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This paper covers the whole of Topic 5. Like a real Cambridge paper, the seven questions range across every sub-topic — catalysts and the active site, temperature and pH, and investigating enzyme activity — and they are deliberately mixed rather than grouped. All three Topic 5 papers do; they differ in the angle they come at it from, not in what they cover.
Question 1 — Liver, Peroxide and the Definition of a Catalyst
Total: 12 marks
Hydrogen peroxide is a toxic substance produced as a by-product of metabolism in every living cell. A student places 2 g of fresh liver into a flask containing 50 cm³ of hydrogen peroxide solution and collects the gas released. Bubbling is vigorous at first and stops completely after four minutes, by which time 46 cm³ of gas has been collected. She then pours off the liquid, leaves the liver in the flask, adds a further 50 cm³ of fresh hydrogen peroxide, and a further 45 cm³ of gas is collected over the next four minutes.
(a)[3]
Name the enzyme involved, name the gas collected, and give the word equation for the reaction it catalyses.
Model Answer — 1(a)
catalase [1]
the gas is oxygen [1]
hydrogen peroxide → water + oxygen [1]
⚠ If you missed marks here: Catalase is one of the enzymes that does not follow the substrate-plus-–ase naming rule, so it has to be learned. In the equation, hydrogen peroxide is the substrate and water and oxygen are the products; candidates who write oxygen on the left have described the reaction running backwards and will struggle with every later part.
(b)[3]
Explain what the result of the second addition of hydrogen peroxide shows about catalase.
Model Answer — 1(b)
the enzyme is not used up / not changed by the reaction [1]
because the same liver catalysed a second, equally large reaction without being replaced [1]
this is the property that makes it a catalyst [1]
⚠ If you missed marks here: The mark is not for saying “it still worked” but for connecting that observation to the definition. Note also what the result rules out: the enzyme had not been denatured, and it had not been consumed — both of which are common explanations offered for why the first reaction stopped.
(c)[3]
Explain why the first reaction stopped after four minutes, and state one further observation that would confirm your explanation.
Model Answer — 1(c)
all the hydrogen peroxide (substrate) had been broken down [1]
so there was no substrate left to bind to the active sites [1]
confirmation: adding fresh hydrogen peroxide restarts the reaction at the original rate, which it could not do if the enzyme had stopped working [1]
⚠ If you missed marks here: A reaction that stops has almost always run out of substrate, never of enzyme. Writing “the catalase was used up” here directly contradicts the answer given in part (b), and examiners notice when two parts of the same question disagree with each other.
(d)[3]
The student repeats the whole investigation using 2 g of liver that has been boiled for five minutes and then cooled. Predict the result and explain it fully.
Model Answer — 1(d)
almost no gas would be collected [1]
boiling denatures the catalase, so the shape of the active site changes [1]
hydrogen peroxide is no longer complementary to the active site, so no enzyme–substrate complexes form and the reaction is not catalysed [1]
⚠ If you missed marks here: Two words decide this answer. Denatured, never “killed” — an enzyme is a protein molecule and was never alive. And active site, because “the enzyme changes shape” on its own describes the event without explaining why the reaction stops. Note also that cooling the liver afterwards makes no difference: denaturation is permanent.
Question 2 — Amylase, Starch and the Meaning of Specificity
Total: 12 marks
Salivary amylase catalyses the breakdown of starch into maltose. A technician sets up four test tubes, each containing 5 cm³ of amylase solution at pH 7 and 35 °C, and adds a different substance to each: tube 1 starch, tube 2 protein (egg albumen), tube 3 cellulose, tube 4 a fat emulsion. After twenty minutes only tube 1 shows any change.
(a)[3]
Using the terms substrate, active site and product, describe what happens in tube 1.
Model Answer — 2(a)
starch is the substrate, and it is complementary in shape to the active site of amylase [1]
the substrate binds to the active site and the reaction takes place there [1]
the product, maltose, is formed and leaves the active site [1]
⚠ If you missed marks here: The question names the three words it wants, so an answer that avoids them cannot score. Substrate goes in and product comes out — describing maltose as binding to the active site reverses the reaction and undermines everything that follows.
(b)[4]
Explain fully why there is no change in tubes 2, 3 and 4.
Model Answer — 2(b)
the active site of amylase has a specific shape determined by the folding of its amino acid chain [1]
protein, cellulose and fat are not complementary in shape to that active site [1]
so they cannot bind to it and no enzyme–substrate complex forms [1]
therefore no reaction is catalysed — this property is called specificity [1]
⚠ If you missed marks here: Cellulose is the interesting distractor here: like starch it is built from glucose, so “amylase works on carbohydrates” predicts the wrong result. Specificity depends on the shape of the whole molecule, not on what it is made of. Answers that say the other molecules are “too big” are refused, since proteases handle those same protein molecules easily.
