Answers · Enzymes

Temperature and enzymes — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. TQ1 · Practice

Model answer. d. The end point is the first drop that stays orange-brown, the colour of iodine on its own: starch is no longer detected.

A full answer includes

I chose d.

Also accept

Watch for

“The end point is when the iodine turns blue-black.” — Blue-black shows that starch is still there.

2. TQ2 · Practice

Model answer. Relative rate = 1000 ÷ 150 = 6.666… = 6.7 (to one decimal place).

Checked automatically

Relative rate (no unit): 6.7 (2 significant figures; within ±0.001)
- Common wrong answer 0.15 : That is 150 ÷ 1000. Divide 1000 by the time: 1000 ÷ 150.
- Common wrong answer 400 : Use the time in seconds, 150 s, not minutes.
- Common wrong answer 6.67 : Round to one decimal place: 6.7.
- Common wrong answer 6.6 : 6.666… rounds up to 6.7.

A full answer includes

1000 ÷ 150 = 6.7

Also accept

Watch for

“Relative rate = time ÷ 1000.” — A shorter time must give a bigger rate, so the time goes on the bottom: 1000 ÷ time.

3. TQ3 · Practice

Model answer. a, b, d and e: the volume of starch, the concentration of amylase, the pH and the time between samples are controlled. The temperature (c) is the independent variable: it is changed on purpose.

A full answer includes

I chose a, b, d and e.

Also accept

a, b, d, e

Watch for

“The temperature must be kept the same.” — Temperature is what you change on purpose; everything else is kept the same.

4. TQ4 · Practice

Model answer. b. Without amylase the starch is not broken down in 5 minutes, so it is the amylase that breaks down the starch in the other tubes.

A full answer includes

I chose b.

Also accept

Watch for

“The control shows the optimum temperature.” — The control has no enzyme: it shows what happens without amylase.

5. TQ5 · Practice

Model answer. As the temperature rises, the amylase and starch molecules gain kinetic energy and move faster. They collide more often, so there are more effective collisions in which starch fits the active site. More enzyme–substrate complexes form every second, so more maltose is made per second and the rate increases.

A full answer includes

Higher temperature: enzyme and substrate molecules have more kinetic energy and move faster.
They collide more often, so there are more effective collisions (substrate fitting the active site).
More enzyme–substrate complexes form each second, so starch is broken down faster.

Also accept

more kinetic energy, more frequent effective collisions, more complexes

Watch for

“Heat gives the enzyme more energy to work harder.” — Describe the molecules: they move faster and collide more often, forming more enzyme–substrate complexes.

6. TQ6 · Practice

Model answer. At 70 °C, well above the optimum, the amylase is denatured: bonds holding its shape break and the active site changes shape. Starch no longer fits the active site, so no enzyme–substrate complexes form. The starch is not broken down, so every drop turns iodine blue-black.

A full answer includes

At 70 °C, far above the optimum, the amylase is denatured.
The active site changes shape.
Starch no longer fits, so no enzyme–substrate complexes form and the starch is not broken down: it stays, so iodine stays blue-black.

Also accept

denatured; active site shape changed; starch does not fit; starch stays

Watch for

“The amylase was killed by the heat.” — Enzymes are not alive; heat changes the shape of the active site.

7. TQ7 · Practice

Model answer. (a) 35 °C: 1000 ÷ 80 = 12.5. 55 °C: 1000 ÷ 125 = 8.0. (b) 45 °C: the shortest time, 50 s, so the highest relative rate (20). It is the best tested temperature; the true optimum lies near it.

Checked automatically

Relative rate at 35 °C (no unit): 12.5 (3 significant figures; within ±0.001)
- Common wrong answer 0.08 : That is 80 ÷ 1000. Divide 1000 by 80.
Relative rate at 55 °C (no unit): 8.0 (2 significant figures; within ±0.001)
- Common wrong answer 0.125 : That is 125 ÷ 1000. Divide 1000 by 125.
Best tested temperature: 45 °C (within ±0.1 °C)
- Common wrong answer 25 : At 25 °C the time was longest, so the rate was lowest.
- Common wrong answer 55 : At 55 °C the rate had already fallen.

A full answer includes

(a) 35 °C: 1000 ÷ 80 = 12.5; 55 °C: 1000 ÷ 125 = 8.0.
(b) 45 °C (shortest time, highest rate).

Also accept

(a) 12.5 and 8.0; (b) 45 °C

Watch for

“The best temperature is the one with the longest time.” — A long time means a slow reaction; the best temperature has the shortest time.

8. TQ8 · Practice

Options as numbered on the pupil sheet: 1 — The molecules gain kinetic energy; 2 — They move faster; 3 — They collide more often; 4 — More collisions are effective: starch fits the active site; 5 — More enzyme–substrate complexes form each second; 6 — Starch is broken down faster.

Model answer. The molecules gain kinetic energy, so they move faster and collide more often. More of the collisions are effective, with starch fitting the active site, so more enzyme–substrate complexes form each second and the rate increases.

A full answer includes

More kinetic energy → molecules move faster → more frequent collisions → more effective collisions (substrate fits the active site) → more enzyme–substrate complexes each second → faster rate.

Also accept

kinetic energy, faster, more collisions, more effective collisions, more complexes, faster rate

Watch for

“Heat makes the enzyme bigger, so more starch fits.” — Below the optimum heat changes the speed of the molecules, not the size of the active site.

9. TT1 · Apply

Model answer. (a) At 37 °C the rate is lower than at 75 °C: the molecules have less kinetic energy, collide less often and form fewer enzyme–substrate complexes. (b) At 95 °C, far above its optimum, the protease is denatured: its active site changes shape, the protein substrate no longer fits, and almost no complexes form.

A full answer includes

(a) Lower at 37 °C: molecules move more slowly and collide less often, so fewer complexes form.
(b) At 95 °C, far above its optimum, the enzyme is denatured: its active site changes shape and the substrate no longer fits.

Also accept

(a) slower at 37 °C, fewer collisions; (b) denatured at 95 °C, active site shape changed

Watch for

“All enzymes are denatured above 40 °C.” — Each enzyme has its own optimum; these bacteria’s enzymes keep their shape at 75 °C.

10. TT2 · Apply

Model answer. (a) In the freezer the enzyme and substrate molecules have very little kinetic energy, so they move slowly and collide rarely: the spoiling reactions are very slow. The enzymes are not denatured, so their active sites keep their shape and they work faster again when the beans thaw. (b) Two minutes in boiling water denatures the enzymes: their active sites change shape, the substrates no longer fit and no enzyme–substrate complexes form. Denaturation is permanent, so cooling, freezing and thawing do not bring the enzymes back.

A full answer includes

(a) Cold gives the molecules little kinetic energy, so they collide rarely and the enzymes work very slowly; but their active sites keep their shape, so they work again when the beans thaw.
(b) Boiling denatures the enzymes: their active sites change shape for good, so the substrates no longer fit, and cooling or thawing does not restore them.

Also accept

(a) cold slows enzymes (less kinetic energy, fewer collisions) but does not denature them; (b) boiling denatures them for good

Watch for

“Freezing kills the enzymes.” — Enzymes are not alive; cold only slows them down, and they work again when warmed.
“Blanching cleans the bacteria off, so the beans keep.” — The question is about the beans’ own enzymes: boiling denatures them.