BIOLOGY WITH HAMZA · MDCAT REVISION
Enzymes
Name: ____________________ Date: ______________
Recall first. Check second. Return to the ideas you missed.
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Learning targets
- Explain how enzymes lower activation energy without changing equilibrium.
- Interpret effects of temperature, pH and substrate concentration.
- Compare competitive and non-competitive inhibition.
Core notes
- Enzymes lower activation energy through an alternative pathway.
- Active sites depend on three-dimensional chemistry and induced fit.
- Saturation limits rate at high substrate concentration.
- Competitive inhibition differs from non-competitive inhibition in site and kinetic effect.
Read, cover and redraw
Substrate collides with the active site.
Specific interactions form an enzyme–substrate complex.
The transition state is stabilised.
Products leave; the enzyme is reusable.
Written recall
1. Why does an enzyme not change the equilibrium position of a reversible reaction?
2. Which observation supports competitive inhibition?
3. Can equal rates imply different mechanisms?
4. Why include an untreated control?
5. Does doubling enzyme double Km?
Apply the idea
1. Which tube has the highest average product-formation rate over the stated first 20 seconds?
| Tube | Product formed in 20 s / μmol |
|---|---|
| P | 8 |
| Q | 14 |
| R | 10 |
- Q
- P
- R
- All have the same rate
My answer: _____ Reason: ____________________________________
2. An inhibitor lowers rate at low substrate concentration, but the inhibited reaction approaches the same maximum rate as the control at very high substrate concentration. Which model fits best?
- Complete irreversible enzyme destruction
- Competitive inhibition
- Removal of all enzyme
- Lowering enzyme concentration to zero
My answer: _____ Reason: ____________________________________
3. An enzyme reaction slows at 5°C, then returns to its earlier rate when warmed to 30°C. What is the strongest conclusion?
- All enzyme molecules were permanently denatured at 5°C
- Warming created a new substrate
- Cooling reduced effective collisions without permanently inactivating the enzyme
- The enzyme was consumed during cooling
My answer: _____ Reason: ____________________________________
Answer key
Cover this section while attempting the questions.
Written recall
1. It lowers the activation barrier for both forward and reverse directions, so equilibrium is reached faster without changing the relative energies of reactants and products.
2. The inhibited reaction can approach the original maximum rate when substrate concentration becomes sufficiently high.
3. Yes. Different kinetic parameters can produce the same rate at one concentration.
4. It establishes activity under the same conditions without inhibitor.
5. No. In the simple model Vmax doubles while Km stays constant.
Applied questions
1. A — Rate is product formed divided by time. The rates are 0.4, 0.7 and 0.5 μmol s⁻¹ for P, Q and R respectively.
2. B — In the ideal competitive model, excess substrate reduces the inhibitor’s effect at the active site, so the same maximum rate can be approached.
3. C — Recovery on warming supports a reversible temperature effect on molecular motion and productive encounters. It does not support permanent denaturation at the lower temperature.