The short answer
How enzymes work, what Km and Vmax mean, and how to read inhibition graphs the GAMSAT way.
Written and checked by GAMSAT tutors — not AI-generated.
Try the reasoning style
We treat forgetting as a failure — a lapse to be patched with reminders and records. Yet a mind that kept everything could not think; it would drown in the undifferentiated noise of every moment it had ever lived. To forget is not so much to lose information as to decide, mostly without our noticing, what was never worth keeping.
The author's argument relies most directly on which unstated assumption?
Pick an option to see how the tutor reasons to the answer — not just whether you were right.
Not quite — the answer is B.
Work backwards from the conclusion: a mind that ‘kept everything’ supposedly ‘could not think.’ That only follows if thinking means leaving most of experience out — so B is the premise the argument quietly rests on. A raises reliability, which the passage never weighs; C contradicts ‘mostly without our noticing’; D smuggles in a claim about intellect the passage never makes. The question rewards finding the hidden premise, not recalling a fact.
Enzymes are biological catalysts. They speed a reaction up by lowering its activation energy () — they do not change how much energy the reaction releases () or where the equilibrium sits. They just get you there faster. Most Section III enzyme questions are really graph-reading and reasoning questions, so focus on what the curves mean.
Km and Vmax in one line
Vmax is the maximum rate when the enzyme is saturated with substrate. Km is the substrate concentration that gives half of Vmax. A low Km means high affinity — the enzyme reaches half-speed with very little substrate.
What else changes the rate
Temperature and pH each have an optimum. Past it the enzyme denatures — the active site loses its shape — so the rate falls away instead of levelling off. Substrate concentration behaves differently: rate climbs until every active site is occupied, then plateaus at Vmax. On a graph that's the giveaway — a peak means temperature or pH, a plateau means saturation.
The trap students fall for
Enzymes do not make a non-spontaneous reaction happen, and they do not shift equilibrium. They lower for both the forward and reverse reaction equally, so they reach the same equilibrium — just sooner. If an option says an enzyme "increases the yield at equilibrium," it's wrong.
Competitive vs non-competitive inhibition
Competitive
- Binds the active site, competing with substrate
- Effect can be overcome by adding more substrate
- Km increases (apparent affinity falls)
- Vmax unchanged
Non-competitive
- Binds elsewhere (allosteric), changing the enzyme's shape
- Adding substrate does NOT relieve it
- Km unchanged
- Vmax decreases
Worked example
An enzyme is studied with and without drug X. With X, the reaction reaches the SAME maximum rate (Vmax) but needs a higher substrate concentration to get to half-speed. What kind of inhibitor is X?
Check yourself
On a rate-vs-substrate graph, a non-competitive inhibitor is added. What happens to the curve?
In the paper this arrives as one table with three or four questions on it, and no sentence telling you which inhibitor is which. Below is that shape. Everything you need is in the columns.
Passage· 3 questions on it
The following is adapted from a report on a bacterial enzyme. The enzyme and the two drugs are not ones you are expected to have met.
| [S] (mmol/L) | Rate, no drug (µmol L⁻¹ min⁻¹) | Rate + P1 | Rate + Q1 |
|---|---|---|---|
| 0.5 | 40 | 17 | 20 |
| 1.0 | 60 | 30 | 30 |
| 2.0 | 80 | 48 | 40 |
| 4.0 | 96 | 69 | 48 |
| 8.0 | 107 | 87 | 53 |
| 20.0 | 114 | 104 | 57 |
| 40.0 | 117 | 111 | 58 |
Q1 of 3 on this passage
Taken together, the three rate columns show that
Q2 of 3 on this passage
What is the value of Km for hydrolase H in the presence of drug P1?
Q3 of 3 on this passage
At [S] = 1.0 mmol/L the rate is 30 µmol L⁻¹ min⁻¹ with either drug. A colleague says this shows the two drugs are equally potent. The best objection is that
Key takeaways
- Enzymes lower activation energy; they don't change ΔG or the equilibrium position.
- Vmax = max rate at saturation; Km = substrate concentration at half-Vmax.
- Low Km = high affinity.
- Temperature and pH have optima (a peak, then denaturation); substrate concentration plateaus at Vmax.
- Competitive inhibitor: Km up, Vmax unchanged (beaten by more substrate).
- Non-competitive inhibitor: Vmax down, Km unchanged.
Practise this with real GAMSAT-style questions
A timed diagnostic, real practice questions with worked explanations, and your first essay marked on our rubric built from ACER's two published criteria.