Focus TG4 KSSM (Biology) Terbitan Penerbitan Pelangi Sdn Bhd

CONTENTS

1Chapter Introduction to Biology and 1 4.4 Lipids 57
Laboratory Rules 4.5 Nucleic Acids 59
SPM Practice 4 61
1.1 Fields and Careers in Biology 2
1.2 Safety and Rules in Biology Laboratory 3 5Chapter Metabolism and Enzymes 63
1.3 Communicating in Biology 4
1.4 Scientific Investigation in Biology 10 5.1 Metabolism 64
11
SPM Practice 1 5.2 Enzymes 64

2Chapter Cell Biology and Organisation 13 5.3 Applications of Enzymes in Daily Life 72

SPM Practice 5 73

2.1 Cell Structure and Function 14 6Chapter Cell Division 76

2.2 Living Processes in Unicellular 6.1 Cell Division 77
Organisms 19

2.3 Living Processes in Multicellular 6.2 Cell Cycle and Mitosis 78
Organisms 21
6.3 Meiosis 83
2.4 Levels of Organisation in Multicellular
Organisms 25 6.4 Issues of Cell Division on Human
Health 87
SPM Practice 2 27
SPM Practice 6 89
3Chapter
Movement of Substances 30 7Chapter Cellular Respiration 92
across the Plasma Membrane

3.1 Structure of Plasma Membrane 31 7.1 Energy Production through Cellular
Respiration 93
3.2 Concept of Movement of Substances 32
Across a Plasma Membrane 7.2 Aerobic Respiration 94

3.3 Movement of Substances Across a 7.3 Fermentation 96
Plasma Membrane in Living Organisms 37

3.4 Movement of Substances Across a SPM Practice 7 101
Plasma Membrane and its Application
in Daily Life 45 8Chapter Respiratory Systems in
Humans and Animals
SPM Practice 3 47 103

4Chapter Chemical Composition in a Cell 51 8.1 Types of Respiratory System 104

8.2 Mechanisms of Breathing 110

4.1 Water 52 8.3 Gaseous Exchange in Humans 114
4.2 Carbohydrates 53
4.3 Proteins 56 8.4 Health Issues Related to the Human 116
Respiratory System

SPM Practice 8 118

iii

9Chapter Nutrition and the Human 12.4 Voluntary and Involuntary Actions 188
Digestive System
121 12.5 Health Issues Related to the 190
Nervous System

9.1 Digestive System 122 12.6 Endocrine System 191

9.2 Digestion 123 12.7 Health Issues Related to the Endocrine
System 194
9.3 Absorption 130
SPM Practice 12 196
9.4 Assimilation 131

9.5 Defaecation 133 1 3Chapter Homeostasis and the Human
Urinary System
9.6 Balanced Diet 133 199

9.7 Health Issues Related to Digestive 137 13.1 Homeostasis 200
System and Eating Habits

SPM Practice 9 140 13.2 The Urinary System 205

1 0Chapter Transport in Humans 143 13.3 Health Issues Related to the 209
and Animals Urinary System

SPM Practice 13 211

10.1 Types of Circulatory System 144 1 4Chapter Support and Movement in
Humans and Animals
10.2 Circulatory System of Humans 150 214

10.3 Mechanism of Heartbeat 155

10.4 Mechanism of Blood Clotting 156 14.1 Types of Skeletons 215

10.5 Blood Groups of Humans 158 14.2 Musculoskeletal System of Humans 215

10.6 Health Issues Related to the Human 160 14.3 Mechanism of Movement and
Circulatory System Locomotion 221

10.7 Lymphatic System of Humans 163 14.4 Health Issues Related to the Human 226
Musculoskeletal System
10.8 Health Issues Related to the Human
Lymphatic System 165 SPM Practice 14 228

SPM Practice 10 167 Sexual Reproduction,
Development and Growth
1 1Chapter Immunity in Humans 171 1 5Chapter in Humans and Animals 231

11.1 Body Defence 172 15.1 Reproductive System of Humans 232

11.2 Actions of Antibodies 175 15.2 Gametogenesis in Humans 233

11.3 Types of Immunity 176 15.3 Menstrual Cycle 236

11.4 Health Issues Related to Immunity 178 15.4 Development of a Human Foetus 239

SPM Practice 11 178 15.5 Formation of Twins 242

1 2Chapter Coordination and Response 15.6 Health Issues Related to the Human 244
in Humans Reproductive System

181 15.7 Growth in Humans and Animals 245

SPM Practice 15 250

12.1 Coordination and Response 182 PRE-SPM MODEL PAPER 254
12.2 Nervous System 183
12.3 Neurones and Synapse 185 ANSWERS 269

iv

8Chapter Respiratory Systems in Humans
and Animals

Coughing, yawning and sneezing are normal reactions in humans that
involve the expulsion of air from the lungs. What are the differences
between coughing, yawning and sneezing?

Coughing expels impurities and bacteria from
the respiratory track. The thoracic cavity will
contract and the pressure inside the lungs
will increase. The air is pushed out with high
pressure and vibrates the vocal cord, thus
producing coughing sound.

CHAPTER FOCUS Sninoohmtaeleetnimooexusyg,ghebntroeafoethxrhinbagolednyco’asrrmnbaoelenlyd.diniTohxreeidrseetfioasnred,
8.1 Types of Respiratory we will yawn to inhale deeper so that
System more oxygen will enter the body and
8.2 Mechanisms of more carbon dioxide can be expelled.
Breathing
8.3 Gaseous Exchange in
Humans
8.4 Health Issues Related
to the Human
Respiratory System

Sneezing is a way that our body uses
to get rid of rashes due to dust
and bacteria. Air is expelled from the
lungs in force at velocity of 160 km
per hour.

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  Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

8.1 Types of Respiratory System Insect Respiratory System

1. The respiratory system is influenced by an 1. Insects such as cockroaches and grasshoppers
organism’s anatomy and physiology. conduct respiration via the tracheal system.

2. Different organisms retrieve oxygen from the 2. Gaseous exchange occurs without involving
atmosphere via different respiratory structures. the circulation of oxygen and carbon dioxide.
Instead, the exchange occurs with simple
3. Unicellular organisms do not have specific diffusion between the respiratory surfaces
systems because of their small size. Gaseous and the tissues of the body via air-filled tubes
exchange occur in the form of diffusion. known as the trachea.

3. Structures in the tracheal system include the
spiracle, trachea, tracheoles and air sacs.

Insect Filled with air to speed up the Trachea • Straight tubes which
Cell body transport of respiratory gases to connect to the spiracle
the tissues and brings air into the
body
Air sacs
• Tracheal wall is
Air sac stiffened by chitin
bands to prevent
Chapter the structure from
Chapter collapsing due to air
pressure

Chitin band

8 Air Spiracles

Tracheole • Small holes on both sides of the thorax
and the abdomen of the insect
• Minute branches that break from the trachea
• Has thin and moist walls to increase efficiency of • Guarded by valves which allow passage of
air
gaseous exchange
• Exist in abundant to increase the total surface • Have hairs that filter out the dust in the air

area for gaseous exchange via simple diffusion

Figure 8.1 Respiratory structures and their functions in a grasshopper

Fish Respiratory System
1. Gaseous exchange occur in fish via the gills which are tissues with feathery threads called filaments.
2. Bony fish have four pairs of gills protected by operculum.
3. Each gill consists of two rows of filaments supported by gill arches.
4. Each filament contains thin discs called lamellae.
5. Apart from being a site for gaseous exchange, gills also filter particulates like sand so they do not

damage the filaments.

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Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

Fish Gills Contains many thin discs called lamellae to increase surface
area for gaseous exchange.

Water enter Operculum
Oxygenated blood Closes to protect the gills

Gill arches
Holds thin filaments in place

Filament

Deoxygenated blood

Lamella Thin membraneous discs filled with blood capillaries
to increase the efficiency of gaseous exchange

Figure 8.2 Respiratory structures and their functions in a fish

Amphibian Respiratory System Chapter
Chapter
1. Amphibians like frogs, salamanders and newts 2. When the frog’s body is fully submerged
can live both on land and in water. Thus, in water, gaseous exchange occurs via the 8
the respiratory structure must be adapted to skin. When it is on land, gaseous exchange
function in both environments. involves both respiratory organs, the skin and
the lungs.

Amphibians Epidermis • Involves a pair of thin sacs to enable respiratory gases to diffuse
in and out easily.
• The skin is thin and Skin
very permeable to • The wall of the lungs is always moist as they are covered with
respiratory gases. fluids. This allows respiratory gases to diffuse into them.

• The skin is always • Lungs are rich with blood capillary networks to circulate
moist due to the layer respiratory gases to body cells.
of mucus which allow
diffusion of respiratory Lungs
gases.
Kulit
• Below the layer of skin
is a network of blood Mucous
capillaries which carry gland
respiratory gases to
body cells.

Figure 8.3 Respiratory structures and their functions in a frog

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  Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

Human Respiratory Structure 3. Apart from the respiratory organ, lungs are
also excretory organs that help eliminate
1. Lungs are the human respiratory structures carbon dioxide from the body.
found in the thoracic cavity.

2. Lungs are protected by the rib cage and
intercostal muscles.

Humans Nasal passage

Trachea • A channel for air to flow in and out
Supported by cartilage to ensure the trachea is • Covered in mucus and fine hairs
always open and can receive air
that trap dust

Chapter Bronchus Alveolus
Chapter Two branches from the trachea that allow
air to flow to the left and right lungs • The lungs contain millions
of alveoli which provides a
Bronchioles large surface area for gaseous
Smaller tubes that branch from the bronchus exchange
and allows air to flow to the alveoli
• Has moist walls that are one cell
Diaphragm thick to allow oxygen to dissolve
A dome-shaped layer of muscle and and diffuse through
fibrous tissue that cover the bottom side
of the rib cage • Large network of blood capillaries
that provide a large surface area
for gaseous exchange

Rib cage Ribs

8 Surrounds and protects the lungs External
intercostal
Intercostal muscles muscle
Wraps the outer and inner sides of the Internal intercostal
rib cage muscle

Figure 8.4 Respiratory structure and functions in humans

EEkxsppeerrimimeennt 82.1

Aim: To study the effect of the increase in total surface area towards diffusion as an analogy in gaseous
exchange

Problem statement: D o small objects have a large total surface area over volume thereby increasing
the substance diffusion rate?

Hypothesis: The smaller the object, the higher the TSA/V and its substance diffusion rate.

Variables:
Manipulated variable: Cube size
Responding variable: Diffusion rate of coloured solution into the cube
Fixed variable: Type of cube and concentration of coloured water

Materials and apparatus: P otato, coloured solution, core borer, sharp knife, tiles, beaker, grid paper,
stopwatch

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Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

Procedure:
1. Potatoes were cut into four cubes with 4 cm, 3 cm, 2 cm and 1 cm sides respectively.
2. The total surface area and volumes for each cube is calculated.
3. The potato cubes are inserted into beakers containing a coloured solution.
4. The time is measured with a stopwatch.
5. The cubes are left in the solution for 30 minutes.
6. The cubes are removed from the beaker when the time is up.
7. Cube with 4 cm sides are cut in two.
8. The coloured area percentage in each potato is estimated with the help of the grid paper.
9. Steps 7 and 8 are repeated with cubes with 3 cm, 2 cm and 1 cm sides.
10. All data is recorded in a table.
11. Step 9 is repeated using a different piece of grid paper to get the average reading.

Observations:

Length of Total surface Volume (cm3) TSA/V Ratio Estimated coloured
sides (cm) area (cm2) area (%)

16 1 6.0 100

2 24 8 3.0 60

3 54 27 2.0 25 Chapter
Chapter
4 96 64 1.5 10

Discussion: 8

1. (a) The cube with the highest TSA/V is the cube with the 1 cm sides.
(b) The cube with the lowest TSA/V is the cube with the 4 cm sides.
(c) The estimated coloured area (%) for the cube with 1 cm sides is 100%.
(d) The estimated coloured area (%) for the cube with 4 cm sides is 10%.

2. (a) The smaller the size, the larger the TSA/V.
(b) The higher the TSA/V the higher the diffusion rate and the larger the coloured area percentage.
(c) Smaller organisms have high TSA/V allowing oxygen to be easily diffused from the atmosphere

into their bodies.

Conclusion:
Hypothesis is accepted. The smaller the object, the higher the TSA/V, and the higher the diffusion rate.

EkAspcteirviimtye8n.12.1

Aim: To study the respiratory structure in insects, fish, amphibian and mammal

A   Cockroach: Tracheae

Procedure:
1. Place a chloroformed dead cockroach on the dissecting board and identify its body parts.
2. Observe and count the number of spiracles on both sides of the cockroach’s thorax and

abdomen.

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  Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals Spiracles
Trachea
3. Cut off the feet and wings of the cockroach.
4. Cut the dorsal area near the spiracles. Be careful to not cut into any

internal organs.
5. Using a magnifying glass, observe the tracheal structure.
6. Draw and label the structures that you can observe.

B   Fish: Gills Figure 8.5 Cockroach respiratory
Procedure: structure

1. Place a fish on the dissecting board.

2. Cut open the operculum from the body of the fish using scissors. Then cut out a piece of the gills.

3. Place the gills in a water-filled petri dish.

4. Identify the gill arches, filaments and lamellae.

5. Take out the gills, remove one gill arch and place it on a piece of white tile.

6. Observe the structure of the gill arch, filament and lamella with a magnifying glass.

7. Draw the gill arch and label the structure.

Operculum

Chapter
Chapter

8

Gill Mouth Filament Gill arch

Figure 8.6 Steps to remove the fish gill

C   Frog: Lungs

Procedure:
1. Place the chloroformed dead frog on the dissecting board front-up and pin the legs in place.
2. Make an incision in the middle of the body below the ventral.
3. Using the scissors, cut the skin of the frog upwards towards the head and across the sides.
4. Remove the skin layer slowly and pin the skin to the dissecting board.
5. Observe the blood vessels below the layer of the skin.
6. Using the scissors, cut the stomach muscles upwards. Identify the glottis and the lungs.
7. Extract the lungs with the scissors and place them on the white tile.
8. Cut out a cross section of the lungs using a blade.
9. Observe the size, shape and texture of the lungs using a magnifying glass. Sketch and label your

observations.

