FORM

4BIOLOGY

KURIKULUM STANDARD SEKOLAH MENENGAH

WRITERS
Gan Wan Yeat
Nor Azlina binti Abd. Aziz
Yusmin binti Mohd. Yusuf
Noor Haniyatie binti Ibrahim

TRANSLATORS
Michelle Harris
Nor Azlina binti Abd. Aziz
Yusmin binti Mohd. Yusuf

EDITORS
Stella Melkion
Mohd Haziq bin Kahar

DESIGNER
Siti Aishah binti Mohd

ILLUSTRATORS
Hazli bin Hashim
Sara Fateha binti Muhd Kamil
Nigel Teoh
Tsen Wee Ling
Chan Khong Wei
Lin Pu Kuan

2019
Book Serial No.: 0177

KPM2019 ISBN 978-967-2250-83-8

First Published 2019
©  Ministry of Education Malaysia

All rights reserved. No part of this book may
be reproduced, stored in any retrieval system or
transmitted in any form or by any means,
electronic, mechanical, photocopying, recording
or otherwise, without prior permission of the
Director General of Education, Ministry of
Education Malaysia. Negotiation is subject to the
calculation of royalty or honororium.

Published for the Ministry of Education Malaysia by:
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KEMENTERIAN
PENDIDIKAN
MALAYSIA

The publishing of this textbook involves
cooperation from various parties. Our
wholehearted appreciation and gratitude goes
out to all involved parties:

•	 Educational	 Resources	 and	 Technology
division.

•	 Committee	 members	 of	 Penyemakan
Naskhah Sedia Kamera, Educational
Resources and Technology Division,
Ministry of Education Malaysia.

•	 Committee	 members	 of	 Penyemakan
Pembetulan Naskhah Sedia Kamera,
Educational Resources and Technology
Division, Ministry of Education Malaysia.

•	 Officers	of	the	Educational	Resources	and
Technology Division and the Curriculum
Development Division, Ministry of
Education Malaysia

•	 Officers	of	the	English	Language	Teaching
Centre (ELTC), Institute of Teacher
Education Malaysia, Ministry of Education
Malaysia.

•	 Chairperson	and	members	of	 the	Quality
Control Panel.

•	 Editorial	 Team	 and	 Production	 Team,
especially the illustrators and designers.

Acknowledgement
iii

Take Note!

Brainstorm!

Malaysian Innovation

Across the Fields

Kerjaya MileniaMillennial Career

STEM Bulletin

Our World of Biology

Biological Lense

Additional information related to theories
and concepts Higher-order thinking questions which

cover applying, analysing, evaluating and
creating questions

Reminders for students when carrying
out activities or experiments

Questions to test pupils’ understanding
at the end of every subtopic

Questions to test pupils’ understanding
at the end of every chapter

A summary of the main concepts

A checklist on the mastery of concepts for
pupils’ reference

Illustrates the application of theories and
concepts learnt in everyday life

Challenges pupils to think critically and
creatively

The successes of Malaysian scientists and
the related scientific and technological
developments in Malaysia

Describes careers related to a concept
or topic

Highlights current developments in the
related fields of science and technology

Shows the connection between biology
and other fields

Suggested activities for pupils to
carry out

This textbook has been written based on the Dokumen Standard Kurikulum dan Pentaksiran (DSKP)
for Biology Form 4, published by the Ministry of Education, Malaysia. It is guided by the concepts
of Scientific Skills, Science Process Skills, Higher Order Thinking Skills, 21st Century Skills and the
STEM teaching and learning approach. The objective of this textbook is to develop well-rounded
pupils and to equip them with the skills needed for the 21st century.

Preface

Special features of this book:

Summary

Self Reflection

Summative Practice

Activity Zone

4.2.3

Learning Standard based on the Form
4 Biology Dokumen Standard Kurikulum
dan Pentaksiran (DSKP)
iv

This book consists of digital resources in the form of Augmented Reality (AR) and
QR codes. These resources can be accessed by scanning the AR or QR codes
which can be found in selected pages, using a smartphone or electronic tablet.

A Fun Learning Experience

Steps in accessing the AR resources

1 3Scan

SnapLearn

1 2

Download
Snaplearn APP

Activate by scanning
this QR Code

Exciting New Experience While Reading

Scan to
experience AR

iOS 7.0 or Android 4.0 and above

3 Scan

SnapLearn

1 2

Download
Snaplearn APP

Activate by scanning
this QR Code

Exciting New Experience While Reading

Scan to
experience AR

iOS 7.0 or Android 4.0 and above

3

Scan

SnapLearn

1 2

Download
Snaplearn APP

Activate by scanning
this QR Code

Exciting New Experience While Reading

Scan to
experience AR

iOS 7.0 or Android 4.0 and above

3

Download the
SnapLearn app onto
your electronic device.

Scan the pages
with the above icon
to access the AR

resources.

2 Scan

SnapLearn

1 2

Download
Snaplearn APP

Activate by scanning
this QR Code

Exciting New Experience While Reading

Scan to
experience AR

iOS 7.0 or Android 4.0 and above

3

AR

Accessing the QR resources
A variety of complementary content such
as videos, additional activities, quizzes and
additional information can be accessed by
scanning the QR codes found in selected pages
using any QR code scanner.

Click on the
button and scan the

above code.

ICT 4.4

Activity: Studying the
effects of changes in
total surface area to
volume ratio (TSA/V)

on diffusion rate.iv

ICT 7.1

Quiz: Test your
understanding of

lipids

ICT 10.1

Video: Aerobic
respiration
5  Metabolism and Enzymes 84
 5.1 Metabolism 86
 5.2 Enzymes 86
 5.3 Application of Enzymes in Daily Life 95

6  Cell Division 98
 6.1 Cell Division 100
 6.2 Cell Cycle and Mitosis 101
 6.3 Meiosis 105
 6.4 Issues of Cell Division on Human
  Health 108

7  Cellular Respiration 112
 7.1 Energy Production through
  Cellular Respiration 114
 7.2 Aerobic Respiration 115
 7.3 Fermentation 117

PHYSIOLOGY OF
HUMANS AND
ANIMALS 125

8  Respiratory Systems in
Humans and Animals 126

8.1 Types of Respiratory System 128
 8.2 Mechanisms of Breathing 132
 8.3 Gaseous Exchange in Humans 136
 8.4 Health Issues Related to the Human
  Respiratory System 138

9  Nutrition and the Human
Digestive System 142

9.1 Digestive System 144
 9.2 Digestion 144
 9.3 Absorption 151
 9.4 Assimilation 153
      9.5  Defaecation   155
     9.6  Balanced Diet 156
     9.7 Health Issues Related to the
  Digestive System and Eating Habits 161

1

2

Title Page  i

Copyright Page and Acknowledgments ii

Preface iii

FUNDAMENTALS OF
BIOLOGY 1

1  Introduction to Biology and
Laboratory Rules 2

1.1 Fields and Careers in Biology 4
 1.2 Safety and Rules in a Biology Laboratory 6
 1.3 Communicating in Biology 11
 1.4 Scientific Investigation in Biology 15

2  Cell Biology and Organisation 20
 2.1 Cell Structure and Function 22
 2.2 Living Processes in Unicellular Organisms 28
 2.3 Living Processes in Multicellular
  Organisms 30
 2.4 Levels of Organisation in Multicellular
  Organisms 37

 3  Movement of Substances

across a Plasma Membrane 44
 3.1 Structure of Plasma Membrane 46
 3.2 Concept of Movement of Substances
  across a Plasma Membrane 49
 3.3 Movement of Substances across a
  Plasma Membrane in Living Organisms 57
 3.4 Movement of Substances across a
  Plasma Membrane and its Application
  in Daily Life 65

 4  Chemical Compositions in

a Cell 70
 4.1 Water 72
 4.2 Carbohydrates 73
 4.3 Proteins 76
 4.4 Lipids 77
      4.5  Nucleic Acids 79

v

Contents
10 Transport in Humans
  and Animals 166
 10.1 Types of Circulatory System 168
 10.2 Circulatory System of Humans 173
 10.3 Mechanism of Heartbeat 180
 10.4 Mechanism of Blood Clotting  183
 10.5  Blood Groups of Humans 185
 10.6 Health Issues Related to the
  Human Circulatory System   187
 10.7  Lymphatic System of Humans 189
 10.8 Health Issues Related to the
  Human Lymphatic System 193

11 Immunity in Humans 198
 11.1 Body Defence 200
 11.2 Actions of Antibodies 204
 11.3 Types of Immunity 205
 11.4 Health Issues Related to Immunity 209

12 Coordination and Response
  in Humans 214
 12.1 Coordination and Response 216
 12.2 Nervous System 220
 12.3 Neurones and Synapse 223
 12.4 Voluntary and Involuntary Actions 226
 12.5 Health Issues Related to the Nervous
  System 230
 12.6 The Endocrine System 233
 12.7  Health Issues Related to the
  Endocrine System 238

13 Homeostasis and the Human
  Urinary System 244
 13.1 Homeostasis 246
 13.2 The Urinary System 252
   13.3 Health Issues Related to
  the Urinary System 259

14 Support and Movement in
  Humans and Animals 264
 14.1 Types of Skeleton 266
 14.2 Musculoskeletal System of Humans 267
 14.3 Movement and Locomotion 272
 14.4 Health Issues Related to the Human
  Musculoskeletal System 277

15 Sexual Reproduction,
  Development and Growth in
  Humans and Animals 284
 15.1 Reproductive System of Humans 286
 15.2 Gametogenesis in Humans 288
 15.3 Menstrual Cycle 293
 15.4 Development of Human Foetus 297
 15.5 Formation of Twins 300
    15.6 Health Issues Related to the
  Human Reproductive System 302
 15.7 Growth in Humans and Animals 303

Glossary 311

References 314

vi
Chapter 1 Introduction to Biology and Laboratory Rules
Chapter 2  Cell Biology and Organisation
Chapter 3  Movement of Substances across a Plasma Membrane
Chapter 4 Chemical Composition in a Cell
Chapter 5 Metabolism and Enzymes
Chapter 6 Cell Division
Chapter 7 Cellular Respiration

This theme
aims to provide a basic

understanding of biology
through scientific investigation
with the inquiry approach. This
theme focuses on cells as basic

living units, the role of the plasma
membrane in cell physiology,
chemical composition in cells

and cell division.

1
THEME FUNDAMENTALS

OF BIOLOGY
Introduction
to Biology and
Laboratory
Rules1

CHAPTER

• What does biology mean?
• What are the fields of study in biology?• What type of careers can you pursue in the fields of biology?
• What are the safety equipment and rules in a biology laboratory?
• How do you communicate in the fields of biology?
• How is knowledge gained through scientific investigation?

Do You KNOW…

What are the effects
of the Industrial
Revolution 4.0 on the
fields of biology?

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1.1 Fields and Careers in Biology

1.1.1 State the meaning of biology.
1.1.2 List the fields of study in biology,

careers and development in related
fields.

1.1.3 Generate ideas about development
in biology fields and contribution of
technology in biology to humans.

1.1.4 Give examples of careers related to
the fields of biology.

1.2 Safety and Rules in Biology Laboratory

1.2.1 Explain self protective equipment
and their functions.

1.2.2 Identify and justify substances that
can be disposed into the sink.

1.2.3 Identify and justify substances that
cannot be disposed into the sink.

1.2.4 Describe methods in managing
biological wastes.

1.2.5 Communicate about steps to
manage accidents in a laboratory.

1.2.6 Conclude safety practices in a
biology laboratory.

1.3 Communicating in Biology

1.3.1 Communicate by constructing
tables based on experimental data.

1.3.2 Plot a suitable graph based on the
data from an experiment.

1.3.3 Sketch biological drawings based
on observations.

1.3.4 Identify body planes, sections and
directional terms in organisms.

1.4 Scientific Investigation in Biology

1.4.1 Design an experiment to solve
a problem using scientific
investigation methods.

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Fields and Careers in Biology
What is Biology?
The word ‘Biology’ comes from the Greek word, ‘bios’ meaning life,
and ‘logos’ meaning study. Biology is a detailed study of living things in
the natural environment. Biology allows us to understand the structure
and function of organisms, as well as the relationships and interactions
between organisms with their environment. Some examples of biological
study areas are shown below in Photograph 1.1.

1.1

Botany
Study of
plant life

Ecology
Study of the
interactions
between
organisms and
their environment

Physiology
Study of the
function and
mechanisms in
organisms

Genetics
Study of
inheritance
and genetic
variation

ICT 1.1

Video: What is
Biology?

(Accessed on 21
August 2019)

PHOTOGRAPH 1.1 Examples of biological research areas

Microbiology

Study of
microorganisms

1.1.1 1.1.2

The three-dimensional
(3D) bioprinting
technique is used to
print and form artificial
tissues and assist in
the field of modern
medical engineering.

Biological Lens

One of Malaysia’s success in technological
innovation is the production of biological scaffold
derived from shellfish for use in dental treatment.
Technological innovation in Malaysia is spearheaded
by researchers from public and private universities
and research institutions.

Malaysian Innovation

4

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Use of microorganisms in
the production of cheese,
soy sauce, tapai and
tempe.

The contribution of biology in everyday life
Biological knowledge has contributed a lot to mankind. Here are some examples of how biology has
contributed to everyday life (Photograph 1.2).

Careers in Biology

In vitro fertilisation, family
planning, plastic surgery and
gene therapy.

MEDICAL

FOOD PRODUCTION

Production of synthetic
vitamins, vaccines, insulin
and synthetic enzymes.

PHARMACEUTICAL

Transgenic crops and animals,
hydroponic and aeroponic
technology.

AGRICULTURE

I want to be a wildlife biologist. Which
biological field should I choose?

PHOTOGRAPH 1.2 Contribution of biology in everyday life

I am using information technology to
produce medicines. What is my profession?

PHOTOGRAPH 1.3  Careers in biology

1.1.3 1.1.4

1 What does biology mean?

2 List three fields of biology and the
development in those fields.

3 Choose one development in the field of
biology and discuss its contributions to
humans.

4 Name five careers that are related to the
fields of biology.

1.1Formative Practice

5

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Safety and Rules in Biology
Laboratory
A biology laboratory is essential for conducting scientific investigations.
To avoid unwanted incidences, we must always adhere to the safety rules
and measures in a biology laboratory. We must also know the personal
protective equipment and understand their functions.

