Topic 5 has nine syllabus statements and, on a good day, you could recite all nine in three minutes. That is exactly why it is dangerous. Challenge papers do not ask you to recite — they hand you a table of oxygen volumes, a graph with two curves crossing, a badly designed method or an enzyme from a hot spring, and ask what it means. Nearly every mark lost in this topic comes from one of six habits: writing that the enzyme was killed; writing that it “changes shape” without naming the active site; explaining the pH curve with kinetic energy, which has nothing to do with pH; saying the enzyme was used up when it was the substrate; treating a cold enzyme as a denatured one; and reading a graph of time as though it were a graph of rate, which reverses every conclusion. Twelve traps, six walkthroughs, six lookalike pairs, a concept map and ten full practice questions below — all aimed at those six habits.
Twelve traps that cost marks on Topic 5 challenge papers. Every one is a sensible-sounding answer that mark schemes refuse.
Six challenge-level questions broken into steps. Try each step yourself before revealing the next one.
From 10 to 30 °C the volume is roughly doubling every ten degrees. A fall to 7 cm³ at 40 °C breaks that trend, and the repeat of 28 cm³ fits it perfectly — so the first 40 °C reading is anomalous. Exclude it from the mean, repeat the measurement, and offer a checkable cause: the water bath had not reached temperature, or the bung leaked.
The rise runs 10 to 40 °C, using the corrected value. Marks: molecules gain kinetic energy; there are more frequent effective collisions between substrate and active site; more enzyme–substrate complexes form per second, so more oxygen is released. Three sentences, three nouns doing the work.
At 60 °C the catalase is denatured; the shape of the active site changes so hydrogen peroxide is no longer complementary to it; no complexes form, so no oxygen is released. Note what you must not write: the enzyme was killed, or the enzyme was used up, or the molecules moved too fast to bind.
The highest measured value is at 40 °C, but with readings ten degrees apart the true optimum lies somewhere between 30 and 50 °C. A strong answer says so and suggests repeating at 2 °C intervals across that range. Claiming “the optimum is exactly 40 °C” asserts more than the data support.
Pepsin peaks at pH 2, trypsin at pH 8. Two enzymes that do the same job to the same substrate, with optima six pH units apart — which is already the answer to the assumption that enzymes work best at pH 7.
pH 7 is far from pepsin’s optimum of 2, so pepsin is denatured: the shape of its active site has changed and protein is no longer complementary to it, so no enzyme–substrate complexes form and no protein is digested. If your sentence contains the words kinetic energy, delete it — pH does not change how fast molecules move.
The stomach contains hydrochloric acid at about pH 2, which is pepsin’s optimum; the small intestine is made alkaline at about pH 8, which is trypsin’s. A single protease could not have its optimum in both places, and would be denatured in one of them. Two enzymes let protein digestion continue all the way along the gut.
Factual: do not say pepsin is “killed” by neutral pH. Logical: do not say trypsin is “better” than pepsin — each has the higher activity in its own range, and the data show trypsin at zero in acid just as clearly as they show pepsin at zero in neutral conditions.
The volume rises steeply at first, then the rise becomes progressively smaller, and after 150 s the volume is constant at 34 cm³. Quote at least one figure and the time at which the line becomes flat — a description with no numbers rarely gets both marks.
Either the hydrogen peroxide has all been used up, or the enzyme has stopped working (for example if it had been denatured by heat released in the reaction). The graph alone cannot separate them, because both produce a line that flattens. Recognising that the data are ambiguous is what the question is really testing.
Add a further volume of hydrogen peroxide to the same yeast. If oxygen is released again at about the original rate, the enzyme was intact and the substrate had run out. If no oxygen is released, the enzyme really had stopped working. Adding more yeast instead would not distinguish them, because fresh yeast would produce oxygen in either case.
24 cm³ ÷ 60 s = 0.4 cm³/s. Two things are being marked: the division, and the unit. In practice the reaction restarts when substrate is added, so the conclusion is that the hydrogen peroxide had been exhausted — a catalyst is not used up.
A protein folds according to the sequence of its amino acids; changing one amino acid can change how the chain folds; the folding determines the shape of the active site; if the active site is a different shape, lactose is no longer complementary to it, so no enzyme–substrate complex forms and no reaction is catalysed.
Only a small part of the surface forms the active site. If the replaced amino acid lies far from it, and the substitution does not disturb the overall folding, the shape of the active site is unchanged and the substrate still fits — so activity is normal.
Denaturation is the loss of shape of a properly folded enzyme, caused by heat or extreme pH. This enzyme was never the right shape: it was assembled incorrectly from the start. Using the word here shows the two ideas have been merged, and the mark scheme is looking for exactly that distinction.
The second altered enzyme is the control that rules out a general explanation such as “any change breaks a protein”. Because it works, the loss of function in the first version must be tied to the position of that particular amino acid — which is the strongest evidence in the stem.
Enzyme X peaks at about 75 °C and enzyme Y at about 37 °C. Both curves have the same characteristic shape — a climb, a peak and a steep collapse — and only their position on the temperature axis differs.
Below the crossing point enzyme Y is the more active; above it enzyme X is. At 37 °C Y is at 100 units while X is at only 18; at 55 °C X is at 55 while Y has fallen to 12. Neither enzyme is simply “better” — each is better in its own range, and quoting both ranges with data is what earns the third mark.
The claim contains something true: X is not denatured at temperatures that destroy Y, because its protein structure is held together more strongly. But it overstates the case — X falls from 100 units to 9 between 75 and 95 °C, which is denaturation happening in the data the student was given.
Enzyme X is very poor at 20 °C (4 units against Y’s 44) because at that temperature it is far to the left of its own optimum, with too little kinetic energy for frequent effective collisions. A shifted curve moves both ends, not just the top.
“A few drops” gives an unknown pH that also drifts as the reaction proceeds [1], so the values on the axis are not the values in the tubes. Buffer solutions of known pH, the same volume in each tube, fix this [1].
Room temperature drifts through a lesson and strongly affects enzyme activity [1], so a difference between tubes might have been caused by temperature rather than pH. All tubes should be held in a water bath at one fixed temperature [1]. This is the fault most likely to change the result, and it is the one candidates most often miss.
A single trial at each pH cannot reveal an anomaly and gives no measure of reliability [1]; repeat three times and take a mean [1]. The volumes and concentrations of starch and amylase are never stated [1]; equal measured volumes of the same solutions are needed, or a tube with more enzyme will finish sooner for reasons that have nothing to do with pH [1].
A tube containing boiled amylase with starch and buffer [1]. The amylase in it is denatured, so the iodine should stay blue-black — showing that the starch disappearing in the other tubes was caused by working amylase and not by anything else in the mixture [1]. Note the difference from part (a): those were control variables, this is a control experiment.
Six pairs that look almost identical and have different answers. The distinction is where the marks live.
Click each node to see how the nine syllabus statements connect into one picture.
Six real student answers. Find the fault before you reveal it.
Ten Cambridge-style challenge questions. Write your answer first, then reveal the model answer and the examiner’s notes.