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Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

(a) Lift the skin using (b) (c)
forceps

Make incisions Make horizontal
upwards towards incisions
the chin

(d) (e) Lift and cut with (f)

a scalpel

Figure 8.7 Steps to remove the frog’s lungs

D   Rat: Lungs

Procedure: Trachea Chapter
Heart Chapter
1. Place the chloroformed dead rat on the dissecting board front-up. Lungs
Diaphragm 8
2. Pin all four limbs in place.
Figure 8.8 Respiratory system
3. Make an incision in the middle of the body below the ventral up structures in a rat
towards the jaw of the rat.

4. Continue the incision all the way to the rectum.

5. Pull the skin and pin it on the dissecting board using forceps.

6. Cut the tissue around the thorax vertically and horizontally so the
thoracic cavity is exposed.

7. Cut and move the ribs aside.

8. Identify the trachea, bronchus and lungs using a magnifying glass.

9. Extract the lungs and place it on the white tile. Make a cross section
of the lung using a blade.

10. Sketch and label your observations.

Discussion:

1. The respiratory organ for the cockroach is the trachea whereas for the fish, it is the gill. Both the
rat and the frog breathe using the lungs. Frogs can also breathe with their moist skin.

2. The respiratory structures for each specimen is as follows:

Specimen Respiratory structures

Cockroach Spiracle, trachea, air sac

Fish Gill arch, filament, lamellae

Frog Lungs, glottis, moist skin

Rat Trachea, bronchus, lungs

3. The respiratory surface for the cockroach is the tracheal, for the fish is the lamellae, for the frog is
the lung and moist skin, and for the rat is the alveoli.

4. The respiratory organs in the cockroach, fish, frog, and rat have large surface areas to increase
diffusion rate.

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  Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

Comparison of human and animal respiratory structures

Table 8.1 Comparison of human and animal respiratory structures

Similarities

• Have specific organs for respiration
• Respiratory surface is always moist, has a high TSA/V ratio and full of blood capillary networks.
• Gaseous exchange occurs via diffusion

Differences

Aspect Humans Insects Fish Amphibians

Respiratory organ Lungs Tracheae Gills Skin and lungs
Respiratory surface Bronchioles Lamellae Skin and lungs
Alveoli
Respiratory structure Spiracle, tracheae, Mouth, mouth cavity, Nostril, mouth, glottis,
Nostrils, nasal cavity, tracheole gills lungs
trachea, bronchus,
bronchiole, alveoli

ChapterCheckpoint 8.1 3. When the volume of the cavity increases,
Chapter the internal air pressure becomes lower than
1. Frogs have two respiratory organs. Name and the atmospheric pressure and air is forced
describe the adaptations found in both organs inwards.
which increase the gaseous exchange efficiency.
4. When the volume of the cavity decreases,
2. Humans have two lungs that contain millions of the internal air pressure becomes higher than
alveoli covered with blood capillaries. State its the atmospheric pressure and air is forced
significance. outwards.

8 8.2 Mechanisms of Breathing Grasshopper Breathing Mechanism

1. The act of breathing in and out is aided by Air sacs
the difference in pressure inside the breathing
cavities of organisms. Trachea Spiracles

2. The change in pressure is produced by
manipulating the volume of the cavity.

Figure 8.9 Grasshopper breathing structure

When breathing in: Abdominal pressure increases Air enters the tracheae and
and the air pressure in the tracheoles via the spiracles
The grasshopper’s abdominal tracheae is reduced
muscles relaxes and the
spiracle valve opens

When breathing out: Abdominal pressure decreases Air exits via the spiracles
and the air pressure in the
The grasshopper’s abdominal tracheae is raised
muscles contracts

Figure 8.10 Grasshopper breathing mechanism

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Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

Fish Breathing Mechanism

When breathing in: Mouth opens When breathing out: Mouth closes

Mouth opens and the Mouth closes and the
base of the mouth is base of the mouth is

lowered raised

The operculum closes Gill The operculum opens Gill

The space in the mouth Operculum Operculum
increases, lowering the close open
pressure in the mouth
The higher pressure The space in the mouth The operculum opens
outside pushes water decreases, raising the and water flows out
into the mouth, carrying pressure in the mouth through the gills
dissolved oxygen into

the mouth cavity

Figure 8.11 Fish breathing mechanism

Frog Breathing Mechanism When breathing out:

When on land, frogs breathe through the nostrils.
When breathing in:

12 3 Chapter
Chapter
The lung muscles 4
The nostrils open, the The nostril closes, contract and the 8
base of the mouth cavity the base of the mouth pressure inside the The deoxygenated air
raises and the glottis abdomen rises mixes with the air in the
lowers and the glottis
closes opens mouth cavity

The volume inside the As the volume inside The glottis opens and The nostrils open
mouth cavity increases, the mouth cavity the deoxygenated air and the glottis opens,
lowering the pressure eliminating the air inside
decreases, the internal exits the lungs
pressure increases. The

glottis opens and air
rushes into the lungs

Figure 8.12 Frog breathing mechanism

Human Breathing Mechanism

Table 8.2 Human breathing mechanism

When breathing in When breathing out

Contracts Outer intercostal muscles Relaxes

Relaxes Inner intercostal muscles Contracts

Moves upwards and outwards Rib cage Moves downwards and inwards

Contracts and flattens downwards Diaphragm Relaxes and arches upwards

Increases Thoracic cavity volume Reduces

Pressure reduces and air is forced in Pressure and air movement Pressure increases and air is forced out

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  Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

Breathing in Breathing out

Air flows in Air flows out

Rib cage moves Ribs Rib cage moves Ribs
upwards and Diaphragm inwards and Diaphragm
outwards when downwards
the external when the inner
intercostal intercostal
muscles contract muscles contract

Diaphragm contracts and Diaphragm relaxes, arches
flattens downwards upwards in a dome shape

Figure 8.13 Human breathing mechanism

EkAspcteirviimtye8n.2.1

Aim: Building a lung model to show the movement of the diaphragm during breathing

Materials and apparatus: T hin rubber sheet, perforated rubber cork, rope, grease, balloons, bell jar,
y-shaped tube and glass tube

Procedure: Glass tube

Chapter Bell jar
Chapter
Balloon Balloon
contracts
8 expands

(a) Thin rubber (b)
sheet

Figure 8.14 Human respiratory system model

1. Build a human respiratory system model as shown in Figure 8.14.
2. Secure a balloon on each of the ends of the Y-shaped tube.
3. Insert the rubber cork into the Y-shaped tube inside the bell jar from below.
4. Apply some grease on the cork so it is airtight.
5. Secure the thin rubber sheet at the base of the bell jar.
6. Tie a piece of rope to the rubber sheet.
7. Pull the rope downwards. Observe the effect on the balloon (Figure 8.14 (a)).
8. Push the rubber sheet upwards and into the bell jar. Observe the effect on the balloon (Figure 8.14 (b)).
9. Complete the table below.

Observations:

Procedure Observations

Rubber sheet pulled downwards Balloons expand

Rubber sheet pushed upwards and into the bell jar Balloons relax

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Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

Discussion:
1. The bell jar model represents the structure in the actual respiratory system in humans as follows:

Model Human respiratory system
Bell jar Thoracic cavity
Balloon Lungs
Rubber sheet Diaphragm
Y-tube
Trachea and bronchus

2. Here is how the model represents the breathing mechanism.

Breathing in mechanism Breathing out mechanism

Model Respiratory system Model Respiratory system

Rubber sheet is Diaphragm contracts Rubber sheet is Diaphragm relaxes
pulled down pushed up
Volume in thoracic
Volume in the bell cavity increases Volume in the bell Volume in thoracic
jar increases
Pressure inside jar reduces cavity reduces
Pressure inside the the thoracic cavity
bell jar reduces Pressure inside the Pressure inside
reduces bell jar increases the thoracic cavity
Air is forced into the
balloon Air is forced into the increases
lungs Chapter
Chapter
Air is forced out into Air is forced out into

the atmosphere the atmosphere 8

Conclusion:

When breathing in, the diaphragm contracts and flattens, the volume in thoracic cavity increases,
the pressure reduces and air is forced into the lungs. When breathing out, the diaphragm relaxes and
arches upwards, the volume in thoracic cavity decreases, the pressure increases and air is forced out
of the lungs.

EkAspcteirviimtye8n.32.1

Aim: To design a model that shows the antagonistic action of intercostal muscles during breathing

Materials: T hin board strip, ice cream sticks, rubber bands, nails

Procedure: Rubber Ice cream
band A stick A
1. Build the rib cage model as shown in Figure 8.15. Ice cream
stick B Nail
2. When each rubber band is stretched, observe and record
the movement of the board strip and ice cream sticks.

Ice cream Board
stick C strip

Rubber
band B

Figure 8.15 Rib cage model

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  Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

Observation: Observation
Procedure Ice cream stick B is lowered
Ice cream stick B is raised
Rubber band A stretched
Rubber band B stretched

Discussion:
1. The rib cage model represents the structure in the respiratory system in humans as follows:

Model Human respiratory system
Rubber band A Inner intercostal muscles
Rubber band B Outer intercostal muscles
Ice cream stick B
Ice cream stick A Sternum
Ice cream stick C Rib cage
Rib cage
Board strips
Spine

Chapter2. (a) When the outer intercostal muscles contract, inner intercostal muscles relaxes, the rib cage is
Chapter pulled upwards and outwards.

(b) This raises the volume in the thoracic cavity.
(c) The pressure in the thoracic cavity becomes lower than the atmospheric pressure.
(d) Air is forced into the lung.

8 3. (a) When the inner intercostal muscles contract, outer intercostal muscles relaxes, the rib cage is
pulled downwards and inwards.
(b) This lowers the volume in the thoracic cavity.
(c) The pressure in the thoracic cavity becomes higher than the atmospheric pressure.
(d) Air is forced out of the lung.

Conclusion:

The movement of inner and outer intercostal muscles causes the air to move out of and into the lungs
during the breathing process.

Checkpoint 8.2 8.3 Gaseous Exchange in Humans

1. How does fish: 1. Gaseous exchange between the lung and
(a) increase the volume in the mouth cavity? blood occur via simple diffusion.
(b) decrease the volume in the mouth cavity?
2. Apart from that, this exchange is aided by the
2. The two opercula on an injured fish are torn. In difference in the partial pressure for oxygen
your opinion, does this injury affect the breathing and carbon dioxide.
process? Explain.

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Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

Table 8.3 The difference between the concentration, partial pressure of oxygen and carbon dioxide at the alveoli
and blood capillary

Alveolus Blood capillaries
High oxygen concentration Low oxygen concentration
Low carbon dioxide concentration High carbon dioxide concentration
High partial pressure for oxygen Low partial pressure for oxygen
Low partial pressure for carbon dioxide High partial pressure for carbon dioxide

Exhaled air Inhaled air

Alveolus wall
(one-cell thick)

Blood in the blood capillary Alveolus Blood in the blood capillary
leading towards the alveolus leading away from the
have a high partial pressure Carbon alveolus have a high partial
for carbon dioxide and low dioxide Oxygen pressure for oxygen and
partial pressure for oxygen. low partial pressure for
carbon dioxide.
Chapter
Red blood Chapter
cell

Figure 8.16 Gaseous exchange across the surface of the alveolus and blood capillaries in the lungs 8

• Gas diffusion occurs from an area of high partial pressure to an area of low partial pressure.

1 • Inhaled air is rich with oxygen causing the partial pressure for oxygen in the alveolus to be higher than the partial

pressure in the blood capillary.

• Thus, oxygen dissolves into the moist alveoli walls and diffuse out from the alveolus into the blood capillary.

2 • Oxygen combines with the haemoglobin in the blood capillaries to form oxyhaemoglobin.
• Oxygen is carried from the lungs to the whole body in the form of oxyhaemoglobin.

Haemoglobin + Oxygen In the lungs Oxyhaemoglobin

In the body
cells

• When it arrives at the body cells, oxyhaemoglobin will break down and release the oxygen it carries due to the

3 lower partial pressure of oxygen.

• The released oxygen will diffuse through the wall of the blood capillaries into the cells. Oxygen is used in cell
respiration.

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  Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

• Cell respiration releases carbon dioxide leading to a higher partial pressure for carbon dioxide inside the cell

4 compared to inside the blood capillary.

• Carbon dioxide diffuses from the body cell to the blood capillary.

5 • Carbon dioxide from the body cell is carried through three ways:
 7% is dissolved in the blood plasma.

 23% combines with the haemoglobin in the red blood cells (erythrocytes) and carried in the form of
carbaminohaemoglobin.

 70% is in the form of bicarbonate ions, HCO3– in the blood plasma
– Carbon dioxide diffuses into the erythrocytes and combine with water to form carbonic acid (H2CO3). This

reaction is catalysed by carbonic anhydrase.

– Carbonic acid, H2CO3 is broken down to bicarbonate ions, HCO3– and hydrogen ions, (H+). Bicarbonate
ions, HCO3– diffuse into the blood plasma and is carried to the lungs.

– Bicarbonate ions, HCO3– diffuse again into the erythrocytes and combines with hydrogen ions, H+ to form
carbonic acid (H2CO3) once again.

– Carbonic acid (H2CO3) break down to carbon dioxide and water.

– The partial pressure for carbon dioxide in the blood capillary is higher than the partial pressure for
carbon dioxide in the alveolus.

– Therefore, carbon dioxide in blood plasma and erythrocytes diffuse through blood capillaries and alveolus
wall into the alveolus.

– Carbon dioxide is exhaled into the atmosphere via the mouth and nose.

Chapter
Chapter
Checkpoint 8.3

8 1. Oxygenated and deoxygenated blood is obtained from a specimen. How do you identify which blood is which?
2. Oxygen from the lungs is carried to the body cells in the form of oxyhaemoglobin that is the combination between
oxygen and haemoglobin. Meanwhile, carbon dioxide from the body cells are carried to the lungs to be eliminated
via three ways that are carbaminohaemoglobin, dissolved in blood plasma and in the form of bicarbonate ions.
What is the need to eliminate this gas from the body?

8.4 Health Issues Related to the Human Respiratory System

1. Chronic Obstructive Pulmonary Disease (COPD) is a disease caused by an obstruction in the respiratory
pathways due to swelling or long-term coughing leading to inflammation of the respiratory pathways
and obstruction of airflow to the lungs.

2. The primary cause of COPD is smoking or long term exposure to polluted air.

3. COPD is not gender-specific and the risks for COPD increases with age.

4. The most common symptom is shortness of breath during physical activities and in the mornings, long-
term caughing as well as wheezing. COPD patients often experience lung failure and inflammation of
the heart.

