who thinks enzymes cannot be regulated picks this. In fact cells control metabolism by controlling the synthesis and activity of specific enzymes, not just by the supply of substrate.
A network of interdependent enzyme-catalysed reactions, each controlled through its own enzyme. — Metabolism is the complex network of interdependent and interacting chemical reactions in an organism. Because each reaction depends on a specific enzyme, control over metabolism can be exerted through these enzymes.
A set of reactions catalysed by a few general-purpose enzymes that act on many substrates. — A student who thinks enzymes are all-purpose catalysts picks this. In fact enzyme specificity means each reaction needs its own enzyme, which is why organisms require many different enzymes.
3 A mutation replaces one amino acid in an enzyme at a position on the surface far from the active site in the folded protein. The enzyme loses most of its activity. What is the best explanation?
Answer and reasoning
The result must be an error, because only the amino acids of the active site itself can affect catalysis — A student who thinks the rest of the enzyme is inert packaging picks this. In fact the whole three-dimensional structure positions and shapes the active site.
The loss of activity shows the enzyme has been denatured, so its chain has broken up into free amino acids — A student who equates any loss of enzyme activity with the polypeptide breaking up picks this. In fact a substitution leaves every peptide bond intact; the chain is complete but folds slightly differently, which is enough to alter the active site.
The change disturbs interactions that hold the enzyme's overall shape, altering the active site — The active site is composed of a few amino acids only, but interactions between amino acids within the overall three-dimensional structure ensure that it has the properties needed for catalysis. A change elsewhere can shift those interactions and distort the active site.
The changed amino acid supplies less energy to the reaction, so the substrate cannot react — A student who thinks enzymes supply energy to substrates picks this. In fact enzymes supply no energy; activity is lost because the active site's shape or chemistry has changed.
4 ATP synthase is embedded in the inner mitochondrial membrane and cannot move freely. How do its substrates, ADP and phosphate, reach its active site?
Answer and reasoning
They move by random molecular motion in the fluid and collide with the fixed active site. — Movement is needed for a substrate and an active site to come together, but movement of the substrate alone is enough. Enzymes immobilized in membranes are reached by substrates moving randomly.
They cannot, so an enzyme fixed in a membrane works only after being released from it. — A student who thinks both partners must move picks this. In fact substrate molecules moving randomly collide with the active site of an immobilized enzyme, which works normally.
The enzyme extends its active site out into the fluid to capture passing substrates. — A student who imagines enzymes reaching for substrates picks this. In fact enzymes do not seek substrates; collisions arise from random molecular motion.
They are carried to the enzyme by a directed flow of fluid in the mitochondrion. — A student who thinks molecules stay still unless carried picks this. In fact molecules in a fluid are in constant random motion, which alone brings them into collision with the active site.
Its peptide bonds are broken, so the polypeptide chain splits into its separate amino acids. — A student who thinks denaturation destroys the molecule picks this. In fact the peptide bonds and primary structure remain intact; only the folding changes.
Its shape alters only briefly and then returns to normal once the conditions are restored. — A student who extends the reversibility of low temperature to denaturation picks this. In fact denaturation is usually permanent; the enzyme does not refold correctly.
Its three-dimensional structure changes, so the active site no longer fits the substrate. — Denaturation disrupts the hydrogen bonds, ionic bonds and other interactions that hold the enzyme's folded shape. The active site loses the shape and properties needed to bind the substrate, so activity is lost.
It is killed, because enzymes are living particles that are destroyed by heat or acid. — A student who takes 'killed' literally picks this. In fact enzymes are protein molecules, not living things; heat or extreme pH changes their shape.
6 Catalase was added to hydrogen peroxide and the volume of oxygen collected was recorded: 0 s, 0 cm³; 20 s, 12 cm³; 40 s, 20 cm³; 60 s, 24 cm³. What is the initial rate of reaction?
Answer and reasoning
0.40 cm³ s⁻¹ — A student who divides the final volume by the whole time (24 / 60) picks this. That is a mean rate over a period in which the reaction had already slowed; the initial rate uses the first interval.
1.7 s per cm³ — A student who divides time by volume (20 / 12) picks this. Rate is amount per unit time, so the volume is divided by the time, not the other way round.
