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1.
The diagrams below represent an enzyme, its substrate and molecules U and V.
Molecule U
Substrate
Enzyme
Molecule V
(a)
Identify whether molecules U or V may be a non-competitive inhibitor.
(1 mark)
(b)
How does a non-competitive inhibitor affect enzyme activity?
(4 marks)
(c)
(d)
2.
A student tries to increase the substrate concentration to reduce the inhibition by the non-competitive inhibitor. Will it
work? Explain your answer.
(2 marks)
A decrease in temperature decreases the rate of the enzyme-controlled reaction. Explain briefly.
(4 marks)
Biological washing powder is an improved type of clothes-washing detergent, with enzymes added to the washing
powder.
(a)
Suggest two types of enzyme that could be present in the biological washing powder.
(2 marks)
(b)
Give two advantages of adding enzymes into washing powder.
(c)
There is a factor that would greatly affect the cleaning power of biological washing powder. Name the factor and suggest
how we can ensure the proper working of the biological washing powder when washing clothes.
(d)
(2 marks)
(2 marks)
The enzymes in biological washing powder are usually packed inside small granules, and are only released when they
come into contact with water. Suggest one advantage of this.
(1 mark)
3.
In an investigation, the rate at which starch was broken down by the enzyme amylase was measured in solutions with
different concentrations of starch. The experiment was carried out at 25ºC and was repeated with mercury ions added to
the starch solutions. The graph shows the results.
11
10
A
9
8
Rate of reaction
(arbitrary units)
B
7
6
5
4
3
2
1
0
0
1
2
3
4
5
6
7
8
9 10
Starch concentration (arbitrary units)
11
12
(a)
What are the products in this enzyme reaction?
(1 mark)
(b)
State which curves represent the results of mercury ion added and non-mercury ion added solutions respectively. Explain
your answers.
(4 marks)
(c)
Why do both curves level off beyond starch concentration of 9 arbitrary units?
(2 marks)
(d)
Sketch on the graph to show the amylase activity when the experiment was carried out at 37ºC in the absence of mercury
ions.
(1 mark)
4.
In order to kill the bacteria during sterilisation, we can attack their metabolic pathways through interfering with their
enzyme activities.
(a)
(i) Most bacteria are temperature and pH sensitive. Suggest how we can kill these bacteria.
(2 marks)
(ii) Explain the mechanisms behind.
(b)
(2 marks)
Besides temperature and pH treatment, we can kill bacteria by inhibiting those enzymes essential in some important
metabolic pathways. The following shows a reaction in the process of respiration in bacteria.
succinate
Enzyme S
fumarate
Malonate is structurally and chemically similar to succinate. Suggest how the addition of malonate can kill bacteria.
(4 marks)
5.
In the production of biofuels, cellulose is first broken down into simple sugars with cellulase. Sugars are then fermented to
produce ethanol, which can be used as biofuels. The following shows the process of biofuel production.
cellulose
cellulase
Reaction 1
glucose
zymase
ethanol
Reaction 2
(a)
Where does the cellulose come from?
(1 mark)
(b)
In which organism can we find zymase that ferments glucose?
(1 mark)
(c)
The optimum pH of reaction 1 is 5, suggest which herbivore’s organ can provide such pH condition.
(1 mark)
(d)
What will be the effect of an increased amount of cellulose on the rate of reaction 2? Please explain.
(2 marks)
(e)
There is heat produced during the process of fermentation. Explain the effects of heat on reaction 2.
(4 marks)
(f)
Suggest another application of the enzyme cellulase.
6.
Read the passage below and answer the questions that follow.
(1 mark)
Toxic soymilk
Soymilk is a healthy drink and contains a lot of nutrients. However, homemade soymilk may
be toxic if it is not heated properly. In soybean, it contains trypsin inhibitor, which inhibits the action
of trypsin – a pancreatic protease. Trypsin inhibitor is a protein that binds to trypsin to form a stable
complex. This inhibits the digestion and hence the absorption of proteins. In order to drink the
healthy soymilk, it is advised to boil the soymilk first.
(a)
(i) To which kind of inhibitor does trypsin inhibitor belong?
(1 mark)
(ii) How does trypsin inhibitor affect the action of trypsin?
(3 marks)
(b)
What would be the optimum temperature and pH of trypsin?
(2 marks)
(c)
Suggest why soybeans contain toxins like trypsin inhibitor.
(1 mark)
(d)
How can boiling minimise the effect of trypsin inhibitor?
(2 marks)
7.
