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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 — Two Enzymes, Two Optima, One Crossing Point
Total: 12 marks
The same reaction is catalysed by enzyme G, taken from a mammal, and by enzyme H, taken from a bacterium living in a hot spring. Both were tested with the same substrate at the same concentration and pH.
temperature / °C
activity of G / units
activity of H / units
15
30
3
25
62
6
37
100
15
50
26
58
65
0
100
80
0
61
95
0
4
(a)[2]
State the optimum temperature of each enzyme, as shown by these data.
Model Answer — 1(a)
enzyme G: 37 °C [1]
enzyme H: 65 °C [1]
⚠ If you missed marks here: Read the optimum straight from the highest value in each column. The temptation is to give a single answer for “the optimum” as though enzymes shared one; two enzymes in the same table with optima 28 °C apart is the whole point of the question.
(b)[3]
The two curves cross between 37 and 50 °C. Explain what this means, using figures from the table.
Model Answer — 1(b)
below about 45 °C enzyme G is the more active and above it enzyme H is [1]
at 37 °C, G is at 100 units while H is at only 15 [1]
at 65 °C, H is at 100 units while G has fallen to 0 [1]
⚠ If you missed marks here: Both ranges are needed, each supported by a figure. A frequent answer here is “enzyme H is better”, which ignores half the table — neither enzyme is superior overall, each is more active within its own temperature range.
(c)[4]
Explain the activity of enzyme H at 15 °C and at 95 °C. Make clear that the two low values have different causes.
Model Answer — 1(c)
at 15 °C the molecules have little kinetic energy [1]
so effective collisions between substrate and active site are infrequent, few complexes form and the rate is low — the enzyme is undamaged [1]
at 95 °C enzyme H has been denatured [1]
the shape of the active site has changed so the substrate is no longer complementary to it and no complexes form [1]
⚠ If you missed marks here: Two values of 3 and 4 units that mean entirely different things. This is also the evidence against the claim that a heat-loving enzyme “cannot be denatured”: enzyme H falls from 100 units to 4 across the top of its own range.
(d)[3]
A student concludes: “These results show that enzyme H cannot be denatured, and that it would be the better enzyme to use in any industrial process.” Evaluate this conclusion.
Model Answer — 1(d)
the first claim is wrong: enzyme H falls from 100 units at 65 °C to 4 units at 95 °C, which shows it is denatured above its optimum [1]
what is true is that H is not denatured at temperatures that destroy G, because its structure is held together more strongly [1]
the second claim is also too broad: at 15 and 25 °C enzyme H is far less active than G, so it would be the poorer choice for a low-temperature process [1]
⚠ If you missed marks here: Evaluate means weigh both sides before judging. Answers that simply agree, or simply write “wrong”, cap themselves at one mark. The idea worth carrying away is that a shifted curve moves both ends: a high optimum is a real disadvantage in the cold.
Question 2 — Product Against Time, at Three Enzyme Concentrations
Total: 12 marks
Three flasks each contain 40 cm³ of the same hydrogen peroxide solution at 25 °C. Flask 1 receives 1 cm³ of catalase solution, flask 2 receives 2 cm³ and flask 3 receives 4 cm³. The total volume of oxygen collected was recorded.
time / s
flask 1 / cm³
flask 2 / cm³
flask 3 / cm³
0
0
0
0
30
9
17
31
60
16
29
42
120
27
40
44
180
35
44
44
240
40
44
44
300
44
44
44
(a)[3]
Calculate the mean rate of oxygen production in the first 30 s for flask 1 and for flask 3, and describe the relationship between the volume of catalase added and the initial rate.
Model Answer — 2(a)
flask 1: 9 ÷ 30 = 0.30 cm³/s [1]
flask 3: 31 ÷ 30 = 1.03 cm³/s (accept 1.0) [1]
the greater the volume of catalase added, the higher the initial rate [1]
⚠ If you missed marks here: Two of these three marks are for arithmetic with a unit, and the unit is dropped constantly. Note that the relationship is not exactly proportional in the data — describing it as “increases with” is safer than claiming it doubles exactly.
(b)[3]
All three flasks eventually produce 44 cm³ of oxygen. Explain why the final volume is the same in every flask.