20 cm³ in 40 s — A student who quotes a reading rather than a rate picks this. A rate must be expressed as a change per unit time, and 20 / 40 would in any case be a mean over a period when the rate had already fallen.
0.6 cm³ s⁻¹ — The initial rate is the gradient of the first, steepest part of the graph: 12 cm³ of oxygen in the first 20 s gives 12 / 20 = 0.6 cm³ s⁻¹.
7 Which statement about intracellular and extracellular enzyme-catalysed reactions is correct? HL
Answer and reasoning
Digestion in the gut is intracellular, because food is broken down inside the cells that line it. — A student who merges absorption with digestion picks this. In fact digestive enzymes are secreted and act in the gut lumen, outside cells; only the products are absorbed.
The Krebs cycle is extracellular, because it breaks down food in the gut to release its energy. — A student who conflates respiration with digestion picks this. In fact the Krebs cycle occurs inside cells, in mitochondria, and its substrates are products of digestion that cells have absorbed.
Extracellular enzymes such as pepsin are made outside cells, which is why they act outside them. — A student who reads 'extracellular' as describing where the enzyme is made picks this. In fact all enzymes are synthesized inside cells; extracellular enzymes are secreted to act outside.
Glycolysis is intracellular; chemical digestion in the gut is extracellular. — Glycolysis and the Krebs cycle take place inside cells and are intracellular. Digestive enzymes are secreted into the gut lumen and act outside cells, so chemical digestion in the gut is extracellular.
8 A resting mammal produces heat continuously, even when it is not shivering and its muscles are relaxed. Why is this heat production inevitable? HL
Answer and reasoning
No metabolic reaction transfers energy with 100% efficiency, so part of the energy of every reaction becomes heat. — Heat generation is inevitable because metabolic reactions are not 100% efficient in energy transfer. Every reaction in every cell releases some heat, and mammals and birds depend on this to maintain a constant body temperature.
Heat is produced only by dedicated reactions, such as those in shivering muscle, which are switched on when cold. — A student who thinks heat comes only from special mechanisms picks this, but the stem says the animal is not shivering. In fact all metabolic reactions release some heat as an unavoidable by-product.
Breaking the bonds in glucose releases energy directly as heat before any of it is transferred to make ATP. — A student who thinks bond breaking releases energy picks this. In fact breaking bonds requires energy; heat arises because the energy released as new bonds form is not all captured in ATP.
Only mammals and birds produce metabolic heat, because in other animals energy transfer has no losses. — A student who takes 'warm-blooded' to mean 'heat-producing' picks this. In fact every organism's metabolism releases heat; mammals and birds are distinctive in relying on it for a constant body temperature.
9 Which statement about non-competitive inhibition at an allosteric site is correct? HL
Answer and reasoning
The inhibitor binds permanently to the allosteric site, so the enzyme is inactivated for the rest of its life. — A student who reads 'non-competitive' as 'irreversible' picks this. In fact binding at an allosteric site is reversible; the enzyme regains activity when the inhibitor leaves.
Any molecule of a suitable size can occupy the allosteric site, and it denatures the enzyme when it does so. — A student who equates inhibition with denaturation picks this. In fact only specific substances bind to an allosteric site, and the conformational change is reversible, not denaturation.
The inhibitor binds reversibly away from the active site, and the resulting shape change stops catalysis. — Only specific substances bind to an allosteric site. Binding causes interactions within the enzyme that lead to conformational changes, altering the active site enough to prevent catalysis; the binding is reversible.
The inhibitor occupies the active site, and adding more substrate displaces it and restores the rate. — A student who applies the competitive mechanism to all inhibitors picks this. In fact a non-competitive inhibitor binds elsewhere, so extra substrate cannot displace it and the maximum rate stays lowered.
10 In bacteria, threonine is converted into isoleucine by a pathway of five enzyme-catalysed steps. What happens when isoleucine is added to a growing culture? HL
Answer and reasoning
Isoleucine binds to the enzyme of the final step, the one that produced it, and blocks that single step only. — A student who reasons from proximity on the diagram picks this. In fact the end product inhibits the first enzyme, which stops the whole pathway and prevents intermediates from accumulating.