PHOTOGRAPH 1.4  Equipment and their functions in a biology laboratory

FUME HOOD LAMINAR FLOW CABINET

1.2.1

1.2
Used when chemicals or
hazardous substances come
into contact with the skin.

EMERGENCY SHOWER STATION

Used to wash the eyes when exposed to
chemicals or hazardous substances.

EYE WASH STATION

To avoid breathing
in hazardous
gases like chlorine,
bromine and
nitrogen dioxide.

Provides a flow of
filtered air for a clean
work area in the
laminar flow cabinet.

Provides an enclosed work space to study materials that have
been (or are likely to have been) contaminated by pathogens.

BIOLOGICAL SAFETY CABINET

6

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Substances that can be disposed into the sink

• Chemicals with pH values 5–9

• Low concentration liquids and solutions that are
harmless to users (sucrose solution, dye solution,
distilled water)

Substances that cannot be disposed into the sink
• Organic solvents (acetone, alcohol, benzene)

• Substances that have a pH value of less than 5 or
more than 9

• Chemicals (acids, greases, oils)

• Solid waste (chemicals, glass, rubber)

• Heavy metals (mercury)

• Volatile substances

• Toxic substances

• Organic waste
(microorganisms, carcases)

• Reactive substances

• Radioactive substances

PHOTOGRAPH 1.5  Personal protective equipment and their functions in the biology laboratory

1.2.1 1.2.2 1.2.3

To handle biological samples

LABORATORY GLOVES

To remove germs on hands

HANDWASH

To protect the skin and clothes from
dangerous chemical splashes

LAB COAT

To prevent injury from glass shards
and chemical spills

LAB SHOES

To protect the eyes from hazardous
chemicals

GOGGLES

To protect against smoke, vapour and dust that
can affect the respiratory system

FACE MASK

Why can certain
substances be
disposed into the
sink?

Brainstorm!

Why can’t certain
substances be
disposed into the
sink?

Brainstorm!

7

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Methods for managing biological waste
After carrying out an experiment, there are wastes that need to be managed with care before disposal.
Methods for managing different biological wastes according to Standard Operating Procedures are
listed in Table 1.1.

TABLE 1.1 Methods for managing biological waste (Photograph 1.6)
Biological

waste
categories

Example Management method

Category A
(sharp wastes)

Sharp instruments such as
syringes, needles, glass,
scalpel and other sharp
instruments that can cause
injuries

Placed into a special bin for sharp material
disposal. This bin does not need to be sterilised.

Category B
(non-sharp
wastes)

Biological solid waste such
as gloves, tissue papers,
petri dishes, plastic culture
containers and hardened agar

Packed first in autoclave resistant biohazard
plastic bags, sterilised in an autoclave for
decontamination, and then placed into a
biohazard bin (Photograph 1.6).  Biohazard
plastic bags cannot be thrown into regular
waste baskets.

Category C
(animal carcases)

Animal carcases, organs and
tissues

Wrapped carefully in absorbent material (such as
tissue papers), packed carefully into a biohazard
plastic bag and frozen.

Category D
(liquids)

Broth culture and liquid
medium such as blood

All biological liquid waste must be
decontaminated by autoclaving before disposal.
Sterilised biological liquid wastes must be
disposed immediately.

Take Note!

• Decontamination by autoclaving
is done at a temperature of 121°C
and pressurised at 15 psi for
20 minutes.

• Biohazard plastic bags that have
been sterilised in an autoclave
and sharp waste bin must be
stored temporarily in a controlled
storage place until the scheduled
time for disposal.

PHOTOGRAPH 1.6 Equipment to
manage biological wastes

Sharp waste bin  Biohazard plastic bag

Biohazard bin

Autoclave

1.2.4 8

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Practices in a biology laboratory
The biology laboratory is a place for learning and research. However, there
are some general safety rules that must always be practised by pupils.

Use a lab coat, gloves, safety shoes and goggles
when appropriate.

CLOTHING ETHICS

• Do not work alone in the laboratory
without supervision.

• Wash your hands after conducting
an experiment.

• Do not bring in irrelevant items into
the laboratory.

• Clean your workstation using disinfectant.

• Dispose wastes according to the
set procedures.

• Do not eat and drink in the laboratory.

• Identify all safety symbols on substances and
equipment before use.

LABORATORY SAFETY RULES

• Stop work immediately and switch off
all nearby power sources.
Unplug appliances.

• Exit the laboratory according to the
emergency exit plan.

• Call the fire and rescue department.

• Do not panic and stay calm.

• Do not turn back to collect
your belongings.

• Assemble at the assembly point.

SAFETY MEASURES FOR FIRES

1.2.5

Accidents in the laboratory
Accidents in laboratories can be caused by carelessness, negligence or lack of skill in
carrying out experiments. Here are the steps in managing accidents in laboratories.

STEPS TO HANDLE GENERAL CHEMICAL SPILLS

1. Inform your teacher.

2. Declare the spill area as a restricted zone.

3. Prevent the chemical spill from spreading using sand.

4. Scoop up the chemical spill using appropriate equipment.

5. Dispose it safely.

STEPS TO HANDLE MERCURY SPILLS

1. Inform your teacher.

2. Declare the spill area as a restricted zone.

3. Sprinkle sulphur to cover the mercury spills.

4. Call the fire and rescue department.

PHOTOGRAPH 1.7 Chemical spills are
stopped from spreading by using sand

1.2.6

Find information on
handling accidents in
laboratories set by the
following agencies:
(i) Malaysian Biosafety

and Biosecurity
Association (MBBA).

(ii) National Institute of
Occupational Safety
and Health (NIOSH).

Activity Zone

9

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Scientific attitudes and noble value practices
when carrying out a scientific investigation:

• Having interest and curiosity towards the
environment.

• Being honest and accurate in recording
and validating data.

• Being diligent and persevering in carrying
out a study.

• Being responsible about the safety of
oneself, others and the environment.

• Appreciating and practising clean and
healthy living.

• Appreciate the contributions of science
and technology.

• Think critically and analytically.

1.2.6

Scientific Attitudes and Noble Values

• Be cautious when handling hot glassware.

• Report any damaged equipment or
glassware to teachers immediately.

• Keep flammable chemicals away from
fire sources.

• Do not touch, taste and smell
chemicals directly.

• Use appropriate gloves when handling
biological specimens.

• Specimens that are not harmful and have
been dissected should be buried or frozen.

• Wash hands with antiseptic detergents
before and after experiments.

• All surfaces and workstations should be
cleaned with disinfectant before leaving
the lab.

HANDLING GLASS AND CHEMICALS HANDLING LIVE SPECIMENS

The following procedures should be followed in the event of an accident:

• Inform your teacher.

• Call the fire and rescue emergency number.

• Remove the victim from the scene.

• Give emergency treatment.

• Make the place of accident as a restricted area.

EMERGENCY HELP

1 State three examples of personal protective
equipment in the lab and their functions.

2 After conducting an experiment, you are
given a task to manage the disposal of
Category B biological waste (non-sharp
wastes). Think of the best way so that
the biological waste does not affect
the environment.

3 An accident has occurred in the laboratory
involving mercury spills. Explain the steps in
handling the mercury spills.

4 Chemicals must be handled with care to
avoid unwanted accidents. List steps for
proper handling of chemicals.

1.2Formative Practice

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Communicating in Biology
Biologists who conduct research need to communicate their research
findings. Communication in biology involves tables, graphs and biological
drawings.

Tables
Tables are used to record and present data. When you conduct an
investigation, the experimental data needs to be recorded in a suitable table.
A table allows you to organise data systematically for easy comparison.
The headings in the table consist of the manipulated variable and the
responding variable with the respective unit.

a. The manipulated variable is placed in the column on the
left side.

b. The responding variable is placed in the column on the
right side.

Table 1.2 shows an example of a table in an experiment on the
effects of different temperatures (manipulated variable)
on the volume of oxygen (O2) released by aquatic plants
(responding variable).

Graph
The relationship between the manipulated variable and
responding variable can be illustrated in the form of a
graph. There are several types of graphs such as line graphs,
bar charts and histograms. The line graph shown in Figure
1.1 is based on the data in Table 1.2.

1.3

1.3.1 1.3.2

Video: How to
draw a graph

(Accessed on 21
August 2019)

ICT 1.2

Temperature
(°C)

Volume of oxygen
released (cm3)

30 2.0

40 14.0

50 12.0

60 6.0

70 3.0

TABLE 1.2 Volume of oxygen released by
aquatic plants at different temperatures

GENERAL METHOD OF
DRAWING A GRAPH

• The responding variable
is represented by the
vertical axis (y-axis) and
the manipulated variable
is represented by the
horizontal axis (x-axis).

• The scaling on the axis
must be uniform.

• Mark the points with an
appropriate symbol such
as ‘x’.

• Title the graph: “Graph
(responding variable) versus
(manipulated variable)”.

FIGURE 1.1 Example of a line graph

Volume of oxygen released versus temperature

Vo
lu

m
e

of
 o

xy
ge

n
re

le
as

ed
 (c

m
3 )

Temperature (°C)

11

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A bar chart presents data using vertical bars
that are not attached to each other.

FIGURE 1.2 Example of a bar chart

FIGURE 1.3 Example of a histogram

METHOD OF DRAWING A BAR CHART

METHOD OF DRAWING A HISTOGRAM

• Draw two axes, the horizontal axis/X
to represent the manipulated variable
and the vertical axis/Y to represent the
responding variable.

• Each width of the bar needs to
be uniform.

• The bar height depends on the
frequency of the data.

• Bar charts are used to make
comparisons between two or more
items at a time.

• Draw two axes where the horizontal axis/X
represents the data that has been divided
into groups according to the appropriate
range while the vertical axis/Y shows
the frequency.

• Insert labels and units on the axes.

• The size or width of the bars must be the
same for all class gap with no spacing
between the bars.

Biological drawings
Biological drawings must be exact to give an accurate representation of an observed specimen.

Characteristics of biological drawings
• Large and accurate; use a sharp pencil and not a colour pencil or pen.
• Not shaded artistically.
• Lines drawn must be clear, clean and continuous. Do not use a ruler to draw the outline of

a specimen.

1.3.2 1.3.3

A histogram shows the visual distribution of
data and the frequency of a value in a data set.

The graph title must be written as
“<Y-axis label> versus <X-axis label>”.

Take Note!

Blood type

N
um

be
r

of
 p

up
ils

Number of pupils versus height

0

2

4

6

8

10

Number of students versus height
N

um
be

r o
f s

tu
de

nt
s

Height
(cm)

11
5

- 1
20

15
6

- 1
60

16
1

- 1
65

12
1

-1
25

12
6

- 1
30

13
1

- 1
35

13
6

- 1
40

14
1

- 1
45

15
1

- 1
55

14
6

- 1
50

Height (cm)

Number of Form 4 pupils versus blood type

N
um

be
r o

f F
or

m
 4

p
up

ils

12

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frontal plane

horizontal plane

sagittal plane

FIGURE 1.4 Example of (a) plan drawing and (b) detailed
drawing of cross section of dicotyledon stem

• Drawings must be labelled. Label lines must point to the correct
structure, no arrowheads, must be straight and not cross each other.

• Drawings must have titles.
• Each structure in the drawing must be placed in the correct position,

and its size should be proportionate to the rest of the structures.
• A plan drawing is a line drawing that shows the outline of a structure.
• A detailed drawing shows the cells in a structure.
• The magnification factor of a drawing must be stated, for example:

= =
Length of drawing in cm

Length of specimen in cm
2.4
1.2

2x

Therefore, the magnification factor is 2x.

(a) Plan drawing

(b) Detailed drawing

Plane, section and direction
In biology, research and observation of an organism’s
structure are done based on the plane, section and
direction. The plane refers to a flat surface of shadow
passing through the body. There are three main planes
used (Figure 1.5):
• sagittal plane (divides the body into right and left parts)
• frontal plane (divides the body into frontal and rear

parts)
• horizontal plane (divides the body into upper and

lower parts)

1.3.4

cortex

phloem

pith

xylem

FIGURE 1.5 Sagittal, frontal and horizontal planes

pith

xylem

phloem

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Many anatomical drawings of animals and plants need to be labelled with
direction. All vertebrates have the same fundamental body plan that is
symmetrical. Direction consists of anterior, ventral, posterior, dorsal,
superior, inferior and lateral. By studying anatomical directions in
biology, you can identify the orientation of the vertebrate that is stated.

1.3.4

FIGURE 1.6 Directional terms in biology

The part that is above all
other parts or towards the
head

Far from the midline or at
the side of the body

Situated below other parts
or towards the feet

SUPERIOR

LATERAL

INFERIOR

superior

inferior

lateral

Cross section divides the structure into
upper and lower portions horizontally while
longitudinal section divides the structure into

left and right portions (Photograph 1.8).

PHOTOGRAPH 1.8
Longitudinal section and
cross section of a red carrot

Longitudinal section

Cross section

Towards the back
part of the body

POSTERIOR

Towards the upper
part of the body

DORSAL

Towards the front
of the body

ANTERIOR

Towards the lower
part of the body

VENTRAL

dorsal

ventral

posterior

anterior

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After you have completed the experiment, you need to write a report to prove that you have
conducted the experiment. If a report is not written, your experiment will be invalid. Here is an
example of an experimental report framework.

Problem statement
What is the effect of a vigorous activity on the pupil’s pulse rate?

Hypothesis
The pupil’s pulse rate increases after a vigorous activity.

Variables
Manipulated: Vigorous activity
Responding: Pupil’s pulse rate
Fixed: Pupil

Materials and apparatus
Stopwatch

Specimen
Pupil

Procedure
 1 Count the pupil’s pulse for one minute before carrying out the activity.

2 The pulse is counted by placing two fingers on the wrist.

3 The pupil is asked to carry out a vigorous activity for three minutes.

4 After the vigorous activity is carried out, count the pupil’s pulse again for one minute.

5 Record the results in the following table.

1.4.1

Activities2.1Activity 1.1   ExperimentStudying the effects of an activity on a pupil’s
pulse rate

Scientific Investigation
in Biology1.4
Scientific investigations prove the validity
of a hypothesis that has been made. The
following are the steps in carrying out
a scientific investigation.

Identifying problems that can be
tested using a scientific investigation

Forming a hypothesis

Identifying and controlling variables
and data collection method

Planning and carrying out a scientific
investigation

1  Collecting data

Interpreting data and
results through scientific
reasoning

Forming a conclusion

Writing a report

2
3

4

5

6

7
8

Measuring pulse rate

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Results

Pupil

Pulse rate (number of pulse beats per minute)

Before the vigorous
activity

After the vigorous
activity

A

B

Discussion
 1 State the relationship between the pupil’s pulse rate with vigorous

activity.