Model Answer — 2(b)
the final volume of oxygen depends on the amount of substrate, and all three flasks contained the same 40 cm³ of hydrogen peroxide [1]
the catalase is a catalyst, so it changes the rate of the reaction but not the amount of product formed [1]
the enzyme is not changed by the reaction, so even 1 cm³ of it eventually converts all the substrate — it simply takes longer [1]
⚠ If you missed marks here: This is one of the sharpest tests of whether the definition of a catalyst is understood. “More enzyme means more product” is intuitive and wrong: more enzyme means the same product, sooner.
(c)[3]
Explain why the line for flask 3 becomes horizontal after 120 s.
Model Answer — 2(c)
all the hydrogen peroxide has been broken down [1]
so no substrate remains to bind to the active sites, and no further product can be formed [1]
the catalase is unchanged: adding fresh hydrogen peroxide would restart the reaction at the original rate [1]
⚠ If you missed marks here: A graph of product against time that flattens means the substrate has run out. “The enzyme was used up” contradicts part (b) of this same question, and “the enzyme was denatured” is the explanation for a different graph entirely — one with temperature on the horizontal axis.
(d)[3]
A student wants to compare the three flasks fairly. Explain why the initial rate over the first 30 s is a better measure than the time taken to reach 44 cm³, and state two variables that must be kept constant.
Model Answer — 2(d)
over the first 30 s the substrate concentration is still close to its starting value, so the rate reflects the enzyme concentration being tested rather than a dwindling substrate supply [1]
variable 1: the volume and concentration of hydrogen peroxide, e.g. 40 cm³ of the same solution in each flask [1]
variable 2: the temperature, held at 25 °C in a water bath (accept pH held constant with a buffer) [1]
⚠ If you missed marks here: Control variables must be named individually and quantified; “keep everything else the same” earns nothing. The first mark is about validity: a measurement taken after the substrate has largely gone tells you more about the starting quantity than about the variable under test.
Question 3 — Three pH Curves on One Pair of Axes
Total: 12 marks
The activities of three enzymes were measured across a range of pH values, each at its own optimum temperature. All values are expressed as a percentage of that enzyme’s maximum activity.
pH
pepsin / %
salivary amylase / %
trypsin / %
1
36
0
0
2
100
0
0
3
58
0
0
5
4
22
0
6
0
78
9
7
0
100
48
8
0
76
100
9
0
20
47
11
0
0
0
(a)[3]
State the optimum pH of each enzyme.
Model Answer — 3(a)
pepsin: pH 2 [1]
salivary amylase: pH 7 [1]
trypsin: pH 8 [1]
⚠ If you missed marks here: Three straightforward marks that are lost only by candidates applying the belief that all enzymes prefer pH 7. Two of these three enzymes contradict it, and pepsin does so by five whole pH units.
(b)[3]
Describe and explain the shape of the salivary amylase curve between pH 5 and pH 9.
Model Answer — 3(b)
the activity rises from 22% at pH 5 to a peak of 100% at pH 7, then falls to 20% at pH 9 — a roughly symmetrical curve [1]
as the pH approaches the optimum the shape of the active site becomes increasingly complementary to starch, so the substrate fits better [1]
more enzyme–substrate complexes therefore form and the rate rises; either side of the optimum the fit is poorer, fewer complexes form and the rate falls [1]
⚠ If you missed marks here: Describe and explain means figures first, then mechanism. The mechanism must be shape and fit throughout: any mention of kinetic energy here is refused, because pH does not change how fast molecules move.
(c)[3]
Explain why the pepsin curve is completely flat at zero from pH 6 to pH 11, rather than continuing to fall gradually.
Model Answer — 3(c)
pH 6 and above is far from pepsin’s optimum of pH 2, so pepsin is denatured [1]
the shape of the active site has changed so protein is no longer complementary to it at all [1]
no enzyme–substrate complexes can form, so the activity is zero rather than merely reduced — and the change is permanent [1]
⚠ If you missed marks here: The pH curve has two regions and they need different language: a graded region near the optimum, where the fit is simply poorer, and an extreme region where the enzyme has been destroyed. A row of zeroes is a result in its own right, not a gap in the data.