(c)[3]
Cellulose, like starch, is built from glucose units. Suggest why amylase nevertheless has no effect on it.
Model Answer — 2(c)
the glucose units are joined together in a different arrangement, so the cellulose molecule has a different overall shape [1]
cellulose is therefore not complementary to the active site of amylase [1]
a different enzyme, with an active site complementary to cellulose, would be required [1]
⚠ If you missed marks here: This is a suggest question, so you are being asked to apply the shape rule to an unfamiliar case rather than recall a fact. The key move is to separate composition from shape: two molecules made of the same units can present completely different shapes to an active site.
(d)[2]
State what would be observed when iodine solution is added to a sample from tube 1 at the start of the investigation and again after twenty minutes.
Model Answer — 2(d)
at the start the iodine turns blue-black, because starch is present [1]
after twenty minutes the iodine remains orange-brown, because the starch has been digested to maltose [1]
⚠ If you missed marks here: Each mark needs both the colour and the reason. “It goes clear” is refused because the iodine was always transparent; the correct description of the negative result is that it stays orange-brown. Reversing the two colours is a common slip that makes every timing in a practical meaningless.
Question 3 — Two Halves of One Curve
Total: 12 marks
The graph of the rate of an enzyme-catalysed reaction against temperature rises from almost zero at 5 °C, reaches its highest value at 42 °C, and has fallen back to zero by 58 °C. The enzyme was obtained from a mammal.
(a)[2]
State what is meant by the optimum temperature of an enzyme, and give its value for this enzyme.
Model Answer — 3(a)
the temperature at which the rate of reaction is highest [1]
42 °C [1]
⚠ If you missed marks here: Optimum means the peak of the curve and nothing more. Definitions such as “the temperature at which the enzyme works” or “the temperature at which it starts to denature” both miss, and the second one describes the wrong side of the graph entirely.
(b)[3]
Explain, in terms of the behaviour of molecules, why the rate increases between 5 °C and 42 °C.
Model Answer — 3(b)
the enzyme and substrate molecules gain kinetic energy and move faster [1]
so there are more frequent effective collisions between the substrate and the active site [1]
so more enzyme–substrate complexes form each second and the rate increases [1]
⚠ If you missed marks here: Three marks, three named phrases. “The particles have more energy” may scrape the first mark and earns nothing else, because the two consequences are what the question is asking for. The word effective matters: only a collision with the active site produces anything.
(c)[4]
Explain why the rate falls to zero between 42 °C and 58 °C.
Model Answer — 3(c)
above the optimum the enzyme is denatured [1]
the bonds holding the folded protein chain are broken, so the chain unfolds and the enzyme changes shape [1]
the shape of the active site changes so it is no longer complementary to the substrate [1]
the substrate can no longer bind, no enzyme–substrate complexes form and the rate falls to zero [1]
⚠ If you missed marks here: Two answers are refused outright here. “The enzyme is killed” — it is a molecule and was never alive. “The molecules move too fast for the substrate to bind” — faster movement never lowers a rate; that sentence comes from trying to use the kinetic energy explanation on the wrong side of the peak.
(d)[3]
A sample of the enzyme is held at 58 °C for ten minutes, then cooled to 42 °C and mixed with substrate. A second sample is held at 4 °C for ten minutes, then warmed to 42 °C and mixed with substrate. Predict and explain both results.
Model Answer — 3(d)
the heated sample shows no activity, because denaturation is irreversible — the chain does not re-fold and the active site does not regain its shape [1]
the cooled sample works normally [1]
because low temperature only reduced the kinetic energy of the molecules; the enzyme was not damaged, so warming restores full activity [1]
⚠ If you missed marks here: Low temperature slows an enzyme down; high temperature destroys it. Because both samples are tested at the same temperature, any difference must come from what happened before the test — which is exactly what this design is built to reveal. Describing the cold sample as “denatured” is the standard error.
Question 4 — Two Proteases, Two Places, Two Optima
Total: 12 marks
Pepsin is a protease produced in the stomach, where the pH is about 2. Trypsin is a protease produced by the pancreas and released into the small intestine, where the pH is about 8. Both catalyse the breakdown of protein into shorter chains and amino acids. A student assumes that all enzymes have an optimum pH of 7.
(a)[2]
State the approximate optimum pH of pepsin and of trypsin.
Model Answer — 4(a)
pepsin: about pH 2 [1]
trypsin: about pH 8 [1]
⚠ If you missed marks here: The optimum of a digestive enzyme matches the region it works in, so the pH values given in the stem are the answer. Candidates who apply the “all enzymes prefer pH 7” rule get both marks wrong, and then cannot explain any of the parts that follow.
(b)[4]
Explain what happens to pepsin when it passes out of the stomach into the small intestine.