5. Example of COPD include asthma, acute bronchitis and emphysema.

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Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

Emphysema

Emphysema involves damage to the
alveoli which has become less elastic and
cannot expand much during breathing.
This leads to shortness of breath. If left
untreated, the alveoli will expand and
burst. This will reduce the gas exchange
surfaces.

Trachea Narrowed Chronic bronchitis
Lungs airway
Chronic bronchitis is a disease
Thickening caused by the infection or Chapter
walls inflammation of the bronchus Chapter
pathways which causes the
Normal Respiratory airway pathways leading to the 8
respiratory of asthmatic patient lungs to become narrow.
Chronic bronchitis patients will
airway experience long term coughs
with a lot of sputum.

Asthma Normal bronchus airway Inflamed bronchus airway

Asthma involves chronic inflammation of the respiratory
pathways. The airways redden, swell and produce
excessive mucus. This leads to the narrowing of the
airways. Asthmatic patients may experience wheezing
as well as shortness of breath on top of difficulties when
doing physical activities.

Checkpoint 8.4

1. (a) What does COPD mean?
(b) What diseases are associated with COPD?
(c) Name the symptoms for COPD?

2. Abu, a Form 3 student in SMK Purnama often have difficulties breathing after Physical Education class. His
teacher brought him to the clinic. The doctor asked if he has problems breathing when he wakes up and often
have coughs with phlegm. In your opinion, does the doctor suspect that Abu has COPD? Explain.

117

  Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

cCONCEPT MAP RESPIRATORY SYSTEMS IN
HUMANS AND ANIMALS

Insect Tracheae system Comparison based on
Fish Gill and Breathing
Amphibian Lungs Mechanism Oxygen
Carbon dioxide
Skin Asthma
Humans Lungs Chronic bronchitis
Emphysema

Chapter SPM Practice 8
Chapter
What is R?
8 Objective Questions 3. Which of the following is the A Trachea
1. Which of the following SPM site for gaseous exchange B Alveolus
SPM happen to reduce the 2016 for human and insect? C Bronchus
2015 amount of carbon dioxide in D Bronchiole
Human Insect
blood after one exercises? 5. Figure 2 shows a respiratory
A Alveolus Tracheole SPM system of an insect.

Heart Breathing B Alveolus Trachea 2016
rate rate J

A Reduced Reduced C Lungs Tracheole Figure 2
What process occurs in J
B Reduced Increased D Lungs Trachea
during gaseous exchange in
C Increased Reduced 4. Figure 1 shows a respiratory insects?
SPM structure. A Osmosis
D Increased Increased B Simple diffusion
2016 C Facilitated diffusion
D Active transport
2. Which of the following are R
SPM adaptations of the frog’s skin Figure 1
2016 for gaseous exchange?

A Moist and rich with blood
capillaries

B Smooth and dry
C Thick and moist
D Folded and thin

118

Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

6. Figure 3 shows the gaseous 8. Figure 4 shows the human 9. Figure 5 shows a tracheae
SPM exchange between body SPM breathing mechanism. SPM system in an insect.
2017 cells and blood capillaries.
2017 2017
Y

ZX RR SS Z

Figure 3 Figure 4 Figure 5
What is the partial pressure What is Z?
Which of the statements is
of carbon dioxide at X, Y true about Figure 4? A Tracheae
and Z? B Spiracle
RS C Muscle
XYZ D Tracheole
A Low High Low A Flattened Dome-
B High High Low diaphragm shaped 10. A group of students climbed
C Low Low High diaphragm SPM to the peak of Mount
D High Low High 2018 Kinabalu. The following are
Outer Outer
7. What structure in the changes that the students
SPM respiratory system of the fish B intercostal intercostal experience.
2017 increases the surface area muscle muscle
P – Breathing rate
for gaseous exchange? contracts relaxes increases
A Filament
B Gill arch C The The volume Q – Oxygen decreases in Chapter
C Operculum volume in in the lungs the bloodChapter
D Base of the mouth the lungs
decrease increase R – The partial pressure
of oxygen into
D The air The air 8
pressure pressure in alveolus decreases
the alveolus
in the Which of the following shows
alveolus is is high the correct order of the
changes that occur?
low A P, Q and R
B R, P and Q
C Q, R and P
D Q, P and R

Subjective Questions
Section A

1. Figure 1 shows part of the human respiratory system.

R
P
Q

Figure 1


119

  Biology Form 4  Chapter 8  Respiratory Systems in Humans and Animals

(a) (i) Name structures P and Q. [2 marks]
(ii) Explain the characteristics for P that is associated with its functions. [3 marks]

(b) (i) A heavy smoker will cough for long periods of time. Explain how this habit reduces the efficiency of
[3 marks]
his respiratory system.

(ii) Explain one disease that the smoker might have. [2 marks]

(c) Suggest a way to increase the ventilation rate when breathing. Explain. [2 marks]

Section B
2. (a) Figure 2.1 shows the human respiratory structure.


Figure 2.1

Explain the specific adaptations for the respiratory structure.

(b) Give a reasonable reason to stay away from smoking.
Chapter [6 marks]
Chapter [10 marks]

(c) A student conducted an experiment to study the breathing rate of an athlete. Figure 2.2 shows a graph

8 depicting the changes in the air volume inside the lungs of an athlete during a period of rest and during
a period of intensive exercise.

2.0

Volume of air (dm3) 1.5

1.0

0.5

Time (second)
0 10 20 30 40 50 60

Resting Intensive training
phase phase

Figure 2.2

Why is the volume of air in the lungs higher during the period of intensive exercise compared to the
[4 marks]
period of rest?

120

1Chapter Introduction to Biology and SPM Practice 1 enhance detergents so that clothes
Laboratory Rules get cleaner. In food technology,
Objective Question food can be modified to make it
Checkpoint 1.1 1. D 2. D 3. A 4. D 5. D taste better and more nutritious.
6. A 7. A 8. B 9. C 10. D Preservation methods can prolong
S1 3 research fields in biology: expiry date of food. Stem cells
• Zoology – The study of animals Subjective Questions have the potential to cure chronic
diseases such as spinal cord injury
which includes structure, physiology, Section A and heart disease.
development and classification. 1. (a) How does physical activity affect (c) Examples of personal protective
• Cytology – The study of cells which equipment that can be worn during
includes structure, cell composition pulse rate? surgery are mask, safety goggles,
and interaction with other cells. (b) As the physical activity increases, gloves and laboratory coat. The
• Anatomy – The study of bodily equipment is worn to prevent from
structure of humans, animals and pulse rate also increases. contacting with spurted blood or
other organisms. (c) (i) Manipulated variable: Types of other body fluid.
S2 • Five examples of biotechnological (d) Determine the problem statement,
products that are available at home: activity make hypothesis, plan investigation,
Clothes detergent, yogurt, cosmetics, determine and control variables,
plastics and fabrics Method to handle: Use different perform experiment, collect data,
types of activities analyse data, interpret data, make
Checkpoint 1.2 conclusion and write report.
(ii) Responding variable: Pulse rate
S1 Three protective equipments and their count
functions:
Method to handle: Count and
• Laboratory coat – Protects clothing record the pulse rate in a minute
from spilt chemical.
(d)
• Gloves – Protects hands from heat
and chemical reactions. 150

• Face mask – Filters fine particles 100 2Chapter Cell Biology and
bigger than 5 micrometres such as Organisation
bacteria, spores, fungi, solid particles
and liquid particles from entering 50 Checkpoint 2.1
respiratory system.
0 Zulfati Sarah S1 Nucleus, cytoplasm and plasma membrane.
S2 Materials that can be discarded into Husna S2 Presence of water in vacuole produces
laboratory sink: 100 ml distilled water,
0.1 M sodium hydroxide Resting Walking Jogging turgor pressure in herbal plants. Turgor
pressure is important for herb to maintain
Materials that cannot be discarded into (e) As the physical activity increase, its shape and prevent it from wilting.
laboratory sink: Concentrated sulphuric pulse rate also increases. This
acid, matchstick shows that more oxygen in the Checkpoint 2.2
blood is pumped from the heart
Checkpoint 1.3 to the rest of the body to produce 1. (a) K: Food vacuole
more energy. L: Nucleus
S1 Vertical axis: Responding variable M: Pseudopodium
Horizontal axis: Manipulated variable (f) Heartbeat N: Plasma membrane
S2 Dorsal view of a fish (b) Structure M is involved in the
Section B
2. (a) (i) Ecology is a scientific movement and capture of food.
Structure M is projected towards
S3 Histogram. It shows frequency study on distribution of its target. Cytoplasm flows into the
distribution of data in ranges value. Its organism, interactions among projected structure M to change
x-axis shows range of marks whereas organisms, and interactions its position. Structure M captures
y-axis shows frequency of pupils. between organisms and their food and subsequently forms food
environment. vacuole.
Checkpoint 1.4 (ii) Nanotechnology is a branch (c) Nucleus of a matured Amoeba
of technology that focuses sp. divides by shrinking along the
S1 Three things that are needed in on manipulation of materials middle. Cytoplasm divides into
the procedure when planning an smaller than 100 nanometres. two parts after the nucleus has
experiment: Data collection, selection of (iii) Longitudinal section is a section completely divided. Finally, two new
apparatus and materials, and planning done along the long axis of a cells are formed.
of procedure to be carried out. structure of organism, organ or
tissue. Checkpoint 2.3
S2 A report shows the flow of research (b) Genetic engineering is used
methodology in the experiment. It also in livestock and agriculture to S1 Smooth muscle cell, cardiac muscle cell
shows how the observations were produce better quality organisms. and skeletal muscle cell.
recorded; the data was analysed and For example, tomato which does
evaluated. In addition, future research to not turn squishy, and remain S2 Meristem cells at the tips of shoot and
be carried out was suggested; and the fresh longer. In medicine, vaccine root require much energy to perform
conclusion was presented to summarise is produced to prevent against active cell division for growth.
the research outcome based on the diseases. Enzymes are produced to
results. Checkpoint 2.4

S1 Cell → Tissue → Organ → System →
Organism

S2 Nose and lungs
S3 Endocrine hormones

269

  Biology Form 4  Answers

S4 Pancreas. Digestive system and as carbohydrate, protein, Checkpoint 3.2
endocrine system. phospholipid and glycoprotein.
For examples, K modifies S1 Osmosis is a process by which
SPM Practice 2 5. C protein to enzyme. L is rough water molecules pass through a
endoplasmic reticulum. L semipermeable membrane from a less
Objective Questions 4. C functions in transporting protein concentrated solute solution (higher
1. C 2. B 3. A synthesised by ribosome to concentration of water molecules) into a
cell surface and packages more concentrated solute solution (lower
Subjective Questions the protein into vessicle to be concentration of water molecules).
transported to other parts of
Section A cell. M is the nucleus. Nucleus S2 (a) Nucleic acid – facilitated diffusion
1. (a) (i) Cell consists of cell wall, consists of chromosomes which (b) Sodium ion – active transport
carry genetic information. This (c) Glycerol – simple diffusion
chloroplast and large vacuole information will determine the S3 (a) In passive transport, movement of
(ii) P: Cell wall characteristics of offspring.
Q: Chloroplast Nucleus also controls all molecules does not require energy
R: Mitochondria activities of a cell. whereas active transport requires
(iii) P is made up of tough cellulose (ii) 1. Palisade mesophyll cell energy.
(b) Movement of molecules by passive
fibres. It provides support to cell contains lots of chloroplasts transport is in the direction (along)
and maintain its shape. and are closely arranged to of its concentration gradient
(b) (i) Amoeba sp. is a single-celled ensure maximum absorption whereas in active transport,
microorganism which performs of light energy. molecules move against its
all functions of a complete living 2. Root hair cell has a large concentration gradient.
organism. surface area to allow maximum S4 Movement of potassium ion by active
(ii) Water diffuses into a vacuole of absorption of water. transport:
a flaccid Amoeba sp. Its vacuole (b) Nutrition • Potassium ion requires ATP energy
enlarges and pushes against Movement of cilia directs food and protein carrier to move against
the plasma membrane until it into oral groove. Food is digested concentration gradient in order to
reaches its maximum size. The by enzymes secreted into food cross plasma membrane.
vacuole contracts and excretes vacuole. Nutrient is absorbed into • Potassium ion attaches to active site
excess water. Amoeba sp. does cytoplasm whereas waste product is of protein carrier.
not burst. excreted out through its anal pore. • ATP dissociates to release phosphate
(iii) Lysosome fuses with food molecule which binds to carrier
vacuole. Lysosome releases Osmoregulation protein (of plasma membrane) and
lysozyme to digest bacteria. Paramecium sp. contains contractile alters the shape of the carrier protein
Digested food is absorbed vacuole which controls entry of to allow potassium ion to cross
by Amoeba sp. whereas excess water into it. Whenever plasma membrane.
undigested food is excreted. contractile vacuole is filled • Carrier protein regains its original
2. (a) with water, it moves to plasma shape.
membrane and excretes excess
Level of cell Structure water. Osmoregulation prevents Checkpoint 3.3
organisation Paramecium sp. from bursting.
S1 (a) Blood
(a) Cell (b) Concentrated salt solution is

(b) Cell Asexual reproduction hypertonic to red blood cells. Water
structure Paramecium sp. performs asexual diffuses out of red blood cell by
osmosis. The red blood cells shrink
(c) Organ reproduction by binary fission. and experience crenation.
Each Paramecium sp. divides (c) Animal cell does not contain cell
once to produce two similar cells, wall. Absence of cell wall does not
which develop into two individual allow red blood cell to maintain its
Paramecium sp. shape. Shape of red blood cell that
experiences crenation:
(d) System 3Chapter Movement of Substances
across a Plasma Membrane S2 Transport of mineral ions into grass
(b) (i) Reproductive system of man. roots of grass is by active transport.
Function: Produce sperms to Checkpoint 3.1 When cell respiration stops, no energy
fertilise ovum. is produced to carry out active transport,
S1 Channel proteins and carrier proteins therefore transport of mineral ions into
(ii) Sperm has tail to facilitate S2 • Main components of plasma root stops.
swimming towards ovum.
membrane are proteins and Checkpoint 3.4
(c) Cell cannot swim towards ovum phospholipids whereas the main
because it does not have energy. components of cell wall are celluloses. S1 (a) Food preservation method
Fertilisation does not occur. • Plasma membrane is semi-permeable (b) Advantages:
whereas cell wall is fully permeable. • Add flavour
(d) Multicellular organism consists of S3 Plasma membrane consists of bilayer • Last longer
many cells. It requires specialised phospholipid with different types of • Looks better
cells to perform specific functions. proteins partially or fully embedded Disadvantages:
Multicellular organism requires in the membrane. The description • Less nutritious
specific systems to fulfill the above is known as the fluid mosaic • High in sugar/salt content
requirements of its body. model of plasma membrane. Plasma
membrane components are always
Section B in free motion and their positions are
3. (a) (i) K is Golgi apparatus. K constantly changing. This explains why
plasma membrane possesses fluidic
functions as a centre for characteristic.
processing, packaging and
transporting substances such

270

Biology Form 4  Answers

S2 Cat faeces caused soil water to become • It is an organic material that can Adhesive forces of water:
hypertonic to cell sap of flower plant. Forces which attract water molecules to
This caused water to diffuse out of the be renewed and easy to obtain
plant and lost a lot of water, and the other substance. The forces that move
plant wilted. If nothing was done, the 3. (a) • It is very soft water up in xylem vessel.
plant would die. A lot of water must be
sprinkled to reduce the concentration • It is very thin Checkpoint 4.2
of minerals dissolved in the soil water.
This would enable water to be absorbed • It consists of two layers of lipid S1 Hydrolysis breaks complex molecule into
into the roots and caused the plant cells simple molecules whereas condensation
to be turgid again, and the plant would • Various types of proteins are combines simple molecules to form
then regain its freshness. complex molecule. Hydrolysis involves
embedded in between the two addition of water molecule whereas
condensation involves removal of water
layers of lipid molecule.