Isoleucine binds to an allosteric site on the first enzyme, reversibly reducing the pathway's activity. — This is feedback inhibition: the end product, isoleucine, binds to an allosteric site on threonine deaminase, the first enzyme of the pathway. Binding is reversible, so the pathway resumes when isoleucine is used up.
The five enzymes are broken down, so the pathway cannot restart after the isoleucine is used up. — A student who thinks feedback inhibition is permanent picks this. In fact the enzymes are unchanged; inhibition is released as soon as the isoleucine concentration falls.
Isoleucine competes with threonine for the active site of the first enzyme, blocking it competitively. — A student who defaults to the active site for any inhibitor picks this. In fact isoleucine does not resemble threonine closely enough to compete; it binds to a separate allosteric site.
Read the ones marked not yet in Learn, then Verify.
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17 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 Catalase in liver cells increases the rate at which hydrogen peroxide breaks down into water and oxygen by a factor of many millions. What is the benefit to the cell of this increase in rate?
Answer and reasoning
Hydrogen peroxide could not be broken down at all without an enzyme to make it possible. — A student who thinks enzymes make reactions possible rather than faster picks this. In fact hydrogen peroxide decomposes slowly on its own; catalase increases the rate enormously.
Toxic hydrogen peroxide is removed fast enough at cell temperature to prevent damage. — The benefit of increasing rates of reaction in cells is that reactions which would be uselessly slow at cell temperature happen fast enough to be useful. Here a toxic substance is destroyed before it can harm the cell.
Catalase supplies the energy that the breakdown of hydrogen peroxide needs. — A student who thinks enzymes provide energy picks this. In fact catalase lowers the activation energy; it does not supply energy to the substrate.
The reaction releases energy that the cell captures as ATP, as in respiration. — A student who thinks every metabolic reaction is an energy-yielding reaction like respiration picks this. In fact the cell does not harvest energy from this reaction; its purpose is to remove a toxic product.
2 A yeast cell synthesizes one amino acid through a pathway of several steps. Why does this pathway require several different enzymes rather than one?
Answer and reasoning
One enzyme could catalyse every step if the substrates were supplied in the correct order. — A student who thinks an enzyme is a general-purpose catalyst picks this. In fact one active site cannot be complementary to several different substrates, so several enzymes are needed.
Each step needs a different amount of energy, and each enzyme supplies a fixed amount of energy. — A student who thinks enzymes supply energy picks this. In fact enzymes lower activation energy and supply no energy; the number of enzymes is set by specificity, not by energy demands.
Enzymes are used up as they work, so several are needed to complete all the steps. — A student who thinks enzymes are consumed picks this. In fact each enzyme is released unchanged after every reaction, so a single enzyme molecule could serve its step indefinitely.
Each step has a different substrate, and an active site fits only one substrate. — Because of enzyme specificity, each enzyme's active site is complementary to one substrate, so each reaction in the pathway needs its own enzyme. This is why living organisms require many different enzymes.
The formation of glycogen from glucose monomers by condensation reactions — Anabolism builds larger molecules from smaller ones; the formation of macromolecules such as glycogen from monomers by condensation is one of the guide's named examples, along with protein synthesis and photosynthesis.
The oxidation of glucose in cell respiration to provide ATP for the cell — A student who classifies respiration as anabolic because it makes ATP picks this. In fact respiration oxidizes glucose to smaller molecules, so it is catabolic; ATP is the energy carrier, not a macromolecule being built.
The digestion of starch in the gut into maltose and then glucose — A student who thinks digestion is anabolic because it produces useful monomers picks this. In fact digestion is hydrolysis of macromolecules into monomers, a catabolic process.
The hydrolysis of ATP to ADP and phosphate, which releases energy — A student who thinks anabolic reactions are the ones that release energy picks this. In fact anabolic reactions require energy; hydrolysis of ATP is a breaking-down reaction that releases it.
4 In the small intestine, maltase converts each maltose molecule into two glucose molecules. How is this reaction classified?
Answer and reasoning
Anabolic; the glucose produced is a useful new molecule that the body can then absorb and make use of — A student who judges by whether the product is useful picks this. In fact the classification depends on the substrate being broken into smaller molecules, which makes it catabolic.