(d)[3]
Compare the general shape of a pH curve with that of a temperature curve, and explain why they differ.
Model Answer — 3(d)
a pH curve is roughly symmetrical about its optimum, whereas a temperature curve rises gradually and then falls very steeply [1]
both sides of the pH curve are caused by the same thing — a change in the shape and fit of the active site, ending in denaturation at the extremes [1]
the temperature curve has two different causes: increasing kinetic energy on the rising side and denaturation on the falling side, which is why the two halves are so unlike each other [1]
⚠ If you missed marks here: This part exists to check that the two graphs have not been merged into one memorised explanation. If you can say why the temperature curve is lopsided, you will never again write about kinetic energy in a pH answer.
Question 4 — From Times to Rates
Total: 12 marks
Amylase and starch were mixed at six temperatures and the time for the iodine test to stop giving a blue-black colour was recorded. The pH was held at 7 with a buffer and all solutions were equilibrated before mixing.
temperature / °C
time for starch to disappear / s
rate = 1000 ÷ time / s⁻¹
10
500
2.0
20
250
4.0
30
125
8.0
40
100
?
50
200
5.0
60
no colour change after 30 minutes
?
(a)[3]
Calculate the missing rate at 40 °C, state the rate at 60 °C, and explain why the reaction is described as faster at 40 °C than at 20 °C even though the recorded number is smaller.
Model Answer — 4(a)
at 40 °C: 1000 ÷ 100 = 10.0 s⁻¹ [1]
at 60 °C the rate is 0, since the reaction never went to completion [1]
a shorter time means a faster reaction: time and rate are inversely related, so 100 s at 40 °C represents a rate two and a half times that of 250 s at 20 °C [1]
⚠ If you missed marks here: The inverse relationship is the hinge of the whole question. Reading the times as though larger meant faster reverses every conclusion in this table — it would make 10 °C the best temperature and 40 °C one of the worst.
(b)[3]
Explain the entry “no colour change after 30 minutes” at 60 °C. State which colour the iodine remained, and why.
Model Answer — 4(b)
the iodine remained blue-black throughout [1]
because the starch was never digested — the amylase had been denatured at 60 °C [1]
the shape of the active site had changed, so starch was no longer complementary to it and no enzyme–substrate complexes formed [1]
⚠ If you missed marks here: “No colour change” is deliberately ambiguous and catches candidates who assume it means the reaction finished instantly. Always ask which colour it stayed: blue-black means the starch is still there, orange-brown means it has gone.
(c)[3]
Using the rate values, describe the pattern between 10 and 30 °C and explain it.
Model Answer — 4(c)
the rate doubles for every 10 °C rise: 2.0, 4.0 then 8.0 s⁻¹ [1]
the molecules gain kinetic energy and move faster, so effective collisions between starch and the active site become more frequent [1]
more enzyme–substrate complexes form each second, so the starch is digested more quickly [1]
⚠ If you missed marks here: The first mark is for spotting the numerical pattern, not merely saying “the rate increases”. Quoting the three values is what turns a description into evidence, and it is exactly the sort of detail that separates a grade 8 answer from a grade 6 one.
(d)[3]
The student concludes that the optimum temperature of this amylase is exactly 40 °C. Assess this conclusion and suggest how the investigation could be improved.
Model Answer — 4(d)
the highest rate measured was at 40 °C, so this is a reasonable estimate [1]
but the readings are 10 °C apart, so the true optimum could lie anywhere between 30 and 50 °C — “exactly 40” claims more precision than the method allows [1]
improvement: repeat at 2 °C intervals between 30 and 50 °C, with three repeats at each temperature and a mean taken [1]
⚠ If you missed marks here: A conclusion may never be more precise than the intervals it was measured at. Notice that the mark scheme credits the student first — a good evaluation says what is reasonable about a claim before saying what is overstated.
Question 5 — Heat First, Test Later
Total: 10 marks
Samples of one enzyme were each held at a different temperature for ten minutes. Every sample was then brought to 30 °C and assayed with the same substrate under identical conditions.
pre-treatment temperature / °C
activity when assayed at 30 °C / %
0
99
20
100
40
98
55
71
65
9
80
0
100
0
(a)[3]
Explain why every sample was assayed at 30 °C rather than at the temperature at which it had been held.
