New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.1/31
Propose a hypothesis (p.
1
-2)
It is the fresh pineapple that causes the jelly to remain in liquid form.
Design and perform an experiment (p.
1
-2)
1 (Answer varies with Ss. The recommended quantity of jelly powder and water is stated on the packet of the jelly powder. Jelly will not set if it is too dilute.)
2 Method II. This makes sure the concentrations of the jelly solutions in different containers are the same.
A Identifying variables
Independent variable
(What will you change?)
The presence of fresh pineapple.
Dependent variable
(What will you measure?)
Whether the jelly set or not.
Controlled variables
(What will you keep constant?)
The amount of jelly solution in each container, the size and shape of the containers, the cooling temperature, the time allowed for setting, etc.
Control
(What is the control in this experiment?)
The jelly without fresh pineapple.
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New Senior Secondary Mastering Biology
Practical workbook answer
B Designing the set-up
Book 1A p.2/31
C Collecting data
1 (Answer varies with Ss.)
2 Repeat the experiment a few more times.
D Risk assessment and safety precautions
1 During the preparation of jelly solution, the hot water may burn our body.
The knife used to cut the pineapple is very sharp and may cut our fingers.
2 Wear a pair of thick gloves when handling hot water.
Handle the knife with care.
Write an experimental report (p.
1
-4)
Objective
To investigate the effect of fresh pineapple on the setting of jelly.
Hypothesis
It is the fresh pineapple that causes the jelly to remain in liquid form.
Apparatus and materials
1 electronic balance
1 refrigerator
2 beakers (500 cm
3
)
1 measuring cylinder (100 cm 3 )
2 plastic containers
2 glass rods
1 knife fresh pineapple hot water jelly powder
Procedure
1 Cut a fresh pineapple into small pieces.
2 Add 50 g of jelly powder and 200 cm
3
of hot water to a beaker. Stir the mixture with a glass rod until all the jelly powders dissolve.
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Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.3/31
3 Pour 100 cm 3 of jelly solution into two containers respectively. Leave the jelly solutions at room temperature for one hour.
4 Put the small pieces of fresh pineapple into one of the containers. Then, refrigerate two containers overnight.
5 Observe any changes of jelly solutions in the two containers on the next day.
Results
The jelly without fresh pineapple set.
The jelly with fresh pineapple does not set.
Analysis and discussion
1 It is used to confirm that the presence of fresh pineapple is the only factor that prevents the jelly from setting.
2 (Answer varies with the design.)
Conclusion
The presence of fresh pineapple prevents the jelly from setting.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.4/31
Results (p.
2
-4)
(Drawings vary with the cells observed. Drawings of human cheek cells and onion epidermal cells are given as examples.)
Questions (p.
2
-4)
1 To allow entry of a suitable amount of light. A dim image may result if there is insufficient light while a faint image may result if the light is too bright.
2 The coarse adjustment knob leads to a larger degree of movement of the body tube.
Any downward movement of the body tube controlled by the coarse adjustment knob may damage the objective or the slide because the distance between the objective and the slide is very small.
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Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.5/31
3
Area of specimen observed (small / large)
Details of specimen
(more / less)
Brightness of image
(bright / dim)
Low power large less bright
High power small more dim
4 Towards the left. An inverted image is formed in the microscope. It moves in a direction opposite to the actual movement of the flatworm.
Results (p.
2
-7)
Questions (p.