Catabolic; a hydrolysis reaction in which a water molecule is used to break the glycosidic bond — Digestion of a larger molecule into monomers is catabolism, and the bond between the two glucose units is broken by the addition of water, which is hydrolysis.
Catabolic; a condensation reaction in which a water molecule is released as the bond is broken — A student who attaches water release to bond breaking picks this. In fact condensation joins monomers and releases water; breaking the bond uses water and is hydrolysis.
Catabolic; energy from ATP is needed to force the two glucose units of the maltose apart — A student who thinks breaking molecules down needs an input of energy picks this. In fact hydrolysis in digestion releases energy overall and does not consume ATP.
5 According to the induced-fit model, what happens when a substrate binds to the active site of an enzyme?
Answer and reasoning
The rigid active site admits only an exactly matching substrate. — A student still using the lock-and-key model picks this. In fact the active site is not rigid: it changes shape as the substrate binds.
The active site closes around a substrate that keeps its shape. — A student who has noticed only the enzyme's movement picks this. In fact the substrate also changes shape on binding, and that change weakens its bonds.
The enzyme seizes the substrate and holds it without any changes. — A student who imagines enzymes actively catching substrates picks this. In fact substrate and active site meet by random collision, and both change shape once bound.
Both the enzyme and the substrate change shape as they bind. — Students should recognize that both substrate and enzyme change shape when binding occurs. The closer fit strains bonds within the substrate, which is part of how catalysis is achieved.
6 Lipase catalyses the hydrolysis of triglycerides but has no effect on starch. What is the reason?
Answer and reasoning
Starch molecules are too large to enter the active site, but any molecule smaller than starch could. — A student who thinks specificity is only a matter of size picks this. In fact many molecules smaller than starch also fail to bind, because the active site must match shape and chemistry.
The active site of lipase is complementary in shape and chemical properties to a triglyceride only. — Enzyme-substrate specificity arises from the structure of the active site: its shape and chemical properties match one substrate, so starch cannot bind in a way that allows catalysis.
Starch requires more energy to break down than lipase is able to supply to the reaction. — A student who thinks enzymes supply energy picks this. In fact enzymes supply no energy; lipase fails to act on starch because starch does not fit its active site.
Lipase seeks out triglyceride molecules in the fluid and ignores starch molecules it meets. — A student who imagines enzymes choosing their targets picks this. In fact lipase collides with all molecules at random; only a triglyceride binds productively to its active site.
7 A student measured the rate of an enzyme-catalysed reaction at six temperatures. The rates, in arbitrary units, were: 10 °C, 2; 20 °C, 4; 30 °C, 8; 40 °C, 15; 50 °C, 9; 60 °C, 1. Which conclusion is supported by these data?
Answer and reasoning
The low rate at 10 °C shows that the enzyme had been denatured by the cold temperature. — A student who treats both ends of the graph the same way picks this. In fact the low rate in the cold is due to slower molecular motion and fewer collisions; the enzyme's structure is unchanged.
The true optimum is 37 °C, the body temperature at which every enzyme works at its best. — A student who assumes all enzymes share the human body's conditions picks this. In fact optima differ between enzymes, and these data show the highest rate at 40 °C.
The fall in rate after 40 °C shows that the substrate had been used up at higher temperatures. — A student who confuses a rate-temperature graph with a product-time graph picks this. In fact each rate was measured with fresh substrate; the fall is caused by denaturation.
The optimum is near 40 °C; above it, denaturation outweighs the faster collisions. — Rate rises with temperature up to 40 °C as molecules gain kinetic energy and collide more often and more energetically. Above the optimum, denaturation of the enzyme reduces the rate sharply.
8 At high substrate concentrations, the rate of an enzyme-catalysed reaction levels off even when more substrate is added. Why?
Answer and reasoning
Every active site is occupied almost all the time, so extra substrate finds no free site to bind. — At saturation the enzyme is working at its maximum rate: as soon as one product leaves an active site another substrate binds. Adding substrate cannot increase the frequency of productive collisions further.
The enzyme molecules have been used up by the reaction, so no more product can be formed. — A student who thinks enzymes are consumed picks this. In fact enzymes are released unchanged; the plateau reflects saturation of active sites, not loss of enzyme.