The following experiment was carried out to study the properties of a washing powder containing a biologically active
substance X.
A
B
film with a layer
of protein coat
washing
powder
solution
boiled
washing
powder
solution
After two hours
protein
coat
absent
protein
coat
present
(a)
Describe the action of substance X on the protein coat.
(2 marks)
(b)
What is the effect of boiling on the biological washing powder solution?
(1 mark)
(c)
Name substance X. Explain why this washing powder can remove egg stain from clothes.
(3 marks)
(d)
It is not advisable to use this washing powder with water at temperatures above 60ºC. Explain why.
(1 mark)
(e)
It is also suggested not to use this washing powder together with bleaching solution. Explain briefly.
(2 marks)
8.
Tom carried out an experiment to investigate the action of an enzyme. The set-up was placed in a water bath at 40°C.
At the beginning
3 minutes later
2 cm3 of
enzyme
solution was
added and
stirred
10 cm3 of
egg white
cloudy
30 minutes later
cloudy
clear
(a)
Name the enzyme.
(1 mark)
(b)
Explain why the solution became clear.
(1 mark)
(c)
Suggest one method to speed up the action.
(1 mark)
(d)
Predict and explain the result if the test tube was placed in a water bath at
(i) 80°C.
(2 marks)
(ii) 2°C.
(2 marks)
9.
Helen carried out an experiment to investigate the action of amylase on starch.
Boiling tube A
Boiling tube B
Boiling tube C
dialysis
tubing
distille
d water
water bath
at 20ºC
Boiling tube
Content inside the dialysis tubing
pH of the content
A
starch + amylase
2
B
starch + amylase
7.5
C
starch + amylase
11
(a)
State the action of amylase on starch.
(2 marks)
(b)
State the tube in which the action was the fastest. Explain your answer.
(2 marks)
(c)
Design a control to avoid the argument that the result is caused by other non-pH factors.
(1 mark)
(d)
Suggest with reason a method to speed up the action.
(2 marks)
(e)
Predict, if the amylase can act on starch, whether the distilled water contains the end product. Explain your answer.
(2 marks)
10.
Read the passage below and answer the questions that follow.
Tenderisation of meat
Enzymes have been utilised by man for thousands of years. One example is the use of enzymes
by some Africans. Much of the meat consumed in these areas is very fibrous and tough. This type of
meat would need to be cooked for hours to break down the fibrous protein, and so make the meat
edible. The people in these areas have learned that after covering the meat with fruit extracts from the
paw-paw tree for several hours, the meat becomes tenderer. They also found that after cooking, the
process of tenderisation stopped and the meat is no longer tender.
(a)
According to the paragraph, name the substrate on which the enzyme in paw-paw works.
(1 mark)
(b)
State the chemical nature of the enzyme.
(1 mark)
(c)
Describe the action of the enzyme on the specific substrate.
(2 marks)
(d)
Explain why cooking stops the tenderisation by enzymes.
(2 marks)
(e)
Complete the following table by filling either ‘more’ or ‘less’:
(2 marks)
Untreated meat
Tenderised meat
Toughness of meat
Cooking time
(f)
Suggest another fruit which has the same effect on meat.
(1 mark)
11.
The diagram below shows how a certain chemical reaction takes place inside the small intestine of a mammal.
+ H2O
+
+
+
D
A
B
C
E
B
(a)
Which molecule is an enzyme?
(1 mark)
(b)
Which are the substrate(s) and product(s) of this reaction?
(4 marks)
(c)
Which part of the enzyme determines the specificity of enzyme action?
(1 mark)
(d)
Draw a simple sketch to show the relationship between enzyme activity and
(i) temperature.
(2 marks)
(ii) pH.
(2 marks)
(i)
(e)
(ii)
Give an example of this type of reaction in the small intestine and state the products of the reaction.
(2 marks)
12.
The graph below shows the results of an experiment demonstrating the breakdown of starch by amylase.
20
Enzyme
activity
(g of starch
10
broken down
per hour)
0
0
10
20
30
40
50
60
Temperature (ºC)
(a)
Describe the changes in the activity of the enzyme when temperature increased from 0ºC to 60ºC?
(3 marks)
(b)
Why did the rate of reaction drop rapidly at temperatures above 40ºC?
(1 mark)
(c)
What is the product of digestion of starch?
(1 mark)
(d)
How would you test for this product?
(2 marks)
(e)
State one factor, other than temperature, that affects the activity of an enzyme.
(1 mark)
(f)
State where amylase can be found in the human body.
