A free GAMSAT Section III data pack: three units, with the reasoning shown
Three original Section III units - an enzyme curve, a titration and a free-body diagram - with six questions and the full reasoning for every option.
75
Section III questions in 150 min
2 min
average per question
3
original units in this pack
6
questions, reasoning shown
Section III does not test what you know about enzymes. It hands you an enzyme, some data about it and about two minutes, then asks whether you can read the data. That is a different skill from the one most candidates spend the final fortnight practising, and it is the one that transfers: 75 questions in 150 minutes, no calculator, and a paper that counts twice in your overall.
The three units below are practice at exactly that. Read each stimulus once, answer both questions at roughly two minutes each, then read the reasoning - which is the part that does the work.
Read every stimulus in the same order
The cheapest habit in Section III is fixing the order in which you take a figure in. Axes, units and scale before the shape of the curve. The shape stated in one plain sentence before you look at the options. A prediction of what must be true before four written options colonise your reading. Candidates who jump straight to the options end up evaluating each one from scratch - four times the work, at two minutes a question.
- 1
Axes first
Quantities, units and scale before the curve.
- 2
Say the shape
One plain sentence: what rises, what flattens.
- 3
Predict
What must be true — before reading any option.
- 4
Then eliminate
Against the stimulus, not against memory.
What this pack is - and what it is not
Every stimulus below is original, built in-house for this pack (AI-assisted drafting, human-reviewed) - not ACER material, and not a reproduction of any published question. That is deliberate: it means each unit can be built around one reasoning move worth practising, and it means nothing here is a copyright problem for you to worry about. The science is standard first-year content. The difficulty is in the reading, which is where the difficulty sits on the real paper too.
About these units
One warning before you start. In Section I the commonest wrong option is truer than the passage. In Section III it is true science the stimulus never gave you - product inhibition, a buffer, an indicator, a safety margin. Real chemistry and real biology, simply absent from the data in front of you. Train yourself to notice the moment you supply a fact the stimulus did not.
The tempting wrong option is real science you were never given
Unit 1 - an enzyme that stops speeding up
Unit 1 · 2 questions
Original · written in-houseAn enzyme measured at fixed enzyme concentration
An enzyme converts a single substrate to product. Initial reaction rate was measured across a range of substrate concentrations, at constant enzyme concentration, pH and temperature. The curve below is the result.
Two point mutants of the same enzyme, Variant A and Variant B, were then measured under identical conditions.
| Enzyme | Vmax (µmol min⁻¹) | Km (mM) |
|---|---|---|
| Wild type | 1.00 | 1.4 |
| Variant A | 1.00 | 4.2 |
| Variant B | 0.45 | 1.4 |
1. The rate curve flattens at high substrate concentration chiefly because
- A.the substrate is being consumed faster than it can be replaced.
- B.essentially every active site is occupied at any instant.
- C.accumulating product begins to inhibit the enzyme.
- D.the enzyme is denatured by prolonged exposure to substrate.
2. Which statement is best supported by the table?
- A.Variant A binds substrate less tightly than the wild type but has the same maximum capacity.
- B.Variant A is the faster enzyme at every substrate concentration.
- C.Variant B binds substrate more tightly than the wild type.
- D.Variant B would reach the wild type's maximum rate if given enough substrate.
Question 1 asks why the curve flattens, and the shape answers it. Rate climbs steeply while substrate is scarce, then stops responding to more substrate at all. The only thing that stops responding is a fixed number of active sites, occupied essentially all of the time (B). A misreads the experiment - these are initial rates measured at each concentration, not one reaction running itself down. C and D are the shape this pack keeps returning to: real mechanisms, product inhibition and denaturation, neither mentioned in the stimulus nor required by the curve. A plateau is what saturation looks like; it is not evidence of damage.