The high concentration of substrate denatures the enzyme, so its active site stops working. — A student who assumes substrate has a damaging limit like temperature picks this. In fact substrate concentration does not change the enzyme's structure.
The amount of energy the enzyme can supply each second has reached its upper limit. — A student who thinks enzymes supply energy picks this. In fact enzymes supply no energy; the limit is the number of active sites available for substrate to bind.
9 Pepsin has an optimum pH of about 2. When it passes with food into the small intestine, where the pH is about 8, its activity falls almost to zero. Which explanation is correct?
Answer and reasoning
The alkaline conditions hydrolyse the peptide bonds of pepsin, breaking it up into free amino acids. — A student who thinks loss of activity means the chain is broken picks this. In fact the peptide bonds are intact; the enzyme's folded shape has been disrupted.
The change in pH alters charges on amino acids, disrupting bonds that hold the active site's shape. — Hydrogen ion concentration affects the charges on amino acids in the enzyme. Away from the optimum, ionic and hydrogen bonds are disrupted, the active site changes shape and at an extreme the enzyme is denatured.
Pepsin has run out of substrate, because protein digestion is completed inside the stomach. — A student who looks for a substrate explanation picks this. In fact protein digestion is far from complete in the stomach; pepsin is inactive at pH 8 because its structure is altered.
At pH 8 the enzyme and substrate molecules move more slowly, so they collide less often. — A student who applies the temperature explanation to pH picks this. In fact pH does not change molecular speed; it acts on the enzyme's structure.
10 A textbook sketch graph shows the rate of an enzyme-catalysed reaction rising steadily with temperature to a sharp peak and then falling steeply. A class measured catalase activity every 5 °C from 5 °C to 70 °C and found a broad plateau between 30 °C and 45 °C rather than a sharp peak. How should the sketch graph be regarded?
Answer and reasoning
As an accurate record of catalase results, which shows that the class must have made an error somewhere. — A student who treats a sketch graph as data picks this. In fact a sketch has no data points or scale; it is a simplified model and cannot overrule real measurements.
As a model that the mismatch has proved to be wrong, so it should be discarded and not used again. — A student who thinks a model is simply right or wrong picks this. In fact a mismatch leads to evaluation and refinement; the general pattern of rise, optimum and fall is still supported.
As a model of the general relationship, which the class data can be used to evaluate and refine. — Generalized sketches of relationships are models. The class results neither prove nor disprove the sketch outright; they are evidence for evaluating it, and here suggest that for catalase the peak should be drawn as a plateau.
As a graph of one specific enzyme that was measured, so it says nothing about catalase. — A student who thinks the sketch records one enzyme's results picks this. In fact it is a generalized model intended to apply to enzymes in general, which is exactly why catalase data can test it.
11 A filter-paper disc soaked in catalase solution sinks in hydrogen peroxide and rises when enough oxygen has formed on it. In 1% hydrogen peroxide the disc took 50 s to rise; in 2% it took 20 s. Which statement about the reaction rate is correct?
Answer and reasoning
Rate is proportional to 1/time, so the rate at 2% (0.05 s⁻¹) is 2.5 times the rate at 1% (0.02 s⁻¹). — A fixed event, the disc rising, takes less time when the reaction is faster, so rate is proportional to 1/time: 1/20 = 0.05 s⁻¹ and 1/50 = 0.02 s⁻¹, a ratio of 2.5.
The rate at 1% is 50 and at 2% is 20, so the rate fell when the concentration was doubled. — A student who reports the measured time as the rate picks this. In fact a shorter time means a faster reaction; rates must be calculated as 1/time before they are compared.
Doubling the concentration doubles the rate, so the rate at 2% is 0.04 s⁻¹, twice that at 1%. — A student who assumes rate is always proportional to substrate concentration picks this and calculates instead of using the data. The measured times give 0.05 s⁻¹, not 0.04 s⁻¹.
The rate cannot be found, because the volume of oxygen produced was not measured in either trial. — A student who thinks a rate must come from an amount of product picks this. In fact the time taken for a fixed change is a valid measure, with rate proportional to 1/time.
12 How does an enzyme affect the energy changes of the reaction it catalyses?
Answer and reasoning
It lowers the energy of the products, so that more energy is released overall than would be without the enzyme. — A student who reads the lower catalysed curve as lower energy throughout picks this. In fact only the peak is lowered; substrate and products are at the same levels with or without the enzyme.