(2 marks)
The graph below shows the effect of pH on the activities of three digestive enzymes found in the human body.
Percentage of enzyme activity
13.
A
100
B
C
80
60
40
20
0
1
2
3
4
5
6
7
8
9
10
11
12
pH
(a)
What conclusion can be drawn from the graph?
(1 mark)
(b)
Which curves would best represent the actions of salivary amylase and pepsin (protease in stomach), respectively?
(2 marks)
(c)
Which parts of the digestive system secrete salivary amylase and pepsin, respectively? Briefly describe the functions of
these two enzymes.
(4 marks)
14. In an experiment investigating the action of a kind of protease on protein, a glass tube of solid egg white was placed in each of
the following labelled test tubes.
Test tube A – with 4 cm3 of 1% protease
Test tube B – with 4 cm3 of 1% protease and several drops of dilute hydrochloric acid
Test tube C – with 4 cm3 of boiled 1% protease and several drops of dilute hydrochloric acid
Test tube D – with 4 cm3 of dilute hydrochloric acid
The test tubes were put into a water bath at 40ºC. The original lengths of egg white were all 30 mm. After one hour, the length of
egg white in each test tube was measured again.
A
B
C
D
water bath at 40℃
egg white
protease
protease
+ dilute HCl
boiled
protease +
dilute HCl
dilute HCl
The length of egg white in each tube after one hour is shown below.
A: 29 mm
B: 25 mm
C: 30 mm
D: 30 mm
(a)
What conclusion can be drawn from the results of the experiment?
(2 marks)
(b)
Explain the results in tubes A, C and D.
(3 marks)
(c)
Explain the following applications of protease in our daily lives.
(i) Biological washing powder contains protease which can wash away the egg stains on clothes.
(2 marks)
(ii) When pineapple juice, which contains protease, is added to a large piece of beef and the beef is left for 30 minutes, it
will become tenderer.
(2 marks)
15. The table below shows the contents in five different test tubes.
Test tubes
Test tube contents
A
B
C
D
E
Whole fat milk (cm3)
5
5
5
5
5
Sodium hydrogencarbonate (cm3)
1
1
1
1
1
Bile salt (cm3)
0
1
0
1
1
Lipase (cm3)
0
0
1
1
1 (boiled)
Water (cm3)
2
1
1
0
0
After one hour, pH paper was used to test the pH of the solutions in the five test tubes. It was found that only solution in test
tubes C and D was acidic.
(a)
Which substance present in the milk was the substrate in this experiment?
(1 mark)
(b)
What was the main purpose of adding sodium hydrogencarbonate solution to each test tube?
(1 mark)
(c)
State the final product which made the solution in test tubes C and D acidic.
(1 mark)
(d)
Compare the acidity of the solution in test tubes C and D. Explain briefly.
(3 marks)
(e)
Explain the results in test tube E.
(2 marks)
(f)
Another test tube, F, contained a mixture of the following.
Whole fat milk (cm3)
5
3
Sodium hydrogencarbonate (cm )
Bile salt (cm3)
Lipase (cm3)
3
Water (cm )
1
1 (boiled)
1
0
The solution was found to be acidic after one hour.
(i) Which test tube could act as a control to test tube F?
(1 mark)
(ii) Is bile salt an enzyme? Explain your answer.
(2 marks)
Answers
1.
(a)
Molecule U
(b)
Non-competitive inhibitor binds to the enzyme at a different place other than the active site.(1 mark)
(c)
(1 mark)
It causes changes to the shape of the enzyme.
(1 mark)
The active site no longer fits the substrates,
(1 mark)
hence it reduces enzyme activity.
(1 mark)
No.
(1 mark)
It is because non-competitive inhibitor does not compete with the substrates for the active sites,
(1 mark)
increasing the concentration of substrates will not reduce the amount of inhibition.
(d)
As temperature decreases, less kinetic energy is supplied to the substrate and enzyme molecules,
(1 mark)
they will move slower.
(1 mark)
Hence the frequency of contact between substrates and active sites decreases,
(1 mark)
the formation of enzyme-substrate complex is less likely formed,
(1 mark)
the reaction rate decreases.
2.
(a)
Any two of the following: (1 mark each)
(2 marks)
protease/ lipase/ amylase
(b)
helps to remove proteins/ lipids/ starch
(1 mark)
causes less damage to fabric
(1 mark)
(c)
water temperature
(1 mark)
(d)
use warm water to wash the clothes
less harm to the users
(1 mark)
(1 mark)
3.
