enzymes

Question 1

Formation of an enzyme-substrate complex increases the rate of reaction.

Explain how.

Answer 1

Reduces activation energy;

Accept ‘reduces Ea’.

2. Due to bending bonds

OR

Without enzyme, very few substrates have sufficient energy for reaction;

Question 2

Lyxose binds to the enzyme.

Suggest a reason for the difference in the results shown in the graph with and without lyxose.

Answer 2

(Binding) alters the tertiary structure of the enzyme ;

Max 1

if lyxose acting as an inhibitor

OR if answer linked to lower rate of reaction

OR if lyxose used an energy source/respiratory substrate

2. (This causes) active site to change (shape);
3. (So) More (successful) E-S complexes form (per minute)

OR

E-S complexes form more quickly

OR

Further lowers activation energy;

Question 3

The scientists grew both strains of fungi on dishes kept at 30 °C. Keeping the dishes at a temperature of 15 °C would affect the results. Use your knowledge of kinetic energy to explain why.

Answer 3

Molecules move at slower speeds;(use slower speed no moves less).

2 diffusion or

2. Decreases rate of diffusion;

4 enzymes.

OR

3. Molecules move at slower speed;
4. Fewer collisions between enzymes and substrates / fewer enzyme-substrate complexes formed;

Question 4

The scientists gave their results as ratios. Explain the advantage of giving the results of this investigation as a ratio.

Answer 4

Allows comparison;
2. Different amounts of fungus added / fungus is different size at start;

Question 5

The catalase produced by the K30 strain of the fungus is mainly an extracellular enzyme. This means that the fungus secretes catalase from its cells into the jelly in the Petri dish.

Describe and explain the evidence from the investigation which shows that the catalase is an extracellular enzyme.

Answer 5

Colourless zone around fungus / colourless zone outside fungus;

2. No fungus growing here / must be enzyme here;

Question 6

In humans, the enzyme maltase breaks down maltose to glucose.
This takes place at normal body temperature.

Explain why maltase:

• only breaks down maltose
• allows this reaction to take place at normal body temperature.

Answer 6

Tertiary structure / 3D shape of enzyme (means);

Accept references to active site

2. Active site complementary to maltose / substrate / maltose fits into active site / active site and substrate fit like a lock and key;

Idea of shapes fitting together

3. Description of induced fit;
4. Enzyme is a catalyst / lowers activation energy / energy required for reaction;

Accept “provides alternative pathway for the reaction at a lower energy level”

5. By forming enzyme-substrate complex;

Question 7

Scientists have investigated the effects of competitive and non-competitive inhibitors of the enzyme maltase.

Describe competitive and non-competitive inhibition of an enzyme.

Answer 7

Inhibitors reduce binding of enzyme to substrate / prevent formation of ES complex;

Max 3 if only one type of inhibition dealt with. Accept maltase and maltose as examples of enzyme and substrate (and others)

Only once, for either inhibitor

(Competitive inhibition),

2. Inhibitor similar shape (idea) to substrate;
3. (Binds) in to active site (of enzyme);

Accept allows max rate of reaction to be reached / max product will eventually be formed

Accept complementary to active site

4. (Inhibition) can be overcome by more substrate;

(Non-competitive inhibition),

5. Inhibitor binds to site on enzyme other than active site;
6. Prevents formation of active site / changes (shape of) active site;

Accept does not allow max rate of reaction to be reached / max product will not be formed

7. Cannot be overcome by adding more substrate;

Question 8

Describe the induced-fit model of enzyme action.

Answer 8

before reaction) active site not complementary to/does
not fit substrate;

2. Shape of active site changes as substrate binds/as
   enzyme-substrate complex forms;

Note. Points 1 and 2 may be made in one statement and ‘complementary’ introduced at any point.

Points 1&2 – active site mentioned once applies for both points

Point 2 – Ignore references to how shape change is caused

3. Stressing/distorting/bending bonds (in substrate leading to reaction);

2

Question 9

explain differences in rate of reaction between line at 37 and line at 25

Answer 9

Initial rate of reaction faster at 37 °C;

2. Because more kinetic energy;
3. So more E–S collisions / more E–S complexes formed;
4. Graph reaches plateau at 37 °C;
5. Because all substrate used up.

Question 10

Suggest how the amino acids at positions 35 and 52 are held close together to form the active site.

Answer 10

idea that amino acid chain folds / tertiary structure;
named bond holding tertiary structure e.g. ionic disulphide hydrogen;