It lowers the activation energy but leaves the energy difference between substrate and products unchanged. — Energy is required to break bonds within the substrate, and the enzyme reduces this activation energy. The energy yield when bonds form in the products is a property of the reaction and is not changed.
It supplies the activation energy, transferring its own energy to the substrate to break the bonds. — A student who thinks the enzyme is an energy source picks this. In fact the enzyme supplies no energy; it lowers the barrier so that more substrate molecules already have enough energy to react.
It releases energy by breaking bonds in the substrate, and this provides the activation energy. — A student who thinks bond breaking releases energy picks this. In fact energy is required to break bonds; energy is released when the bonds of the products are made.
13 In an energy diagram for a reaction, the substrate is at 40 kJ mol⁻¹ and the products are at 15 kJ mol⁻¹. The highest energy reached during the reaction is 90 kJ mol⁻¹ without an enzyme and 60 kJ mol⁻¹ with the enzyme. For the enzyme-catalysed reaction, what are the activation energy and the overall energy yield?
Answer and reasoning
activation energy 60 kJ mol⁻¹; energy yield 25 kJ mol⁻¹ — A student who reads the peak's value from the axis as the activation energy picks this. Activation energy is measured from the substrate level, so it is 60 − 40 = 20 kJ mol⁻¹.
activation energy 20 kJ mol⁻¹; energy yield 45 kJ mol⁻¹ — A student who reads the energy released as the fall from the peak to the products (60 − 15) picks this. That fall includes the activation energy that was put in; the yield is substrate minus products, 25 kJ mol⁻¹.
activation energy 20 kJ mol⁻¹; energy yield 25 kJ mol⁻¹ — Activation energy is the rise from substrate to the highest point: 60 − 40 = 20 kJ mol⁻¹ with the enzyme. The energy yield is substrate minus products: 40 − 15 = 25 kJ mol⁻¹, the same as without the enzyme.
activation energy 50 kJ mol⁻¹; energy yield 25 kJ mol⁻¹ — A student who thinks the enzyme supplies energy rather than lowering the barrier keeps the uncatalysed value, 90 − 40 = 50 kJ mol⁻¹. With the enzyme the highest point is 60 kJ mol⁻¹, so the activation energy is 20 kJ mol⁻¹.
14 What distinguishes the Calvin cycle and the Krebs cycle from glycolysis? HL
Answer and reasoning
The cycles regenerate all of their intermediates and so produce no net product, whereas glycolysis produces pyruvate as its end product. — A student who thinks a cycle makes nothing new picks this. In fact substrates enter and products leave each cycle: the Krebs cycle releases carbon dioxide and reduced NAD, and the Calvin cycle releases triose phosphate.
In the cycles an intermediate used in the first step is regenerated in the last, whereas glycolysis runs from glucose to pyruvate in a chain. — A cyclical pathway regenerates one intermediate with each turn: RuBP in the Calvin cycle and oxaloacetate in the Krebs cycle. Glycolysis is linear, converting glucose through a series of intermediates to pyruvate with nothing regenerated.
The cycles use up their starting compound on every turn, so it must be supplied continuously from outside, unlike the glucose used in glycolysis. — A student who applies linear-pathway logic to cycles picks this. In fact the starting compound of a cycle is what is regenerated; it is glycolysis whose starting substrate, glucose, is consumed.
The cycles take place in the gut, where food is broken down to release its energy, whereas glycolysis takes place inside cells. — A student who conflates the Krebs cycle with digestion picks this. In fact both cycles are intracellular: the Krebs cycle in mitochondria and the Calvin cycle in chloroplasts.
15 The rate of an enzyme-catalysed reaction was measured at six increasing substrate concentrations, without and with inhibitor X. Without X the rates were 20, 35, 50, 58, 60 and 60 units. With X they were 8, 18, 33, 48, 57 and 60 units. What type of inhibitor is X? HL
Answer and reasoning
Non-competitive, because X lowers the rate at most concentrations, which shows that it is bound irreversibly. — A student who thinks non-competitive means irreversible picks this. In fact an irreversible or non-competitive inhibitor would lower the maximum rate; here the maximum is fully recovered.