2
-8)
1 To flatten the specimen so that they can be seen easily in one plane of focus for the objective lens. To prevent the objective lens from getting dirty by touching the specimen or the mounting medium. To prevent the specimen from drying out because of evaporation. To protect the specimen from being damaged.
2 Cell membrane, nucleus and cytoplasm.
3 Anywhere inside the cell.
4 No. / Yes, they have small vacuoles.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.6/31
Results (p.
2
-12)
Questions (p.
2
-13)
1 Cell wall, cell membrane, nucleus, cytoplasm, chloroplast and granule.
2 Near the side of the cell.
3 No. Not all of them contain chloroplasts or chlorophyll. Only those with chloroplasts are green.
4 Similarities: Both of them have a nucleus, a cell membrane and cytoplasm.
Differences:
1 Plant cells are often larger than animal cells.
2 Plant cells have a definite shape while the animal cells do not.
3 Plant cells have a thick cell wall and some have chloroplasts. Animal cells do not have cell walls or chloroplasts.
4 Plant cells usually have a large vacuole while animal cells do not.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.7/31
Results (p.
2
-15)
Size
Genetic material
Nuclear membrane
Organelles bounded by a double membrane
(e.g. mitochondria)
Endoplasmic reticulum
Prokaryotic cell
Usually smaller
DNA lying free in the cytoplasm
Absent
Absent
Absent
Eukaryotic cell
Usually larger
DNA enclosed in the nucleus
Present
Present
Present
Questions (p.
2
-16)
1 Actual size of E. coli =
14 cm
60 000
= 0.00023 cm = 2.3 μ m
13 cm
Actual size of Guinea pig bone marrow cell =
12 500
= 0.00104 cm = 10.4 μ m
Guinea pig bone marrow cell is larger than E. coli.
2 Cell X is a eukaryotic cell because it has a true nucleus.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.8/31
Results (p.
3
-2)
Set-up
Experimental
Control
Change in liquid level in the capillary tube
Rises
Lowers until it reaches the liquid level of water in the beaker
Questions (p.
3
-3)
1 Its small bore gives a more obvious change in liquid level.
2 Sucrose solution on the outside of the tubing will affect the result. Rinsing the tubing ensures no such sucrose solution is present.
3 a There is a net water movement from distilled water to sucrose solution. b Osmosis. c Differential permeability.
4 After a certain period of time, the force produced by the weight of the liquid column balances the force developed by the water potential gradient.
5 a The liquid level will rise faster and higher. b The liquid level will drop and the tubing will eventually shrink.
Conclusion (p.
3
-3)
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.
Results (p.
3
-6)
Set-up Change in liquid level in the thistle funnel
A Rises
B Lowers until it reaches that of water in the beaker
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Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.9/31
Questions (p.
3
-6)
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 the distilled water to concentrated sucrose solution through the differentially permeable animal tissues by osmosis. The volume of liquid inside the thistle funnel increases and the liquid level rises.
Conclusion (p.
3
-7)
Living animal tissues are differentially permeable. When solutions with different water potential are separated by living animal tissues, osmosis takes place.
Results (p.
3
-9)
Concentration of sodium chloride solution
0%
Appearance of the red blood cells
Many red blood cells swell and burst.
0.45%
0.9%
1.35%
Few red blood cells swell and burst.
Red blood cells are normal.
Few red blood cells shrink and become wrinkled.
1.8% Many red blood cells shrink and become wrinkled.
Questions (p.
3
-9)
1 0.9% sodium chloride solution is isotonic to the red blood cells. At this concentration, the red blood cells appear normal. It shows that there is no net movement of water into or out of the cells because there is no difference in water potential between the cells and its surrounding.
