Enzymes

Questions............................................................................................................1

HKALE 1995 Biology Paper I...............................................................................1




HKALE 1999 Biology Paper I B2.........................................................................5


HKALE 1996 Biology Paper I.............................................................................10



Questions

HKALE 1995 Biology Paper IHKALE 1995 Biology Paper I8. (a) An experiment was conducted to investigate the effect of temperature on the

activities of amylases I and II. These enzymes were obtained from the same type

of tissue of vertebrate animals I and II respectively. Similar pH, enzyme

concentration and substrate concentration were used. The experimental data

obtained are recorded in the following table :

Temperature/℃ Time for complete digestion of 1 unit of substrate / minute

Amylase I Amylase II

10 2.0 5.0

15 1.6 3.5

20 1.5 2.5

25 1.6 2.0

30 2.0 1.6

35 4.0 1.4

40 10.0 1.5

(i) Plot a graph to demonstrate the effect of temperature on the activities of

amylases I and II.

(5 marks)

(ii) Explain the results for amylase I. (2 marks)

(iii) What are the optimum temperatures for amylase I and amylase II ? What

thermoregulatory ability would the data for amylase I suggest for vertebrate

animal I ? (2 marks)

(b) The graph below shows the effects of two chemicals, A and B, on the activity of

amylase I. Similar pH, temperature, enzyme concentration and chemical

concentration were used in the experiment.

(i) Compare and contrast the effects of chemical A and chemical B on the

activity of amylase I.

(3 marks)

(ii) Explain how each chemical, A and B, exerts its effect on amylase I. (3 marks)

(Total : 15 marks)

Suggested Solution…

HKALE 1996 Biology Paper IHKALE 1996 Biology Paper I10. (a) A student performed an experiment to monitor the progress of a simple enzyme-

catalysed reaction involving one substrate and one product. He prepared replicate

reaction tubes. In each tube, the concentration of only one reaction component,

either the substrate or product, was measured. The results of three selected tubes

are shown as curves A, B and C below :

(i) Identify which measured component, substrate or product, is represented by

each curve.

(1 marks)

(ii) It was realized that the reaction mixture in one of the three tubes had been

wrongly prepared. This resulted in a different reaction condition.

(1) State which curve represents the result of an error in the preparation.

Give a reason for your choice. Suggest two possible mistakes in the

student's preparation of this reaction mixture. (3 marks)

(2) What evidence shows that the other two curves represent identical

reaction conditions ? (2 marks)

(iii) With reference to curve C,

(1) calculate the rate of enzyme reaction at the 3rd minute, given that 100%

concentration is equivalent to 100 mmoles of the measured component

(Show the readings you take from the graph in your calculation.) ; (2

marks)

(2) compare and explain the difference in enzyme reaction rate at the 3rd

and 15th minute. (2 marks)

(iv) Curve A will finally level off and will not reach 100 %. Explain this

phenomenon. (11 marks)

(b) Explain what is meant by the 'active site' of an enzyme. (1 marks)

Total : 15 marks


HKALE 1997 Biology Paper IHKALE 1997 Biology Paper I2. Compare and contrast the characteristics of competitive and non-competitive inhibitors

on enzyme activity.

(4 marks)


HKALE 1998 Biology Paper IHKALE 1998 Biology Paper I6. Give one application of enzyme and explain its effect. (2 marks)


HKALE 1999 Biology Paper I B2HKALE 1999 Biology Paper I B212. The following reaction is catalysed by the enzyme succinate dehydrogenase :

(a) State three parameters that can be used to determine the rate of this reaction. (3 marks)

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(b) The graph below shows the effect of succinate concentration on the rate of reaction under

optimum pH and temperature :

The curve flattens out at X. Give two explanations for this observation based on the

mechanism of enzymatic reaction. (2 marks)

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(c) Malate and succinate are metabolic intermediates in the Krebs cycle. They have the

following structural formulae :

(i) Suggest what effect malate might have on the rate of reaction catalysed by

succinate dehydrogenase. (1 mark)

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(ii) The positions of malate, succinate and fumarate in the Krebs cycle are shown in the

following flowchart :

Malate can play a role in regulating the rate of reactions in the Krebs cycle. Explain

the possible mechanism by which malate can achieve such a control. (4 marks)

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Total : 10 marks

HKALE 1995 Biology Paper IHKALE 1995 Biology Paper I8. (a) (i) Refer to the graph shown below

－ title of graph

－ properly scaled and fully labelled X and Y axes with correct units

1－ 1 curve labelled or keyed

－ correctly plotted points (correct conversion for 1/t) for the 2

curves (1 marks for each curve) (if students plot time VS

temp : title () properly scaled and fully labelled axis with

correct unit ().)

Deduct mark if 1-2 points are plotted wrongly, and if

extra points are shown. No mark if > 2 points are plotted

wrongly.