Non-competitive, because a competitive inhibitor would have kept the rate low at every substrate concentration. — A student who thinks a competitive inhibitor permanently blocks the active site picks this. In fact competitive inhibition is reversible and is overcome by high substrate concentration, as the data show.
Either type, because raising the substrate concentration overcomes both competitive and non-competitive inhibitors. — A student who thinks extra substrate defeats any inhibitor picks this. In fact substrate cannot displace a non-competitive inhibitor from an allosteric site, so recovery of the maximum rate identifies X as competitive.
Competitive, because at high substrate concentration the inhibited rate reaches the uninhibited maximum. — With X the rate is lower at low substrate concentration but climbs to the same maximum of 60 units. Substrate outcompetes the inhibitor for the active site at high concentration, the defining feature of competitive inhibition.
16 Statins are drugs prescribed to reduce blood cholesterol. How do they act? HL
Answer and reasoning
They resemble the substrate of an enzyme in the cholesterol synthesis pathway and bind reversibly to its active site. — Statins are competitive inhibitors of HMG-CoA reductase, an enzyme in the pathway that synthesizes cholesterol. Because they resemble the substrate they compete for the active site, reducing the rate of cholesterol production.
They hydrolyse cholesterol molecules that are already circulating in the blood into products that are excreted. — A student who assumes 'lowering cholesterol' means removing it picks this. In fact statins are enzyme inhibitors that reduce cholesterol synthesis; they do not break cholesterol down.
They bind permanently to the active site of the enzyme, so cholesterol synthesis stops for the rest of life. — A student who thinks a competitive inhibitor blocks the site for good picks this. In fact binding is reversible, which is why statins must be taken continuously to keep cholesterol low.
They denature the cholesterol-synthesizing enzyme so that it can no longer fold into its working shape. — A student who equates inhibition with denaturation picks this. In fact statins bind reversibly to an intact enzyme's active site; the enzyme is unchanged when the statin leaves.
17 Penicillin inhibits transpeptidases, the bacterial enzymes that cross-link peptidoglycan in the cell wall. Which statement correctly describes both the action of penicillin and how bacteria become resistant to it? HL
Answer and reasoning
Penicillin competes reversibly with the substrate for the active site; bacteria resist it by producing extra substrate that outcompetes the drug molecules. — A student who assumes every active-site inhibitor is competitive picks this. In fact penicillin binds covalently and irreversibly, so extra substrate cannot displace it; resistance requires a changed enzyme.
Penicillin binds to an allosteric site on transpeptidase and inactivates it permanently; resistant bacteria have a transpeptidase that lacks the allosteric site. — A student who assumes that any irreversible inhibitor must be acting non-competitively at an allosteric site picks this. In fact penicillin binds at the active site and changes it chemically; resistance comes from an altered active site that still cross-links peptidoglycan but no longer binds penicillin.
Penicillin bonds irreversibly to the active site, changing it chemically; a changed transpeptidase that penicillin cannot bind confers resistance. — Mechanism-based inhibition is a consequence of chemical changes to the active site caused by the irreversible binding of an inhibitor. Resistance is conferred by a mutant transpeptidase whose altered active site still cross-links peptidoglycan but does not bind penicillin.
Bacteria exposed to penicillin alter their own transpeptidase during their lifetime and pass this acquired change to their offspring. — A student who thinks exposure causes the change picks this. In fact the altered transpeptidase arises by mutation; penicillin selects the bacteria that already carry it, and they pass it on.
That was your twenty minutes. Real practice on C1.1 is past-paper questions marked against the mark scheme.
What the exam asks of C1.1
Paper 1A tests definitions: catalyst, active site, denaturation, saturation. Paper 1B gives rate graphs against temperature, pH or substrate concentration and asks you to describe the relationship and explain it with collision theory. Paper 2 uses *outline*, *explain* and *compare*: explain induced fit, explain why rate plateaus, compare competitive with non-competitive inhibition. At HL, expect *explain* on feedback inhibition using isoleucine, and on penicillin and resistance.
Compiled from the IB Biology guide (first assessment 2025, updated May 2026 for 2028) and our question bank · Specialist review in progress. How these pages are made ·
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