2 0% and 0.45% sodium chloride solutions are hypotonic to the red blood cells. At these concentrations, the red blood cells swell and burst. It shows that the water potential of the red blood cells is lower than that of the sodium chloride solution. Water enters the red blood cells by osmosis.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.10/31
3 1.35% and 1.8% sodium chloride solutions are hypertonic to the red blood cells. At these concentrations, the red blood cells shrink and become wrinkled. It shows that the water potential of the red blood cells is higher than that of the sodium chloride solution. Water leaves the red blood cells by osmosis.
Conclusion (p.
3
-10)
When red blood cells are put in a hypotonic solution, water enters the cells by osmosis.
The red blood cells swell and finally burst.
When red blood cells are put in a hypertonic solution, water leaves the cells by osmosis.
The red blood cells shrink.
Results (p.
3
-12)
In concentrated sucrose solution
In less concentrated sucrose solution
In very dilute sucrose solution
Questions (p.
3
-13)
1 To prevent the evaporation of sucrose solution, which may change its water potential and affect the results. This also provides a flat surface for observation and keeps the objective lens of the microscope clean.
2 The cytoplasm swells up gradually until the cell membrane presses tightly against the cell wall.
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Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.11/31
3 No. This is because the concentration of the content / water potential of each cell varies.
Conclusion (p.
3
-13)
When the surrounding fluid has a lower water potential than the plant cells, water leaves the cells by osmosis. The cells finally become plasmolyzed and flaccid. When the water potential of the surrounding fluid increases, water will enter the cells by osmosis. The cytoplasm expands and the cells become turgid.
Results (p.
3
-15)
Liquid inside the beaker
Distilled
Water
Initial weight (g)
Strip
1
Strip
2
Strip
3
Average
Final weight (g)
Strip
1
Strip
2
Strip
3
Average
2.81 2.76 2.80 2.79 3.08 3.15 3.07 3.10
10% sucrose solution
2.76 2.87 2.83 2.82 2.79 2.80 2.81 2.80
Percentage change in weight (%)
11.11%
-0.71%
20% sucrose solution
2.84 2.83 2.85 2.84 2.57 2.62 2.55 2.58 -9.15%
Questions (p.
3
-16)
1 Osmosis cannot takes place across the potato peel because the peel is impermeable to water. Any peel left on the potato strips will affect the result.
2 To prevent the evaporation of water which may change the concentration of the liquids in the set-ups and affect the results.
3 To absorb the surplus water on the surface of the potato strips which may increase the weight of the potato strips and affect the results.
4 To minimize the water loss from the potato strips by evaporation. Any water loss will decrease the weight of the potato strips and affect the results.
5 The potato strips in distilled water become heavier. This is because the water potential of distilled water is higher than that of the potato cells. Water enters the potato strips by osmosis.
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New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.12/31
6 The weight of the potato strips in 10% sucrose solution changes very slightly. This is because the water potential of 10% sucrose solution is nearly the same as that of the potato cells. There is almost no net movement of water into or out of the potato strips.
7 The potato strips in 20% sucrose solution become lighter. This is because the water potential of 20% sucrose solution is lower than that of the potato cells. Water leaves the potato strips by osmosis.
Conclusion (p.
3
-17)
Living plant tissues are differentially permeable. When living plant tissues are placed in solutions with different water potential, osmosis takes place.
Results (p.
3
-19)

