Free samples
Sample questions, fully worked.
Attempt each one before revealing the answer. What you are judging is not the question, it is the explanation underneath it.
A competitive inhibitor is added to an enzyme-catalysed reaction that was previously running normally. Assuming substrate concentration can be increased without limit, what happens to the apparent Km and to Vmax?
- Km increases; Vmax is unchanged
- Km decreases; Vmax is unchanged
- Km is unchanged; Vmax decreases
- Km increases; Vmax decreases
- Both are unchanged
Correct answer: A, Km increases, Vmax unchanged
A competitive inhibitor binds the active site and competes directly with substrate. Because the competition is reversible, flooding the system with enough substrate will always out-compete the inhibitor, so the maximum achievable rate is unaffected, Vmax is unchanged. But you now need more substrate to reach half that maximum, so the apparent Km rises. Lower apparent affinity, same ceiling.
Why the others fail
BA falling Km means higher apparent affinity. Inhibition does the opposite.
CThis is the signature of a non-competitive inhibitor, which binds elsewhere and cannot be out-competed by substrate.
DBoth changing points towards uncompetitive inhibition, where Km and Vmax fall together.
EDescribes no inhibition at all.
At standard temperature and pressure, what volume is occupied by 8.0 g of oxygen gas? Take the molar volume at STP as 22.4 L/mol and the relative atomic mass of oxygen as 16.
- 2.8 L
- 5.6 L
- 11.2 L
- 22.4 L
- 44.8 L
Correct answer: B, 5.6 L
Oxygen gas is diatomic: O2, so its molar mass is 2 × 16 = 32 g/mol, not 16. That gives 8.0 / 32 = 0.25 mol. Volume is then 0.25 × 22.4 = 5.6 L. The whole question turns on one word in the stem: "oxygen gas" rather than "oxygen atoms".
Why the others fail
A2.8 L corresponds to 0.125 mol, a molar mass of 64, as if the gas were O4.
CThe intended trap. Using 16 g/mol gives 0.5 mol and 11.2 L, the answer for atomic oxygen, which is not what was asked.
D22.4 L is one full mole; that would need 32 g, four times what you were given.
E44.8 L is two moles, 64 g of O2.
A bag contains 4 red marbles and 6 blue marbles. Two marbles are drawn one after the other without replacement. What is the probability that both are red?
- 2/5
- 4/25
- 2/15
- 1/6
- 6/25
Correct answer: C, 2/15
The first draw is 4 red from 10 marbles: 4/10. Because the marble is not replaced, the second draw is from a changed bag, 3 red remaining out of 9 total: 3/9. Multiply: (4/10) × (3/9) = 12/90 = 2/15. The phrase "without replacement" is the entire question; miss it and every number shifts.
Why the others fail
A2/5 is just 4/10 · the first draw only, ignoring the second.
BThe main trap. (4/10)² = 4/25 is the answer with replacement, where the bag resets between draws.
D1/6 does not arise from this setup at all.
E6/25 comes from using the blue count somewhere it does not belong.
All questions are original, written from the published syllabus. They are not reproduced from any past paper.
The full bank holds 500 questions.
Every one explained to the same depth, weighted to the real paper.
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