• It is dynamic / can move freely S2 Hydrochloric acid is used to break
chemical bonds in non-reducing sugar.
• It is a strong structure / not easily Sodium hydrogen carbonate powder is
used to neutralise the pH of solution.
broken
Checkpoint 4.3
(b) • Example: glycerol / fatty acids
S1 Dipeptide is a product of condensation
SPM Practice / Vitamin A/D/E/K // Dissolved of two amino acids whereas polypeptide
is produced from hundreds of amino
Objective Questions 3 gases such as oxygen and carbon acids.
1. C 2. B 3. A
6. A 7. B 8. D 4. C 5. A dioxide S2 Involves in growth, assists in movement,
9. A 10. C • When concentration of R formation of haemoglobin, enzyme,
hormone and antibody
increases, its movement rate
Checkpoint 4.4
Structure Questions across membrane also increases
• Movement of R is by simple S1 Carbon, hydrogen and oxygen
S2 Condensation of one molecule glycerol
Section A diffusion
1. (a) (i) Q: Bilayer phospholipids • R dissolves in plasma membrane and three molecules of fatty acids
R: Glycerol S3 Advantages of steroid:
S: Fatty acids and crosses it • Can effectively treat inflammation
• Provides energy
(ii) • Semipermeable • Along the direction of Disadvantages of steroid:
• Cause increase in body weight
• Fluid characteristics concentration gradient • It is a type of drug/stimulant which
• Size of molecule R is very small
(iii) • Simple diffusion • The movement process of R is causes addiction.
S4 No. Overweight is caused by excessive
• Oxygen / CO2 continuous until the concentration
(b) • Facilitated diffusion of R in the cell equals the fatty tissue in the form of adipose tissue
• Carrier protein / Pore protein concentration of R outside the cell that are kept below the dermis / skin
• From a region of higher to lower layer. Cholesterol is in the blood.
// dynamic equilibrium is achieved.
concentrations // down the Checkpoint 4.5
(c) 5% sucrose solution for 30 minutes:
S1 Stores and transfers genetic information
concentration gradient • Shape and structure of cell S2 DNA and RNA
• DNA is a double stranded whereas
• Amino acid binds to protein carrier remains the same
• Carrier protein changes shape RNA is single stranded.
• 5% sucrose solution is isotonic to • DNA carries genetic information
and moves molecule across
cell sap whereas RNA copies genetic
information from DNA for protein
plasma membrane • Rate of water absorption in and synthesis.

(c) • Active transport out of the cell is the same by
• Carrier protein
osmosis

• Against concentration gradient • There is no net water movement
• Requires energy (ATP)
• Have active sites across plasma membrane
• Binds with iodine ion
• Moves the ion across cell by 20% sucrose solution for 30 minutes:
• Shape and structure of cell
going against concentration
becomes flaccid
gradient • Water diffuse out from cell into
2. (a) (i) P: Fatty acids and glycerol
Q: Pore protein environment
(ii) P: F orms the structure of cell • Vacuole shrinks and becomes

membrane smaller
Q: Passage in plasma membrane • Cytoplasm / plasma membrane is

for movement of water-soluble pulled away from cell wall
• 20% sucrose solution is
molecules and ions
hypertonic to cell sap
• Cell experiences plasmolysis

(iii) Very small size Distilled water for 30 minutes: SPM Practice 4
(b) • Soluble nutrient absorbs through • Cell becomes turgid
• Distilled water is hypotonic Objective Questions 4. D 5. A
the pore of structure Q, whereas 1. A 2. B 3. B 9. C 10. D
nutrient binds to structure R and towards cell sap 6. C 7. A 8. B
• Water diffuse into cell
alters it, then allows the nutrient to • Cell vacuole expands and

enter or exit the cell. becomes bigger Subjective Questions
• Q only allows water-soluble • Cytoplasm / plasma membrane
Section A
molecules to pass through pushes towards cell wall 1. (a) X: Sucrose
whereas R allows both water- Z : Lactose
(b) • Y is rich in lipid / fats / oil
soluble and water-insoluble • Excess fats will be converted to

molecules to pass through. 4Chapter cholesterol
• Which is suspended in blood /
(c) • Method 2 Chemical Compositions in
• Concentrated salt solution is a Cell stick to lumen wall of blood vessel
• Increases risk of heart attack
hypertonic to cell sap of potato
(c) (i) • Involves condensation process
• Water diffuse out of potato to the Checkpoint 4.1 • Removes one molecule of water
environment • Produces a more complex
S1 Water dissolves many substances
• Potato cells lack water S2 Cohesive forces of water: molecule
• Potato becomes crispy when fried Forces which attract water molecules to
(d) • Coconut husk is not packed /
water molecules. The forces that move
contains air spaces in it
water in xylem vessel.

271

  Biology Form 4  Answers

(ii) • Involves hydrolysis process S2 Metabolism is a process that occurs • Enzyme is released from Golgi
• Addition of one molecule of only in living cells such as cells found in apparatus in the form of secretory
live animals, fungi and mosses.
water vessicle
• Produces simpler molecules S3 Yes. The body of a person with higher
(d) • Benedict solution is an indicator metabolic rate burns food at a higher • Secretory vessicle moves to
rate to produce energy than a person plasma membrane and diffuses
for reducing sugar with a lower metabolic rate. Therefore, a
• Presence of reducing sugar forms person with higher metabolic rate tends out of cell
to have lesser body fats and weigh less
brick red precipitate than a person with lower metabolic rate. (b) • Amount of substrate molecules
• No change when sucrose is boiled increase
Checkpoint 5.2
with Benedict solution because • More substrate can bind to active
sucrose is not a reducing sugar S1 Intracellular enzyme and extracellular sites of enzyme
enzyme. Intracellular enzyme is
Section B produced inside cell and is used within • More catalytic activities occur
2. (a) • Water can absorb a lot of heat the cell whereas extracellular enzyme is • Higher rate of reaction
produced inside cell but is secreted out (c) • Adding lipase enzyme: Remove
energy without increasing its to be used outside the cell.
temperature. fats stains
• Maintain human body temperature S2 Pancreas secretes lipase to digest fats.
at optimum temperature. When pancreas is damaged, lipase will • Adding protease enzyme: Remove
• Chemical reactions in human not be produced. Therefore, fats will not protein stains
body are at maximum level. be digested.
• E.g. Cell respiration / enzymatic • Adding amylase enzyme:
activities S3 Transport vessicle moves protein Removes starch stains
(b) • Cohesive forces are forces that released from endoplasmic reticulum
attracts water molecules together. to Golgi apparatus whereas secretory • Clothes washed with the enzymes
• Cohesive forces help to pull other vessicle carries enzyme released by are cleaner
water molecules up through xylem Golgi apparatus and secretes the
vessel. enzyme out of the cell. 2. (a) • As enzyme concentration
• Adhesive forces are forces that increases, more enzyme
enable water molecules to stick to S4 High temperature denatures enzyme
wall of xylem vessel. whereas change in pH only alters shape molecules are available
• Adhesive forces pull water and charges found on active sites of
molecules up at the edges of enzyme. • It increases the chances of
xylem vessel. substrate molecules that attach to
• Both cohesive and adhesive Checkpoint 5.3
forces move water up xylem active sites of enzymes
vessel. S1 Protease, lipase and lactase hydrolyse
• Both forces help to maintain cell protein, fats and lactose found in milk • Increase in amount of enzyme-
turgidity of leaves. to produce instant infant formula, a substrate complexes
• Both forces provide mechanical digested form of milk/food because
support to the stem of the tree. infant cannot produce the digestive • Increase in catalytic activity by
(c) • Energy source enzymes yet. enzyme
• Growth
• To produce new cells / replace S2 Amylase hydrolyses starch found in • Higher rate of reaction
damaged cells fruits to sucrose which is sweet. • Increase in products of reaction
• Actin and myosin function in • If concentration of enzyme
muscle contraction SPM Practice 5
• To produce DNA continues to increase, rate of
• To regulate chromosome structure Objective Questions
/ determine genetic characteristics 1. B 2. B 3. B 4. B 5. C reaction reaches constant;
• To produce hormones 6. A 7. C 8. B 9. A 10. D
• To regulate internal environment • because the amount of substrate
• To produce haemoglobin Subjective Questions is limited
• To produce enzymes
• To produce antibodies Section A • No increase in the amount of
1. (a) P: Nucleus enzyme-substrate complexes
5Chapter Metabolism and Enzymes Q: Rough endoplasmic reticulum (ER)
R: Ribosome • No increase in the rate of reaction
Checkpoint 5.1 S: Golgi apparatus (b) • Enzyme is the lock
T: Transport vessicle • Substrate is the key
S1 (a) Catabolism and anabolism U: Secretory vessicle • Conformations (shape) of both
(b) Catabolism is the breakdown • DNA found in nucleus carries
lock and key must correspond to
of complex molecule to form genetic information
simpler ones whereas anabolism • Genetic information is copied by one another to enable formation
is the condensation of simple
molecules which are the products RNA of enzyme-substrate complex
of catabolism, to form complex • RNA diffuses out from nucleus
molecule. (c) • Factor: pH
(c) Catabolism: Breaking down of and attaches to ribosome • Enzyme is very sensitive to
glucose during respiration • Genetic codes are used to
change in pH
Anabolism: Condensation of synthesised protein
fatty acids and glycerol to form • Protein is transported by rough • which may alter conformation of
phospholipids enzyme
endoplasmic reticulum (ER)
• Rough ER releases transport • such as alteration of conformation
of enzyme active sites
vessicle
• Transport vessicle diffuses into • Change of pH also alters charges
on substrate
Golgi apparatus
• Protein is modified, processed • Substrate conformation will no longer
correspond to active site of enzyme
and packaged to form enzyme
• No formation of enzyme-substrate
complex

• No catalytic activity

6Chapter Cell Division

Checkpoint 6.1

S1 Karyokinesis is the division of a cell
nucleus during mitosis or meiosis
whereas cytokinesis is the division of
cytoplasm, and both types of division of
a cell produce two new cells.

S2 6

272

Biology Form 4  Answers

S3 Gamete (iii) For growth / increase in length Metaphase
Somatic cell or height of plant • Chromosomes are arranged at
equatorial plane
Body cell Reproductive cell (c) (i) Tissue culture • Spindle fibres are fully formed
(ii) • A plant is genetically the same
Consists Consists to its parent plant Anaphase
of diploid of haploid • A new plant that is produced • Both sister chromatids of each
number/2 sets of number/1 set of does not show any genetic chromosome separate at
chromosomes chromosomes variation centromere
• A new plant possesses the • Sister chromatids are pulled
Checkpoint 6.2 same characteristics as its towards opposite poles of cell
parent plant in terms of level during shortening of spindle
S1 DNA replication occurs. Each of resistance towards diseases fibres
chromosome replicates to produce two / pest
identical chromatids. Telophase
(d) (i) The cells have been exposed • Both sets of chromosomes
S2 G1 phase determines whether the to carcinogenic substance / reach opposite poles of cell
cell cycle will be continued. If the radioactive rays. It causes • Chromosomes unwind / look
environment is not conducive for growth, mutation to gene that control like fine threads
cell will not enter synthesis phase. cell cycle. The cells then • Spindle fibres disappear //
Instead, cell will enter G0 phase, where undergo repeated uncontrolled a new nuclear membrane is
cell does not divide. mitosis. formed to enclose each set of
chromosomes // formation of
S3 Prophase → Metaphase → Anaphase (ii) Radioactive rays / chemotherapy new nucleolus
→ Telophase can be used to kill cancer
cells. When cancer cells die, (b) • In cytoplasm, actin filaments
S4 Cytokinesis they cannot divide by mitosis contract to cleave cytoplasm by
S5 • Increase number of cells during anymore. pulling plasma membrane inwards

growth 2. (a) Anaphase I • A cleavage furrow is formed at the
• Replace injured/damaged/dead cells (b) A pair of homologous chromosomes equatorial plane of cell
• Regeneration
separate and moves towards • The cleavage furrow occurs
Checkpoint 6.3 opposite poles gradually until two daughter cells
(c) Because it possesses centriole and are formed
S1 30 no cell wall
S2 Crossovers involve exchange of (d) 2 7Chapter
(e) (i)
DNA segments between non-sister
chromatids. Crossovers produce new Metaphase Metaphase I Cellular Respiration
combination of genes in chromosome
/ various genetic combinations / (ii) Chromosomes are arranged Checkpoint 7.1
different types of gametes in terms of at the equatorial plane
genetic combinations. During random during metaphase whereas S1 Food is made up of complex molecules
fertilisation, zygote possesses different homologous chromosomes are such as carbohydrate, protein and
combinations of diploid chromosomes. arranged at the equatorial plane lipid which are rich in chemical energy.
Therefore, variations occur among during metaphase I. During catabolism, bonds of complex
individuals/organisms within the same molecule are broken down to simpler
species. (iii) Create genetic variations among molecules to release energy.
daughter cells by independent
Checkpoint 6.4 assortment of homologous S2 Glucose
chromosomes during metaphase
S1 Cancer cell is due to uncontrolled mitosis. I. Checkpoint 7.2
Mutation occurs in gene which controls
cell cycle. Cancer also occurs when Section B S1 Glucose + Oxygen → Carbon dioxide +
normal cell undergoes transformation and 3. (a) (i) P: Telophase; Q: Metaphase; R: Water + Energy
become cancerous.
Prophase; and S: Anaphase S2 Presence of oxygen breaks down/
S2 Down syndrome is a genetic disease (ii) R, Q, S, P oxidises glucose molecule completely to
when each somatic cell consists of 47 (iii) Prophase release all energy stored in it.
chromosomes instead of the normal 46.
Individual with Down syndrome has slit • Chromosomes condense and S3 During aerobic respiration, energy
eyes / short body / mental retardation. coil stored in glucose is released as ATP.
Breaking down of glucose begins with
SPM Practice 6 • Chromosomes look shorter / glycolysis in cytoplasm. Glucose is then
thicker / can be seen under a oxidised in mitochondria where energy
Objective Questions 4. C 5. A light microscope is released in stages. Each stage is
1. B 2. D 3. D 9. B 10. C catalysed by enzymes.
6. A 7. C 8. A • Each chromosome consists
of a pair of sister chromatids Checkpoint 7.3
Subjective Questions joined at centromere
S1 Fermentation of lactic acid in muscle
Section A • Spindle fibres begin to form in cell:
1. (a) Phase P is cytokinesis. Formation of between centrioles
Glucose → Lactic acid + Energy
cell plate in the middle of cell. Cell • Each pair of centrioles moves
plate expands outwards until it fuses towards opposite poles of cell Fermentation of alcohol in yeast:
with plasma membrane. A new zymase
cell wall is formed, and two new • Nucleolus disappears // nuclear
daughter cells are produced. membrane disintegrates Glucose → Ethanol + Carbon
(b) (i) Mitosis dioxide + Energy
(ii) Apical meristem (tip of shoot or
S2 The breakdown of glucose is incomplete
tip of root) // lateral meristem // without the presence of oxygen, and
cambium some energy is still found in lactic acid /
ethanol molecule.