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.13/31
Questions (p.
3
-19)
1 Amoeba takes in food particles by phagocytosis: when Amoeba gets close to the food particles, pseudopodium starts to form to surround the food particles. The whole food particles are finally engulfed by the Amoeba .
2 Phagocytosis is important for:
1 the nutrition of some single-celled organisms, e.g. Amoeba engulfs food particles by phagocytosis;
2 body defence against diseases, e.g. in humans and other mammals, certain white blood cells engulf harmful microorganisms by phagocytosis.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.14/31
Results (p.
4
-3)
Sample hydrogen peroxide + liver extract distilled water + liver extract hydrogen peroxide + distilled water
Glowing splint relights
+
–
–
Questions (p.
4
-3)
1 This increases the surface area for reactions.
2 Grinding action produces heat. The high temperature resulted may denature any enzyme present in the tissues.
3 The gas given off is oxygen.
4 a It is a control to show that no oxygen is given off from the liver extract. b It is a control to show that no oxygen is given off from the hydrogen peroxide.
5 Liver extract reacts with hydrogen peroxide to produce oxygen.
6 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, can be used in a further experiment. If boiled liver extract has no catalytic action, it is more likely that the reaction is catalysed by an enzyme.
7 Yes. For the three test tubes, only one variable (the sample) is changed at a time, other variables (e.g. the volume and temperature of hydrogen peroxide, liver extract and distilled water) are kept constant.
Conclusion (p.
4
-4)
The breakdown of hydrogen peroxide is catalysed by the liver extract, probably by an enzyme in the liver tissues. Nevertheless, further experiments should be done to confirm this.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.15/31
Results (p.
4
-7)
Temperature (°C) Time for disappearance of blue-black colour (min)
0
20
The blue-black colour does not disappear.
40
60
80
100
(Results vary with the origin of amylase.)
The blue-black colour does not disappear.
Questions (p.
4
-7)
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 temperatures. Its activity increases with temperature and is the 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 the denatured amylase will not restore even when it is cooled.
5 Measuring the rate of appearance of maltose molecules.
Conclusion (p.
4
-8)
Amylase is inactive at low temperatures. Its activity increases with temperature until it reaches a maximum. Afterwards the activity decreases and stops.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.16/31
Results
(p.
4
-10)
Tube
A
B
C
D
E
F pH
3.2
4.0
5.2
6.0
7.0
8.0
Depth of brick-red precipitate settled
(mm)
(Results vary with the origin of invertase.)
Questions
(p.
4
-11)
1 To ensure the invertase has sufficient time to catalyse the breakdown of sucrose into glucose and fructose.
2 (Answer depends on results.)
3 a No or less brick-red precipitate will be formed. This is because extremely low pH will denature the invertase and reduce the enzyme activity. b No or less brick-red precipitate will be formed. This is because extremely high pH will denature the invertase and reduce the enzyme activity.
4 Weighing the precipitate formed. / Using an arbitrary system of ‘+’ to denote the relative amount of precipitate.
Conclusion
(p.
4
-11)
Invertase is most active in the acidic medium (pH 5.2) and less active in the neutral and alkaline medium.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.17/31
Results
(p.
4
-13)
Tube
A
Sample that contains copper(II) sulphate solution
Depth of brick-red precipitate settled
B silver nitrate solution (Answer varies with Ss.)
(mm)
C distilled water
Questions
(p.
4
-14)
1 It is a control to show that the activity of invertase is slowed down or stopped by the inhibitor.
2 No or less brick-red precipitate is formed in tubes A and B because copper(II) ions and silver ions are inhibitors of invertase. They slow down or stop the activity of invertase.
Brick-red precipitate is formed in tube C because the activity of invertase is not affected by any inhibitor.
3 Whether an inhibitor is competitive or non-competitive can be found out by increasing the substrate concentration of the reaction medium. The reaction rate will be increased in case of competitive inhibition, but not in case of non-competitive inhibition.
Conclusion
(p.
4
-14)
Copper(II) ions and silver ions are inhibitors of invertase. They slow down or stop the activity of invertase.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.18/31
Results
(p.
4
-17)
Well
A
B
C
D
E
Sample
Pineapple juice
Kiwi fruit juice
Papaya juice
Guava juice
Distilled water
Diameter of clear zone
(number of squares on graph paper)
(Results vary with Ss.)
Questions
(p.
4
-17)
1 It is a control to show that the formation of the clear zones is due to the fruit juices.
2 Proteases in the fruit juices break down the white milk protein nearby. Therefore, the white colour of the milk disappears and 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
(p.
4
-18)
Pineapple, kiwi fruit, papaya and guava contain proteases that can break down proteins, but their activities differ from one another.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.19/31
Design and perform a fair test
(p.
4
-20)
1 Protease. / Lipase.
2 By mixing the washing powder with distilled water well.
3 (Answer varies with Ss. The recommended quantity of washing powder and water is stated on the packet of the washing powder. This ensures the washing powder works in the best conditions.)
A Identifying variables
Independent variable
(What will you change?)
The brand of the washing powder.
Dependent variable
(What will you measure?)
Controlled variables
(What will you keep constant?)
Time for the disappearance of the food stains, diameter of the clear zones on a milk-agar plate, etc.
Temperature, pH, amounts of
Brand A and Brand B washing powders, areas of food stains, etc.
B Designing the set-up
(Answer varies with Ss.)
C Collecting data
1 (Answer varies with Ss.)
2 Provide the optimum temperature and pH for the enzyme to work.
3 Repeat the experiment a few more times.
D Risk assessment and safety precautions
1 (Answer varies with the design.)
2 (Answer varies with the design.)