Model Answer — 5(a)
it keeps the assay conditions constant, so temperature during the measurement cannot affect the result [1]
the only difference between the samples is the pre-treatment they received [1]
so any difference in activity must be caused by lasting damage to the enzyme, not by the molecules moving faster or more slowly during the test [1]
⚠ If you missed marks here: This is a question about experimental design as much as about enzymes. If each sample had been assayed at its own temperature, a low reading could mean either damage or simply little kinetic energy — and the experiment would be unable to tell them apart.
(b)[4]
Describe and explain the pattern shown by these results.
Model Answer — 5(b)
pre-treatment up to 40 °C has almost no effect — activity stays at 98–100% [1]
above 40 °C activity falls sharply, reaching 0% by 80 °C [1]
the fall occurs because the enzyme is denatured: the protein chain unfolds and the shape of the active site changes [1]
the substrate is no longer complementary to the active site, so no enzyme–substrate complexes form — and the loss is permanent, since the samples were cooled before testing and did not recover [1]
⚠ If you missed marks here: The 0 °C row is the control that proves cold causes no damage, and it is often overlooked. Notice that this table cannot show the optimum: it measures how much enzyme survived, not how fast it works at each temperature.
(c)[3]
A student says these results show the optimum temperature of the enzyme is 20 °C. Explain why this conclusion cannot be drawn from these data, and describe what experiment would be needed instead.
Model Answer — 5(c)
every sample was assayed at 30 °C, so the table shows how much enzyme survived each pre-treatment, not how fast the enzyme works at each temperature [1]
the 20 °C sample was simply undamaged, like the 0 and 40 °C samples, so no optimum can be identified from these figures [1]
to find the optimum, the enzyme should be mixed with substrate and the rate measured at each temperature, using a range of temperatures with repeats [1]
⚠ If you missed marks here: This is the hardest part of the paper because it asks what the data actually measure. The clue is in the method: an experiment in which every measurement is made under identical conditions cannot possibly reveal how those conditions affect the rate.
Question 6 — How Much Catalase Is in a Potato?
Total: 12 marks
Discs of equal size were cut with a cork borer from four plant tissues and from a boiled potato. Each set of discs was added to 25 cm³ of hydrogen peroxide at 25 °C, and the volume of oxygen collected in 60 s was recorded. Three repeats were taken and the mean calculated.
tissue
mean oxygen in 60 s / cm³
range of repeats / cm³
potato
21.0
20–22
celery
6.0
5–7
apple
13.5
13–14
carrot
9.0
8–10
boiled potato
0.3
0–1
(a)[3]
Calculate how many times greater the mean rate for potato is than for celery, express the celery value as a percentage of the potato value, and state what the results suggest about the tissues.
Model Answer — 6(a)
21.0 ÷ 6.0 = 3.5 times greater [1]
(6.0 ÷ 21.0) × 100 = 28.6% (accept 29%) [1]
the tissues contain different concentrations of catalase, with potato containing the most of those tested [1]
⚠ If you missed marks here: Show the working for both calculations; a bare number with no method loses the mark if it is wrong. The third mark is for a conclusion in terms of enzyme concentration — saying “potato is better” describes the result without interpreting it.
(b)[3]
Explain the purpose of the boiled potato, and explain its result.
Model Answer — 6(b)
it acts as a control experiment, showing that the oxygen released by the other tissues was produced by the enzyme rather than by anything else in the plant material [1]
boiling has denatured the catalase, so the shape of its active site has changed and hydrogen peroxide is no longer complementary to it [1]
no enzyme–substrate complexes form; the small volume of 0.3 cm³ is consistent with a little hydrogen peroxide breaking down on its own [1]
⚠ If you missed marks here: The third mark rewards taking the 0.3 cm³ seriously instead of rounding it to nothing. Note also the distinction being tested: this is a control experiment, an extra tube designed to rule out a rival explanation, not a control variable.
(c)[3]
State three variables that had to be controlled in this investigation, and for each explain briefly why.