273

  Biology Form 4  Answers

S3 Anaerobic respiration in plant produces Release lots of Release a little • Helps to prevent it from
ethanol, carbon dioxide and energy energy energy collapse / damage
whereas anaerobic respiration in animal
produces lactic acid and energy. Produce 38 Produce 2 molecules • Ensure that the air passage is
always open
SPM Practice molecules of ATP of ATP
(b) (i) • Smoking causes alveolar
7 Produce carbon Produce lactic acid surfaces to dry and induce
dioxide, water and and energy in human coughing.
Objective Questions: 5. A energy muscle; or ethanol,
1. D 2. C 3. B 4. D 10. C carbon dioxide and • Reduce efficiency of gaseous
6. B 7. B 8. C 9. C energy in yeast and exchange (between alveolus
plant and blood capillaries.)
Subjective Questions:
Glucose is oxidised Glucose is not • Decrease oxygen intake into
Section A lungs.
1. (a) (i) P: Oxygen Q: Carbon dioxide completely oxidised completely
(ii) To break down / oxidise (ii) • Bronchitis / lungs cancer
Occur in Occur in cytoplasm • Cigarette smoke causes
glucose completely and release mitochondria
all energy stored in glucose dryness to respiratory
molecules. 8Chapter Respiratory System in passages / cigarette smoke
(iii) Muscle cell carries out anaerobic Humans and Animals contains tar which is a type of
respiration. Glucose molecule carcinogen.
is partially broken down to Checkpoint 8.1 (c) • Taking a deep breath
lactic acid. Energy produced is • Breath in more air / oxygen
less than that produced during S1 Lungs and skin. Both the organs are
aerobic respiration. moist and have large surface areas Section B
(b) (i) Volume of dough increases / as well as filled with network of blood 2. (a) • Lungs consist of many alveoli:
dough rises. Zymase in yeast capillaries.
dissociates glucose to ethanol provide large surface area and
and carbon dioxide. Presence S2 Surface of alveolus is where gaseous increase rate of diffusion of gases
of carbon dioxide increases exchange takes place. The presence of / oxygen / carbon dioxide
volume of dough. many alveoli increases surface area for • Surface of alveolus is moist:
(ii) Zymase denatured in high gas absorption, therefore increases the facilitate diffusion of oxygen /
temperature. Yeast respiration efficiency of gaseous exchange. carbon dioxide
does not occur. • Alveolus is covered with blood
(iii) Production of wine and beer capillaries: facilitate transport of
oxygen
Section B Checkpoint 8.2 • Capillary wall is only one-celled
2. (a) • Aerobic respiration begins with thick: facilitate diffusion of
S1 (a) By lowering the floor of buccal cavity respiratory gases
glycolysis in cytoplasm. (b) By raising the floor of buccal cavity (b) • Tar in cigarette smoke
• During aerobic respiration, S2 Yes. Volume of buccal cavity cannot • sticks to walls of alveolus /
bronchus
glucose is oxidised completely in be controlled. Air pressure in buccal • Heat of cigarette smoke
the presence of oxygen to release • Causes alveolus to dry / lose its
all energy stored in the molecule cavity cannot be controlled. Interruption moist characteristic
/ a large amount of energy is • Acid in cigarette smoke
released / 2898 kJ energy. of movement of water which contains • damages respiratory passage
• Carbon dioxide and water are • Tar in cigarette smoke
produced as waste products. dissolved oxygen. • Causes lung cancer
• The process is controlled by a • Blocks surface of alveolus
chain of complex biochemical Checkpoint 8.3 • Nicotine
reactions which are catalysed • Causes addiction
by respiratory enzymes in S1 Oxygenated blood is bright red whereas • Cigarette smoke contains carbon
mitochondrion. deoxygenated blood is dark red. monoxide
• Some of the energy is released as • Competes with oxygen to bind to
heat energy. S2 Carbon dioxide is acidic. If its content is haemoglobin
• The remaining energy is used to high in the body, blood pH will decrease (c) • Require more oxygen
synthesise ATP molecules from • To increase rate of respiration
ADP and inorganic phosphate. and interfere with processes in the body. • Supply more energy
(b) • Yeast contains zymase that • More carbon dioxide must be
catalyses conversion of glucose to Checkpoint 8.4 removed
ethanol, carbon dioxide and energy. • To maintain blood pH
• The process is known as alcohol S1 (a) Chronic obstructive pulmonary
fermentation by anaerobic disease is caused by obstruction 9Chapter Nutrition and Human
respiration in yeast. Digestive System
• Carbon dioxide that is released of respiratory passage due to
during fermentation by yeast helps
to rise dough. swelling or inflammation caused by
• Room temperature / optimum
temperature increases rate of persistent coughing.
reaction of zymase.
(c) (b) Asthma, bronchitis and emphysema
(c) Shortness of breath, coughing with

phlegm and breathing difficulty

S2 Yes. Coughing with phlegm and breathing
difficulty are symptoms associated with

COPD

SPM Practice 8 Checkpoint 9.1

Objective Questions 4. D 5. B S1 Stomach contains acid and enzymes
1. D 2. A 3. A 9. B 10. C to digest food. Its muscle contract
6. C 7. A 8. C
periodically to churn food so that food
Subjective Questions
can be better digested.
Aerobic Anaerobic Section A Q: Bronchus
respiration respiration 1. (a) (i) P: Trachea S2 • Duodenum accepts bile from gall
bladder to emulsify fats.
Occur in the Occur in the absence (ii) • Strengthened by cartilaginous
• It also accepts amylase, trypsin and
presence of oxygen of oxygen ring in the shape of C lipase from pancreas to hydrolyse

starch, polypeptides and fats.

274

Biology Form 4  Answers

S3 Ileum: to increase the movement period (iii) 3 1 0Chapter Transport in Humans and
of chyme in ileum so as to maximise Exchange of gases in alveolus Animals

absorption of nutrients.

Checkpoint 9.2 Mineral ions move from ground Checkpoint 10.1
water which is hypotonic into cells
S1 Oily or fatty food of root hair S1 Complete: Oxygenated blood and
• Secretion of bile juice / salt stops deoxygenated blood are completely
• Fats cannot be emulsified A drop of blue ink is added into a 3 separated in the blood circulation.
S2 No / less secretion of intestinal juice that glass of water which causes the
water to turn blue Incomplete: Oxygenated blood and
contains maltase, sucrase, lactase and deoxygenated blood are mixed in some
parts of the circulation.
erepsin. Digestion of maltose, sucrose,
S2 Mixing of oxygenated blood and
lactose and dipeptides stops / rate of deoxygenated blood. Blood transported
to body cells has lesser oxygen content.
digestion decreases.
Checkpoint 10.2
Checkpoint 9.3 (b) (i) Rice: Starch
Milk: Lactose S1 When heart stops beating, the individual
S1 Blood capillaries absorb water-soluble dies.
molecules such as glucose, amino acids (ii) • Rice is high in starch
and vitamin C whereas lacteal absorbs • Glucose is the basic unit of S2 (a) Platelet
fats-soluble molecules such as fatty (b) Leucocyte
acids and glycerol as well as vitamins A, carbohydrate S3 Oxygenated blood and deoxygenated
D, E and K. • Glucose is a type of simple
blood will be mixed. Lesser oxygen
S2 Vitamin B and C are absorbed by blood sugar will be transported to body cells. Less
capillaries whereas vitamins A, D, E and (c) • Benedictʼs solution is the indicator energy will be produced, and the
K are absorbed by lacteal. body cannot perform active physical
for reducing sugar exercises.
Checkpoint 9.4 • A test is considered positive when
Checkpoint 10.3
S1 Channel the end-product molecules of its blue colour solution is changed
digestion from small intestine to liver. to brick red precipitate S1 Can produce own impulse for muscle
• Sucrose is a non-reducing sugar contraction and relaxation without
S2 Carry out deamination process to stimulus from nervous system.
produce urea which is excreted as urine Section B
2. (a) Gastritis S2 Sinoatrial node impulse causes
or sweat. • Gastric juice corrodes wall of contraction to both atria whereas
atrioventricular node impulse causes
Checkpoint 9.5 empty stomach contraction to both ventricles.

S1 The longer faeces stay in large intestine, • Gastric ulcer S3 Right atrial upper wall → whole atrial
more water will be reabsorbed from wall → Bundle of His muscle fibres →
Obesity Purkinje fibres → whole ventricular wall
faeces and make it harder. Hardening of • Fats accumulate at arterial wall
• High blood pressure / stroke // S4 Shrinking of skeletal muscles weakens
faeces causes constipation. muscle contractions. Less / no pressure
cardiovascular disease on vein. Blood pressure in vein will be
S2 Fibres mix with faeces to increase water reduced. Difficulty in transporting blood
content of faeces. Soft faeces facilitate Anorexia nervosa to heart.
• Psychological problem / having
defaecation. S5 Heart is in thoracic cavity of chest.
the impression that he/she is fat Pulse is stronger in chest than arm
Checkpoint 9.6 • Malnutrition
Checkpoint 10.4
S1 Steamed fish does not contain oil that Muscle dysmorphia
increases cholesterol level in blood. • Obsess in muscle building S1 The scab is removed too early. Cells
• Diet taken is insufficient to provide below damaged tissue of wound have
S2 Imported fruits take a longer period not divided completely to heal the
to reach Malaysia. High temperature energy wound.
(b) (i) Chicken burger
destroys some vitamin C of imported S2 Interruption of blood circulation occurs in
(ii) • A 16 years old pupil requires a the leg region, and not at the heart.
fruits. lot of energy
S3 Haemophilic patient has lesser
S3 Lack of vitamin C causes scurvy. Fresh • Chicken burger contains coagulating factor in blood as compared
fruits contain high vitamin C content. high contents of fats and to blood of a healthy person.
carbohydrate
Checkpoint 9.7 Checkpoint 10.5
• Protein for growth
S1 Have appropriate diet and perform • However, chicken burger is S1 To ensure that the blood type of donor is
physical exercises. suitable for recipient.
high in salt
S2 Yes. Typhoid fever is spread by • Can cause high blood S2 Only blood type O. The mother’s blood
Salmonella typhi bacteria found contains anti-A and anti-B antibodies.
in faeces, which is transmitted by pressure Blood will coagulate if the mother
houseflies to food and drinks. • High fats receives blood from other blood types.
• Can increase blood
SPM Practice 9 S3 Rhesus negative recipient will produce
cholesterol antibodies to destroy rhesus antigen of
• and leads to heart attack donor’s blood.

Objective Questions OR
1. D 2. B 3. B 4. B 5. A
6. A 7. C 8. D 9. C 10. C • A 16 years old pupil requires a
lot of energy
Subjective Questions
• Bread is rich in carbohydrate
Section A • Provides energy
1. (a) (i) Simple diffusion • Vegetables contain a lot of
(ii) The crystal molecules diffuse
fibres
from higher concentration area • To avoid constipation
to lower concentration area • Less fats
until a dynamic equilibrium is • To prevent obesity
achieved.