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.20/31
Write an experimental report (p.
4
-22)
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 conduct the experiment by using a milk-agar plate (see Practical 4.5).
Another method is using two test tubes containing equal volumes of boiled egg white cubes.
Add the two washing powder solutions into the test tubes and compare the rate of dissolving of the egg white cubes.
Results
(Answer varies with Ss.)
Analysis and discussion
1 (Answer depends on results.)
2 No. We should consider the price of each brand.
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.)

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Results
(p.
5
-5)
A Test for glucose using Clinistix paper
Sample
Original colour
Clinistix paper
Final colour
Glucose solution Pink Purple
Distilled water Pink Pink
B Test for reducing sugars using Benedict’s test
Sample
Mixture of Benedict’s solution and sample
Observable change
Glucose solution Brick red precipitate is formed
Distilled water No observable change
C Test for starch using iodine test
Sample
Original colour
Iodine solution
Final colour
Glucose solution Brown Blue-black
Distilled water Brown Brown
D Test for lipids using grease spot test
Sample
Cooking oil
Presence of translucent spot after drying?
Before immersing into organic solvent
Yes
After immersing into organic solvent
No
Egg white solution
Distilled water
Yes
No
Yes
Not applicable
Book 1A p.21/31
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New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.22/31
E Test for proteins using Albustix paper
Albustix paper
Sample
Egg white solution
Original colour
Yellow
Final colour
Blue-green
Distilled water Yellow
F Test for vitamin C using DCPIP solution
Sample
Yellow
Original colour
DCPIP solution
Final colour
Vitamin C solution Blue Colourless
Boiled vitamin C solution
Distilled water
Blue
Blue
Blue
Blue
Questions
(p.
5
-6)
1 No. This is because these food tests are qualitative tests for showing the presence of certain food substances. They are not quantitative tests.
2 The water bath has a better control over the temperature and can prevent bumping of the mixture.
3 As the red colour of the blood will mask the results of the Benedict’s test, the blood sample should be diluted with distilled water first. Alternatively, the blood sample should be centrifuged and the plasma collected is used for the Benedict’s test.
4 The translucent spot caused by lipids is permanent. On the contrary, the translucent spot caused by water disappears as water evaporates.
5 Lipids but not proteins dissolve in organic solvent. Thus, the translucent spot caused by lipids disappear whereas the one caused by proteins remains on the filter paper.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.23/31
6 a Boiling destroys the reducing property of vitamin C. Thus, only the vitamin C solution that has not been boiled can reduce the blue DCPIP solution and decolourize it. b Do not boil or overheat fruits and vegetables.
7 No. Glucose, like vitamin C, is reducing and will decolourize DCPIP solution no matter vitamin C is present in the sample or not.
Results (p.
5
-12)
Food sample
Glucose
Reducing sugars
Starch
(Results vary with the types of foods tested.)
Lipids Proteins Vitamin C
Question
(p.
5
-12)
(Answer depends on the types of foods tested.)
Design and perform a fair test
(p.
5
-14)
1 Yes. Accurate measurement is necessary because a comparison of vitamin C content in different fruits and vegetables is needed.