Model Answer — 6(c)
the size and number of discs, cut with a cork borer and a ruler, so that the surface area and mass of tissue exposed are the same [1]
the volume and concentration of hydrogen peroxide, e.g. 25 cm³ of the same solution, so the amount of substrate available is the same [1]
the temperature, held at 25 °C in a water bath (accept pH held with a buffer, or the time of collection), because it affects the rate of an enzyme-catalysed reaction [1]
⚠ If you missed marks here: Each mark requires a named variable and a reason. “Keep everything else the same” and “to make it a fair test” collect nothing, however confidently they are written. Quantities strengthen the answer at almost no cost in time.
(d)[3]
The ranges of the repeats are given in the last column. Explain what these tell you about the results, and suggest one further improvement to the investigation.
Model Answer — 6(d)
the repeats agree closely — for example apple varies by only 1 cm³ — so the results are reliable [1]
reliability is not the same as validity: close agreement shows the method was consistent, not that it measured the right thing [1]
improvement: repeat with tissue from several different potatoes, celery sticks and so on, because individual specimens vary in enzyme concentration (accept measuring the initial rate over a shorter interval) [1]
⚠ If you missed marks here: The distinction between reliability and validity is examined directly here and is worth learning as a pair of sentences. The improvement mark needs something specific to this investigation; “do more repeats” is too weak when three consistent repeats have already been taken.
Question 7 — One Graph, Four Claims
Total: 10 marks
A student investigates a protease using a suspension of powdered milk, which is cloudy. As the protein is digested the suspension becomes clear, and the time taken for a cross drawn on paper to become visible through the tube is recorded. Results at pH 7: 20 °C, 180 s; 30 °C, 90 s; 40 °C, 45 s; 50 °C, 60 s; 60 °C, the suspension was still cloudy after 20 minutes.
(a)[3]
Explain why the suspension becomes clear, and state why the time taken can be used as a measure of enzyme activity.
Model Answer — 7(a)
the protease breaks down the large, insoluble protein molecules that scatter light into small soluble products [1]
so the suspension becomes clear and the cross becomes visible [1]
a shorter time means a faster reaction, so time is inversely related to the rate and can be converted using 1 ÷ time [1]
⚠ If you missed marks here: The third mark is the one that matters for the rest of the question. Candidates who forget the inverse relationship read this table backwards and identify 20 °C, the slowest condition, as the optimum.
(b)[4]
Four students each make a claim about these results. State whether each is supported by the data, giving a reason. Claim 1: the optimum is 40 °C. Claim 2: the enzyme works twice as fast at 30 °C as at 20 °C. Claim 3: at 60 °C the reaction was too slow to see. Claim 4: at 50 °C the enzyme has begun to denature.
Model Answer — 7(b)
Claim 1 — partly supported: 40 °C gives the shortest time, but with 10 °C intervals the true optimum lies between 30 and 50 °C [1]
Claim 2 — supported: 180 s falls to 90 s, so the rate has exactly doubled [1]
Claim 3 — not supported: the enzyme has been denatured at 60 °C, so the reaction is not merely slow but has stopped altogether [1]
Claim 4 — supported: the time rises from 45 s to 60 s, so the rate has fallen despite the higher temperature, which indicates denaturation has begun [1]
⚠ If you missed marks here: Four short judgements, each needing a reason. The one most often missed is claim 3: “too slow to see” sounds cautious but is a different biological statement from “denatured”, and the collapse from 60 s to nothing at all is far too abrupt for a simple slowing.
(c)[3]
Suggest two reasons why this method might give less precise results than collecting a gas, and state one advantage it has.
Model Answer — 7(c)
judging when the cross “becomes visible” is subjective, and different observers would choose different moments [1]
the end-point is gradual rather than sharp, so the reaction time cannot be pinpointed (accept: the cloudiness of the milk suspension may vary between tubes) [1]
advantage: it needs no gas-collecting apparatus, is quick to set up, and works for a reaction that produces no gas at all [1]
⚠ If you missed marks here: Evaluating a method means being specific about where the uncertainty enters. “Human error” is never credited; “the moment at which the cross becomes visible is a matter of judgement” is the same idea stated in a way an examiner can award.
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