Author: Philip Colls
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Enzyme booklet revision questions    NAME:
Q1. (a)     Describe the induced-fit model of enzyme action.
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(b)     A scientist investigated the hydrolysis of starch. He added amylase to a suspension of starch and measured the concentration of maltose in the reaction mixture at regular intervals. His results are shown in the graph below.
 
Determine the rate of the reaction at 10 minutes. Show how you obtained your answer.
Rate of reaction ____________________ mg dm^−3 min^−1
(2)
(c)     Explain the results shown in the graph.
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(d)     A quantitative Benedict’s test produces a colour whose intensity depends on the concentration of reducing sugar in a solution. A colorimeter can be used to measure the intensity of this colour.
The scientist used quantitative Benedict’s tests to produce a calibration curve of colorimeter reading against concentration of maltose.
Describe how the scientist would have produced the calibration curve and used it to obtain the results in the graph.
Do not include details of how to perform a Benedict’s test in your answer.
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____________________________________________________(3) (Total 9 marks)
Q2.
A technician investigated the effect of temperature on the rate of an enzyme-controlled reaction. At each temperature, he started the reaction using the same volume of substrate solution and the same volume of enzyme solution.
The figure below shows his results.

                Time after start of reaction / s
(a)     Give one other factor the technician would have controlled.
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(1)
(b)     Calculate the rate of reaction at 25 °C.

 
Answer ____________________
(2)
(c)     Describe and explain the differences between the two curves.
(5)
(Total 8 marks)
Q3.
(a)     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.

(5)
(b)     Scientists have investigated the effects of competitive and non-competitive inhibitors of the enzyme maltase.
Describe competitive and non-competitive inhibition of an enzyme.

(5)
(Total 10 marks)
Q4. A student investigated the effect of chewing on the digestion of starch in cooked wheat.
He devised a laboratory model of starch digestion in the human gut. This is the method he used.
1.      Volunteers chewed cooked wheat for a set time. The wheat had been cooked in boiling water.
2.      This chewed wheat was mixed with water, hydrochloric acid and a protein-digesting enzyme and left at 37 °C for 30 minutes.
3.      A buffer was then added to bring the pH to 6.0 and pancreatic amylase was added. This mixture was then left at 37 °C for 120 minutes.
4.      Samples of the mixture were removed at 0, 10, 20, 40, 60 and 120 minutes, and the concentration of reducing sugar in each sample was measured.
5.      Control experiments were carried out using cooked wheat that had been chopped up in a blender, not chewed.
(a)     What reducing sugar, or sugars, would you expect to be produced during chewing?
Give a reason for your answer.
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(2)
(b)     In this model of digestion in the human gut, what other enzyme is required for the complete digestion of starch?
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(c)     What was the purpose of step 2, in which samples were mixed with water, hydrochloric acid and pepsin?
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(d)     In the control experiments, cooked wheat was chopped up to copy the effect of chewing.
Suggest a more appropriate control experiment. Explain your suggestion.
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(e)     The figure below shows the student’s results.
 
                               Incubation time / minutes
Explain what these results suggest about the effect of chewing on the digestion of starch in wheat.
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___________________________________________________(3) (Total 9 marks)
Q5. A student investigated the effect of lipase concentration on the hydrolysis of lipids. He took a beaker containing a suspension of lipids. He placed a pH probe attached to a data logger into the beaker. After 5 minutes, he added the lipase solution. The data logger recorded the pH. The apparatus used is shown in the diagram below.
 
(a)     The student did not add a buffer to the lipase solution. Explain why.
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(b)     Give two variables the student would have controlled in this investigation.
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(c)     Give the suitable control for this investigation.
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The data logger recorded the pH. The graph below shows what happened after he added the lipase solution.
 
(d)     Draw a tangent on the graph and use it to calculate the rate of change at 5 minutes.

Rate of change at 5 minutes = _________________ pH minute^−1
(2)
(e)     Explain the results shown in the graph.
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(2)
(f)      The student repeated the experiment with a higher concentration of lipase solution.
Describe and explain the results you would expect him to get.
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(3)
(Total 11 marks)
Q6.
A student investigated the effect of substrate concentration on the initial rate of an enzyme-catalysed reaction.
She added 10 cm^3 of an enzyme solution to 10 cm^3 of substrate solutions of different concentrations. At 30-second intervals, she tested samples of each mixture for the presence of substrate.
•        A – in the absence of an inhibitor.
•        B – with a competitive inhibitor added to the substrate solution.
•        C – with a non-competitive inhibitor added to the substrate solution.
Her results are shown in the graph below.
 