2 Fruits or vegetables that have juices very pale in colour. The pale colour of the juices will not mask the decolourization of DCPIP solution.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.24/31
A Identifying variables
Independent variable
(What will you change?)
The type of fruits and vegetables.
Dependent variable
(What will you measure?)
Controlled variables
(What will you keep constant?)
The number of drops of sample needed to decolourize
1 cm
3
of DCPIP solution.
Amount of each type of fruits and vegetables used to extract the juices, volume and concentration of DCPIP solution used with each sample, temperature of sample and DCPIP solution, etc.
B Designing the set-up
(Answer varies with Ss.)
C Collecting data
1 (Answer varies with Ss.)
2 Use less DCPIP solution or a more dilute DCPIP solution.
3 Take any dilution factor of the juices into consideration in the comparison of vitamin C content.
Repeat the experiment with more samples from the same types of fruits and vegetables.
D Risk assessment and safety precautions
1 The knife used to cut fruits and vegetables is very sharp and may cut our fingers.
2 Handle the knife with care.
Write an experimental report
(p.
5
-16)
Objective
To compare the vitamin C content in different fruits and vegetables.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.25/31
Apparatus and materials
10 test tubes
1 test tube rack
10 droppers
1 measuring cylinder (10 cm
3
)
1 mortar and pestle
1 knife
1 filter funnel filter paper or fine muslin
0.02% DCPIP solution distilled water fruits and vegetables (e.g. orange, lemon, cabbage)
Procedure
1 Cut the fruit or vegetable into small pieces.
2 Put the small pieces into a mortar and grind with a small known quantity of cool distilled water (if necessary) using the pestle.
3 Squeeze the ground materials through several layers of pre-moistened fine muslin or filter them through a filter paper to remove the debris. Collect the juice extracted.
Skip this step if a fine and fairly colourless suspension is obtained.
4 Put 1 cm 3 of DCPIP solution in a test tube.
5 Use a dropper to add the juice, drop by drop, to the
DCPIP solution until the solution is decolourized. Record the number of drops of juice added.
6 Repeat steps 4 and 5 with the juices extracted from different fruits and vegetables. If the decolourization is too quick (i.e. the juice is too concentrated), dilute the juice by a known volume of distilled water and repeat steps 4 and 5 again. Take this dilution factor into consideration in the comparison of vitamin C content.
7 Repeat steps 4 and 5 with distilled water. It is a control.
Results
Sample
Lemon
Orange
Cabbage
Distilled water
Number of drops of sample needed to decolourize
1 cm 3 of DCPIP solution
15
12
19
Cannot decolourize DCPIP solution
Analysis and discussion
1 (Answer depends on the results.)
2 (Answer depends on the results.)
3 Reducing property.
4 a The vitamin C content will decrease as vitamin C will be oxidized by the air.
b The vitamin C content will decrease as vitamin C will be destroyed by high temperatures.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.26/31
5 The vitamin C in the juices will be oxidized by the air.
The observation of complete decolourization of DCPIP solution is subjective, especially when the juices are coloured.
There are other reducing substances, e.g. glucose and fructose, in the juices.
6 Fruits or vegetables that have juices very dark in colour cannot be used. The dark colour of the juices masks the decolourization of DCPIP solution.
Conclusion
(Answer depends on the results.)

