Certificate Biology - New Mastering Basic Concepts
Chapter 3
Suggested answers to Activity Books Ch 3
Cell activities and organization
Practical 3.1 Demonstration of the catalytic action of enzymes
(Bk 1, p.26)
Results
Sample
Gas given off
Glowing splint relighted
hydrogen peroxide + liver extract
+
+
distilled water + liver extract
-
-
hydrogen peroxide + distilled water
-
-
(Bk 1, p.27)
Questions
1
This increases the surface area for reactions.
2
Grinding action generates heat. The high temperature resulted may denature any enzyme present
in the tissues.
3
The gas given off is oxygen.
4
It is a control to show that no oxygen is given off from the liver extract.
5
It is a control to show that no oxygen is given off from the hydrogen peroxide.
6
Liver extract reacts with hydrogen peroxide to produce oxygen.
7
No. This experiment only shows that the breakdown of hydrogen peroxide is speeded up by the
liver extract. Boiled liver extract, instead of fresh liver extract, is used in a further experiment.
If boiled liver extract has no catalytic action, it is more likely that the reaction is catalyzed by
an enzyme because enzymes are denatured by boiling.
8
Yes. For the three test tubes, only one variable (the sample) is changed at a time, the others (e.g.
the volume and temperature of hydrogen peroxide, liver extract and distilled water) are kept
constant.
Conclusion
The breakdown of hydrogen peroxide is catalyzed by the liver extract, probably by an enzyme in the
liver tissues. Nevertheless, further experiments should be done to confirm this.
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Certificate Biology - New Mastering Basic Concepts
Suggested answers to Activity Books Ch 3
Practical 3.2 Investigation of the effect of temperature on enzyme
activity
(Bk 1, p.29)
Results
Temperature (C)
Time for disappearance of blue-black colour (min)
0
The blue-black colour does not disappear.
20
(Results vary with the origin of amylase.)
40
(Results vary with the origin of amylase.)
60
(Results vary with the origin of amylase.)
75
(Results vary with the origin of amylase.)
100
The blue-black colour does not disappear.
(Bk 1, p.30)
Questions
1
To ensure that the amylase and starch solutions inside the tubes reach the respective
temperatures before the reaction starts.
2
To prevent the changing of the condition of a mixture by any residue in the dropper.
3
Amylase is inactive at low temperature. Its activity increases with temperature and is highest at
60 °C. Afterwards the activity decreases and stops at 100 °C. With a rise in temperature, the
kinetic energy of amylase and starch molecules increases. They collide and react more
frequently. As the temperature increases further, the active sites of amylase become distorted
(i.e. the enzyme is denatured) and the reaction rate decreases. At 100 °C, all amylase is
denatured and no reaction takes place.
4
a
Starch will be digested and blue-black colour will disappear. This is because the inactive
amylase will resume its activity with an increase in temperature.
b
Starch will not be digested and blue-black colour will remain. This is because the activity of
denatured amylase will not restore even when it is cooled.
5
By measuring the rate of appearance of maltose molecules.
Conclusion
Amylase is inactive at low temperature. Its activity increases with temperature until it reaches a
maximum. Afterwards the activity decreases and stops.
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Certificate Biology - New Mastering Basic Concepts
Suggested answers to Activity Books Ch 3
Practical 3.3 Investigation of the effect of pH on enzyme activity
(Bk 1, p.34)
Results
Filter paper disc
Diameter of clear zone (mm)
C1
12
Extent of clearness
(clear, slightly milky, milky, unchanged)
Slightly milky
C2
9
Milky
C3
15
Clear
C4
0
Unchanged
C5
0
Unchanged
C6
0
Unchanged
Questions
1
They are controls. C4 checks whether the results observed are due to the enzyme alone. C5
checks whether an acidic medium affects starch. C6 checks whether an alkaline medium affects
starch.
2
To provide a temperature at which amylase works well.
3
They indicate the amount of starch left, which in turn indicates enzyme activity.
4
Alkaline medium.