(a)     Explain the results without inhibitor (curve A) shown in the graph.
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(2)
(b)     The graph shows that the maximum initial rate of reaction (V_max) when a competitive inhibitor was present (curve B) is different from that when a non-competitive inhibitor was present (curve C).
Explain this difference.
(4)
(c)     The Michaelis constant (K_m) is the substrate concentration at which the initial rate of reaction is half its maximum value (V_max).
How could you use the Michaelis constant to determine the type of inhibition occurring in an enzyme-catalysed reaction?
Use information from the graph to support your answer.
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____________________________________________________(1) (Total 7 marks)
Q7. A biochemist isolated a protease from a bacterium. He investigated the effect of temperature on the rate of hydrolysis of a protein by this protease. He measured the mass of protein hydrolysed in 5 minutes at each temperature.
The results are shown in the table below.

Temperature / °C

Mass of protein hydrolysed / g

Rate of hydrolysis /
_______________

5 | 0.48 |
10 | 1.11 |
15 | 1.23 |
20 | 1.05 |
30 | 0.78 |
45 | 0.12 |
(a)     Process the data in the table. Plot the processed data on the graph paper.
 
(4)

(b)     A student concluded from a graph of the data in the table that the bacterium lives at 15 °C.
Does the data support the student’s conclusion? Give reasons for your answer.
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(4)
(c)     Suggest two variables the biochemist controlled when investigating the effect of temperature on the rate of breakdown of a protein by the protease.
1. _________________________________________________________________
2. _________________________________________________________________
(1)
(Total 9 marks)
Q8.
(a)     Formation of an enzyme-substrate complex increases the rate of reaction.
Explain how.
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(b)     A scientist measured the rate of removal of amino acids from a polypeptide with and without an enzyme present. With the enzyme present, 578 amino acids were released per second. Without the enzyme, 3.0 × 10^–9 amino acids were released per second.
Calculate by how many times the rate of reaction is greater with the enzyme present. Give your answer in standard form.

Answer = ____________________ times faster
(2)
Another scientist investigated an enzyme that catalyses the following reaction.
ATP → ADP + Pi
The scientists set up two experiments, C and L.
Experiment C used
•        the enzyme
•        different concentrations of ATP.
Experiment L used
•        the enzyme
•        different concentrations of ATP
•        a sugar called lyxose.
The scientists measured the rate of reaction in each experiment. Their results are shown in the graph.

(c)     Calculate the rate of reaction of the enzyme activity with no lyxose at 2.5 mmol dm^–3 of ATP as a percentage of the maximum rate shown with lyxose.
Answer = ____________________ %
(2)
(d)     Lyxose binds to the enzyme.
Suggest a reason for the difference in the results shown in the graph with and without lyxose.
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(3)
(Total 9 marks)