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.27/31
Questions
(p. 6-2)
1
A Mouth
C Stomach
E Pancreas
G Ileum
I Liver
K Rectum
2 A

B

C

D

G

J

K

L
3
Structure
A
C
D
B
D
F
H
J
L
Oesophagus
Duodenum
Appendix
Caecum
Colon
Anus
Nutrition process involved
Ingestion, digestion
Digestion
Digestion, absorption
G Digestion, absorption
L Egestion
4 The salivary glands, the pancreas and the liver.
Design and perform an experiment
(p.
6
-4)
1 Proteins.
2 37°C. To simulate the body temperature at which pepsin works well.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.28/31
A Identifying variables
Independent variable
(What will you change?)
Dependent variable
(What will you measure?)
The solutions in the test tubes.
The disappearance of egg white cubes.
Controlled variables
(What will you keep constant?)
Number and size of egg white cubes, temperature and total volume of solution, etc.
Control
(What is the control in this experiment?)
A test tube with an egg white cube, hydrochloric acid and distilled water. A test tube with an egg white cube, sodium carbonate solution and distilled water. A test tube with an egg white cube and distilled water.
B Designing the set-up
(Answer varies with Ss.)
C Collecting data
1 (Answer varies with Ss.)
2 Provide a higher temperature, use smaller egg white cubes instead of a large one, raise the egg white cube from the bottom of the test tube by using a toothpick, etc.
3 Repeat the experiment a few more times.
D Risk assessment and safety precautions
1 The knife used to cut the hard-boiled egg is very sharp and may cut our fingers.
Dilute hydrochloric acid is corrosive and dilute sodium carbonate solution is irritant.
2 Handle the knife with care. Wear disposable gloves. Wash hands thoroughly with liquid soap and water after the experiment.
Write an experimental report
(p.
6
-6)
Objective
To investigate the action of pepsin in protein digestion.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.29/31
Apparatus and materials
6 test tubes
1 test tube rack
1 measuring cylinder (10 cm
3
)
1 knife
1 water bath dilute hydrochloric acid dilute sodium carbonate solution distilled water pepsin solution
1 hard-boiled egg
Procedure
1 Cut the egg white of the hard-boiled egg into six small cubes of length 0.5 cm and put one in each of the six test tubes.
2 Add the following solutions to the test tubes.
Test tube
A
B
C
D
E
F
Solution
5 cm
3
pepsin solution + 5 cm
3 dilute hydrochloric acid
5 cm
3
pepsin solution + 5 cm
3
dilute sodium carbonate solution
5 cm
3
pepsin solution + 5 cm
3
distilled water
5 cm
3
hydrochloric acid + 5 cm
3
distilled water
5 cm
3
sodium carbonate solution + 5 cm
3
distilled water
10 cm
3
distilled water
3 Leave the test tubes in a water bath at 37°C overnight.
4 Observe and note any changes in the size and appearance of the egg white cube in each test tube.
Results
Test tube
A
B
C
D
E
F
Observation
The egg white cube disappears.
The egg white cube remains intact.
The egg white cube remains intact.
The egg white cube remains intact.
The egg white cube remains intact.
The egg white cube remains intact.
Analysis and discussion
1 Stomach.
2 Protease.
3 Peptides.
4 Positive. Pepsin is a protein and it changes the colour of the Albustix paper.
5 (Answer varies with the design.)
Conclusion
Pepsin digests proteins in an acidic medium.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
Book 1A p.30/31
Results
(p.
6
-10)
Appearance
Mixture of oil and bile salt solution
An emulsion is formed.
Mixture of oil and distilled water
Two layers of liquids can be seen: oil on the top and water at the bottom. The two liquids do not mix.
Question
(p.
6
-10)
Water cannot break down oil into small droplets as the bile salt solution does. Therefore, no emulsion is formed and two layers of liquids can be seen.
Conclusion
(p.
6
-10)
Bile salts can break down lipids into small droplets. It is an emulsifying agent.
Results (p.
6
-12)
Starch Reducing sugars
At start – –
After one hour – +
Questions (p.
6
-12)
1 Starch solution on the outside of the tubing will affect the result. Washing the tubing ensures no such starch solution is present.
2 Less water allows a higher concentration of starch or reducing sugar molecules for easy detection.

Oxford University Press 2009

New Senior Secondary Mastering Biology
Practical workbook answer
3
Part of the model
Dialysis tubing
Part of the human body
Wall of the small intestine
Book 1A p.31/31
Water surrounding the tubing Blood
Starch and amylase mixture in the tubing
Mixture of undigested food and digestive enzymes in the small intestine
4 a Reducing sugar (maltose) is found. b Amylase digests starch into maltose. Maltose molecules are small enough to pass through the tubing and diffuse into the water outside the tubing, whereas the large starch molecules are retained inside the tubing.
5 Through digestion, food substances are broken down into small molecules that can diffuse through the intestinal wall / epithelium into the blood for use in our body.
6 Maltose molecules are not small enough to pass through the small intestine.
The small intestine can absorb digested food by active transport but the dialysis tubing cannot.
The small intestine can secrete enzymes but the model cannot.
The small intestine shows peristalsis but the model does not.
There are many types of food molecules in the small intestine apart from starch.
The food molecules have to pass through more than one layer of cells instead of only one layer of tubing.
The blood is enclosed in blood vessels.
7 Diffusion rate of the reducing sugar molecules can be increased by stirring the surrounding water and using a water bath at a higher temperature.
More concentrated solutions of starch and amylase can be used to speed up the reaction rate.

Oxford University Press 2009