2 Explain why the pulse rate increases after doing a vigorous activity.

Conclusion
Based on the results, state if the hypothesis can be accepted or rejected.

Report
Prepare a complete report based on the experiment that was carried out.

Make sure the pupil
is seated quietly
when the pulse rate
is taken.

Take Note!

Make sure pupils
with health issues
such as breathing
and cardiovascular
problems do not
take part in this
activity.

Take Note!

1.4.1

1 The table below shows the results of an experiment to study the effects of substrate concentration
on biochemical reactions.

Test tube Starch

concentration
(%)

Time to complete starch
hydrolysis (minute)

Rate of hydrolysis of
starch (% minute-1)

A 1.0 2.5 0.4

B 2.0 3.0 0.7

C 3.0 3.5 0.9

D 4.0 4.0 1.0

(a) Based on the table above, identify the manipulated variable and responding variable.

(b) Draw a graph of rate of starch hydrolysis versus starch concentration using the correct scale
and units.

2 State five characteristics that a biological drawing must have.

3 State the three main planes that are used for observation of the structure of an organism.

4 List the steps in a scientific investigation.

1.3Formative Practice

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Have you mastered the following concepts?
• Meaning of biology
• Areas of study in biology, careers and progress in the related fields
• Development in the fields of biology and contribution of biological technology to

humans
• Careers related to biology
• Personal protective equipment and their functions
• Substances that can be disposed into the sink
• Substances that cannot be disposed into the sink
• Methods of managing biological waste
• Steps to manage accidents in the laboratory
• Practices in a biology laboratory
• Tables based on data from experiments conducted
• Appropriate graphs based on data from experiments conducted
• Biological drawings based on observations made
• Body planes, sections and directional terminology in organisms
• Experiment to solve problems using the scientific investigation method

Fields and Careers
in Biology

Safety and Rules
in the Biology

Laboratory
Communicating in

Biology

Scientific
Investigation in

Biology

INTRODUCTION TO BIOLOGY AND LABORATORY RULES

• Meaning of
biology

• Developments in
biology

• Contribution of
biology in life

• Careers in
biology

• Personal protective
equipment and
their functions

• Methods of
managing
biological wastes

• Safety practices
in the biology
laboratory

• Tables
• Graphs
• Bar charts
• Histograms
• Biological drawings
• Planes, sections

and directions

Steps in a
scientific
investigation

Summary

Self Reflection

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1 Why are you not allowed to wear slippers while in the laboratory?

2 Give the meaning of

(a) cross section

(b) longitudinal section

3 Discuss methods of biological waste management.

4 Describe the stages involved in a scientific investigation.

5 (a) An experiment was conducted to demonstrate the hydrolysis of starch by the amylase
enzyme. Experiment apparatus provided are as in Figure 1. After an hour, samples from
each test tube were tested using iodine test.

(i) Identify the problem statement for this experiment.

(ii) State the appropriate hypothesis for this experiment.

(iii) Identify the manipulated variable, responding variable and constant variable.

(b) A pupil spilled hydrochloric acid while conducting an experiment. What action should
be taken by the pupil?

FIGURE 1

-1
0

0
10

20
30

40
50

60
70

80
90

10
0

11
0

1 ml starch solution +
1 ml 0.1% amylase

1 ml starch solution +
1 ml distilled water

water bath
(37ºC)

thermometer A B

Summative Practice 1

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Essay Questions

6

(a) Figure 2 shows an accident in a biology laboratory due to two pupils playing with fire
while conducting an experiment. The accident caused one of the boys’ shirt to catch
fire. Explain what the victim's friend should do.

(b) Azlin is conducting an experiment involving the use of a thermometer. Accidentally, she
drops the thermometer on the floor and the thermometer breaks. What steps does
Azlin need to take in handling the spilled mercury waste? Explain your answer.

(c) Plan an experiment to study the effect of sugar on the growth of yeast.

Enrichment

7 The emergence of Industrial Revolution 4.0 has begun to change the way we work and
the way we live. Various industries are experiencing drastic changes and at the same
time transforming jobs related to those industries. In your opinion, what is the effect of the
Industrial Revolution 4.0 to everyday life? Justify your opinion.

8 If you are given an assignment by your teacher to collect information on the development of
tissue culture technique in Malaysia, suggest the steps that need to be taken to complete
your assignment.

Complete answers are
available by scanning the
QR code provided

FIGURE 2

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Which cell
has the longest
life span?

• What are the components of plant cells and animal cells?
• What is the life process of the unicellular organism?
• What is the relationship between specialised cell structures and their functions?
• What are the levels of

organisation in multicellular organisms?

Do you KNOW...

20

Cell Biology and
Organisation2

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2.1 Cell Structure and Function

2.1.1 Prepare microscope slides of animal and plant
cells.

2.1.2 Identify the structures of animal and plant
cells based on observations through a light
microscope.

2.1.3 Analyse the components of animal and plant
cells as seen on micrographs.

2.1.4 State the main functions of components of
animal and plant cells as seen on micrographs.

2.1.5 Compare and contrast components of animal
and plant cells.

2.2 Living Processes in Unicellular Organisms

2.2.1  Conceptualise living processes in unicellular
organisms such as Amoeba sp. and
Paramecium sp.

2.2.2 Deduce living processes in unicellular organisms
as seen through a light microscope.

2.3 Living Processes in Multicellular Organisms

2.3.1  Correlate the uniqueness of specialised cell
structures with their functions in multicellular
organisms.

2.3.2  Identify specialised cells in multicellular
organisms.

2.3.3 Analyse the density of certain organelles with
the functions of specialised cells in multicellular
organisms.

2.3.4 Describe the effects of deficiency, absence or
failure in the function of an organelle of certain
cells in multicellular organisms.

2.4 Levels of Organisation in Multicellular Organisms

2.4.1 Make a sequence of levels of organisation in
multicellular organisms.

2.4.2 Identify cells, tissues or organs in an organ
system.

2.4.3 Communicate about organ systems in
multicellular organisms with their main functions.

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2.1.1 2.1.2

Cell Structure and Function2.1  You may still recall the shape of cells you learned in Form 1. Much like
the bee hive which is made up of hexagonal-shaped units, all living
things are also made up of cells that are combined together. Cells are the
basic units of all living things. Let us examine the structure of animal
cells and plant cells through a light microscope.

Activities2.1Activity 2.1 ObservationPreparing and examining slides of plant cells

Materials
One big onion, iodine solution, distilled water and filter paper

Apparatus
Glass slides, cover slips, light microscope, scalpel, a pair of forceps, dropper and mounting needle

Procedure
 1 Cut a piece of scale leaf from a big onion (a).

2 Remove the translucent epidermis from the inside of the scale leaf using a scalpel.

3 Add a drop of distilled water at the centre of the slide and lay the onion epidermis on the water
(b).

4 Using a mounting needle, cover the onion epidermis with a cover slip by placing it at a 45°
angle to the slide and slowly press it down. The temporary slide of this specimen is named as
wet mount. Make sure there are no air bubbles trapped in the wet mount slide (c) and the
onion epidermis is not folded.

5 Add a drop of iodine solution on one side of the cover slip. Place the filter paper at the opposite
end of the cover slip to draw the iodine solution to absorb and stain the entire onion epidermis.
This technique is known as the irrigation technique (d).

epidermis
bawang besar

setitis air

forseps
epidermis

larutan iodin

(a) (b) (c) (d)

kaca penutup
jarum tenggek

(c) (d)

epidermis
bawang besar

setitis air

forseps
epidermis

larutan iodin

(a) (b) (c) (d)

kaca penutup
jarum tenggek

cover slip
mounting needle iodine solution

filter paper

Preparation of plant cell slide

Air bubbles that are
trapped under the
cover slip can be
removed by gently
tapping the cover
slip with the tip of a
sharp pencil.

Take Note!(a) (b)

epidermis
bawang besar

setitis air

forseps
epidermis

larutan iodin

(a) (b) (c) (d)

kaca penutup
jarum tenggek

epidermis
bawang besar

setitis air

forseps
epidermis

larutan iodin

(a) (b) (c) (d)

kaca penutup
jarum tenggek

a pair of forceps
epidermis

glass
slide

drop of distilled
water

epidermis
one big onion

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Activities2.1Activity 2.2 ObservationPreparing and examining slides of animal cells

Materials
Animal cell (cheek cell), methylene blue solution, distilled water, filter paper and
toothpick

Apparatus
Glass slides, cover slips, light microscope, dropper and mounting needle

Procedure
 1 Gently scrape the inside of the cheek with a blunt tooth

pick.

2 Transfer the scrapping into a drop of water on a clean
glass slide and cover with the cover slip.

3 Stain the cheek cells with methylene blue solution using
the irrigation technique.

4 Examine the cheek cells through a light microscope.
Draw the cheek cells that can be observed. Record the
magnification power.

Discussion
 1 What are the components of the cell that can be observed through the light microscope?

2 Does the human cheek cell have a permanent shape?

3 Can you see dark blue spots on the prepared cheek cell?

4 What are the similarities between animal cells and plant cells? Which component is present
in the onion cell but is not present in the cheek cell?

Conclusion
Suggest a suitable conclusion for this observation.

2.1.1 2.1.2

Using the same slide preparation method, prepare a slide of cheek cells (slide of animal cells) using
a methylene blue solution.

6 Use a piece of filter paper to absorb excess iodine and distilled water.

7 Observe the slide through a light microscope with a low-power objective lens and then with a
high-power objective lens.

8 Draw the epidermal cells structure that can be observed and label it completely. Record the
magnification power.

Discussion
 1 What are the components of the cell that can be observed through the light microscope?

2 Does a plant cell have a fixed shape?

3 Name another type of plant that can be used in this activity to replace the onion.

Conclusion
Suggest a suitable conclusion for this observation.

Preparation of animal cell slide

methylene blue solution filter paper

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mitochondrion

centriole

Golgi
apparatus

plasma membrane

ribosome

rough
endoplasmic
reticulum

smooth
endoplasmic
reticulum

lysosome nucleus
nucleolus

cytoplasm

• Small cylindrical components that exist in pairs in
animal cells

• Made up of complex
arrangement of microtubules

• Does not exist in plant cells

Function:

Forms spindle fibre during cell division in animal cells

• Consists of a stack of parallel flattened sacs that are
coated by a single cell membrane

• New membrane is added at one end of the Golgi
apparatus and vesicles bud off from the other end.

Function:

Processes, modifies, packs and
transports chemicals such as protein,
carbohydrate and glycoprotein
(combination of carbohydrate and
protein)

CENTRIOLE

GOLGI APPARATUS

• It is rod-shaped or spherical

• It consists of two layers of membranes, which are the smooth outer
membrane and folded inner membrane

• Contains enzymes that play a role in cellular respiration

Function:

• A site that generates energy through the glucose oxidation process during
cellular respiration

• Energy released in the form of ATP molecules (adenosine triphosphate) to be used by the cells

MITOCHONDRION (PLURAL: MITOCHONDRIA)

• Outer membrane that surrounds the entire content of cell

• Made of proteins and phospholipids

• Thin and elastic film

• Partially permeable

Function:

• Separates content of cell from the external environment

• Controls movement of substances into and out of the cell

• Allows exchange of nutrients, respiratory gases and waste materials between cells and their
surroundings

PLASMA MEMBRANE

outer environment

plasma
membrane

cytoplasm

FIGURE 2.1 Animal cells

2.1.3 2.1.4

Components of animal and plant cells and their functions

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• Largest component in the cell

• Spherical, compressed and enclosed in a
nuclear membrane with many pores

• The nucleus contains chromosomes, nucleolus
and nucleoplasm.

Function:

• Controls all cell activities

• Has chromosomes that contain
deoxyribonucleic acid (DNA). DNA determines
the cell characteristics and metabolic function.

• Consists of a system of interconnected folded flattened sacs

• Endoplasmic reticulum membrane is continuous with the nuclear membrane.

• There are two types of endoplasmic reticulum:

– Rough endoplasmic reticulum has ribosomes attached to the surface

– Smooth endoplasmic reticulum does not have ribosomes

Function:

• The transport system within the cell

• Provides a wide surface for enzyme attachment and biochemical
reactions

• The rough endoplasmic reticulum transports proteins synthesised by
ribosomes.

• The smooth endoplasmic reticulum synthesises and transports
glycerol and lipids, and carries out the detoxification of drugs and
metabolic by-products.

NUCLEUS (PLURAL: NUCLEI)

• Small, compact and spherical granules

• Consists of protein and ribonucleic acid
(RNA)

• Ribosomes are present on the surface of
the rough endoplasmic reticulum or exist
freely in the cytoplasm.

Function:

• Site for protein synthesis

RIBOSOME

ENDOPLASMIC RETICULUM

rough endoplasmic reticulum

smooth endoplasmic reticulum

ribosome

nucleoplasmnuclear
membrane

nucleolus
chromatin

• Small spherical sac enclosed in a single membrane

• Contains hydrolytic enzymes

Function:

• Hydrolyses complex organic molecules such as protein, nucleic
acid and lipid

• Breaks down bacteria and components of damaged cells

LYSOSOME

lysosome

2.1.3 2.1.4 25

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• A strong and rigid outer layer

• Made from cellulose fibre

• Fully permeable

Function:

• Maintains the shape of plant cells

• Provides mechanical support to
plant cells

2.1.3 2.1.4

• Consists of a jelly-like
medium that contains
components of the
suspended cells

• Contains organic
compounds (such
as protein, lipid and
carbohydrate) and
inorganic compounds
(such as potassium ions)

Function:

Acts as a medium for
biochemical reactions in cells

CYTOPLASM

cytoplasm

• Oval shaped

• Consists of two layers of membrane

• Contains chlorophyll pigments in the
grana that give plants a green colour

Function:

Chlorophyll absorbs
sunlight and converts
it to chemical
energy during
photosynthesis.

CHLOROPLAST

CELL WALL

• Liquid-filled sac, which is the cell sap.

• A vacuole is surrounded by the tonoplast
membrane.

• Young plant cells have many small vacuoles while
mature plant cells have a large vacuole.

• The vacuole in animal cells is small.

• Cell sap contains water, organic acids, sugars,
amino acids, enzymes, mineral salts, oxygen,
carbon dioxide and metabolic by-products.