(Bonus : mark if student can distinguish 'reaction rate' and 'enzyme

activity'. Activities of enzyme can be expressed as arbitrary

unit- of substrate over time, to be indicated on Y axis.)

Effect of temperature on the activities of amylase I and II

(correct conversion for 1/t. For marker's reference in checking accuracy of

points)

Temperature /C Rate of react on 1/t / min-1

Amylase I Amylase II10 0.5 0.215 0.625 0.2920 0.67 0.425 0.625 0.530 0.5 0.62535 0.25 0.7140 0.1 0.67

(Possible Max. 5)

(ii) As temperature increases from 10 C to 20 C, the kinetic

energy () of the molecules increases and there is increased

collision () between enzyme and

1

substrate molecules (to form the enzyme substrate complex

which subsequently breaks down to form the products). As

temperature increases further, the active site of the enzyme is

progressively destroyed / the enzyme undergoes thermal

denaturation (1), and the reaction rate declines.

1

(2)

(iii) The optimum temperature for amylase I is around 20C ()

while that for amylase II is the value shown on the candidates

graph ().

1

It is possible that vertebrate animal I is poikilothermic / cold

blooded animal.

1

(2)

OR

The thermoregulatory ability cannot be deduced owing to

insufficient data.

Deduct mark for spelling

mistake.

(b) (i) Both chemicals A and B inhibit the activity of amylase I.

1

The inhibitory activity of A is competitive / can be overcome

by increasing the substrate concentration but the inhibitory

activity of B is non-competitive / cannot be overcome by

increasing substrate concentration.

2

(3)

OR

Contrast the effect at high substrate concentration (1) and at low

substrate concentration (1).

(ii) Chemical A competes with the substrate for the active site ()

on amylase I and produced inhibitory effect at low substrate

concentration (). Chemical A is structurally similar to the

substrate ().

max. 1

However, when substrate concentration increases, more

substrate molecules are available to compete for the active

site on amylase I.

1

Thus inhibitory effect is overcome at high substrate

concentration.

1

Chemical B binds with enzyme in such a way that it reduces

the catalytic properties of the enzyme.

(Total : 15 marks)

(Possible max. 15 marks)back to top

HKALE 1996 Biology Paper IHKALE 1996 Biology Paper I10. (a) (i) Curve A = product ()

Curve B = product ()

Curve C = substrate ()

(1)

(ii) (1) Curve B () because it indicates a slower rate of reaction ()

when compared to the reaction rates indicated by curves A

and C ()

Possible errors are : (any two, 1mark each)

(1)

－ lower enzyme concentration was used / less enzyme

was added

－ lower substrate concentration was used / less substrate

was added

－ possible contaminant(s) / inhibitor(s) was introduced

into the reaction mixture (any other acceptable answer, do not

accept temperature difference as an answer)

(2)

(3)

(2) The remaining two curves / curves A and C show a

corresponding conversion to the product from the substrate

(1) according to (with respect to) their % concentration changes (1) at

various time points ().

(2)

OR Rate of formation of the product as shown in curve A is the

same as the rate of disappearance of the substrate as shown in curve C

(2) throughout the time points ().

(iii) (1) 8 mmoles min -1(1) N.B. No mark for no unit

Readings (1)

(2)

(2) At 3rd minute : faster rate of enzyme reaction (1) / or at 15th

minute : slower rate of enzyme reaction

(1)

At 3rd minute, substrate concentration is higher () / at 15th

minute, substrate concentration is lower, at 3rd minute

chance of collision between enzyme and substrate molecules is higher

() / at 15th minute chance of collision between enzyme and substrate

molecules is less, at 15th minute rate is limited by product inhibition ().

(1 )

(2)

(iv) The reaction is reversible (). End product effect pushes the

reaction backwards and equilibrium is reached (1)

(1)

(b) It is a special region on the enzyme molecule with a specific

configuration () that can bind with the substrate to form an enzyme － substrate complex (). Reaction occurs at the active site ()

(1)

(Total : 15 marks)

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HKALE 1997 Biology Paper IHKALE 1997 Biology Paper I2. Both lower the activity of enzymes () and their actions are reversible () / no

permanent damage to the enzyme.

1

Competitive inhibitor － inhibitor molecules have similar structure as the

substrate molecule (), compete with the substrate for the active sites on the

enzyme molecule (), inhibition effect overcome by increased substrate

concentration ().

1

Non-competitive inhibitor molecules attack the enzyme molecule not at the

active site (), but changes the conformation of the enzyme () prevents the

active site from catalyzing reactions, inhibition effect not overcome by

increased substrate concentration ().

1

Q2 = 4 marks

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HKALE 1998 Biology Paper IHKALE 1998 Biology Paper I6. meat tenderizer (1)

( accept correct alternatives)

contains protease () which breaks down protein in the meat(), renders the

meat more tender()

(2)

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Enzymes

Documents

low substrate

high substrate

increased

15th minute

3rd minute

substrate

competitive

suggested