Question 2 moves to the table, and its whole point is that Km and Vmax move independently. Variant A has the wild type's ceiling (Vmax 1.00) but needs three times the substrate to reach half of it (Km 4.2 against 1.4) - weaker binding at unchanged maximum capacity (A). B is tempting until you test it at the bottom of the range, where a higher Km makes A the slower enzyme, so "faster at every concentration" fails exactly where the difference lives. C reads Variant B's unchanged Km as a change. D is the true-in-general trap: more substrate does raise B's rate, but only towards 0.45, because Vmax is a ceiling and no amount of substrate lifts a ceiling.
Answer key
Unit 2 - a titration with one number worth extracting
Unit 2 · 2 questions
Original · written in-house20.0 mL of hydrochloric acid, titrated with 0.100 mol L⁻¹ sodium hydroxide
A 20.0 mL sample of hydrochloric acid of unknown concentration is titrated with sodium hydroxide of concentration 0.100 mol L⁻¹, and the pH is recorded after each addition. The equivalence point is marked on the curve.
Also recorded: the pH rises by about one unit between 0 and 20.0 mL of added base, and by more than four units between 24.0 mL and 26.0 mL.
3. The concentration of the hydrochloric acid is closest to
- A.0.025 mol L⁻¹
- B.0.080 mol L⁻¹
- C.0.125 mol L⁻¹
- D.0.250 mol L⁻¹
4. The best explanation for the difference in pH change between those two intervals is that
- A.the neutralisation reaction speeds up as the equivalence point approaches.
- B.pH is logarithmic, and near equivalence a small further addition removes most of the acid still present.
- C.sodium chloride formed early in the titration buffers the solution.
- D.the indicator changes colour across that interval.
Question 3 is arithmetic wearing chemistry's coat, and the curve hands you the only number you need: equivalence at 25.0 mL. That is where moles of base added equal moles of acid present, so 25.0 mL × 0.100 mol L⁻¹ = 2.50 mmol, and those 2.50 mmol were sitting in 20.0 mL of acid - 2.50 ÷ 20.0 = 0.125 mol L⁻¹ (C). Option B is the same two volumes divided the wrong way round. A reports the mole count as though it were a concentration. D doubles, for a reaction that is one to one. Not one wrong option needs bad chemistry; each needs one careless step, which is what the question is for.
Question 4 asks why the curve has that shape at all. Before equivalence you are removing acid from a large excess, so each addition changes the hydrogen-ion concentration by a small factor; within a millilitre or two of equivalence you are removing almost all of what remains, and because pH is a logarithm that same small addition moves it by a very large factor (B). A confuses reaction rate with equilibrium position - the neutralisation is fast throughout. C invents a buffer: sodium chloride is the salt of a strong acid and a strong base and buffers nothing. D borrows an indicator the stimulus never mentions, and an indicator reports a pH change rather than causing one.
Answer key
Unit 3 - a block that is not accelerating
Unit 3 · 2 questions
Original · written in-houseA 4.0 kg block pushed across a level bench
A constant horizontal force F is applied to a 4.0 kg block resting on a level bench. After a brief acceleration the block settles to a constant velocity of 0.60 m s⁻¹, which it keeps for as long as F is applied. Take the frictional force on the moving block as constant at 6.0 N.
The diagram shows every force acting on the block: the normal force N, the weight mg, the applied force F and friction f.
5. While the block moves at constant velocity,
- A.F must be greater than f, because the block is moving.
- B.F and f are equal in magnitude and opposite in direction.
- C.N must be greater than mg, because the block is being pushed forward.
- D.the net force on the block is forward but small.
6. F is now doubled, with everything else unchanged. The block's acceleration is closest to
- A.0 m s⁻²
- B.1.5 m s⁻²
- C.3.0 m s⁻²
- D.4.5 m s⁻²
Question 5 is Newton's first law wearing a picture. The block is moving, and moving is what makes candidates reach for option A: surely a push that produces motion must beat friction. It had to once, to start the block. It does not have to now - constant velocity means zero acceleration, which means zero net force, which means F and f are equal in size and opposite in direction (B). C is a real force claim answering a question nobody asked, and it is false here besides: the push is horizontal, so the bench still supports exactly the weight. D contradicts the word constant.