(a)
maltose
(1 mark)
(b)
Mercury ion-added condition: B
(1/2 mark)
Non-mercury-ion added condition: A
(1/2 mark)
Mercury ions bind to the amylase molecules and is irremovable.
(1 mark)
Some of the enzymes have their active sites permanently deformed and lose their catalytic functions.
(1 mark)
(c)
The remaining intact enzymes keep action and the maximum rate of reaction is lowered.
(1 mark)
Enzyme is a limiting factor/ Substrate is no longer limiting.
(1 mark)
Active site of enzyme is saturated with substrates.
(1 mark)
(d)
(1 mark)
1
37ºC, without mercury
1
1
09
ions
A
8
Rate of
reaction
(arbitrary
units)
B
7
6
5
4
3
2
1
00
4.
(a)
1
2
3 4 5 6 7 8 9 1
Starch concentration (arbitrary
0
units)
1
1
1
2
(i) Heat sterilisation/Use of high temperature
Use of strong acid/strong base
(ii) High temperature or extreme pH alters the shape of the enzyme’s active site.
This causes denaturation of enzyme and the enzyme loses its catalytic functions.
(b)
(1 mark)
(1 mark)
(1 mark)
(1 mark)
Since melonate has the structure and chemistry similar to that of succinate,
it competes with succinate molecules for the active site of enzyme S,
this prevents the formation of enzyme-substrate complexes and so the reaction rate decreases.
(1 mark)
(1 mark)
This hinders the reaction and thus the respiration process.
(1 mark)
Bacteria may die due to lack of energy.
(1 mark)
(Bonus mark: very high dose of melonate should be used to kill bacteria)
5.
(a)
plant materials/ crops/ plant cells
(1 mark)
(b)
yeast
(1 mark)
(c)
stomach
(1 mark)
(d)
An increased amount in substrate cellulose will increase the amount of glucose produced.
(1 mark)
An increase amount in substrate glucose will increase the rate of reaction 2.
(1 mark)
(e)
Heat will increase the temperature which will increase the chance of enzyme-substrate complex formation. (1 mark)
It will increase the rate of reaction 2 at first.
(1 mark)
As the temperature increases above the enzyme’s optimum temperature, the shape of enzyme starts to change and the
enzyme loses its catalytic function.
(f)
(1 mark)
The rate of reaction 2 decreases afterwards.
(1 mark)
making stonewashed jeans/ clarifying fruit juice or releasing fruit juice in fruit cells
(1 mark)
6.
(a)
(i)
non-competitive inhibitor
(ii) Trypsin inhibitor binds to trypsin and change the shape of its active site.
(b)
(1 mark)
(1 mark)
Substrate no longer fits the active site and thus the inhibitor reduces trypsin activity.
(1 mark)
Trypsin then cannot digest protein into peptides and amino acids.
(1 mark)
Optimum temperature: 37ºC
(1 mark)
Optimum pH: 8
(1 mark)
(c)
This can prevent herbivores from ingesting/consuming them.
(1 mark)
(d)
Trypsin inhibitor is protein in nature.
(1 mark)
Boiling provides high temperature which leads to the denaturation of trypsin inhibitor.
(1 mark)
Substance X breaks down / digests the solid protein coat
(1 mark)
into soluble substances/amino acids.
(1 mark)
7.
(a)
(b)
Substance X loses its function upon boiling (making it unable to break down the protein coat)/ is denatured.
(c)
X is an enzyme / protease.
(1 mark)
(1 mark)
This washing powder contains enzyme / protease that digests the proteins in egg stains to small molecules/ amino acids
(1 mark)
which are soluble in water / can be removed by water.
(1 mark)
(d)
The enzyme will be denatured at temperatures above 60ºC.
(1 mark)
(e)
Bleaching solution is alkaline.
(1 mark)
The activity of the enzyme in the washing powder may be inhibited at this high pH value.
(1 mark)
(a)
protease
(1 mark)
(b)
The protein in egg white was broken down in the presence of protease.
(1 mark)
(c)
increase the concentration of the enzyme
(1 mark)
(d)
(i)
(1 mark)
8.
The solution would remain cloudy because
at a temperature higher than 40°C, the enzyme would be denatured and could not catalyse the hydrolytic reaction.
(1 mark)
(ii) The solution would remain cloudy because
at low temperature, the enzyme would become inactive.
(1 mark)
(1 mark)
9.