5
In both cases, no clear zone will be observed. This is because low temperature (0 °C)
inactivates the enzyme and high temperature (100 °C) denatures the enzyme.
Conclusion
Amylase from mung beans is most active in the alkaline medium, less active in the neutral medium
and the activity is unobservable in the acidic medium.
Practical 3.4 Design an investigation of the enzyme activities of
different washing powders
(Bk 1, p.35)
Propose a hypothesis
1
(Answer depends on the kinds of washing powder provided.)
2
a
(Answer varies with Ss.)
b
(Answer varies with Ss.)
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Certificate Biology - New Mastering Basic Concepts
Suggested answers to Activity Books Ch 3
(Bk 1, p.36)
Design and perform an experiment
1
Amylase that works on starch. / Proteases that work on proteins.
2
Temperature, pH, concentrations of enzyme and substrate.
3
Temperature, pH, amounts of Brand A and Brand B washing powders, areas of food stains, etc.
4
The brand of the washing powder.
5
Time for the disappearance of the food stains, diameter of the clear zones on a starch-agar plate
or milk-agar plate, etc.
6
(Answer varies with the design.)
7
(Answer varies with Ss.)
8
Provide the optimum temperature and pH for the enzyme to work.
9
Repeat the experiment a few more times.
10
Do not conduct the experiment at a temperature or pH that will denature the enzyme.
(Bk 1, p.37)
Write an experimental report
Objective
(Answer varies with Ss.)
Apparatus and materials
(Answer varies with Ss.)
Procedure
Ss can carry out the experiment in a number of ways. Ss may follow the procedure in Practical 3.3,
but dip the filter paper discs into the two washing powder solutions instead of the enzyme solutions at
different pH. Ss may also conduct the experiment by using a milk-agar plate (see Practical 3.5).
Another method is by the use of two test tubes containing equal volumes of coagulated egg white
cubes. Add the two washing powder solutions into the test tubes and compare the time for the
disappearance of the egg white cubes.
Results
(Answer varies with Ss.)
(Bk 1, p.38)
Analysis and discussion
1
(Answer depends on results.)
2
(Answer varies with Ss.)
3
It is because enzyme activity increases at a higher temperature.
4
It is because proteins in silk and wool will be broken down by the proteases.
5
(Answer varies with the design.)
Conclusion
(Answer varies with Ss.)
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Certificate Biology - New Mastering Basic Concepts
Suggested answers to Activity Books Ch 3
Practical 3.5 Investigation of protease activities in different fruit
juices
(Bk 1, p.40)
Results
(Results vary with Ss.)
Questions
1
It is a control to show that the formation of the clear zones is due to the fruit juices.
(Bk 1, p.41)
2
Proteases in the fruit juices break down the white milk protein nearby. Therefore, the clear
colour of the agar is shown around the wells containing fruit juices.
3
(Answer depends on results.)
4
It is because the proteases in pineapple are denatured by the high temperature during the canning
process.
5
The proteases in fresh pineapple can break down the proteins in beef steak. Leaving beef steak in
contact with slices of pineapple for half an hour allows enough time for the enzymes to work.
Conclusion
Pineapple, kiwi fruit, papaya and guava contain proteases that can break down proteins,
but their activities differ from one another.
Practical 3.6 Study of osmosis in non-living material
(Bk 1, p.43)
Results
Set-up
Change in liquid level in the capillary tube
Experimental
Rises
Control
Lowers until it reaches the liquid level of water in the beaker
Questions
1
Its small bore gives a more obvious change in liquid level.
2
Sucrose molecules on the outside of the tubing will affect the result. Rinsing the tubing ensures
no such sucrose molecules are present.
3
a
There is a net water movement from distilled water to sucrose solution.
b
Osmosis.
c
Differential permeability.
a
The liquid level will rise faster and higher.
b
The liquid level will drop and the tubing will eventually shrink.
4
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Suggested answers to Activity Books Ch 3
Conclusion
When sucrose solution is separated from distilled water by a dialysis tubing, osmosis takes place and
there is a net movement of water molecules from distilled water to sucrose solution.