Function:

• Water is absorbed into the
vacuole plant cell and the
cell becomes turgid.

• In unicellular animals, the
vacuole contracts during
osmoregulation, osmosis
and excretion.

VACUOLE

vacuole
tonoplast

cytoplasm

cell wall

plasma
membrane

cytoplasm

nuclear membrane

vacuole

chloroplast

nucleolus

nucleus

endoplasmic
reticulum

Golgi appratus

cell wall

plasma membrane

mitochondrion

chloroplast

FIGURE 2.2 Plant cell

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2.1.5

Compare and contrast the
components of animal cells
and plant cells
You have learned about the components in animal
and plant cells. What are the similarities and
differences between the components in animal
and plant cells?

• Has a fixed shape

• Has a cell wall

• Has chloroplasts

• Has a large vacuole

• Stores carbohydrate in the form of starch

• Does not have a centriole

Both cells are made of nucleus, cytoplasm, plasma membrane, Golgi apparatus, mitochondrion,
endoplasmic reticulum and ribosomes.

• Does not have a fixed shape

• Does not have a cell wall

• Does not have chloroplasts

• No vacuole/if present, it is small

• Stores carbohydrate in the form of
glycogen

• Has centrioles

PLANT CELLS ANIMAL CELLSDIFFERENCES

SIMILARITIES

PHOTOGRAPH 2.1  Structure of plant and animal cells through a light microscope

2.1.5

2.1  Formative Practice

1 State one structural difference between
the rough endoplasmic reticulum and the
smooth endoplasmic reticulum.

2 Why is the use of an electron microscope
important in studying cells?

3 Chloroplast is found in some plant cells
only. Give one example of a plant cell that
does not have chloroplast and give your
reasons.

4 Compare and contrast animal cells and
plant cells.

cytoplasmnucleus
plasma
membrane

cytoplasm

nucleus

cell wall

plasma membrane vacuole

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2.2.22.2.1

Living Processes in
Unicellular Organisms2.2
Unicellular organisms are made up of only one cell. However, this
cell is a complete unit of life like multicellular organisms. Unicellular
organisms carry out all life processes: respiration, movement, nutrition,
responding to stimulus, reproduction, growth and excretion. Protozoans
are the simplest form of unicellular organisms. Examples of protozoa
are Amoeba sp. and Paramecium sp. Let us look at the life processes of
Amoeba sp. and Paramecium sp.

• Amoeba sp. constantly changes its shape when
it encounters obstacles. Amoeba sp. moves by
extending out its pseudopodium (false feet).
This is followed by the flow of cytoplasm into
the extended pseudopodium.

• Paramecium sp. moves using rhythmic cilia
beats.

MOVEMENT

Amoeba sp. and Paramecium sp. respond to
stimuli such as chemicals, touch or bright light
by moving away from the stimuli.

RESPONDING TO STIMULI

Exchange of oxygen and carbon dioxide
gases occur through the plasma membrane by
simple diffusion on the surface of the cell.

RESPIRATION

Ameoba sp. and Paramecium sp. grow by
synthesising new cytoplasm.

GROWTH

ICT 2.1

Activity: Studying
the life processes of

unicellular organisms

vakuol makanan

endoplasma

ektoplasma

nukleus

membran
plasma

vakuol mengecut

pseudopodium

food

pseudopodium

1

2

3
4

lysosome

food vacuole

nutrients absorbed

digested
food

FIGURE 2.3 Nutrition in Amoeba sp.

Amoeba sp. moves towards food by
extending its pseudopodium to trap food
particles by phagocytosis (Figure 2.3). For
Paramecium sp. the presence of cilium beat
helps transfer food particles into the oral
groove.

The food vacuole is combined with lysosome.
The food particles are hydrolysed by the
enzyme lysozyme in the lysosomes.

The nutrients are absorbed into the
cytoplasm.

Undigested food is discharged when the
Amoeba sp. moves. Undigested food in the
Paramecium sp. is discharged through the
anus.

NUTRITION

1

2

3

4

undigested
food

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2.2.22.2.1

1 What is the function of the
contractile vacuole in a
Paramecium sp.?

2 How does the  Amoeba sp.
move?

3 How does the Amoeba sp.

reproduce when the
environment is unfavourable?

4 Predict what will happen to
the contractile vacuole if the
Paramecium sp. is placed in a
concentrated salt solution.

2.2  Formative Practice

PHOTOGRAPH 2.2 Amoeba sp.

• Waste such as carbon dioxide and ammonia are
removed by diffusion.

• As Amoeba sp. and Paramecium sp. live in
freshwater environments, water will diffuse by
osmosis and fill the contractile vacuole.

• When the vacuole expands to the maximum size,
contraction occurs and water is excreted from
time to time.

• This process is called osmoregulation.

EXCRETION

vakuol makanan

endoplasma

ektoplasma

nukleus

membran
plasma

vakuol mengecut

pseudopodium

food vacuole

pseudopodium

nucleus

plasma membrane

contractile
vacuole

PHOTOGRAPH 2.3
Paramecium sp.

Conduct an
experiment using
a light microscope
to examine the
living processes of
unicellular organisms.

Activity Zone

oral groove
anus

contractile
vacuole

cilium

nucleus

• When the conditions
are suitable and there
is plenty of food,
Amoeba sp. and
Paramecium sp. will
reproduce via
asexual reproduction
that is binary fission
through mitosis.

• However, when the
environmental
conditions
are not suitable,
such as dry
conditions, low
temperature and
food shortage,
the Amoeba sp.
forms spores that
will only germinate when the environment
improves.

• For Paramecium sp., sexual reproduction,
that is conjugation occurs when
environmental conditions are not suitable.

REPRODUCTION

Conjugation

Binary fission

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2.3.1

Living Processes in
Multicellular Organisms2.3
There are various types of cells in multicellular organisms which are
different in size, shape and arrangement. The cell structure correlates
to its function, and each type of cell is specialised to carry out a specific
function. Photographs 2.4 and 2.5 show several types of specialised cells
found in humans and plants as well as how these cells are adapted to
carry out their respective functions.

• Arranged as
multinuclear striated
fibres

• Contract and relax to
generate movement

MUSCLE CELL

• Has a long tail to enable it to
swim towards the ovum in
the Fallopian tube

• The head carries a set of
chromosomes from the male

SPERM CELL

• Long and thin in
shape

• Functions in sending
nerve impulses

NERVE CELL

• Can change shape

• Can change shape

• Functions in
destroying pathogens

WHITE BLOOD
CELL

• Thin and flat cells

• Coats the surface
of organs such as
the digestive tract

EPITHELIAL CELL

• Does not contain a
nucleus

• Shaped as a
biconcave disc

• Functions to optimise
transportation of oxygen

RED BLOOD CELL

PHOTOGRAPH 2.4 Several types of human cells that undergo specialisation to carry out
different functions

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2.3.1

• Long cylindrical tubes arranged
from end to end

• Transports organic materials from
leaves to storage organs

such as fruits

SIEVE TUBE ELEMENT

• Long, continuous hollow tube

• Functions in transporting
water and mineral salts
from the roots to the
other parts of the
plant

XYLEM VESSEL

• Has a long projection which
adds surface area for the
absorption of water and
mineral salts

ROOT HAIR CELL

ICT 2.2

Activity: Observe plant
tissue and animal tissue

slides through a light
microscope

Cross section
of leaf

• Consists of long cylindrical cells,
arranged vertically and close to
each other

• Contains high chlorophyll density

• This arrangement allows
maximum absorption of
sunlight for photosynthesis

PALISADE MESOPHYLL CELL

• Cells are loosely arranged with lots of
air space in between

• Large air space allows exchange of
gas from the inside of the leaves to the
palisade mesophyll cells

SPONGY MESOPHYLL CELL

PHOTOGRAPH 2.5 Certain types of plant cells
undergo specialisation to carry out different functions

• Modified lower epidermal cells
with the thicker cell wall on the
inner side

• Controls the opening and
closing of the stoma. Stoma
is the opening that allows
the exchange of oxygen and
carbon dioxide

GUARD CELL

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Cell organisation in humans
You have learned that multicellular organisms (such as humans) consist of different types of cells
with specific functions. How are these cells arranged to form a complex organism?
Tissues are a group of cells that have the same structure and function and are arranged together to
carry out a specific function. In organisms, tissues can be classified into four different types which
are epithelial tissue, muscle tissue, nerve tissue and connective tissue.

Epithelial tissue
Epithelial tissue covers the outer surface (skin) and hollow surfaces in the body (digestive tract
and respiratory tract) (Figure 2.4). The epithelial tissue structure is adapted based on its function.
For example, the epithelial tissue on the skin protects against infections, injuries, chemicals and
dehydration. Epithelial tissues that coat the trachea have projections like hair, known as cilia
(singular: cilium).

2.3.1 2.3.2

Epithelial tissue covers
the small intestine

Epithelial tissue covers the surface of the
trachea and bronchus

FIGURE 2.4 Types of epithelial
and muscle tissues

Epithelial tissue covers the surface of
lungs, body cavities and blood vessels

Epithelial tissue covers the surface
of the mouth and oesophagus

Epithelial tissue lines tubules, glands
and kidney ducts

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PHOTOGRAPH 2.6  Nerve tissue

Muscle tissue
There are three types of muscle tissue: smooth muscle (found in the digestive tract, blood vessel,
urinary tract and reproductive tract), skeletal muscle (found in legs and hands) and cardiac muscle
(found in the heart wall) (Figure 2.4).

2.3.1 2.3.2

Nerve tissue
Nerve tissue is made of neuron or
nerve cells (Photograph 2.6). Each
neuron consists of a cell body and
nerve fibre called dendrite and axon.
Nerve tissues can detect stimuli and
then send information in the form of
an electrical signal (nerve impulses)
to the muscles or glands. Nerve tissue
regulates and controls body activity.

Contraction and relaxation of smooth muscle
enable involuntary activities such as peristalsis
along the digestive tract.

Skeletal muscle is involved in controlled movement.
Skeletal muscles contract and relax to generate
movement in bones and limbs.

Cardiac muscle builds walls of the heart that
contract to pump blood to the whole body.
Cardiac muscle contraction is involuntary.

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Connective tissue
The connective tissue consists of various types of tissues and fibres
(Figure 2.5). This tissue is distributed all over the body and has many
functions. One of its functions is to link the organs.

These tissues form tendons
and ligaments. The tendon
connects bones and muscles
while the ligaments connect
bones to bones.

It links the epithelial
tissue to the tissue
below it, and fixes the
organs in their positions.

Connective tissues
keep fat under the skin
dermis and the surface
of all main organs.

ADIPOSE TISSUE

collagen fibres

nucleus

Bone forms the body
frame and protects the
internal organs.

BONE

FIGURE 2.5 Connective tissue

ICT 2.3

Activity: Discuss the importance
of cell specialisation in

multicellular animals and plants

Blood plays a functional
role in regulation,
transportation and

protection.

BLOOD TISSUE

FIBROUS CONNECTIVE
TISSUE

LOOSE CONNECTIVE
TISSUE

2.3.22.3.1

white
blood cell

red blood
cell

CARTILAGE

Cartilage encloses
bone tips to prevent
the bone from
wearing out.

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Tissue organisation in plants
Tissue organisation in plants is summarised in Figure 2.6.

PHOTOGRAPH 2.7
Meristem tissue at the tip
of the root

FIGURE 2.6 Tissue organisation in plants

PHOTOGRAPH 2.8
Different types of ground
tissues

Meristem tissue

Lateral
meristem

tissue

Dermal
tissue

Vascular
tissue

Ground
tissue

Apical
meristem

tissue

Permanent tissue

PLANT TISSUE

Xylem tissue

Phloem
tissue

Parenchyma tissue

Collenchyma tissue

Sclerenchyma tissue

Functions in giving support to young, non-woody stems
(herbaceous plants).

COLLENCHYMA TISSUE

Functions to store starch, protein and water.
This tissue can also carry out photosynthesis.

PARENCHYMA TISSUE

Functions in providing support and mechanical
strength to all mature parts of the plant.

SCLERENCHYMA TISSUE

2.3.22.3.1
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2.3.3

Density of certain cell components and specialised cell
functions

Since the functions performed by cells are different, some cells have a higher density of certain
cell components. The density of a cell component in a particular cell is related to the specific
function of the cell. Table 2.1 provides examples of cells that have a higher density of certain cell
components.

TABLE 2.1 Relationship between cell component density with specialised cell functions

Types of cell
Cell component found in

abundance
Function

Sperm cell

Mitochondrion

Requires a lot of energy to swim
towards the uterus and Fallopian
tube to fertilise the secondary
oocytes

Muscle cell such as flight
muscle cells in insects and
birds

Requires a lot of energy to
contract and relax to enable
movement and flight

Plant meristem cell Requires a lot of energy to carry
out active cell division process
to produce new cells

Palisade mesophyll cell
Chloroplast

Absorbs more sunlight to
carry out the process of
photosynthesisSpongy mesophyll cell

Pancreatic cell

Rough endoplasmic reticulum

Golgi apparatus

Increases synthesis and
secretion of digestive enzymes

Goblet cell in intestinal
epithelium and respiratory
tract

Produces mucus

Liver cell • Metabolises carbohydrates

• Detoxification of drugs and
poisons

The xylem functions in
transporting water and

mineral salts from the roots
to other parts of the plant.

Ligneous xylem tissue wall provides
support and mechanical strength to the plants.

XYLEM TISSUE

The phloem functions in
transporting organic matters
such as sucrose from the leaves
to all parts of the plant.

PHLOEM TISSUE

Vascular tissues are made up of xylem tissue and phloem tissue.

VASCULAR TISSUExylem vessel

sieve tube

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2.3.4 2.4.1

PHOTOGRAPH 2.9 Tissues that form the heart organ

A group of different tissues
combine to form organs.
Organs perform special
functions as a result of the
combination of tissues that
form the organ. For example,
the heart organ is composed
of epithelial tissue, cardiac
muscle tissue, connective
tissue and nerve tissue. The
epithelial tissue fills up space
in the heart. Cardiac muscle
tissue functions in pumping
blood to the rest of the body.
Connective tissues such
as the blood, connect the
systems in the organ. Nerve
tissue regulates the rhythm
of the heartbeat.