Mark schemes
Q1.
(a)     1.      (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 max
(b)     1.      Tangent to curve drawn;
Tangent drawn at about 10 minutes
2.      Value in range of 8 to 11;
1 mark only for correct answer
2
(c)     1.      (Rate of) increase in concentration of maltose slows as substrate/starch is used up
OR
High initial rate as plenty of starch/substrate/more E-S complexes;
Reject ref. to amylase being used up
2.      No increase after 25 minutes/at end/levels off because no substrate/starch left;
Accept ‘little’
Ignore references to substrate a limiting factor
2
(d)     1.      Make/use maltose solutions of known/different concentrations 
(and carry out quantitative Benedict’s test on each);
2.      (Use colorimeter to) measure colour/colorimeter value of each 
solution and plot calibration curve/graph described;
Axes must be correct if axes mentioned, concentration on x-axis and colorimeter reading on y-axis
3.      Find concentration of sample from calibration curve;
3
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Q2.
(a)     Concentration of substrate solution / of enzyme solution / pH.
1
(b)     1.      2.5 / 0.04;
1 mark for correct value
2.      g dm^–3 minute^–1 / g dm^–3 s^–1;
1 mark for related unit
2
(c)     1.      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.
Allow converse for correct descriptions and explanations for curve at 25 °C
5
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Q3.
(a)     1.      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;
Accept idea that binding stresses the bonds so more easily broken
Do not award point 5 simply for any reference to E-S complex
5
(b)     1.      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;
5 max
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Q4.
(a)     1.      Maltose;
2.      Salivary amylase breaks down starch.
2
(b)     Maltase.
1
(c)     (Mimics / reproduces) effect of stomach.
1
(d)     1.      Add boiled saliva;
2.      Everything same as experiment but salivary amylase denatured.
2
(e)     1.      Some starch already digested when chewing / in mouth;
2.      Faster digestion of chewed starch;
3.      Same amount of digestion without chewing at end.
Accept use of values from graph
3
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Q5.
(a)     Student was measuring change in pH
OR
Buffer would maintain a constant pH.
1 max
(b)     1.      Volume of suspension of lipids;
2.      Concentration of suspension of lipids;
3.      Volume of lipase solution;
4.      Temperature;
2 max
(c)     Boiled lipase solution;
1
(d)     –0.34 = 2 marks
0.34 = 1 mark
2
(e)     1.      Fatty acids produced;
2.      Curve levels off as all substrate used up.
accept the lower pH inactivates / denatures the enzyme
2
(f)      1.      Faster fall in pH and levels off at same point;
2.      More enzyme = substrate complexes formed;
3.      Same amount of fatty acids produced / product
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Q6.
(a)     1.      Increases because more enzyme-substrate complexes formed;
Neutral; more collisions
2.      Levels off because all enzyme molecules involved in enzyme-substrate complexes (at a given time)
1. and 2. Accept ES
2. Reject enzymes are used up
OR
Levels off because no free active sites (at a given time)
OR
Levels off because enzyme (concentration) is limiting factor.
2
(b)     1.      Competitive inhibitor binds to active sites of enzyme but non-competitive inhibitor binds at allosteric site / away from active site;
2.      (Binding of) competitive inhibitor does not cause change in shape of active site but (binding of) non-competitive does (cause change in size of active site);
3.      So with competitive inhibitor, at high substrate concentrations (active) enzyme still available but with non-competitive inhibitor (active) enzymes no longer available;
4.      At higher substrate concentrations likelihood of enzyme-substrate collisions increases with competitive inhibitor but this is not possible with non-competitive inhibitor;
4
(c)     Reaction with non-competitive inhibitor has the same value of K_m as with no inhibitor / value is 5 (g dm^–3) / reaction with competitive inhibitor has higher K_m value than with no inhibitor / value is 7 (g dm^–3).
1
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Q7.
(a)     1.      IV on x axis and DV on y axis and both axes on linear scales;
2.      Axes labelled clearly and with correct units separated from variable by solidus or in brackets;
3.      All rates calculated correctly;
4.      Points plotted correctly and joined by ruled lines and no extrapolation;
4
(b)     Yes:
1.      Expect optimum temperature of enzyme to be same
OR
Similar to temperature where bacterium lives;
2.      Optimum temperature for enzyme (appears to be around) 15 °C;
No:
3.      Need data from more temperatures (between 10 °C and 20 °C);
4.      Data for only isolated enzyme
OR
Isolation may affect activity;
4
(c)     1.      Initial / starting substrate concentration
2.      Enzyme concentration
3.      pH.
Any 2 for 1 mark
1 max
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Q8.
(a)     1.      Reduces activation energy;
Accept ‘reduces E_a’.
2.      Due to bending bonds
OR
Without enzyme, very few substrates have sufficient energy for reaction;
Accept ‘Due to stress/pressure/tension on bonds’ OR ‘Due to weakening bonds’.
Ignore references to ‘breaking bonds’.
2
(b)     1.93 × 10^11;;
Allow 1 max for
578/3.0 × 10^–9
1.93 × 10^x when x ≠11
Correct answer with incorrect standard form e.g. 19.3 × 10^10
Accept any number of significant figures ≥2, if rounding correct (1.926^• × 10^11). Same principle applies to one max answers.
2
(c)     31.4;;
Allow 1 max for
0.44 and 1.4
32.8
33.1
30
29.3
Accept any number of significant figures ≥2, if rounding correct (31.4284714).
Same principle applies to 1 max answers.
32.8 = Both readings at 2.5 mmol dm^–3 (0.44/1.34)
33.1 = Both readings at 2.5 mmol dm^–3 (0.44/1.33)
30 = Incorrect reading for C (0.42/1.4)
29.3 = Incorrect reading for C (0.41/1.4)
2
(d)     1.      (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;
Accept ‘acts as a co-enzyme’
Accept description for E-S complexes.
3
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