Practical 3.7 Study of osmosis in living plant cells
(Bk 1, p.45)
Results
In concentrated
sucrose solution
In less concentrated
sucrose solution
In very dilute
sucrose solution
Questions
1
To prevent the evaporation of sucrose solution, which may change its water potential and affect
the results. This also keeps the objective lens of the microscope clean.
2
The cytoplasm swells up gradually until the cell membrane presses tightly against the cell wall.
3
No. This is because the concentration of the content / water potential of each cell varies.
Conclusion
When the surrounding fluid has a lower water potential than the plant cells, water leaves the cells. The
cells will eventually plasmolyze and become flaccid. As water potential of the surrounding fluid
increases, water will enter the cells again. The cytoplasm expands to fill the whole cells. The cells
then become turgid.
Practical 3.8 Study of osmosis in living plant tissues
(Bk 1, p.47)
Results
Potato
cup
Condition
X
Filled with distilled water
Y
Filled with concentrated sucrose solution
Z
Boiled, filled with concentrated sucrose solution
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Change in liquid level
in the potato cup
No change
Rises
No change
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Certificate Biology - New Mastering Basic Concepts
Suggested answers to Activity Books Ch 3
Questions
1
The skin of the potatoes is impermeable to water. Peeling off the skin exposes the potato cells for
the experiment.
2
To prevent evaporation of water, which may change the concentration of the liquids in the
set-ups and affect the results.
3
No change is observed in the liquid level of potato cup X. Though living potato cells are
differentially permeable, no water potential gradient exists between the both sides of the cup.
Therefore osmosis does not occur.
4
The liquid level in potato cup Y rises. Since living potato cells are differentially permeable and a
water potential gradient exists between the both sides of the cup, osmosis occurs. Water moves
from the distilled water (higher water potential) outside the cup to the sucrose solution (lower
water potential) inside the well.
5
No change is observed in the liquid level of potato cup Z. Though a water potential gradient
exists between the both sides of the cup, boiling the potato kills the cells and the cell membranes
lose their differential permeability. Therefore, osmosis does not take place and sucrose
molecules diffuse across the fully permeable cell membranes.
(Bk 1, p.48)
Conclusion
Living plant tissues are differentially permeable. When solutions with different water potential are
separated by living plant tissues, osmosis takes place. However, the differential permeability of cell
membranes will be lost if the tissues are dead, i.e. cell membranes are fully permeable to all solutes.
Therefore, no osmosis can take place across dead tissues.
Practical 3.9 Study of osmosis in living animal tissues
(Bk 1, p.49)
Results
Set-up
Change in liquid level in the thistle funnel
A
Rises
B
Lowers until it reaches that of water in the beaker
Questions
1
Set-up B is a control. It shows that any change in liquid level in set-up A is due to the
concentrated sucrose solution.
2
Distilled water has a higher water potential than concentrated sucrose solution, so there is a net
movement of water from distilled water to concentrated sucrose solution through the
differentially permeable animal tissues by osmosis.
Conclusion
Animal tissues are differentially permeable. When solutions with different water potential are
separated by animal tissues, osmosis takes place.
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Certificate Biology - New Mastering Basic Concepts
Suggested answers to Activity Books Ch 3
Practical 3.10 Examination of different levels of organization
in organisms
(Bk 1, p.50)
Results
Level
Flowering plant (e.g. Hibiscus)
Mammal (e.g. rat, human)
Tissue
Examples:
(by examining the leaf under
the microscope)
 vascular tissue
 epidermal tissue
Examples:
 muscle
 nerve
 connective tissue
Organ
Examples:
 root
 stem
 leaf
 flower
Examples:
 heart
 stomach
 intestine
 lung
System
Examples:
 transport system
(vascular tissues in root,
stem, flower and fruit)
Examples:
 digestive system
 respiratory system
 excretory system
 reproductive system
STS connection 3.1 Applications of enzymes in commercial
products and industrial processes
(Bk 1, p.52)
Task 1
1
The bag was made from a sheep’s stomach and contained rennin. It was this enzyme that caused
the coagulation of milk.