2.4  Levels of Organisation in
Multicellular Organisms

connective tiss
ue

cardiac muscle tissue

nerve tissue

epithelial tissue

All cell components perform their respective functions to ensure that the cells function at the
optimum level. The collaboration of all the components of the cells is similar to a factory with
different parts and each with its specific function. What happens to the cell if there is a deficiency,
absence or failure in any of the cell components such as lysosomes, mitochondrion, chloroplasts
or ribosomes?
Failure of the mitochondrion function or a mitochondrion disjunction can cause stunted growth,
weak muscles, hearing and vision problems. Tay-Sachs is a hereditary disease caused by the failure
of enzymes to produce in the lysosomes. Tay-Sachs patients will experience stunted growth and
mental retardation.

1 State the components of cells found in high
density in cells that secrete protein products.

2 Briefly describe the structure and
function of the nerve tissue.

3 Explain why the palisade mesophyll cell
contains a lot of chloroplast.

4 Fat cells have three adaptive characteristics
to allow the storage of fat. Describe the three
characteristics.

2.3  Formative Practice

heart

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2.4.2 2.4.32.4.1

Each multicellular organism is made up of many
organs. Different organs work together to carry
out a specific function to form an organ system.

There are 11 main organ systems in the human
body (Figure 2.7), which are the respiratory
system, digestive system, blood circulatory
system, lymphatic system, nervous system,
integumentary system, endocrine system,
skeletal system, muscular system, urinary
system and reproductive system.

• Ovary, uterus, Fallopian tube, vagina
and cervix

Main function

• Produces ovum and female sex
hormones

FEMALE REPRODUCTIVE SYSTEM

• Testes, prostate gland and penis

Main function

• Produces sperm and male sex
hormone

MALE REPRODUCTIVE SYSTEM

• Spleen, lymph nodes and
lymph vessels

Main function

• Maintains balance of bodily
fluids and prevents infectious
diseases

LYMPHATIC SYSTEM

• The endocrine
gland that secretes
hormones

Main function

• Coordinates body
activities with the
nervous system

ENDOCRINE SYSTEM

• Skeletal muscles, smooth muscles and cardiac
muscles

Main function

• Contracts and relaxes to produce movements in
different parts of the body

MUSCULAR SYSTEM

FIGURE 2.7 Main organ systems in the
human body

• Trachea, nose, lungs and diaphragm

Main function

• Exchange of oxygen and carbon dioxide gases
between the body and external environment

RESPIRATORY SYSTEM

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2.4.22.4.1 2.4.3

• Brain, spinal cord and peripheral nerves

Main function

• Detects and sends information in the body, as
well as coordinates body activities

NERVOUS SYSTEM

• Mouth, oesophagus, stomach, liver, pancreas,
small intestine and large intestine

Main function

• Digests food into a simpler form for easy
absorption

DIGESTIVE SYSTEM

• Bone, cartilage, ligament and tendon

Main function

• Supports the body, protects the internal
organs and provides a base for muscle
adhesion

SKELETAL SYSTEM

• Heart, artery, vein and blood capillary

Main function

• Transports nutrients, respiratory gases and
waste products

BLOOD CIRCULATORY SYSTEM

• Kidney, ureter, urethra and bladder

Main function

• Eliminates waste products such as urea and
uric acid from the body

URINARY SYSTEM

• Skin

Main function

• Protects the body from physical injury,
infection and dehydration

INTEGUMENTARY SYSTEM

Cell

Tissue

Organ

System

Multicellular
organism

All the organ systems
are coordinated and
collaborated as one
complete organism.
The sequence of the
organisation of cells
in a multicellular
organism is shown in
Figure 2.8.

System

Multicellular organism

Cell

Tissue

Organ

FIGURE 2.8 The sequence
of the organisation of cells
in human beings

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2.4.32.4.2

The plant system is divided into the shoot
system and the root system (Figure 2.9).
• The shoot system consists of stems, leaves,

shoots, flowers and fruits.
• Stems and twigs are support systems that

support the leaves at a vertical position to
allow maximum absorption of sunlight
during photosynthesis.

• Flowers are involved in the pollination
process.

• The root system consists of all roots in a
plant that function in absorbing water and
mineral salts as well as providing support
for plants.

FIGURE 2.9  Two main systems in plants

root
system

shoot
system

leaf
fruit

flower

stem

root

2.4  Formative Practice

1 State arrangement of sequence in the
organisation of cells in a multicellular
organism.

2 The skin is the largest organ in the body.
Why is skin classified as an organ?

3 Stems, twigs and flowers are organs in
a plant. Describe how stems, twigs and
flowers function in the shoot system.

4 Predict what will happen if the organisation
level of cells does not exist in an organism.

Cell Structure and
Function

Living Processes in
Unicellular Organisms

Living Processes in
Multicellular Organisms

Levels of
Organisation

CELL BIOLOGY AND ORGANISATION

Amoeba sp. and
Paramecium sp.
• Movement
• Response to

stimuli
• Nutrition
• Respiration
• Growth
• Excretion
• Reproduction

Relationship
between specific
cell structure
and cell function

Cell  Tissue
 Organ

System
Organism

Effect of the absence of
cell components to the cell

Relationship
between cell
component
density and cell
function

Cell components
• Nucleus •  Mitochondrion
• Ribosome      •  Centriole
• Golgi apparatus •  Chloroplast
• Cell membrane  •  Cell wall
• Vacuole     •  Cytoplasm
• Lysosome
• Rough endoplasmic reticulum
• Smooth endoplasmic reticulum

Summary

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1 Why is chloroplast only found in plant cells and not in animal cells?

2 Explain the asexual reproduction of Amoeba sp.

3 An aphid is a small insect which sucks sap in plants. An aphid puts its mouth in the stem
of a plant and sucks its nutrients. Name the liquid obtained from the plant and explain why
aphids are mostly found in the daytime.

4 Figure 1 shows a plant cell.

PP

State the characteristics of P and its function.

Have you mastered the following important concepts?
• Structure and function of animal cells and plant cells
• Similarities and differences between animal cells and plant cells
• Living process of unicellular organism
• Living process of multicellular organism
• Relationship between cell component density in a cell and its specific function
• Effect of the absence or failure of the cell component function in multicellular organisms
• Sequence of levels of organisation in multicellular organisms
• Organ systems with its main functions in multicellular organisms

Self Reflection

FIGURE 1

Summative Practice 2

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5 X is a digestive organ consisting of cells with a high concentration of rough endoplasmic
reticulum for protein digestion. What is X? Explain your answer.

6 Plant cells undergo process X to form xylem vessels. Name process X. Explain how xylem
vessels are specialised in transporting water and mineral salts.

7 Arif is a farmer. He uses a weed killer to control the weed population growth on his farm.
Weed killers can stop the transportion of certain minerals in plants through the hair root
tissue. Explain why.

8 Explain how an Amoeba sp. can survive in fresh water that is hypotonic to the cytoplasm of
the organism.

9 Figure 2 shows one cell that can be seen through an electron microscope.

(a) (i) Name structure K.
   (ii) State the function of structures K and N.

(b) Explain how L mantains the turgidity of the
 cell.

10 State the differences between the following:
  (a) meristematic tissue with plant epidermal

tissue,
  (b) cell and tissue,
  (c) epithelial tissue and muscle tissue.

11 Figure 3 shows the structure of a leaf.

(a) Name the cells labelled P and R.

(b) State the functions of cells P and R.

(c) Explain the role of cell Q in increasing the rate of photosynthesis.

(d) S contains two types of tissue.

(i) Name the two tissues.

R

Q

P

S

FIGURE 3

N

L

K

FIGURE 2

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(ii)  Explain the adaptive features of tissues in S that help in the transportation of materials in
plants.

Essay Questions

12 Using an example of a unicellular organism, explain the five living processes of this
organism.

13 Figure 4 shows a unicellular organism that lives in a fresh water pond.

FIGURE 4

(a) Explain the function of X in osmoregulation.

(b) Predict what will happen if a cell does not have a Golgi apparatus. Explain your
answer.

(c) For each of the following cell, explain the relationship between structure and function.

(i)  (ii)

Enrichment

14 Many cancers start with the epithelial tissue. This includes lung cancer, colon cancer and
skin cancer. What are the two characteristics of this tissue that can cause cancer?

15 The damage to the cartilage disc between the vertebrae of the spine is difficult to repair
  by their own cells. What is the latest technology that can be used to develop
  new cartilage tissues?

X

cell P cell Q

Complete answers are
available by scanning the
QR code provided

FIGURE 5

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What is the
association of carrier
protein on plasma
membrane with
genetic diseases
such as hyperkalemic
periodic paralysis
(HYPP) in horses?

• What is the fluid mosaic model?
• What are the characteristics of substances that are able to move across the plasma membrane?

• What are the examples of passive and active transport?
• How do you apply the concept of movement of substances in everyday life?

Do you KNOW...

Movement of
Substances
Across the Plasma
Membrane3

CHAPTER

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3.1 Structure of Plasma Membrane

3.1.1 Justify the necessity of movement
of substances across a plasma
membrane.

3.1.2 Describe the components of a
plasma membrane and its function
based on the fluid mosaic model.

3.1.3 Draw and label the components of
a plasma membrane based on the
fluid mosaic model.

3.1.4 Describe the permeability of a
plasma membrane.

3.2 Concept of Movement of Substances
Across a Plasma Membrane
3.2.1 State the characteristics of

substances that are able to move
across a plasma membrane in
these aspects:

• size of molecules
  • polarity of molecules
  • ionic charge
3.2.2 Conduct experiments to study the

movement of substances across a
selectively permeable membrane
by using:

• Visking tubing
  • simple osmometer
3.2.3 Describe by using examples

movement of substances across a
plasma membrane:

• passive transport
  • active transport
3.2.4 Compare and contrast passive

transport and active transport.

3.3 Movement of Substances Across a
Plasma Membrane in Living
Organisms

3.3.1 Explain by using examples the
process of passive transport in
organisms.

3.3.2 Explain by using examples the
process of active transport in
organisms.

3.3.3 Define solution:
  • hypotonic solution
  • hypertonic solution
  • isotonic solution

3.3.4 Design an experiment to
study the effects of different
concentrations of solution on
animal and plant cells.

3.3.5 Communicate about the effects
of hypotonic, hypertonic and
isotonic solutions on cell based
on movement of water molecules:

• animal cell
  • plant cell

3.4 Movement of Substances Across a
Plasma Membrane and its Application
in Daily Life

3.4.1 Conduct an experiment to
determine the concentration of
cell sap of a plant tissue.

3.4.2 Correlate the concentration of
cell sap in a plant tissue with the
phenomenon of plant wilting.

3.4.3 Explain by using examples the
application of the concept of
movement of substances across
a plasma membrane in daily life.

3.4.4 Communicate about reverse
osmosis in water purification

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glycolipid

phospholipid
bilayer

carrier
proteins

CYTOPLASM

EXTRACELLULAR

glycoprotein

channel protein

cholesterol

A nonpolar
tail which is
hydrophobic

A polar
head
which is
hydrophilic

Structure of Plasma Membrane
The necessity of movement of substances
across a plasma membrane
Living cells require substances from the external environment to carry
out living processes. At the same time, the metabolic processes in cells
produce waste that need to be disposed of from the cells. What are the
substances required by cells and the waste products of cells? How do these
substances move in and out of cells?
Cells must allow some substances to move into and out of the cells to
maintain the living processes. The movement of substances into and
out of the cell is regulated by the plasma membrane. Before you learn
about how substances move across the plasma membrane, let us look at
its structure.

Structure of the plasma membrane
In Chapter 2, you have learned that each cell is enclosed in a thin plasma
membrane which separates the living cell from its environment.
According to the fluid mosaic model, protein molecules float within
the phospholipid bilayer, forming a
mosaic pattern that changes frequently.
Figure 3.1 shows the plasma membrane.

3.1
ICT 3.1

Activity:
To discuss the need for

movement of substances
across the plasma membrane

3.1.23.1.1 3.1.3

Draw a model
of the plasma
membrane
and its related
components.

Activity Zone

FIGURE 3.1
Plasma membrane

Phospholipid

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Each phospholipid molecule consists of:
•  A polar head  which is hydrophilic (attracted to water)
•  A nonpolar tail  which is  hydrophobic (repels water)
The heads of the phospholipid molecule in the outer layer face the
extracellular fluid, whereas the phospholipid heads in the inner layer
face the cytoplasm. The tails of the phospholipid molecule of the two
layers face each other.

There are various types of protein molecules embedded partially or
fully within the membrane. The protein molecules are widely dispersed
between the phospholipid bilayer. The protein molecules with channels
or canals are known as channel proteins, whereas the protein molecules
that function as carriers are called carrier proteins (Figure 3.2).

3.1.33.1.2

Some proteins and lipids have carbohydrate chains attached to them,
known as glycoprotein and glycolipid respectively (Figure 3.1).
Glycoprotein and glycolipid act as receptors to hormones such as
insulin, stabilise the membrane by forming hydrogen bonds with water
and act as antigens for cell identification.
There are also cholesterol molecules found between the phospholipid
molecules (Figure 3.1). Cholesterols make the phospholipid bilayer
stronger, more flexible and less permeable to water-soluble substances
such as ions. The phospholipid bilayer, proteins and cholesterols are not
static but form a dynamic and flexible structure. This contributes to the
‘fluid’ characteristic of the plasma membrane and makes the plasma
membrane more flexible.

FIGURE 3.2  Models of channel  and carrier proteins

moleculeExtracellular Extracellular

molecule

pore

channel protein carrier protein

Cytoplasm Cytoplasm

There were many
plasma membrane
models proposed by
scientists. However,
the fluid mosaic
model proposed by
S. J. Singer and G. L.
Nicholson in 1972 is
the accepted model
by scientists today.

Biological Lens

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Permeability of a plasma membrane
A membrane is said to be permeable to a substance if the membrane
allows the substance to pass through it freely. On the contrary, a
membrane is impermeable if a substance is unable to move across it.
The plasma membrane is a selectively permeable membrane. What
does selectively permeable membrane mean? Selectively permeable
membrane only allows free movement of certain molecules across it, and
prevent or limit the movement of other substances.
The plasma membrane has a selectively permeable property due to its
building structure. The phospholipid bilayer and protein determine
the membrane permeability towards specific substances. We will learn
about the characteristics of substances that are able to move across the
plasma membrane in the next unit.

ICT 3.2

Video: Fluid mosaic
model of a plasma

membrane

3.1.4

Biochemists who
specialise in cell
and molecular
biology can work in
the pharmaceutical
industry,
biotechnology and
forensics.