(a)
(b)
Starch is broken down
(1 mark)
in the presence of amylase to reducing sugars or maltose.
(1 mark)
B,
(1 mark)
because amylase works best at pH 7.5.
(1 mark)
(c)
Same set-up but the dialysis tubing contains starch and distilled water.
(1 mark)
(d)
The temperature of the water bath is kept at 40ºC.
(1 mark)
This is the optimum temperature at which the enzyme works most efficiently.
(1 mark)
OR
Amylase with a higher concentration is used.
Higher concentration of enzyme increases the rate of reaction.
(e)
Yes,
(1 mark)
because the end products inside are small enough to diffuse through the dialysis tubing to the distilled water outside.
(1 mark)
10.
(a)
The substrate is protein.
(1 mark)
(b)
The enzyme is protein in nature.
(1 mark)
(c)
The enzyme speeds up the breakdown of protein to amino acids.
(2 marks)
(d)
Enzyme is denatured when cooked at high temperatures.
(1 mark)
The denatured enzyme cannot carry out the breakdown process.
(1 mark)
(e)
Untreated meat
Tenderised meat
Toughness of meat
tougher
less tough
Cooking time
longer
shorter
(1/2 mark each)
(2 marks)
(f)
pineapple
(1 mark)
11.
(a)
molecule B
(1 mark)
(b)
substrates: -molecule A and water
(2 marks)
products: -molecules D and E
(2 marks)
active site
(1 mark)
(c)
(d)
(i)
(ii)
Enzyme
activity
Enzyme
activity
pH
Temperature (ºC)
(2 marks)
(e)
(2 marks)
Digestion of sucrose by sucrase
(1 mark)
The products are glucose and fructose.
(1 mark)
OR
Digestion of lipid by lipase
The products are fatty acids and glycerols.
OR
Other reasonable answers
12.
(a)
The enzyme activity increased from 0ºC to 40ºC.
(1 mark)
The optimum temperature of the enzyme was 40ºC.
(1 mark)
The enzyme activity decreased from 40ºC to 60ºC.
(1 mark)
(b)
High temperatures caused denaturation and loss of functions in enzymes.
(1 mark)
(c)
maltose
(1 mark)
(d)
Heat it with Benedict’s solution.
(1 mark)
If an orange or a brick red precipitate is formed, it indicates that the product is maltose (reducing sugar).
(1 mark)
(e)
pH/ substrate concentration/ inhibitor concentration
(1 mark)
(f)
It can be found in the mouth cavity and also in the small intestine.
(2 marks)
(a)
Each type of enzyme works best at its own optimum pH.
(1 mark)
(b)
Action of salivary amylase: curve B
(1 mark)
Action of pepsin: curve A
(1 mark)
Salivary amylase is secreted by the salivary glands.
(1 mark)
It can digest starch into maltose.
(1 mark)
Pepsin is secreted by the stomach.
(1 mark)
It can digest proteins into peptides.
(1 mark)
This kind of protease can digest protein in the presence of dilute HC l/ in acidic condition
(1 mark)
but boiled protease is ineffective in the digestion of protein.
(1 mark)
13.
(c)
14.
(a)
(b)
(c)
Tube A – Protease was the most effective in acidic condition. It alone could not digest protein effectively. (1 mark)
Tube C – Protease was denatured by boiling. It could not digest protein though dilute HCl was added.
(1 mark)
Tube D – Dilute HCl alone could not digest protein.
(1 mark)
(i) Protease in the biological washing powder digests protein in egg stain into simpler forms
(1 mark)
which are soluble in water and can be washed away by water.
(ii) Beef is rich in fibrous protein.
Protease in pineapple juice digests the fibrous protein in beef into simpler forms, making it tenderer.
(1 mark)
(1 mark)
(1 mark)
15.
(a)
fats
(1 mark)
(b)
To provide a slightly alkaline condition for the lipase to work.
(1 mark)
(c)
fatty acids
(1 mark)
(d)
Solution in test tube D is more acidic than that in test tube C, because
(1 mark)
the bile salts in D emulsified (broke down the fat into small droplets) the fat and increased the surface area for the action
(e)
(f)
of lipase.
(1 mark)
With a larger surface area, the enzyme reaction rate was faster in D and more fatty acids were formed.
(1 mark)
The shape of the active site of the lipases was altered after boiling.
(1 mark)
The substrates could not fit into the active site and the enzyme could not function.
(1 mark)
(i) Test tube D
(1 mark)
(ii) No
(1 mark)
It will not be denatured after boiling.
(1 mark)