When a question asks how an enzyme pattern affects a larger process, describe the pattern with figures, explain it using enzyme shape or collisions, then say what that does to the whole process. The third step is the one students leave out.
This lesson sits in integrated biological reasoning and builds on the transport chain in the previous lesson. Confirm the exact wording for your exam year on the Cambridge subject page.
What are the three steps?
- Describe: name the optimum, give two values, and describe how the rate changes either side.
- Explain: below the optimum, molecules move slowly and collide less often. Above it, the active site changes shape (denaturation) and the substrate no longer fits.
- Apply: say what the enzyme normally does in the larger process, such as breaking starch into sugar in digestion, and what happens to that process at the unfavourable condition.
Worked example (invented data)
A student investigates an enzyme that breaks down starch. The figures are invented for this lesson.
| Temperature (°C) | Starch broken down (mg per minute) |
|---|---|
| 20 | 4 |
| 30 | 8 |
| 40 | 15 |
| 50 | 3 |
Question: Describe the pattern, explain it, and suggest what would happen to digestion if a person’s gut stayed at 50 °C.
Step 1, describe: the rate rises from 4 to 15 mg per minute between 20 °C and 40 °C, then drops to 3 mg per minute at 50 °C. The optimum is near 40 °C among these values.
Step 2, quantify the rise: from 20 °C to 30 °C the rate went 4 to 8, so it doubled (an increase of 100%). From 30 °C to 40 °C it rose by 7 mg per minute, so (15 − 8) ÷ 8 = 0.875, an increase of 87.5%.
Step 3, quantify the fall: from 40 °C to 50 °C the rate fell by 12 mg per minute. 12 ÷ 15 = 0.80, a decrease of 80%.
Step 4, explain: up to 40 °C, more kinetic energy means more collisions between enzyme and starch. At 50 °C the active site loses its shape, so starch no longer fits.
Step 5, apply: if the enzyme were denatured in the gut, starch would be broken down much more slowly. Less glucose would be absorbed into the blood, so less would reach cells for respiration.
The mistake to watch for
Mistaken answer: “At 50 °C the enzyme is killed, and enzymes get used up so the rate falls.”
Enzymes are proteins, not living things, so they cannot be killed. They are not used up either, because they are catalysts and are unchanged at the end. The rate falls because the active site is denatured.
A better answer: “The active site changes shape, so the substrate cannot bind, and fewer reactions happen.”
Check yourself
1. An enzyme works at 6 units at 25 °C and 12 units at 35 °C. Calculate the percentage increase.
Show answer
(12 − 6) ÷ 6 = 1.0, so 100% increase.
2. The rate at the optimum is 20 units. At a higher temperature it falls to 4 units. Calculate the percentage decrease.
Show answer
(20 − 4) ÷ 20 = 0.80, so 80% decrease.
3. An enzyme from the stomach has an optimum of pH 2, but the small intestine is about pH 8. Explain why the stomach enzyme would work poorly in the small intestine.
Show answer
At pH 8 the shape of the active site would change (denature), so the protein substrate would not fit and the rate would be very low.
Where this leads next
Continue with linking inheritance and variation without overclaiming. Mixed questions are in the practice set, and the scientific investigation critic helps with data checking.
Moving from describing a graph to explaining its effect is a skill our teachers practise one-to-one in Co-ordinated Sciences tuition.