Millennial Career

3.1Formative Practice

1 What are the two main components of the plasma membrane?

2 Explain the meaning of fluid mosaic model.

3 Predict what will happen to the plasma membrane if there is no
cholesterol.

4 Explain the role of phospholipid and protein in ensuring the
permeability of the plasma membrane.(Accessed on 21 August 2019)

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H2O

H2O O2

O2

CO2

CO2

Concept of Movement of
Substances Across a Plasma
Membrane

3.2

lipid soluble molecule

large molecule

small molecule and ion
small nonpolar molecule

channel
protein

ICT 3.3

Video: Movement of molecules
and ions across the plasma

membrane

3.2.1

The characteristics of substances that are
able to move across a plasma membrane
There are three common factors that determine whether a molecule can pass
through a plasma membrane, which are molecule size , polar molecule and
ionic charge.

FIGURE 3.3  Movement of substances across a plasma membrane

Nonpolar molecules

Examples:

• Fatty acid

• Glycerol

• Fat soluble vitamins (A, D, E, K)

• Steroid compounds

LIPID-SOLUBLE SUBSTANCES LIPID INSOLUBLE SUBSTANCES

• Polar molecules (Example: water)

• Nonpolar molecules (Examples:
oxygen, carbon dioxide)

• Ion (Examples: K+, Na+, Ca2+,
Mg2+)

SMALL MOLECULE AND ION

Examples:

• Glucose

• Amino acid

LARGE MOLECULE

CHARACTERISTICS OF MOVEMENT OF
SUBSTANCES ACROSS A PLASMA MEMBRANE

carrier protein

Extracellular

Cytoplasm

(Accessed on 21 August 2019)

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Problem statement
Does the size of dissolved particles affect the movement of substances across a selectively
permeable membrane?

Hypothesis
Only small molecules are able to diffuse through a selectively permeable membrane while a large
molecule cannot diffuse through a selectively permeable membrane.

Variables
Manipulated: Molecule size
Responding: Presence of molecules in the Visking tubing and beaker
Fixed: Surrounding temperature and soaking time

Materials
Benedict’s solution, 1% starch suspension, iodine solution, 30% glucose solution, distilled water,
Visking tubing (12 cm) and thread

Apparatus
Beaker, test tube, Bunsen burner, dropper and measuring cylinder

Procedure
 1 Soak the Visking tubing in water for 5 minutes to soften it. Make a knot and tie one end of the

Visking tubing with thread to prevent leaking.

2 Fill the Visking tubing with the 10 ml glucose solution and 10 ml starch suspension. Tie one end
of the Visking tubing tightly with a thread. The solution colour is recorded.

3 Rinse the outside of the Visking tubing with distilled
water.

4 Add 400 ml of distilled water in a beaker.

5 Place the Visking tubing in the beaker and leave it for
30 minutes.

6 After 30 minutes, remove and transfer the
Visking tubing into a dry beaker.

7 (a) Conduct the iodine test for the solutions in the
Visking tubing and in the beaker. Put 2 ml of each solution in separate test tubes and add
1 ml of the iodine solution. Observe the colour produced.

(b) Conduct the Benedict’s test on the solutions in the Visking tubing and in the beaker. Put
2 ml of each solution into separate test tubes and add 1 ml of Benedict’s solution. Heat the
solutions in a water bath for 5 minutes and record any changes in colour.

Conduct the following experiment to show that the movement of substances across a
selectively permeable membrane depends on the size of the molecule.

3.2.2

Activities2.1Activity 3.1 Experiment
Studying the movement of substances
across a selectively permeable membrane

Visking tubing
10 ml glucose
solution +
10 ml starch
suspension

400 ml
distilled
water

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Results

Contents Iodine test Benedict’s test

Visking tubing 10 ml glucose solution +
10 ml starch suspension

Beaker 400 ml distilled water

Discussion
 1 What molecule is found in (a) Visking tubing  (b) beaker?

2 What are the inferences that can be made based on the (a) size of the starch molecule (b) size of
the glucose molecule compared to the pore size in the Visking tubing?

3 What are the similarities between a Visking tubing and a plasma membrane?

Conclusion
Is the hypothesis accepted? Suggest a suitable conclusion.

3.2.2

Problem statement
How do the water molecules permeate across selectively permeable membranes?

Hypothesis
Water molecules permeate from an area of high water potential to an area of low water potential.

Variables
Manipulated: Time
Responding: Increase in the level of sucrose solution in a capillary tube
Fixed: Concentration of sucrose solution

Materials
30% sucrose solution, Visking tubing (12 cm), thread and distilled water

Apparatus
Retort stand with a clamp, 25 cm capillary tube, syringe, ruler, 50 ml beaker, marker pen, scissors
and stopwatch

Procedure
 1 Cut a Visking tubing (12 cm).

2 Soak the Visking tubing in water for 5 minutes to soften it.

3 Tie one end of the Visking tubing tightly using thread to form a bag.

4 Fill the Visking tubing with the 30% sucrose solution using the syringe.

Activities2.1 Experiment
Studying the movement of substances across a
Visking tubing using a simple osmometer

Activity 3.2

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5 Tightly tie the other end of the Visking tubing to a capillary tube.

6 Rinse the outer part of the Visking tubing with distilled water.

7 Clamp the capillary tube vertically to the retort
stand.

8 Immerse the Visking tubing in a beaker
containing distilled water.

9 With a marker pen, mark the level of sucrose
solution in the capillary tube at the beginning
of the experiment.

10 Measure and record the level of sucrose
solution in the capillary tube for every
2 minutes from the previous level for
12 minutes.

11 Plot a graph of the level of sucrose solution
(mm) against time (minutes).

Results

Time (minutes) 0 2 4 6 8 10 12

Level of sucrose solution (mm)

Discussion
 1 Based on the graph, explain the relationship between the levels of sucrose solution and time.

2 What caused the change in the levels of sucrose solution?

3 What inferences can be made regarding the size of the sucrose molecules and water, and the
pore size of the Visking tubing?

4 Predict the results if the experiment is repeated using a Visking tubing filled with distilled water,
and a beaker filled with a 30% sucrose solution.

Conclusion
Is the hypothesis accepted? Suggest a suitable conclusion.

The movement of substances across a plasma membrane occurs through
passive and  active transport.

Passive transport
An analogy of passive transport is like riding a bicycle downhill. The
process does not involve the use of energy. Simple diffusion, osmosis
and facilitated diffusion are examples of passive transport.

Do you know that
an egg membrane
can be used to
replace the Visking
tubing? Can you
make an egg
osmometer?

Our World of Biology

capillary tube

distilled water
Visking tubing
30% sucrose
solution

initial level of
sucrose solution

3.2.352

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keupayaan air tinggi
kepekatan bahan
terlarut rendah

keupayaan air rendah
kepekatan bahan
terlarut tinggi

selectively permeable membrane
water molecule

sucrose
molecule

3.2.3

Simple diffusion
Simple diffusion is the movement of molecules
or ions from an area of high concentration to an
area of low concentration. The moving molecules
are said to move down the concentration
gradient until a dynamic equilibrium is achieved.
This may occur with or without the presence of
a plasma membrane. Lipid soluble molecules
(fatty acids and glycerol), oxygen and carbon
dioxide diffuse through the phospholipid bilayer
as simple diffusion (Figure 3.4).

FIGURE 3.4  Simple diffusion through the phospholipid
 bilayer

Osmosis
Osmosis is a passive transport process that is similar to diffusion but it involves only water molecules.
Osmosis refers to the net movement of water molecules from an area of high water potential  (low
solutes concentration) to an area of  low water potential (high solutes concentration) randomly
through a selectively permeable membrane. The selectively permeable membrane is permeable to
water but impermeable to some solutes such as sucrose molecules (Figure 3.5). The same situation
occurs in cells through the phospholipid bilayer (Figure 3.6).

Facilitated diffusion
Lipid-insoluble molecules such as ions, large molecules such as amino acids and glucose are unable
to pass through the phospholipid bilayer. These substances move across the membrane with the
aid of transport proteins (carrier or channel proteins). This process is known as facilitated
diffusion. Facilitated diffusion does not require energy because the transport proteins transport
molecules down a concentration gradient. The process continues until a dynamic equilibrium
is achieved when the concentration of molecules is the same at both sides of membranes.

FIGURE 3.6  Osmosis through the phospholipid bilayerFIGURE 3.5  Osmosis

high water potential low water potential Achieved dynamic
equilibrium

osmosis

high water potential
water molecule

low water potential

phospholipid
bilayer

high
concentration

low
concentration

phospholipid
bilayer

oxygen
Solutes:

carbon dioxide
fatty acids
glycerol

Simple diffusion

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3.2.3

CHANNEL PROTEINS form channels or canals to
allow small-sized solutes and ions to diffuse across
the plasma membrane. Channels have specific internal
characteristics that only allow specific ions to pass
through it.

CARRIER PROTEINS have specific sites and can
only bind to a specific molecule. For example, glucose
molecules can only bind to the specific site of a glucose
carrier protein.

FIGURE 3.7  Facilitated diffusion through channel and
carrier proteins

Active transport
Movement of substances across a plasma membrane may occur by active transport. The
characteristics of an active transport are as follows:
• The movement of molecule or ion substances across a plasma membrane occurs against a

concentration gradient.
• It requires energy from ATP (adenosine triphosphate) molecules generated during cellular

respiration.
• It requires specific carrier protein with specific sites to bind with certain molecules or ions.

• Carrier proteins also possess receptors to bind with ATP molecules. Carrier proteins change
shape when a phosphate group attaches to it. As a result, molecules or ions move across a
membrane (Figure 3.9).

glucose

specific site

carrier protein

Extracellular
glucose
concentrations
are higher
than in
cytoplasm.

Carrier proteins
return to their
original shape
and are ready to
transport other
molecules.

Glucose binds
with the specific
site of the
carrier proteins.

Carrier proteins
change shape to
allow glucose
molecules to pass
through the carrier
proteins and to enter
the cell.

Extracellular

Cytoplasm

FIGURE 3.8  Facilitated diffusion through carrier proteins

channel protein carrier proteinchannel protein

glucose

carrier protein

ion

The process of the movement of glucose molecules across a plasma membrane occurs
through a facilitated diffusion (Figure 3.8).

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4 5 6

Na+

Na+

Na+

Na+

Na+

Na+

Na+

Na+

Na+Na+

Na+

Na+

Na+
Na+

P
ATP

ADP

Na+Na+

Na+ Na+

Na+

Na+

Na+

K+

K+

K+

K+K+

K+

K +

K +
K +

K +
K +

K+

K+
K+

K+

K+K+K+ P

Three sodium ions bind to the
carrier protein.

Two potassium ions from outside the cell
bind with a carrier protein. The phosphate
group leaves the carrier protein.

ATP molecules (adenosine triphosphate)
decompose into ADP (adenosine
diphosphate) and P (phosphate). The
phosphate group is bound to the carrier
protein.

Extracellular fluid:
High Na+ concentration

Extracellular fluid:
Low K+ concentration

Cytoplasm:
High K+ concentration

Cytoplasm:
Low Na+ concentration

The loss of the phosphate group restores
the original shape of the carrier protein.

Phosphate bond provides energy and
changes the shape of the carrier protein,
then sodium ion is transported through the
carrier protein out of a cell.

Potassium ion is transported through
the carrier protein into the cell.

1 2 3

P

3.2.3

FIGURE 3.9  Sodium-potassium pump

Active transport results in the accumulation or excretion of molecules or ions in the cell. Carrier
proteins involved in active transport are known as pumps.
For example, in animal cells, the carrier proteins that transport sodium ions to extracelullar,
and potassium ions into the cell are called sodium-potassium pump. The mechanism of the
sodium-potassium pump is described in Figure 3.9.

FIGURE 3.10  Proton pump

H+

H+

H+

H+

H+

H+

ATP

Cytoplasm:
High pH
Low concentration
of hydrogen ions

Extracellular fluid:
Low pH
High concentration
of hydrogen ions

Carrier protein

+

+

+

+

+There is another type of pump known as the
proton pump. For example, the proton pumps
that are found on the epithelial cells lining the
stomach cavity. Proton pump causes the acidity
of the stomach contents. Energy from the ATP
enables the hydrogen ion to be transported by
the carrier proteins (proton pump) towards the
extracellular fluid. This causes an accumulation
of the hydrogen ion and acid production in the
stomach cavity. The proton pump mechanism is
described in Figure 3.10.

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PASSIVE
TRANSPORT

ACTIVE
TRANSPORT

Energy is not
required

Occurs
following the

concentration of
the gradient

Occurs until
a dynamic

equilibrium is
achieved

Requires
energy

Occurs
against the

concentration
of the gradient

There are
accumulation

and disposal of
molecules or

ions

Moving a
substance across

a membrane

FIGURE  3.11  Similarities and differences between passive and active transports

3.2.4

Occurs through a
selectively permeable

membrane

1 Give some examples of substances that are able to pass through the phospholipid bilayer.

2 Describe how sodium ions are transported out of the cell.

3 A scientist conducts an experiment by soaking the roots of a plant in a solution which contains
mineral salts at different levels of oxygen concentrations. Explain why the mineral salts
concentration in the roots increased when the oxygen concentration in the solution increases by
10%.

3.2Formative Practice

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Active and passive transport in living
organisms
Where do passive and active transport occur in  living organisms?

Movement of Substances
Across a Plasma Membrane
in Living Organisms

3.3
Passive transport in organisms occurs during:

• gaseous exchange between an alveolus and
a blood capillary through simple diffusion
(Figure 3.12)

• reabsorption of water occurs by osmosis
through the renal tubule in the kidney

• absorption of water by a plant root hair cell by
osmosis (Figure 3.13)

• absorption of fructose molecule in the villus by
facilitated diffusion

Active transport in organisms occurs during:

• absorption of glucose and amino acids in the
villus

• reabsorption of glucose through the renal
tubule in the kidney

• transport of sucrose from a leaf to a phloem
tissue

• absorption of mineral ions by a plant root hair
cell (Figure 3.13)

water diffuses by osmosis
from soil water into the
root hair cells

mineral ion absorbed into the
hair root cell through active
transport

rootssoil grains
root hair

water

mineral
salts

FIGURE 3.13  Absorption of water by osmosis and mineral
ions through active transport

FIGURE  3.12  Gaseous exchange at the alveoli occurs by
simple diffusion

3.3.1 3.3.2

blood from
pulmonary veins alveolus

blood from
pulmonary artery

oxygenated
blood

blood
capillary

carbon dioxide molecules diffuse from the
blood capillaries entering the alveoli according
to the concentration gradient

deoxygenated
blood

air movement

oxygen
molecules
diffuse from
the alveolus
into the blood
capillaries
according to the
concentration
gradient

O2 CO2

a layer of water

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3.3.3

Isotonic, hypotonic and hypertonic solutions
Generally, there are three types of solutions: isotonic, hypotonic and hypertonic. As you have
learned, diffusion of water occurs by osmosis from an area of high water potential to an area of
low water potential across a plasma membrane. The explanation on each solution is summarised in
Table 3.1.