Model Answer — 4(b)
the pH rises to about 8, which is far from pepsin’s optimum of about 2 [1]
pepsin is denatured [1]
the shape of its active site changes so protein is no longer complementary to it [1]
no enzyme–substrate complexes form, so pepsin no longer digests protein — and the change is permanent [1]
⚠ If you missed marks here: The full chain is worth four marks and takes three sentences. Two things sink answers here: the word “killed”, and stopping at “pepsin stops working”, which restates the question. Note that the change is irreversible, so pepsin does not resume if conditions become acidic again.
(c)[3]
Suggest why the human digestive system produces two different proteases rather than one.
Model Answer — 4(c)
each enzyme is only active close to its own optimum pH [1]
the stomach is strongly acidic and the small intestine is alkaline, so no single protease could be active in both [1]
having a protease matched to each region allows protein digestion to continue throughout the gut [1]
⚠ If you missed marks here: This is a suggest question, so the marks are for reasoning rather than recall. The strongest answers make the impossibility explicit: an enzyme with an optimum of pH 2 is denatured at pH 8, so it could not simply be reused further down.
(d)[3]
A student explains the shape of a graph of enzyme activity against pH by writing: “As the pH increases towards the optimum the molecules gain kinetic energy, so there are more collisions and the rate rises.” Explain why this answer would gain no credit, and give the correct explanation.
Model Answer — 4(d)
pH does not affect the kinetic energy of molecules — that is the effect of temperature [1]
as pH approaches the optimum the shape of the active site becomes increasingly complementary to the substrate, so the substrate fits better [1]
more enzyme–substrate complexes therefore form and the rate rises; beyond the extremes the enzyme is denatured [1]
⚠ If you missed marks here: This is the commonest zero-mark sentence in the whole topic, and it is written by candidates who understand everything else. The temperature and pH graphs look alike, so the explanation gets copied across. The syllabus itself is the giveaway: kinetic energy appears in the temperature statement and is deliberately absent from the pH one.
Question 5 — Built Wrong, Not Broken
Total: 10 marks
An enzyme consists of a single chain of 214 amino acids folded into a precise shape. Researchers produce two altered versions. In version A, amino acid 96 — which lies within the active site — is replaced with a different amino acid. In version B, amino acid 12 — which lies on the far side of the molecule — is replaced. Version A has no measurable activity. Version B behaves exactly like the original enzyme. Both versions are the same length as the original and are produced in normal amounts.
(a)[4]
Explain why version A has no activity.
Model Answer — 5(a)
the sequence of amino acids determines how the protein chain folds [1]
the folding determines the shape of the active site [1]
with a different amino acid in the active site, the shape of the site is altered [1]
the substrate is no longer complementary to it, so no enzyme–substrate complex forms and no reaction is catalysed [1]
⚠ If you missed marks here: The four marks are four links in one chain, and they must appear in order. A very common answer here is “the enzyme has been denatured” — but denaturation is damage done to a correctly folded enzyme by heat or extreme pH. This enzyme was assembled with the wrong shape from the start and was never damaged at all.
(b)[3]
Suggest why version B behaves exactly like the original enzyme.
Model Answer — 5(b)
amino acid 12 is not part of the active site and lies far from it [1]
the replacement does not alter the folding in the region of the active site, so the shape of the active site is unchanged [1]
the substrate still fits, complexes form as before and the rate is normal [1]
⚠ If you missed marks here: Most of an enzyme molecule is scaffolding: only the small region that binds the substrate must be exactly right. Version B is a built-in control — because it works, the failure of version A cannot be blamed on “any change breaks a protein”, and must be tied to the position of that particular amino acid.
(c)[3]
Explain why it would be incorrect to describe version A as denatured.
Model Answer — 5(c)
denaturation is the loss of the correct three-dimensional shape of an enzyme that was originally folded correctly [1]
it is caused by high temperature or extreme pH breaking the bonds that hold the fold [1]
version A was never correctly shaped — it was assembled with the wrong amino acid, so it was built with a faulty active site rather than damaged [1]
⚠ If you missed marks here: Examiners set this part specifically to see whether two ideas have been merged. A useful test: could the molecule ever have worked? If yes and it stopped, that is denaturation. If it never worked, the fault is in its construction.
Question 6 — Reading a Results Table Properly
Total: 12 marks
A student investigates the effect of temperature on catalase using discs cut from one potato. She records the volume of oxygen collected in 60 s, taking three repeats at each temperature.
temperature / °C
repeat 1 / cm³
repeat 2 / cm³
repeat 3 / cm³
mean / cm³
10
6
5
6
5.7
20
12
13
12
12.3
30
24
23
25
24.0
40
31
9
32
24.0
50
14
15
14
14.3
60
0
0
0
0.0
(a)[2]
Identify the anomalous result and give one possible cause.