275

  Biology Form 4  Answers

Checkpoint 10.6 (c) • Chemical substances in cigarette Similarities
smoke deposit on walls of arteries
S1 Cardiovascular system is an alternative to form plaques • Both are transport media
name for blood circulation system • Both contains nutrients and waste
that consists of heart, blood and • Lumen of blood vessel becomes
blood vessels. This system transports narrower / constriction of arteries substances
essential substances to body cells and / blockage of lumen
removes toxic waste substances out of Differences
the body. • Blood flow stops
• Blood vessels get swollen Blood Lymphatic fluid
S2 To ensure that all cells, tissues and
organs function properly. (d) (i) • Platelets gather at wound Closed Type Open
• Thrombokinase is released
S3 Angina is caused by narrowing of • Thrombokinase converts Two directions Direction One direction
arterial lumen whereas heart attack is (tissue to heart)
caused by blockage of heart arteries. prothrombin to thrombin
Blood can still flow through narrow • Thrombin converts fibrinogen Transport Function Transport
artery, but it stops flowing in blocked glucose / fatty acids
artery. Therefore, tissues will be more to fibrin amino acids / and glycerol /
severely damaged during heart attack. • Fibrin forms a mesh on top of vitamin B and vitamins A, D,
C / oxygen E and K
Checkpoint 10.7 wound / hormone /
• To trap blood cells enzyme / urea
S1 Fluid that flows in lymphatic vessel is • To form scab
known as lymph. Lymph returns useful Accept any three of the above Capillaries,
substances such as water, minerals and
hormones to maintain optimal blood statements that are in correct Heart, arteries, lymph vessels,
circulation system. sequence veins and
capillaries Structure lymph ducts,
S2 Plasma proteins and erythrocytes have (ii) • Haemophilia lymph nodes,
fixed shapes. Both cannot diffuse out at • Injection of clotting factor
arterial-capillary ends to form interstitial 2. (a) (i) P: Carotid artery Q: Vena cava and thymus
fluid and lymphatic fluid. (ii) Stroke / atherosclerosis
(b) • Transport deoxygenated blood glands
Checkpoint 10.8
from the whole body to heart (b) (i) • Examples of nutrients are
Disrupt contents and function of blood since • Consists of bicuspid valve to glucose
blood contents that diffuse out from blood
capillaries into intercellular spaces cannot prevent backflow of blood • Blood glucose enters arterial
diffuse back into blood circulation. (c) • Create a new passage for blood end which has high pressure

SPM Practice 10 5. D flow to bypass the blocked • Glucose diffuses through
10. B artery / arteriole on surface of capillary wall into spaces
Objective Questions 4. B heart caused by deposition of between cells
1. D 2. B 3. B 9. C cholesterol
6. B 7. B 8. D • Use artery from another part of • Glucose diffuses into cells
the body // femoral artery along the direction of glucose
Subjective Questions • Blocked artery / arteriole prevents concentration gradient
blood flow to cardiac muscles and
Section A Body causes muscular heart / cardiac (ii) • Blood vessel is not elastic
1. (a) cells failure • Plaque blocks part of blood
(d) (i) • Himalaya is a high altitude
Lungs vessel
region where its partial oxygen • Makes it difficult for blood to
pressure is low
• An individual must have more flow through
red blood cells to transport • And causes varicose vein
more oxygen • Less oxygen supply to leg
(ii) • Cell N is a crescent shaped
cell M causes
• Gene mutation / chemical
changes in gene structure 1 1Chapter Immunity in Humans
• Alters genetic code / code for
Body synthesis of amino acids Checkpoint 11.1
cells
Section B S1 Fever is a body protective response
(b) (i) 3. (a) (i) • Blood pressure is high at against infection. Increase in body
temperature above normal body
(ii) • Carry out regular physical arteriole-capillary end temperature stimulates production of
exercises • Some substances in blood are more antibodies. In addition, the rise
in body temperature kills and prevents
• Reduce intake of fatty food filtered out (except red blood multiplication of pathogen.
cells) into intercellular spaces
(between cells) to form tissue S2 The third line of defense is the specific
fluid. defense mechanism that involves
• Blood pressure is low at antibody production by lymphocyte.
capillary-venule end Lymphocyte must first recognise the
• Causes substances in tissue antigen before it can produce antibody
fluid to diffuse into venule against it.
• Some fluid (10%) remain in
intercellular spaces to form
tissue fluid.
(ii) Fluid Q is lymph

276

Biology Form 4  Answers

Checkpoint 11.2 SPM Practice 11 Provide immediate Takes a long time
S1 (a) S: Monocyte / Neutrophil / Phagocyte protection for the antibodies to
T: Lymphocyte Objective Question 4. C 5. D be produced
1. C 2. A 3. D 9. B
(b) Undigested 6. D 7. B 8. B

Pathogen substance Subjective Questions Stops when breast Provides long term

Section A feeding stops protection
1. (a) • Neutralisation
Neutrophil • Antibody binds to bacterial toxin Note:
• To neutralise / inactivate the Corresponding differences = 2 marks
S destroys pathogen by Non corresponding differences = 0 mark
phagocytosis. S surrounds and toxin. (b) • Acquired Immunodeficiency
engulfs pathogen. Lysozyme fuses (b) (i) Statement X
with phagosome. Lysozyme digests Syndrome
pathogen. Digested substance will • Acquire natural active • HIV infection
be absorbed by S. Undigested immunity • Weakens body immune system
substance will be excreted. • By destroying T-lymphocytes
• Measle virus infection • By disabling B-lymphocytes
(c) Antibody stimulates body to produce
(d) Specific antibodies so that they cannot produce
antibodies
Checkpoint 11.3 • Until the antibody • When body is infected with HIV
concentration exceeds level of • Immune system cannot function
S1 (a) X: Artificial active immunity immunity. • Body will be infected with various
Y: Artificial passive immunity types of pathogens
(b) • The antibodies produced • Avoid having multiple sex partners
remain in blood of the person • Avoid drug addiction and sharing
Similarities injection needle
(ii) Statement Y • Avoid having sex with person
• Acquire artificial active infected with AIDS
immunity
• Both are involved in body defense • Receive injection of measle 1 2Chapter Coordination and Response
system / fighting infection vaccine in Humans
• Body is stimulated to produce
• Both immunities involve injections antibodies
• Both require antibodies • Booster injection increases
antibody production
Differences • Until the concentration
exceeds the immunity level
Immunity X Immunity Y • The antibodies produced Checkpoint 12.1
remain in the blood of the
Body produces Body receives person S1 Components involved are stimulus,
antibodies antibodies from receptor, integration centre, effector and
outside (c) (i) Neutrophil / Monocyte response.
(ii) Phagocytosis
Receive injection of Receive injection of (iii) S2 External stimulus; e.g.: environmental
temperature, light intensity, sound
vaccine antiserum
Internal stimuli; e.g.: blood sugar level,
Does not show Shows immediate blood pH, blood osmotic pressure
immediate immune immune response
response S3 When photoreceptors of retina detect
light, nerve impulse will be produced
Lasted for a long Lasted only for a Section B and sent to central nervous system,
period short while 2. (a) then nerve impulse will be transmitted to
effector for appropriate response
Injection is given Similarities
after an individual is S4 To ensure the survival of organisms.
Vaccination is given infected or at high • Protects body against pathogenic infection
before an infection risk of acquiring the • Antibodies are produced outside of the Checkpoint 12.2
infection.
body S1 Central nervous system consists of brain
and spinal cord; peripheral nervous
(c) • In immunity X, the first injection Differences system consists of cranial nerves and
/ inoculation produces low spinal nerves.
concentration of antibodies. Method 1 Method 2
The second injection increases S2 The person will die because medulla
concentration of antibodies until oblongata regulates body activities that
above the immunity level. As are important for life.
such, a person is protected from
the pathogenic infection. Natural passive Artificial active Checkpoint 12.3
immunity immunity
• In immunity Y, the concentration of S1
antibodies from the first injection
decreases rapidly. If the infected Mother’s milk is Vaccine is Sensory Motor Relay
person has not recovered, a introduced into introduced into neuron neuron neuron
second injection is given again to baby’s body baby’s body
raise the antibody concentration.
Antibodies are found Antibodies will be Carry nerve Carry nerve Receive
in mother’s milk produced by the impulse impulse of and transmit
body from relay neuron nerve
receptor and transmit impulse
Checkpoint 11.4 Provide protection or sensory it to effector from
until the baby can organ to sensory
HIV can be transmitted by blood. Drug use its immune Antibodies remain central neuron
addict uses needle to inject drug into his/her system to produce in the body nervous to motor
body. This contaminates the needle with HIV. its own antibodies system neuron
HIV is transmitted to other addicts if they
share the contaminated needle.

277

  Biology Form 4  Answers

Found in Found in Found in Subjective Questions Section B
dorsal root ventral root central 3. (a) (i) • When nerve impulse reaches
of spinal of spinal nervous Section A
nerve nerve system 1. (a) (i) X: S ensory neuron / afferent synaptic knob, vesicle that
contains neurotransmitters
Possess Possess Possess neuron moves towards presynaptic
short axon long axon various Y: Motor neuron / efferent neuron membrane.
types of (ii) • Vesicle membrane fuses with
Cell body Cell body shapes presynaptic membrane and
is in the is at the XY neurotransmitters are released
middle of terminal of into synapse.
axon axon Function Transmit nerve Transmit nerve • Neurotransmitters diffuse
impulse from impulse from across synapse and bind to
receptor to spinal cord to specific protein receptors at
spinal cord effector postsynaptic membrane.
• New nerve impulse is initiated
S2 Impulse transmission in myelinated Structure Body cell is in Body cell is at and moves along receiver
neuron is faster than in unmyelinated the middle of the terminal of neuron or effector.
axon axon
neuron. (ii) • Alcohol slows down function
of central nervous system /
Checkpoint 12.4 Possess short Possess long slows down transmission of
nerve impulse to brain.
S1 To protect human from injuries and axon axon
dangerous situations . • Alcohol slows down reflex
(iii) Nerve impulse from receptor action and interferes
S2 Pain receptor in skin receives pain is transmitted to spinal cord. coordination, therefore causes
stimulus. Nerve impulse is triggered and However, his nerve impulse slurred speech / inability to
transmitted to spinal cord by sensory cannot be transmitted from control body balance
neuron. Sensory neuron transmits spinal cord to effector. His
impulse to relay neuron in grey matter hand biceps muscles could not • Long term consumption of
of spinal cord. Relay neuron sends contract // his hand was not alcohol causes damage to
impulse to motor neuron. Impulse is pulled away from hot object. cells in brain and liver.
transmitted to effector (arm muscles) by
motor neuron. Bicep muscles contract (b) To avoid serious injuries • Damaged brain leads
and pull the hand from the sharp nail. (c) (i) Motor neuron / efferent neuron to mental problems
(ii) Q is node of Ranvier. Q allows such as hallucination /
Checkpoint 12.5 unconsciousness.
impulse to jump from one node
S1 Alzheimer disease is due to failure of to another node, and speeds up • Damaged liver is associated
memory and thinking caused by death transmission of impulse. with liver cirrhosis / liver
of brain neurons. (iii) Speed of impulse becomes cancer.
slower
S2 Drug increases or decreases / inhibits (iv) Multiple sclerosis • Alcohol is acidic and can
action of neurotransmitter at synapses 2. (a) (i) P: Pituitary gland cause stomach ulcer / gastric
in brain. Therefore, brain cannot receive Q: Pancreas problem.
or send / transmit impulse to effector (ii) Pituitary gland secretes
to initiate a response towards any hormone that influence the (b) • Eye is the receptor which detects
stimulus. Alcohol is a relaxant that slows effects of other endocrine the stimulus, which is the robber.
down impulse transmission in brain. glands.
This slows down reflex action and also (b) (i) Insulin and glucagon • Nerve impulse is transmitted by
causes confusion and inability to control (ii) When level of blood glucose sensory neuron from receptor to
body balance. increases, insulin will be brain where it is interpreted.
secreted to convert excess
Checkpoint 12.6 glucose to glycogen, which is • Brain makes decision that the
stored in liver. robber is dangerous.
S1 Pituitary gland secretes some hormones (c) The symptom is called gigantism.
that control other endocrine glands. Excessive growth hormone is • Brain transmits nerve impulse via
secreted during childhood, which motor neuron to effectors, which
S2 • Both systems contain targeted tissues causes the individual to grow are muscles and adrenal gland.
and organs exceptionally tall.
(d) • Adrenal gland secretes
• Both systems produce responses adrenaline.
towards stimulus
• Adrenaline increases rate of
Checkpoint 12.7 Nervous system Endocrine system heartbeat and rate of breathing to
bring more oxygen to body.
S1 Excess antidiuretic hormone (ADH) Consists of nerves Consists of
causes detention of water in body. and neurons endocrine glands • Adrenaline also converts glycogen
Insufficient ADH causes diabetes in the liver into glucose.
insipidus where much urine is produced Transmission of Transmission of
frequently. information in information in the • Blood pressure increases to
the form of nerve form of hormone ensure more oxygen and glucose
S2 Hypothyroidism is caused by lack of impulse are sent to muscle tissues.
tyrosine hormone and reduces body Hormone is
metabolic rate and also rate of heart Nerve impulse transported to • These activities increase rates of
beat whereas hyperthyroidism is caused is transmitted to targeted organ in metabolism and cell respiration.
by excess tyrosine which increases targeted organ by blood circulation
metabolic rate and also rate of heart neurons • More energy is produced to fight
beat. Response or run away from the robber.
Response generated is slow
generated is fast 1 3Chapter Homeostasis and Human
Targeted area is Urinary System
SPM Practice Targeted area is widespread
12 5. C localised Checkpoint 13.1
Objective Question 10. C
1. C 2. C 3. A 4. B S1 Latent energy of body is used to
6. D 7. A 8. D 9. B evaporate sweat. Sweat evaporation

provides cooling effect to body.