2
Under hot weather, the kinetic energy of the milk protein and rennin molecules increased and the
molecules moved faster. This increased the chance they hit each other and formed the
enzyme-substrate complexes. Coagulation of milk was therefore speeded up.
3
The movement of the camel which produced a stirring effect.
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Certificate Biology - New Mastering Basic Concepts
Activity 3.1
Suggested answers to Activity Books Ch 3
Phillips head screwdrivers and straight blade
screwdrivers
(Bk 1, p.53)
Task
1
The unscrewing of strips represents a catabolic reaction, whereas the reassembling of strips
represents an anabolic reaction.
2
Straight blade screwdrivers because their shape fits that of the screws.
3
Enzymes are specific in their actions.
4
No. It is because its shape no longer fits that of the screws.
5
Enzymes are denatured at high temperatures and they lose their functions.
Exercise 3
(Bk 1, p.54)
Multiple-choice questions
1
C
2
C
3
A
4
A
7
C
8
A
9
D
10
D
5
C
6
D
(Bk 1, p.55)
Structured questions
1
a
Anabolism (0.5m)
It shows a building-up process. (0.5m)
b
W: enzyme (0.5m)
X: substrate (0.5m); Y: substrate (0.5m)
Z: product (0.5m)
c
i
It serves as a biological catalyst which speeds up the reaction. (1m)
ii
Denatured at high temperature / temperature and pH sensitive / reusable / specific in
action (Any 2, 1m each)
i
pH (0.5m) and temperature (0.5m).
ii
Change in pH: It alters the shape of enzyme’s active site hence, affects the activity of
enzyme. (1m)
Temperature above the optimum temperature: It alters the shape of enzyme’s active
site (0.5m). Denaturation will occur if extremely high temperature is reached (0.5m).
Temperature below the optimum temperature: The enzyme will be inactivated. (1m)
d
(Bk 1, p.56)
2
a Enzymes are biodegradable, thus cause less pollution to the environment. (1m)
Enzymes are specific in action, thus pure or specific substances can be produced. (1m)
Enzymes are reusable, thus the cost of production can be reduced. (1m)
b
Enzymes are made by living cells (0.5m) and able to speed up chemical reactions (0.5m).
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Certificate Biology - New Mastering Basic Concepts
Suggested answers to Activity Books Ch 3
c
i
ii
Protease / Lipase (Any 1, 1m)
Silk and wool are animal fabrics that are mainly made up of proteins (0.5m). They will
be digested by protease (0.5m).
Cotton is plant fabrics that are mainly made up of cellulose (0.5m). It will not be
digested by protease (0.5m).
d
No. (1m)
It is because the enzymes will be denatured at such a high temperature (60 °C). (1m)
(Bk 1, p.57)
3
a Cell membranes detach from the cell walls / Cells become plasmolysed. (1m)
b
i
The cells are immersed in a hypertonic solution. (1m)
ii
There is a net water movement from the cells to the solution outside by osmosis (1m).
It is because the water potential inside the cells is higher than that of the solution
outside the cells (1m).
c
Cell membrane (1m)
d
The cells remain unchanged (1m). It is because the differential permeability of the cell
membrane is destroyed during boiling (1m).
e
Place the tissue in a dry place. (1m)
f
The cells remain unchanged (1m). It is because the rigid cell walls prevent the net
movement of water from the solution outside into the cells (1m).
(Bk 1, p.58)
4
a Tissues: a group of similar cells working together to perform one or a few particular
function(s). (1m)
Cells: the basic building unit of organisms. (1m)
Organs: different tissues group together to form an organ to carry out more specialized
functions. (1m)
Systems: several organs and tissues work together to carry out a number of functions in a
coordinated way. (1m)
b
cells  tissues  organs  systems (1m)
c
Suggested answer:
Cell: cardiac muscle cell (1m)
Tissue: cardiac muscle (1m)
Organ: heart (1m)
System: circulatory system (1m)
d
Division of labour can be achieved among different tissues, organs and systems (1m) so
that different body activities can be carried out at optimum conditions (1m).
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