Question 6 pays you for having answered Question 5. Because the block was moving at constant velocity, F was 6.0 N - equal to the friction. Double it and 12.0 N acts against an unchanged 6.0 N, so the net force is 6.0 N and the acceleration is 6.0 ÷ 4.0 = 1.5 m s⁻² (B). A treats the old state as permanent. C divides the whole applied force by the mass, with friction quietly dropped. D adds friction to the push instead of subtracting it - the sign error a rushed free-body diagram invites.
Answer key
The five shapes these six questions used
Six questions, eighteen wrong options, five recurring shapes. Naming the shape is faster than re-deriving each option from scratch, and the list is short enough to carry into the exam.
| Shape | How it appeared here | The tell |
|---|---|---|
| Science the stimulus never gave you | "product inhibition", "the indicator" | A real mechanism, absent from the data |
| True in general, capped here | "enough substrate would reach the same rate" | A number in the table sets a ceiling |
| Right values, wrong operation | "0.080 mol L⁻¹" | The two correct numbers, combined backwards |
| The intuitive story | "it is moving, so the push must exceed friction" | Everyday sense standing in for a law |
| A real claim, unasked | "the normal force exceeds the weight" | Answers a question the stem did not pose |
Learn the shape, not the topic
How to run the pack
Do not do all three in one sitting. One unit, cold, at about two minutes a question; then read the reasoning and write down the shape of whichever wrong option tempted you - not the topic, the shape. A fortnight of that produces a list of about five entries, and it is a better revision document than another chapter of notes. If you want more units of this kind, our free 15-minute practice test puts six questions from both reasoning papers in front of you at exam pace with the reasoning worked through after each answer - and inside Aptavia, timed practice with a stored explanation on every question is free to use.
- 1
One unit, cold
~2 minutes a question, no second look.
- 2
Answer, then read
The reasoning, option by option.
- 3
Name the shape
Of whichever wrong option tempted you.
- 4
Next unit tomorrow
Spaced beats binged for this skill.
Reading a figure well is not a talent. It is an order of operations, practised until it is boring.
The units in this pack were written in-house for demonstration (AI-assisted drafting, human-reviewed) and are not ACER material. GAMSAT® is a registered trademark of ACER. ACER is not affiliated with, and does not endorse, Aptavia.
Key takeaways
- Section III supplies the science — the marks are in reading the data you are given, not in recalling more of it.
- Fix the order on every figure: axes and units, then the shape in one sentence, then a prediction. Options last.
- A plateau is saturation, not damage — and Km and Vmax move independently of each other.
- At the equivalence point, moles of base added equal moles of acid present: one line of arithmetic gives the concentration.
- Constant velocity means zero net force, however fast the object is moving.
- Five distractor shapes covered all eighteen wrong options in this pack — name the shape rather than re-deriving each option.
Frequently asked questions
Are the questions in this pack real GAMSAT questions?
No. All three units were written in-house for this pack (AI-assisted drafting, human-reviewed). They are not ACER material and not reproductions of any published question - each one was built around a specific reasoning move.
How long should a GAMSAT Section III question take?
About two minutes on average: the paper is 75 questions in 150 minutes, sat without a calculator. Work each unit here at roughly that pace rather than untimed.
Do I need to memorise enzyme kinetics for Section III?
The stimulus supplies the data you reason from - Unit 1's curve and table give both Vmax and Km. What transfers is reading a saturating curve and noticing that Km and Vmax move independently, not recalling values.
What is the commonest wrong-answer shape in Section III?
True science the stimulus never gave you - product inhibition, a buffer, an indicator. Five shapes covered all eighteen wrong options in this pack, and naming the shape is faster than re-deriving each option.
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