278

Biology Form 4  Answers

S2 Carbon dioxide will dissolve in blood • Blood plasma is filtered out from S2 Importance of skeleton:
plasma to form carbonic acid. Carbonic glomerulus into Bowmanʼs capsule • Provides support and shape to body
acid then break down to form hydrogen by the high hydrostatic pressure • Protects internal organs such as
ions and hydrogen carbonate ions. High due to the smaller diameter of
increase in hydrogen ion reduces pH of efferent arteriole as compared to brain, lungs and heart
blood. afferent arteriole.
Checkpoint 14.2
S3 During active physical exercises, body • Filtered fluid is called glomerular
cells respire actively to produce lots of filtrate. S1
energy. The cells require more oxygen
and glucose for cell respiration. This • Glomerular filtrate contains water, Cervical Thoracic
reduces level of blood glucose. In glucose, amino acids, urea and vertebrae vertebrae
addition, cell respiration produces a other small molecules.
lot of heat energy. The heat produced Short spinous Long spinous
is transmitted throughout the body by • Glomerular filtrate does not process process
blood circulation system and causes contain red blood cells and
body temperature to increase. plasma proteins. Small centrum Large centrum

Checkpoint 13.2 (b) • Meat and egg contain high Short and wide Long transverse
contents of protein / amino acids.
S1 To move all useful substances back into transverse process process
blood circulation for body cells to use. • Excess amino acids will be
deaminated in liver and converted Possess a pair of No transverse
S2 (a) A little but concentrated urine to ammonia / urea / uric acid. transverse foramina foramina // Does
(b) A lot but diluted urine for vertebral not have passage
S3 Drug will be secreted into distal • Urea will be transported to kidney arteries to pass for vertebral
and excreted as urine. through // Possess arteries to pass
convoluted tubules during production of passage for through
urine. • Increase in urea concentration in vertebral arteries to
urine. pass through
SPM Practice 13
(c) • When a person drinks too
Objective Questions 4. B 5. D little water / sweat too much, S2 Cartilage reduces friction between
1. A 2. D 3. D 9. C 10. A the osmotic pressure will be bones during movement and absorbs
6. C 7. B 8. B increased. shock during movement. Synovial fluid
functions as lubricant to reduce friction
Subjective Questions • Osmoreceptor in hypothalamus between bones and provides nutrients
detects increase in osmotic for cartilage.
Section A pressure and nerve impulse is
1. (a) (i) X: Pituitary gland sent to pituitary gland. S3 Pair of muscles that acts
Y: Adrenal gland antagonistically, when one set of
(ii) Gland X which is pituitary • Pituitary gland is stimulated to muscles contracts the other set relaxes,
secrete ADH into blood circulation. and vice versa.
gland will secrete less ADH.
• ADH increases water permeability S4 Elbow joint: humerus, ulna and radius
Distal convoluted tubules and of walls of distal convoluted tubule Knee joint: femur, tibia and fibula
and collecting duct.
collecting tubules will be less Checkpoint 14.3
• More water will be reabsorbed
permeable to water / less water into blood capillaries. S1 There will not be any movement.
Movement only occurs when one set of
will be absorbed into blood • This will result in production of muscles contract and the other relaxes,
capillaries. Gland Y which is more concentrated but lesser and vice versa.
volume of urine.
adrenal gland will be stimulated S2 Wing possesses aerofoil shape and the
• When a person drinks too much bones are hollow filled with air sacs.
to produce more aldosterone. water, blood osmotic pressure will
be lowered / decreased Checkpoint 14.4
Distal convoluted tubules and
• Osmoreceptor in hypothalamus S1 Osteoporosis is a disease that causes
collecting tubules will be more detects decrease in osmotic bones to become brittle / porous and
pressure. easily broken. Causes:
permeable to salt / more salt
• Adrenal gland is stimulated to • Oestrogen was no longer secreted in
will be absorbed into blood produce aldosterone. menopausal women

capillaries. Lots of diluted urine • Aldosterone increases salt • Lack of calcium and phosphorous in
permeability and decreases water diet
will be produced. permeability of walls of distal
(iii) Gland Y which is adrenal gland convoluted tubule and collecting S2 Bent body posture for long hours while
tubule. sitting in front of computer causes
will be stimulated to secrete backbones to bend and congested
• Pituitary gland is less stimulated abdomen. This cause shoulders and
adrenaline to speed up the to secrete ADH. back muscles to get tired.

conversion of glycogen to • Walls of distal convoluted tubule
and collecting tubule will be less
glucose. Metabolic rate will permeable to water.

be increased to release more • This will result in production of
diluted but high volume of water.
energy.

(b) (i) Decrease in diffusion rate of
excreted substance such as

water, salt and urea into dialysis

fluid. This is due to the reduction

of surface areas of dialysis

tubules that is straightened. 1 4Chapter

(ii) Excess water in the body could Support and Movements in
not be excreted / reabsorption Humans and Animals

of water did not occur. Water SPM Practice 14

was accumulated in the body Checkpoint 14.1

especially at the leg and ankle S1 Leech possesses internal body fluid Objective Questions
which functions as hydrostatic skeleton 1. A 2. D 3. C 4. D 5. A
which caused the swelling. 6. C 7. B 8. B 9. B 10. D
that provides support and maintain its
Section B
2. (a) • Ultrafiltration takes place body shape.

279

  Biology Form 4  Answers • N is lumbar vertebrae S3 Spermatogenesis Oogenesis
• N is the biggest and strongest Occurs in testis
Subjective Questions Occurs in
vertebrae Produce 4 sperms ovary
Section A • N has large centrum to bear the
1. (a) Vertebra X has two passages / Produce one
weight of lower back of the body primary oocyte
transverse foramens for vertebral (c) • This man might have and three polar
arteries to pass through. Vertebra bodies
Y possesses a long spinal process osteoarthritis.
which projects to the back. • This is because of his age and his Checkpoint 15.3
(b) Provide surface for muscles
attachment. job as a football coach involves S1 Follicle-stimulating hormone (FSH),
(c) Slight movement at the joints. frequent use of joints Luteinizing hormone (LH), Oestrogen,
(d) • Control body balance while • The cartilage becomes thinner
and the bones ends meet during Progesterone
standing, sitting and moving in movement of joints.
various directions. • Since the treatment is not able S2 • Stimulates ovulation
• Can support 10 times the body to solve the problem he should • Stimulates development of corpus
weight of a person. undergo surgery to replace
(e) (i) Can cause injury to joints, the damaged knee joint with a luteum in ovary
stainless steel and polyethylene
soft tissues and bones. Body joint. • Stimulates ovary to secretes
weight is supported at front progesterone
edges of both feet. Lifting of 1 5Chapter Sexual Reproduction,
heel bone causes calf muscles Development and Growth S3 Menopause is caused by reduction of
to shorten. Excessive pressure in Humans and Animals FSH, LH, oestrogen and progesterone
will be applied onto knees
and backbones because high in the woman body
heel shoes have small surface
areas to support the entire body Checkpoint 15.1 Checkpoint 15.4
weight.
(ii) A person can become slightly S1 Structure Function S1 Fallopian tube
hunch due to backbones S2 To ensure corpus luteum produces
alignment problem. Ovary Produces ovum and
secretes oestrogen and appropriate amount of oestrogen and
Section B progesterone
2. (a) • Human skeleton consists of progesterone during first trimester of

axial skeleton and appendicular pregnancy.
skeleton.
• Axial skeleton consists of skull, Checkpoint 15.5
rib / thoracic cage and vertebral
column. S1 Identical twins and unidentical twins
• Appendicular skeleton // vertebral
column supports body weight // Fallopian Moves secondary oocyte S2 Identical twins Unidentical
provides places for joining rib twins
bones, pelvic girdles and surfaces tube to uterus with the aid of
for attachment of back muscles
and neck so as to provide support cilia
to the body.
• Appendicular skeleton consists of Uterus Structure where embryo One ovum is Two ova are
pectoral girdles, arm bones, pelvic attaches and develops fertilised by one fertilised by
girdles and leg bones. into foetus sperm to produce two sperms to
• Appendicular skeleton enables one zygote produce two
movement such as shifting zygotes
positions and performing other Cervix Secretes mucous that
physical activities. assists movement of
• Bones are joined by ligaments at sperms towards Fallopian Embryo that is Embryo that is
joints. tube formed divides formed does not
• Joints allow movement of bones into two divide into two
and body.
• Surface of bone (at joint) is Vagina Accepts penis during
covered with cartilage and sexual intercourse and
synovial fluid to reduce friction function as delivery Checkpoint 15.6
between bones. passage during childbirth
• Tendon joins muscle to bone. S1 Inability to produce children
• Tendon is a strong, non-elastic S2 The woman will have difficulty getting S2 • Testis could not produce sperms
tissue that can withstand pregnant. • Low sperm count
pressure. • Blocked sperm duct
• Skeletal muscles are arranged in Checkpoint 15.2 S3 • Tumour in uterus
antagonistic pair that contracts • Malfunction of uterus
and relaxes alternatively to move S1 Gametogenesis ensures survival of a • Damaged or blocked Fallopian tube
bones during body movement. species and avoid it from extinction.
(b) • M is thoracic vertebra. Checkpoint 15.7
• M has long spinous process S2 • Primordial germ cell (2n) undergoes
projected downwards. repeated mitosis to produce S1 Growth is the changes that occur in an
• M provides attachment sites for spermatogonium (2n). organism that increases number of cells,
muscles / ligament and attached volume, size and mass.
to ribcage to form thoracic cavity. • Spermatogonium (2n) develops and
produce primary spermatocyte (2n). S2 Dry mass is the mass of an organism
after removal of all its water content.
• Primary spermatocyte (2n) undergoes Organism must be killed and dried at
meiosis I to produce secondary very high temperature (e.g. 70°C in
spermatocyte (n). oven for plant) until its mass remains
constant. This method is suitable for
• Secondary spermatocyte (n) determining dry mass of plant.
undergoes meiosis II to produce
spermatid (n). S3 Lag phase, exponential phase,
stationary phase, maturation phase,
• Spermatid (n) differentiates to produce senescence phase and death phase.
sperm (n).

280

Biology Form 4  Answers

SPM Practice 15 (c) • Before birth, primordial germ cell Differences
(2n) undergoes repeated mitosis
Objective Questions to produce oogonium (2n). Human being Insect
1. C 2. D 3. B 4. A 5. C
6. D 7. C 8. A 9. B 10. B • Oogonium (2n) develops and S-shaped (Sigmoid) Ladder-like / Step-
produce primary oocyte (2n)
shape graph like shape graph
• Primary oocyte (2n) undergoes
Subjective Questions meiosis but stops at prophase I. There are six There are two
phases in its phases in its
Section A • Secondary oocyte (n) undergoes growth curve growth curve
1. (a) (i) P: Seminal vesicle meiosis II and stops at metaphase
Q: Prostate gland II. No distinct vertical Five distinct vertical
R: Sperm duct or horizontal lines lines and 5 distinct
S: Testis • Matured Graafian follicle moves to differentiate each horizontal lines to
(ii) T: Holds and protects testis close to surface of ovary and phase indicate each instar
U: Copulating organ which releases secondary oocyte into stage
Fallopian tube. The process is
ejaculates sperms into vagina known as ovulation. Absence of point Presence of five
(b) (i) X: Fallopian tube Y: Uterus to indicate zero zero growth (on 5
• If fertilisation occurs, secondary growth horizontal lines)
(ii) • Produces secondary oocyte oocyte will complete meiosis II to
produce ovum (n) and three polar
• Secrete or estrogen and bodies (n). If fertilisation does not
occur, endometrium sheds and
progesterone becomes menstrual blood. Absence of rapid Presence of (five)
growth rapid growth
(iii) Sperm cannot fertilise ovum and (d)
zygote will not be produced. Presence of
ecdysis
The woman will have difficulty Absence of ecdysis

getting pregnant.
2. (a) (i) P: FSH Q: LH
Spermatogenesis Oogenesis Human cells Mitosis only occurs
(ii) Oestrogen: Repairs and are constantly during ecdysis
Occurs in testis Occurs in ovary undergoing mitosis
thickens endometrium

Progesterone: Stimulates Produce four Produce one ovum Absence of Absorption of air
thickening of endometrium sperms and three polar bodies absorption of air to during ecdysis to
increase size increase size
(iii) Ovulation cannot occur and The smallest cell The largest cell in
corpus luteum cannot be formed in human body human body

in ovary. Complete Meiosis II will only
meiosis be completed if
(b) (i) Umbilical cord arteries - carry fertilisation takes PRE-SPM MODEL PAPER
deoxygenated blood and place.
Paper 1
nitrogenous waste substances 1. D 2. A 3. B 4. D 5. A
6. C 7. B 8. A 9. D 10. A
from foetus to placenta. 11. C 12. B 13. C 14. A 15. C
16. B 17. C 18. C 19. C 20. C
Umbilical cord veins – carry 4. (a) • Pituitary gland secretes follicle- 21. C 22. B 23. D 24. A 25. B
oxygenated blood and nutrients stimulating hormone (FSH) to 26. C 27. B 28. B 29. B 30. A
stimulate development of follicle in 31. B 32. B 33. A 34. A 35. A
from placenta to foetus. ovary. 36. B 37. D 38. D 39. B 40. A
41. A 42. A 43. A 44. A 45. B
(ii) Function as endocrine organ • Primary oocyte in primary follicle 46. C 47. D 48. C 49. A 50. C
that secretes oestrogen and will develop into secondary oocyte
in Graafian follicle. Paper 2
progesterone after the first
• Developing follicle releases Section A
trimester / fourth month of oestrogen. 1. (a) Cell P is a plant cell because it

pregnancy. • Oestrogen stimulates development contains chloroplast.
of follicle and thickening of (b) (i) P: Palisade mesophyll cell
Section B endometrium. Q: Sperm cell
3. (a) Meaning of gametogenesis: (ii) Palisade mesophyll cell contains
• Gametogenesis is the process • When oestrogen rises to a high
enough level, it stimulates pituitary many chloroplasts as shown
which produces gametes / gland to secrete LH. in the histogram. Sperm cell
reproductive cells contains many mitochondria and
• LH level which peak on day-13 no chloroplast as shown in the
Importance of gametogenesis: stimulates ovulation on day-14 to histogram.
• Increase offspring to ensure release secondary oocyte from (c) (i)
Graafian follicle.
survival of species Rough
• Produce variations in subsequent • Remaining follicular tissue will endoplasmic
form corpus luteum. reticulum
generations through cross
linkages during meiosis • Corpus luteum secretes Golgi
• Maintain the diploid status of cells progesterone that stimulates apparatus
in offspring thickening and increase of blood
(b) • Spermatogenesis is a process of vessels in endometrium. Chloroplast
sperm production that occurs in
the seminiferous tubules • If fertilisation does not occur, LH Mitochondria
• Primordial germ cell (2n) level drops and causes corpus
undergoes repeated mitosis to luteum to break down. This
produce spermatogonium (2n) reduces secretion of oestrogen
• Spermatogonium (2n) and progesterone.
develops and produce primary
spermatocyte (2n) • Endometrium sheds and
• Primary spermatocyte (2n) menstruation occurs.
undergoes meiosis I to produce
secondary spermatocyte (n) (b)
• Secondary spermatocyte (n)
undergoes meiosis II to produce Similarities (ii) Cell of salivary gland
spermatid (n) (d) (i) Smooth endoplasmic reticulum
• Spermatid (n) differentiates to • Height of human and length of instar (ii) Liver cell
produce sperm (n). increases with time.

• Both shows zero growth during maturity
phase.