TABLE  3.1 Isotonic, hypotonic and hypertonic solutions
Concept Definition Explanation

Isotonic
solution

Solution A and B
have the same
concentrations of
solutions.

Solutions A and B are
isotonic toward each
other. No net movement
of water.

Hypotonic
solution

Solution A has a low
solutes concentration
and high water
potential.

Solution A is hypotonic
to solution B. Water
diffuses from solution A
to B by osmosis.

Hypertonic
solution

Solution A has a high
solutes concentration
and low water
potential.

Solution A is hypertonic
to B. Water diffuses
from solution B to A by
osmosis.

water molecule

A

A B

B

A B

selectively permeable membrane

water molecule

A

A B

B

A B

selectively permeable membrane
water
molecules

solutes

soluble
substances

A

A B

B

A B

selectively permeable membrane

soluble
substances

Conduct an
experiment to
observe the effect
of osmosis on
an egg plasma
membrane.

Activity Zone

selectively permeable membrane

selectively permeable membrane

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The effects of hypotonic, hypertonic and
isotonic solutions on animal cells and
plant cells
The cell cytoplasm contains solutes such as glucose and mineral
salts. The fluid found in the cell is known as intracellular fluid.
Each cell is surrounded by extracellular fluid.
Therefore, the cell will always go through the osmosis process and
movement of water across the plasma membrane, depending on the
water potential of the extracellular fluid and intracellular fluid.

3.3.4 3.3.5

• When red blood cells are placed in a
hypotonic solution, water will diffuse
into the cells by osmosis, causing
the cells to swell and finally burst.

• This is because the plasma membrane is too thin to withstand the
osmotic pressure built up in the cells.

• The burst of red blood cells is known as haemolysis.

EFFECT OF HYPOTONIC
SOLUTION

• Water diffuses into and out of the
cell by osmosis at the same rate.

• No net movement of water
across the plasma membrane.

• The cells maintain their normal
shape.

EFFECT OF ISOTONIC
SOLUTION

THE EFFECTS
OF HYPOTONIC,
ISOTONIC AND
HYPERTONIC

SOLUTIONS ON
ANIMAL CELLS

• When red blood cells are placed in a
hypertonic solution, water will move out of
the cells by osmosis.

• This will cause the cells to
shrink.

• The red blood cells
are said to undergo
crenation.

EFFECT OF HYPERTONIC SOLUTION

AR

The term haemolysis is
only used for the burst
of red blood cells.

Biological Lens

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3.3.4 3.3.5

THE EFFECTS
OF HYPOTONIC,
ISOTONIC AND
HYPERTONIC

SOLUTIONS ON
PLANT CELLS

• When plant cells are placed in a hypotonic solution, water will diffuse
into the vacuoles by osmosis.

• This will cause the vacuoles to expand and push the cytoplasm
and plasma membrane against the cell wall.

• In this condition, the cells are said to be turgid.

• Plant cells do not burst because the cell wall is rigid and strong.

• Turgor pressure is important to plant cells because it gives support
and maintains the shape of cell.

• Cell turgidity causes guard cell to swell to allow the stomata to remain
open for photosynthesis.

EFFECT OF HYPOTONIC SOLUTION

• When plant cells are placed in a hypertonic
solution, water will diffuse out of the
vacuoles by osmosis.

• Vacuoles and the cytoplasm will shrink,
causing the plasma membrane to be pulled
away from the cell wall.

• This process is known as  plasmolysis.
Plasmolysis causes leaves and stems to
bend downwards and wilt.

• Plasmolysed plant cells can regain
their turgidity if the cells are
returned to a hypotonic solution
immediately.

• The cells are said to undergo
deplasmolysis.

EFFECT OF HYPERTONIC SOLUTION

• When the sap of the plant cell and extracellular
solution are isotonic, water potential is the same.

• The movement of water diffusion in and out of
cells is the same.

• Cells become flaccid.

EFFECT OF ISOTONIC SOLUTION

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3.3.4

Activities2.1Activity 3.3 Experiment
Studying the effects of hypotonic, hypertonic
and isotonic solutions on animal cells

Problem statement
What is the effect of hypotonic, hypertonic and isotonic solutions on animal cells?

Hypothesis
• Hypotonic solution causes the cell to explode.
• Hypertonic solution causes the cell to shrink.
• Isotonic solution causes the cell to remain normal.

Variables
Manipulated: Concentration of surrounding solution
Responding: Chicken blood cells condition
Fixed: Surrounding temperature and types of cell

Materials
Fresh chicken blood, 0.15 M and 0.50 M sodium chloride solutions, filter
paper and distilled water

Apparatus
Light microscope, glass slide, dropper, mounting needle and cover slip

Procedure
 1 Prepare four slides and label as A, B, C and D.

2 Place a drop of fresh chicken blood on slide A and put a cover slip onto it.

3 Observe the shape of the red blood cells under a light microscope.

4 Put a drop of distilled water on slide B and cover with a cover slip.

5 Add a drop of blood on the edge of one side of the cover slip. At the same time, place a filter
paper at the opposite side of the cover slip to spread the blood beneath the cover slip.

6 Examine the slide under a light microscope and draw the shape of the red blood cells in the
table below.

7 Repeat steps 4 to 6. Replace the distilled water with 0.15 M sodium chloride solution (slide C)
and 0.50 M sodium chloride solution (slide D).

Results
 Slide Observation Labelled drawing of cell shape

A (red blood cells)

B (red blood cells in distilled water)

C (red blood cells in 0.15 M sodium
chloride solution)

D (red blood cells in 0.50 M sodium
chloride solution)

Discussion
 1 Discuss and explain the results obtained for each slide.

2 Which solution is hypotonic, hypertonic and isotonic to the red blood cells?

3 What happened to the red blood cells in slides B and D?

Conclusion
Is the hypothesis accepted? Suggest a suitable conclusion.

blood

filter paper

Red blood cells
from the chicken
are kept in sodium
chloride solution to
prevent blood clot.

Take Note!

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Problem statement
What is the effect of different solution concentrations on plant cells?

Hypothesis
• Hypotonic solution causes the plant cells to become turgid.
• Hypertonic solution causes the plant cells to

undergo plasmolysis.
• Isotonic solution causes the plant cells to become

flaccid.

Variables
Manipulated: Concentration of surrounding solution
Responding: Plant cells condition
Fixed: Types of plant cells

Materials
0.50 M and 1.0 M sucrose solutions, distilled water, filter paper and an onion bulb

Apparatus
Light microscope, glass slide, cover slip, small knife, mounting needle, dropper and a pair of
forceps

Procedure
 1 Prepare three slides and label as A, B and C.

2 Peel off the epidermal layer (skin) from the inner surface of an onion scale leaf and place it in a
drop of distilled water on slide A. Use a cover slip to cover the glass slide.

3 Observe the epidermal cells of the onion scale leaf under the microscope. Draw and label
your observation.

4 Peel off another epidermal layer (skin) of the onion scale leaf and put it on slide B. Cover with
a cover slip. Add a drop of 0.50 M sucrose solution on the edge of one side of the cover
slip. Spread the solution under the cover slip by placing a filter paper at the opposite side
of the cover slip. Examine the epidermal cell under the microscope. Draw and label your
observation.

5 Repeat step 4 by replacing 0.50 M sucrose solution with 1.0 M sucrose solution (slide C).

6 Flow excess distilled water through the onion scale leaf on slide C. Observe the cells under
the microscope. Draw and label your observation.

Results

Slide Observation Labelled drawing of cell shape

A (distilled water)

B (0.5 M sucrose solution)

C (1.0 M sucrose solution)

Slide C after being flowed with
excess distilled water

Activities2.1Activity 3.4 Experiment
Studying the effects of hypotonic, hypertonic
and isotonic solutions on plant cells

3.3.4

sucrose
solution filter paper

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Discussion
 1 Discuss and explain the results obtained for each slide.

2 Which solution is hypotonic, hypertonic and isotonic to plant cells?

3 What happened to the plant cells in slide C?

Conclusion
Is the hypothesis accepted? Suggest a suitable conclusion.

2.1Activity 3.5 Experiment
Studying and determining the concentration of
extracellular solution which is isotonic to the
cell sap of a plant

Problem statement
What is the concentration of the extracellular solution which is isotonic to the cell sap of potatoes?

Hypothesis
Concentration of the extracellular solution which is isotonic to the potato cell sap does not change
the mass of the potato cell sap.

Variables
Manipulated: Concentration of sucrose solution
Responding: Percentage of difference in the mass of potato strips
Fixed: Soaking period, surrounding temperature and type of potato

Materials
Potato, distilled water, 0.1 M, 0.2 M, 0.3 M, 0.4 M, 0.5 M, 0.6 M sucrose solutions and filter paper

Apparatus
Test tube, cork borer, knife, forceps, ruler, 50 ml beaker, measuring cylinder, test tube rack and
electronic weighing scale

Procedure
 1 Seven petri dishes are prepared and labelled as A, B, C, D, E, F and G.

2 Fill each beaker with the following solutions:

Beaker A:  distilled water  Beaker E: 0.4 M sucrose solution
 Beaker B: 0.1 M sucrose solution Beaker F: 0.5 M sucrose solution
 Beaker C: 0.2 M sucrose solution Beaker G: 0.6 M sucrose solution
 Beaker D: 0.3 M sucrose solution
 3 A medium-sized cork borer is pressed into a potato.

4 The potato strip is then removed from the cork borer.

5 Next, the potato strip is cut to 50 mm in length.

6 Steps 3 to 5 are repeated to obtain 6 more cylinder of potato strips of the
same length.

7 Wipe the potato strips using filter paper and each is weighed to get the initial mass.

8 Pour the solutions in beaker A to G into the test tube A to G, respectively. Each potato strip is
soaked completely in each test tube.

Do not hold the
potato with your
hands while cutting.
You should use a
cutting board as a
base.

Take Note!

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9 After soaking for 30 minutes, remove each potato strip from the respective test tube and wipe
dry with a filter paper. Weigh each strip to get the final mass.

10 The results are recorded. Plot a graph of the percentage of difference in mass against the
concentration of sucrose solution.

Results

 Test

tube
Beaker

Mass of potato strips (g) Difference in
mass (g)

Percentage of
difference in

mass (%)Initial mass Final mass

A Distilled
water

B
0.1 M
sucrose
solution

Discussion
 1 What is the texture and condition of each potato strip after being soaked in different

concentrations of sucrose solutions?

2 How do you determine the concentration of sucrose solution which is isotonic to the potato sap
cell from your graph?

3 Based on the experiment, discuss the process of osmosis in different concentrations of sucrose
solutions.

Conclusion
Is the hypothesis accepted? Suggest a suitable conclusion.

3.4.1

1 Explain what will happen to the following cells when they are soaked in a solution with a higher
concentration of solutes compared to the intracellular fluid:

(a) red blood cells

(b) plant cells

2 Explain the effect of the hypotonic solution on red blood cells.

3 Explain the effect of the isotonic solution on plant cells.

4 Explain the following situations based on the osmosis concept:

(a) Why do hawkers spray water on their fruits and vegetables?

(b) Explain what happens when sugar is sprinkled on strawberries.

3.3Formative Practice

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Phenomenon of plant wilting
Excessive use of fertilisers may cause wilting in plants. Dissolved
fertilisers will cause soil water to be hypertonic to the sap cell of roots.

Consequently, water will diffuse by osmosis from the roots’ cell
sap to the soil, and cells will become plasmolysed. Cells in plants

will recover once they are watered. However, if the period of
plasmolysis is prolonged, wilted plants will eventually die.

3.4.2

Wilted plants

You have definitely eaten canned pickles, salted fish and
fruits before. How are the concepts of osmosis and

diffusion applied in food preservation?
You can also apply the concept of movement

of substances across a plasma membrane by
conducting Activity 3.6.

Movement of Substances
Across a Plasma Membrane
and Its Application in Daily Life

3.4

Activities2.1Activity 3.6
Project

Applying the concept of movement of
substances across a plasma membrane

Materials
Egg, banana, cabbage, a variety of fruits, gelama fish, sugar, salt, vinegar and boiled water

Apparatus
Knife, bottle and oven

Procedure
 1 Conduct the activity in a group.

2 Apply the concept of movement of substances across a plasma
membrane in food production.

3 Use local raw materials and market the products at schools.

4 Among the food products that you can prepare are colourful
preserved eggs, different flavours of smoked banana, salted fish,
pickles and coloured cabbage.

A group of researchers in
Malaysia were successful
in developing Sil-RH
Membrane Distillation.
This membrane is made
from paddy chaff and
can be used for the
desalination process.

Malaysian Innovation

Fresh plants

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water-insoluble
drugs are
transported in the
hydrophobic area

The concept of movement of substances across a plasma membrane has many
applications in our daily lives, for example:

Saline solutions, normally used in
medicine, are isotonic solution to the
blood plasma. It contains 0.85–0.90 g
sodium chloride per 100 ml.

Rehydration drinks such as oral
rehydration salts help to
recover loss of water and
electrolytes in individuals
with diarrhoea.

water-soluble drugs
are transported in
the hydrophilic area

Isotonic drinks help
athletes to recover
loss of water and
electrolytes such
as potassium and
sodium through
perspiration.

pressure is
applied

salt water
goes in

foreign particles and
microorganisms

selectively permeable
membrane

water molecules

fresh water
comes out

Reverse osmosis is a technology commonly
used to extract fresh water from seawater using
the desalination process. In a reverse osmosis
equipment, pressure is applied to push the
seawater through a semi-permeable membrane.
The membrane allows water molecules to pass
through it but not foreign particles, salt and
microorganisms. As a result, only pure fresh water
is released.

Study the reverse
osmosis process in
water purification.

Activity Zone

ICT 3.4

Video: Liposome application
(Accessed on 21 August 2019)

Liposomes are vesicles that contain aqueous
solution surrounded by a phospholipid bilayer
membrane. Liposomes are used to protect drugs
or active substances taken orally from being
destroyed by gastric juices. This way, drugs can
reach the target cells.