Model Answer — 6(a)
the value of 9 cm³ at 40 °C (repeat 2) is anomalous, because it disagrees sharply with the other two repeats at that temperature and breaks the rising trend [1]
a possible cause: the bung or the syringe leaked, or the water bath had not reached temperature, or timing began late [1]
⚠ If you missed marks here: An anomaly is a point that breaks an established pattern, not simply a value you dislike. Both marks are available only if you say which pattern it breaks, and offer a cause that is specific and checkable — “human error” is never credited.
(b)[3]
Recalculate the mean at 40 °C, explain why your value differs from the one in the table, and state what the student should do next.
Model Answer — 6(b)
corrected mean = (31 + 32) ÷ 2 = 31.5 cm³ [1]
the anomalous value has been excluded, and including it dragged the mean down to 24.0 cm³ [1]
she should repeat the measurement at 40 °C to replace the discarded reading [1]
⚠ If you missed marks here: You must show the exclusion explicitly — the arithmetic alone earns nothing if the anomaly is still in it. Notice how much the conclusion depends on this: with the printed mean, 30 and 40 °C appear identical and the optimum is invisible.
(c)[3]
Using the corrected data, state the best estimate of the optimum temperature and explain how confident the student can be in it.
Model Answer — 6(c)
the highest mean is at 40 °C [1]
the true optimum lies somewhere between 30 and 50 °C, because readings were taken only every 10 °C [1]
to locate it more precisely she should repeat at smaller intervals, for example every 2 °C between 30 and 50 °C [1]
⚠ If you missed marks here: A conclusion may never be more precise than the intervals it was measured at, so “the optimum is exactly 40 °C” overstates the evidence. “Narrow the intervals near the peak” is a standard improvement mark and is available in almost every investigation of this type.
(d)[4]
Explain the results at 20 °C and at 60 °C. Your answer must make clear that the two results have different causes.
Model Answer — 6(d)
at 20 °C the molecules have relatively little kinetic energy [1]
so effective collisions between hydrogen peroxide and the active site are infrequent, few complexes form and the rate is low — but the enzyme is undamaged [1]
at 60 °C the catalase has been denatured [1]
the shape of the active site has changed so hydrogen peroxide is no longer complementary to it, no complexes form and no oxygen is produced [1]
⚠ If you missed marks here: Both rows show a low rate and they mean completely different things. The instruction “must make clear that the causes are different” is the examiner telling you that a merged explanation will be capped. Describing the 20 °C enzyme as denatured is the specific error being hunted here.
Question 7 — Enzymes in a Washing Machine
Total: 10 marks
Biological washing powder contains a protease and a lipase. The manufacturer recommends a wash at 40 °C and states on the packet that a 90 °C wash gives no additional benefit on protein or grease stains. The packet also advises soaking bloodstained fabric in cold water before washing, rather than in hot water.
(a)[3]
Name the substrate of each enzyme and explain why two different enzymes are included rather than one.
Model Answer — 7(a)
protease acts on protein; lipase acts on fats and oils (lipids) [1]
each enzyme has an active site complementary in shape to only one type of substrate [1]
so a single enzyme could not break down both types of stain — this is specificity [1]
⚠ If you missed marks here: Both substrates are needed for the first mark, so naming only one loses it. The explanation must go through the active site: “because they do different jobs” restates the observation without giving the reason behind it.
(b)[3]
Explain why a 90 °C wash gives no additional benefit on these stains.
Model Answer — 7(b)
at 90 °C the enzymes are denatured [1]
the shape of the active site changes so protein and fat molecules are no longer complementary to it [1]
no enzyme–substrate complexes form, so the enzymes contribute nothing and the wash relies on the detergent alone [1]
⚠ If you missed marks here: The third mark is the one candidates miss: some cleaning still occurs, and it is not enzymic. Answers that predict a worse result at 90 °C than at 40 °C overlook the detergent entirely, which the packet has quietly told you about.
(c)[4]
Explain why bloodstained fabric should be soaked in cold water rather than hot water before washing, and suggest why the manufacturer does not simply recommend a cold wash for everything.
Model Answer — 7(c)
blood contains proteins; hot water denatures them, changing their shape so that they bind tightly into the fabric [1]
a denatured protein is harder for the protease to remove, so the stain becomes fixed [1]
a cold wash is not recommended generally because at low temperature the enzymes have little kinetic energy [1]
effective collisions between substrate and active site are infrequent, so the stains are removed very slowly; 40 °C is a compromise close to the optimum but below the temperature at which the enzymes denature [1]
⚠ If you missed marks here: This part asks you to apply denaturation to the stain rather than the enzyme, which is unfamiliar and worth reading twice. The final mark rewards seeing 40 °C as a deliberate compromise: warm enough for a useful rate, cool enough to leave the active sites intact.
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