281

  Biology Form 4  Answers

2. (a) (i) Meiosis • Glomerular filtrate contains urea • Undigested material moves
(ii) very slowly and is assisted by
waste material, water, glucose, peristalsis.

amino acids and mineral salts • Reabsorption of water and
minerals in colon produce
• When the filtrate flows through faeces, a semi-solid waste.

renal tubules, useful substances • Faeces contain dead cells
/ waste materials such as
are reabsorbed by blood bile pigments and toxic
substances.
capillaries which surround the
(ii) • Lack of protein
Accept any one of renal tubules • Lack of protein in children

the figures above • Glucose and amino acids are can cause marasmus and
kwashiorkor.
(b) (i) P: Chiasma reabsorbed by active transport • Amino acids in protein are
Q: Tetrad / Bivalent needed for growth and other
whereas water is reabsorbed by body functions in children.
• Symptoms of marasmus
osmosis include sudden and extreme
loss in body weight.
(ii) (c) • Dialysis fluid contains mineral • Kwashiorkor causes loss in
body tissues, decrease in
salts to ensure that blood does brain mass, bones clearly
seen beneath skin, poor
not lose mineral salts by diffusion. digestion and protruded navel.
• Vitamin B12 is used to form
(d) • By diffusion red blood cells.
• Food lack of vitamin B12
• Urea concentration is higher causes anaemia in children.
• The affected children will be
(iii) Process that occurs in P in blood of patient than urea lethargic, experiences frequent
is known as crossing over. fainting, have pale skin and
Crossing over enables exchange concentration in dialysis fluid. often feel tired.
of genetic materials between (b) • Unbalanced diet // Diet which
homologous chromosomes. • Urea and other waste materials does not contain seven classes of
food in an appropriate ratios
(c) (i) Down syndrome will diffuse out of blood into • Excessive lipids / fats …
(ii) Male • Causes cardiovascular disease
(iii) Down syndrome is caused by dialysis fluid through cellophane such as obesity
• Excess carbohydrate
an addition of one chromosome membrane. • Causes obesity / increase in blood
to the existing pair of glucose
chromosome-21 in zygote. (e) • Dialysis fluid is isotonic to blood • Adequate amount of protein
(iv) Flat face, slanted eyes and • Is needed for normal growth
protruding tongue plasma of a normal person. • Lack / no vitamin
(Accept any one of the characteristics) • Causes vitamin deficiency
3. (a) (i) P: Simple diffusion • Change in osmotic pressure of diseases such as night blindness
Q: Facilitated diffusion / rickets / osteomalacia / anaemia
(ii) P: O xygen / Fatty acid / Glycerol patient / pellagra / beri-beri. Note: Any
Q: Glucose / Amino acid appropriate example of vitamin
R: Sodium ion / Potassium ion • Osmosis takes place deficiency
(iii) • Lack / no fibres
• Molecules diffuse between dialysis • Causes sickness related to
QR insufficient minerals / stunted
fluid and blood (flows in dialysis growth / muscle cramps / heart or
Similarities kidney failure / dental decay
tubing) 7. (a) (i)
5. (a) X: Glucagon Y: Insulin
(b) • Organ P is pancreas, which
secretes hormone X known as

glucagon

• Glucagon stimulates conversion of
glycogen to glucose in organ Q,

which is liver

• Blood glucose level increases

back to normal

(c) (i) • Urine of diabetic patient

Both processes involve carrier protein contains glucose

• Kidney fails to reabsorb all

Differences glucose from glomerular

filtrate

Movement of Movement of • Glucose is secreted into urine
molecules down molecules against
concentration concentration (ii) • Add 2 to 3 drops of Benedictʼs
gradient gradient
solution into 2 ml urine sample

in a test tube

Do not require ATP Require ATP / • Incubate the test tube in a

/ energy energy boiling water bath

The process results • Brick red precipitate will be
in accumulation
The process of excretion of observed if there is glucose in
continues until a substances from
dynamic equilibrium cell the urine
is achieved
(d) • Immediately swallow a glucose

tablet or drink glucose solution.

(iv) Plasma membrane is thin • Excessive dose of insulin / Organism P Organism Q
and semi-permeable allows hormone Y will empty glucose in

only small molecules to pass blood

through. • Taking glucose will provide Similarities

(b) • The concentrated salt solution is immediate energy to the diabetic
hypertonic to fish cells / Fish cells
patient • Both have closed blood circulation
are hypotonic to concentrated salt • Blood flows in blood vessels
Section B
solution Differences
6. (a) (i) • After the nutrients were
• Water diffuses out of the cell by absorbed in the small Single blood Double blood
osmosis intestine, the remaining circulation system circulation system
intestinal content moves to
• Microorganisms are unable to colon. Blood flows through Blood flows through
reproduce / survive heart only once in heart twice in one
• The intestinal content one complete cycle complete cycle
4. (a) P: Proximal convoluted tubule consists of a mixture of water,
Q: Loop of Henle undigested food (fibres) and
(b) • Reabsorption process waste materials.

282

Biology Form 4  Answers

Heart has two Heart has four • Menstrual blood will be discharged 9. (a) (i) • Reflex action is an automatic
chambers chambers response
Day 7 – 13
Heart has one Heart has two atria • Follicle cells / tissues secrete • Which is instantaneous (very
atrium and one and two ventricles fast) / involuntary
ventricle oestrogen
Has septum • Oestrogen helps to repair • Without being controlled by
No septum brain / involuntary action
Oxygenated blood endometrial wall / uterine layer
Oxygenated blood flows from lungs to • Oestrogen inhibits pituitary gland • Which involves only skeletal
flows from gills to the rest of the body muscles and spinal cord
the rest of the body from secreting FSH
Deoxygenated • Oestrogen stimulates pituitary Importance
Deoxygenated blood flows from • Prevent a person from serious
blood flows from the rest of the body gland to secrete LH
the rest of the body to lungs injury
to gills Day 14 • because reflex action happens
• LH stimulates ovulation
• Graafian follicle will release instantaneously
(ii) • When a tendon below knee is
secondary oocyte
• LH will stimulate formation of knocked, quadriceps muscles
(ii) • Name of disease: Hole in tighten
heart / Adult congenital heart corpus luteum • Stretch receptor in quadriceps
defects muscles is stimulated
Day 15 – 28 • Nerve impulse is initiated at
• Place: At heart septum // • Corpus luteum will secrete sensory neuron
Muscular wall that separates • Sensory neuron transmits
right ventricle and left ventricle progesterone and oestrogen nerve impulse to motor neuron
in spinal cord via dorsal root
• Oxygenated blood and • Progesterone will sustain thickening of spinal nerve
deoxygenated blood will mix of endometrial wall / uterine layer • Nerve impulse crosses
(a little) synapse between sensory
• Corpus luteum will shrink and neuron and motor neuron
• Decrease in supply of reduce its progesterone secretion • Motor neuron transmits nerve
oxygenated blood from heart impulse to effector, that is
to the rest of body // Low • Endometrial wall / uterine layer will quadriceps muscles via ventral
blood pressure that flows from shrink root of spinal nerve
the heart • The stimulated quadriceps
• Menstrual cycle will start again muscles contract which
• Body cells receive less oxygen (b) • Fluid X is amniotic fluid causes the leg to kick forward
• Reduced in rate of respiration • It acts as a cushion / absorbs
(iii) • New neurones cannot be
// less ATP energy will be shock from external pressure produced to replace damaged/
produced dead neurones
• Organism Q will be weak / • It acts as lubricant for body of
easily fainted // can have only foetus such as preventing fingers • Which cause less production
limited physical activity of neurotransmitter / dopamine
(b) • The infant acquires natural and toes from growing together
passive immunity through • Nerve impulse crosses
mother’s milk. (webbing) synapse at a very slow pace
• Mother’s milk contains antibodies
to fight infection caused by • Contains antibodies to protect • Therefore, impulse
pathogens // diseases such foetus from infections transmission slows down
as diarrhoea / stomach ache /
asthma / inflammation of lungs / • Allow foetus to float in the • Cerebral arteries harden
infection of lungs // reduce risk of amniotic sac • Muscular function is inefficient
lungs inflammation
• Early milk of mother / colostrum • Allow foetus to move freely in the / muscles harden / muscles
contains a lot of antibodies sac weaken
• Antibodies produced by mother • Trembling movement /
cannot be maintained for long // • Allow muscles and bones of problem with balance
will disintegrate foetus to develop well / problem with body
• The antibody level in infant blood coordination
will be below the immunity level • Amniotic fluid consists of water (b) • When chased by a goose /
• Antibody action is specific to the from mother’s body and urine of stimulus is received / fear is
pathogen developed
• The produced antibody is not foetus • Nerve impulse is transmitted from
effective against diseases such as receptor to brain
hepatitis (c) Smoking • Brain interprets information
• Therefore, infant must be • Cigarette smoke contains • Nerve impulse is transmitted from
vaccinated brain to adrenal gland
• Infant will then acquire artificial nicotine, carbon monoxide and • Adrenal gland is stimulated to
active immunity produce adrenaline
• Lymphocytes are stimulated to carcinogenic substances • Which increases heart pulse rate
produce antibodies against the / blood pressure / blood flow to
virus • Which can diffuse from mother’s muscles
8. (a) Day 1 – 6 blood across placenta into blood • Which increases breathing rate /
• Pituitary gland secretes FSH glucose level
• FSH stimulates development of vessel of foetus. • Which increase metabolic rate
ovarian follicle • More energy is produced
• Endometrial wall / uterine layer • Cause premature birth / stillbirth • To fight or to run away from the
disintegrates / death of baby during childbirth / goose
• Nerve impulse is transmitted to
birth handicap / underweight baby skeletal muscles to initiate running
away from goose
Drug
• Drugs such as cocaine /

methamphetamine / heroine
• Contains dangerous chemicals
• Which increases the chances of

having baby who is deformed
/ brain damaged / pre-matured
/ underweight / stillbirth (death
during pregnancy)

Alcohol

• Alcohol is transferred from
mother’s blood to blood of foetus

• Interfere with foetus growth /
foetal deformities / miscarriage /

stillbirth

• Cause foetal alcohol syndrome
(FAS)

283

  Biology Form 4  Answers

Paper 3 Volume of fruit juice that is (b) (i) 1. The volume of lemon juice 2. What is the effect of pH on a
1. (a) used to decolourise 1 ml of used to decolourise 1 ml mixture / pepsin-albumin?
DCPIP solution is 1.5 ml
Type DCPIP solution (ml) 3. D oes pH affect activity of enzyme /
of fruit 2. The volume of pineapple pepsin?
juice 1.5 juice used to decolourise 1 ml
DCPIP solution is 2.5 ml 4. Is acidic pH medium optimum for
Lemon 2.0 activity of enzyme / pepsin?
(ii) 1. L emon juice contains the
Mango 2.5 highest concentration of • Hypothesis
vitamin C 1. Pepsin activity is highest at acidic
Pineapple
2. Pineapple juice contains medium of pH 2 / 3
the lowest concentration of 2. A t pH 3, pepsin converts cloudy
vitamin C
albumen suspension to a clear
(c) solution faster than at pH 7 or pH 8
3. pH influences pepsin activity in
Variable Method to handle the variable converting albumen suspension to
a clear solution
Manipulated variable: Use different types of fruit juices (such as lemon juice, 4. A ctivity of pepsin in converting
Type of fruit juice mango juice and pineapple juice) albumen suspension to a clear
solution depends on pH
Responding variable: Use a syringe to record the volume of fruit juice used
• Variables
Volume of fruit juice used to to decolourise 1 ml of DCPIP solution // Calculate 1. M anipulated variable: pH
2. Responding variable: Activity of
decolourise 1 ml of DCPIP percentage of concentration of vitamin C by using the
enzyme / pepsin / cloudiness of
solution // percentage of vitamin C formulae: mixture / condition of mixture
3. C ontrolled variable: Concentration
concentration Percentage of vitamin C in fruit juice (%) = / volume of albumen (suspension) /
pepsin (solution)
Volume of 0.1% ascorbic asid
Volume of fruit juice used × 0.1% • List of apparatus and materials
Apparatus:
Concentration of vitamin C in fruit juice (mg) = Bunsen burner, beaker, thermometer,
Volume of 0.1% ascorbic asid
wire gauze, tripod stand, stopwatch,
Volume of fruit juice used × 1.0 mg cm–3 test tubes, measuring cylinder,
syringe, dropper
Controlled variable: Fix the same volume of DCPIP at 1 ml for each of the
Volume of DCPIP solution three experiments. Materials:
Albumen (suspension), 1% pepsin
(d) 1. Lemon juice contains the highest percentage of vitamin C as compared to mango
juice and pineapple juice (solution), distilled water, ice cubes,
0.1 M hydrochloric acid, 0.1 M sodium
2. Volume of lemon juice that is used to decolourise 1 ml DCPIP solution is the least hydroxide solution, pH paper
as compared to the volumes of mango juice and pineapple juice
• Experimental procedure
(e) (i) 1. L abel three test tubes as A, B and C.
2. Fill each test tube with 5 ml
Type of fruit juice Volume of fruit juice Percentage of Concentration of
used to decolourise 1 vitamin C (%) vitamin C (mg cm–3) albumen suspension with a syringe.
Lemon ml of DCPIP solution 3. A dd the following solutions to each
Mango 0.007 0.667
Pineapple (ml) 0.005 0.5 of the three test tubes:
0.004 0.4 A: 1 ml of 0.1 M hydrochloric acid +
1.5
1 ml 1% pepsin solution
2.0 B: 1 ml distilled water + 1 ml 1%

2.5 pepsin solution
C: 1 ml 0.1 M sodium hydroxide + 1
(ii) Lemon juice • There will be less ascorbic acid /
vitamin C in the mango juice that ml 1% pepsin solution
Mango juice had been left overnight 4. D ip a piece of pH paper into each
Pineapple juice
• Because the vitamin C had been test tube and record its pH.
(f) • Lemon juice contains the highest oxidised by the oxygen dissolved 5. Immerse all three test tubes into
concentration of vitamin C as in the juice.
compared to mango juice and a water bath set at 37°C for 20
pineapple juice (h) • Acid ascorbic content in fruit juice minutes using a stopwatch.
• Can be determined by measuring 6. Observe the appearances of the
• This is because lemon juice mixtures in the beginning and after
contains the most amount of the volume of fruit juice required to 20 minutes in a table
ascorbic acid decolourise 1 ml DCPIP solution 7. R epeat steps 1 – 6 to get average
• And the acid ascorbic content in reading.
• Therefore, lemon juice requires a fruit juice is influenced by the • Presentation of data
the least volume to decolourise 1 types of fruit juices
ml of DCPIP solution (i) Appearance

(g) • More than 2.2 ml of mango juice Apparatus Material of mixture Appearance
will be required to decolourise 1
ml DCPIP solution • Syringe with • Mango juice Test pH at the of mixture
needle • Pineapple juice tubes beginning after 20
• Lemon juice
• Reagent bottle • Acid ascorbic A of the minutes
• Beaker • DCPIP solution B
C experiment

2. • Problem statement:
1. W hat is the effect of PH on enzyme

/ pepsin?

284