3.4.3 3.4.4

1 Explain why green chillies are preserved using vinegar and sugar.

2 Pickled pineapples are prepared by soaking the pineapple pieces
in a thick sugar solution. State two advantages and disadvantages
of this method as compared with keeping fresh pineapples.

3 Ariana had diarrhoea after eating stale food. Suggest something
that can help her to recover.

4 A liposome is a vesicle that contains an aqueous solution that is
surrounded by a phospholipid bilayer membrane. Explain the use
of liposomes in everyday life.

3.4Formative Practice

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MOVEMENT OF SUBSTANCES ACROSS A PLASMA MEMBRANE

• Fluid mosaic
model

• Plasma
membrane
consists of
carrier proteins,
channel proteins,
phospholipid
bilayer,
glycoprotein,
glycolipid and
cholesterol

• Passive and active
transport in living
organisms

• Isotonic, hypotonic
and hypertonic
solutions

• Effect of isotonic,
hypotonic and
hypertonic solutions
on animal cells and
plant cells

• Phenomenon of
plant wilting

• Application of
movement of
substances
across a plasma
membrane in
daily life

• Reverse osmosis
process in water
purification

Concept of
Movement of

Substances Across a
Plasma Membrane

Movement of
Substances Across a
Plasma Membrane in

Living Organisms

Structure of Plasma
Membrane

• Passive
transport

– Simple
diffusion

– Osmosis
   – Facilitated

diffusion

• Active
transport

Have you mastered the following important concepts?
• The necessity of movement of substances across a plasma membrane
• Components of a plasma membrane and its function based on the fluid mosaic model
• Permeability of a plasma membrane
• The characteristics of substances that are able to move across a plasma membrane
• Passive and active transport
• Hypotonic, hypertonic and isotonic solutions
• Effect of different solution concentrations on plant cells and animal cells
• Application of the concept of movement of substances across a plasma membrane in

daily life

Movement of
Substances Across

a Plasma Membrane
and its Application in

Daily Life

Summary

Self Reflection

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1 State the meaning of passive transport.

2 The characteristic of a membrane decides if the molecule can move across the membrane.
Explain why the plasma membrane is impermeable to many types of molecules.

3 Explain why food is preserved using a thick sugar solution.

4 (a)   Explain the role of contractile vacuole in controlling the water equilibrum in an Amoeba sp.

(b) Predict what will happen when the Amoeba sp. is transferred to the sea. Explain your
answer.

5 (a) Figure 1 shows one component cell that is found in the cell.

(i) Name the component of the cell in Figure 1.

(ii) Label X and Y.

(b) In an experiment, the palisade mesophyll is soaked
in filtered water for a period of time. Explain what
will happen to the cell.

6 Figure 2 shows a type of substance movement across a
plasma membrane. Explain why calcium ion in Figure 2
is taken to extracellular.

7 A research on the effect of the concentration of different salt solutions on human red blood
cells was conducted. A droplet of blood was dropped into three test tubes A, B and C
containing different solution contents. Test tube A contained 4.5% sodium chloride, test
tube B contained filtered water and test tube C contained 0.85% sodium chloride. All three
test tubes were set aside for 30 minutes.

(a) Explain the observed solution of each test tube for 30 minutes.

(b)  Describe your observation on test tubes A and B.

(c)  Draw the shape of the red blood cell for each test tube when viewed through the
microscope.

(d)  Suggest a suitable concentrated solution to keep red blood cells. Give justifications for
your answer.

Summative Practice 3

X:Y:

Extracellular

Intracellular

X
Y

FIGURE 1

FIGURE 2

Extracellular

Intracellular

Ca2+ PATP

ADP
ATP site

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Essay Questions

8 State the similarities and differences between simple diffusion and osmosis.

9 (a) Some housewives soak their vegetables in salt solution before cooking to get rid of
pesticides and worms.

(i) Explain why vegetables that are soaked too long in salt solution wither.

(ii) Suggest a way to revive the turgidity of the vegetables.

(b)  Explain in detail how the molecules of amino acid are moved across a plasma
membrane.

(c)  Cosmetic cream containing lyposome is said to be more effective in skin care. Give your
assessment on this statement.

Enrichment

10 You work in a company that produces drinks for different groups of athletes. The suggested
carbohydrate concentration for the drinks are as follows: A gymnast:1–3%; middle distance
runner: 6–8%; and footballer:10–12%.

Give justifications for the concentration of drinks that you produce.

11 The rate of movement of material P and material S across a plasma membrane is shown in
Figure 3 and 4.

(a) Give an example of material P and explain how material P can move across a membrane
plasma.

(b) Compare the movements of material P and material S across the membrane plasma.

FIGURE 3

Rate of movement of P (mg s–1)

Concentration of material P (unit)

0 5 10
10–4

10–3

10–2

10–1

1

P

Ra
te

of
mo

ve
me

nt
of

P
(u

nit
 s-1

)

Concentration of material P (unit)

FIGURE 4

Movement rate of S (mg s–1)

Concentration of material S (unit)

0 5 10
10–4

0.1

0.2

0.3

0.4

S

Mo
ve

me
nt

ra
te

of
S

(u
nit

s-1
)

Concentration of material S (unit)

Complete answers are
available by scanning the
QR code provided

P

S

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How can the
knowledge
of biological
elements and
molecules be
used for biological
control of fresh
water snails?

• What is the chemical composition in a cell?
• What are the properties of water and its importance in a cell?
• What are the elements found in carbohydrates, proteins, lipids and nucleic acids?
• How are polymers produced from monomers?
• What is the importance of carbohydrates, proteins, lipids and nucleic acids in a cell?

Do You KNOW…

Chemical
Composition
in a Cell4

CHAPTER

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4.1 Water

4.1.1 Describe the properties
of water molecule.

4.1.2 Correlate the properties
of water with its
importance in the cell.

4.2 Carbohydrates

4.2.1 List the elements of
carbohydrate.

4.2.2 Explain the types of
carbohydrates:

•  monosaccharides.
•  disaccharides.

•  polysaccharides.
4.2.3 Conceptualise

the formation and
breakdown of:

•  disaccharides.
•  polysaccharides.

4.2.4 Write and explain the
word equation for
the formation and
the breakdown of
disaccharides.

4.2.5 Justify the importance
of carbohydrates in cell.

4.3 Proteins

4.3.1 List the elements
of proteins.

4.3.2 Conceptualise
the formation and
breakdown of
dipeptides and
polypeptides.

4.3.3 Write and explain the
word equation for
the formation and
the breakdown of
dipeptides.

4.3.4 Justify the importance
of proteins in a cell.

4.4 Lipids

4.4.1 List the elements in lipids.
4.4.2 Explain the main types

of lipids.
4.4.3 Describe the formation

and the breakdown of
a triglyceride.

4.4.4 Write and explain the
word equation for
the formation and
the breakdown of a
triglyceride.

4.4.5 Justify the importance
of lipids in cell and
multicellular organisms.

4.5 Nucleic acids

4.5.1 List the elements in
nucleic acids.

4.5.2 Explain the structure of
nucleotides:

• nitrogenous base
   • ribose or

deoxyribose sugars
   • phosphate
4.5.3 Describe the structure

of the nucleic acids:
   • deoxyribonucleic

acid (DNA)
   • ribonucleic acid

(RNA)
4.5.4 Justify the importance

of nucleic acids in cells:
   • carrier of hereditary

information.
   • production of

proteins.
4.5.5 Describe the formation

of chromosomes from
DNA and proteins.

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Water
In Form Two, you have learned briefly about water and organic
compounds. Examples of organic compounds are carbohydrates,
proteins, lipids and nucleic acids. What is the function of organic
compounds and water in the cells of an organism?

Properties of water and its importance in a cell

4.1

FIGURE 4.2 Cohesive force and adhesive force inside the
xylem tube

4.1.1 4.1.2

H
H

H

H

O

O hydrogen bond

• Water is an inorganic compound consisting of the
hydrogen (H) and oxygen (O) elements.

• Water molecules are polar molecules because shared
electrons between oxygen and hydrogen will be attracted
towards oxygen which is more electronegative (δ–).

• This polarity produces hydrogen bonds and allows water
to act as a universal solvent (Figure 4.1).

• The universal solvent properties of water allow solutes
such as glucose and electrolytes to be transported
through the plasma membranes into cells for biochemical
reactions.

• Water molecules are attached to each other
through a cohesive force.

• At the same time, water molecules are also
attached to other surfaces through adhesive
force.

• Both forces produce the capillary action which
allows water to enter and move along narrow
spaces, such as in the xylem tube.

• Water has a high specific heat capacity of
 4.2 kJ kg-1 °C-1.
• This means that 4.2 kJ of heat energy is required
 to raise the temperature of one kilogram of water
 by 1°C.
• Water absorbs a lot of heat energy with a small rise in

temperature. This characteristic is very important to
maintain the body temperature of organisms.

POLARITY OF WATER

COHESIVE FORCE AND ADHESIVE FORCE
OF WATER

SPECIFIC HEAT CAPACITY OF WATER

FIGURE 4.1 Hydrogen bond
between water molecules

xylem vessel

water

adhesive
force

cohesive
force

1 What chemical bonds are
broken when water changes
from liquid to water vapour?

2 Why is water known as polar
molecule?

3 State the meaning of adhesive
force and cohesive force.

4 Explain how perspiration helps
to lower the body temperature.

4.1Formative Practice

ICT 4.1

Quiz: Test your understanding
of water

How do aquatic
animals live in frozen
sea water?

Brainstorm!
PHOTOGRAPH 4.1
A polar bear in the
sea covered with ice

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4.2.1 4.2.2

4.2  Carbohydrates
Organic compounds are chemical compounds that contain carbon
elements. Large and complex compounds form macromolecules. Most
macromolecules are polymers comprising small molecules known as
monomers (building blocks). Carbohydrates, proteins and nucleic acids are
polymer molecules of organic compounds.
Carbohydrates are important as a source of energy and the basic structure of
some organisms. Carbohydrates are organic compounds consisting of the
elements carbon (C), hydrogen (H) and oxygen (O) in the ratio 1:2:1 and
with the chemical formula (CH2O)n.

Types of carbohydrates
There are three main types of carbohydrates, which are:
• monosaccharides (simple sugars)
• disaccharides
• polysaccharides (complex sugars)

Monosaccharides
Monosaccharides are carbohydrate monomers, which are the simplest
carbohydrate units. Monosaccharides can combine to form polymers
through a condensation reaction. Most monosaccharides taste sweet, can
form crystals and dissolve in water.
Examples of monosaccharides:
• Glucose is a sugar found in plants such as rice and wheat as well as fruits

such as grapes. Glucose is the most commonly found monosaccharide
and most polysaccharides are formed from this sugar.

• Fructose is the sugar found in honey and sweet fruits.
• Galactose is found in milk.
Monosaccharide has the reducing power, which is the ability to transfer
hydrogen (or electron) to other compounds. This is called the
reducing process. When the monosaccharide is heated in
Benedict’s solution, the monosaccharide will reduce the
blue copper (II) sulphate to a brick red precipitate of
copper (I) oxide which is not soluble in water. All
monosaccharides give this reaction and it is known as
reducing sugars.

PHOTOGRAPH
4.2  Foods rich in
carbohydrates

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Disaccharides
Disaccharide molecules are formed when two simple sugar molecules
(monosaccharides) combine through condensation to form a
disaccharide unit. This process involves the removal of a water
molecule.

Disaccharides can also be broken down to their monosaccharide units
through hydrolysis and the addition of one water molecule.

Examples of disaccharides include sucrose, maltose and lactose
(Photograph 4.3). Lactose and maltose are reducing sugars while
sucrose is a non-reducing sugar.

PHOTOGRAPH 4.3 Examples of disaccharides and its sources

Sucrose is found in sugar cane,
sweet fruits and sugar beet

Lactose in milk
Maltose is found in
 grains

4.2.34.2.2 4.2.4

Design an experiment
to ascertain the
presence of reducing
sugar and non-
reducing sugar
(sucrose).

Activity Zone

Maltose + water       Glucose + glucose

Sucrose + water      Glucose + fructose

Lactose + water      Glucose + galactose

hydrolysis

hydrolysis

hydrolysis

Glucose + glucose  Maltose + water

Glucose + fructose  Sucrose + water

Glucose + galactose  Lactose + water

condensation

condensation

condensation

Sucrose is a non-
reducing sugar
because it does not
reduce copper (II)
sulphate solution. If
sucrose is hydrolysed
(disintegrated) into units
of monosaccharides
(glucose and fructose)
earlier, the Benedict’s
test will show a positive
result.

Biological Lens

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4.2.3 4.2.5

TMK 4.2

Polysaccharides
Polysaccharides are sugar polymers
consisting of monosaccharide monomers.
Similar to disaccharides, polysaccharides
are formed through the condensation process and involves hundreds of
monosaccharides to form long molecular chains. Polysaccharides are
not soluble in water due to their large molecular size. Polysaccharides
neither taste sweet nor crystallise.
Polysaccharides can also disintegrate through hydrolysis with the help
of dilute acids, boiling and enzyme action. Polysaccharides play various
roles in organisms.

Importance of carbohydrates in cells
• As a source of energy, for example glucose
• As a food reserve, for example glycogen in animal cells and starch

in plant cells
• As a support structure, for example cellulose in the plant cell

wall

ICT 4.2

Quiz: Test your
understanding of

carbohydrates

Chitin, a type of
polysaccharide, is used
as a surgical thread.
The advantage is that
chitin will decompose
after the sewn wound
heals.

Our World of Biology

1 Name the elements in carbohydrates.

2 List the main types of carbohydrates.

3 Give examples of reducing sugars
and non-reducing sugars.

4 Explain why sucrose is a
non-reducing sugar.

4.2Formative Practice

Cellulose forms the main
structure of the plant cell wall.

Glycogen is the
main storage of
polysaccharide found in
muscle cells and animal
liver cells.

Starch is the main
storage of polysaccharide
in plants. Starch is also
found in chloroplasts.
Source: grains, potatoes
and legumes.

plant cell
wall

PHOTOGRAPH 4.4
Examples of polysaccharides

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4.3

Each dipeptide can be broken down into an amino acid through
hydrolysis.
There are about 20 types of amino acids present naturally.  Various types
of polypeptide molecules can be formed from the 20 types of amino
acids. This is because each type of protein differs in terms of the amino
acid sequence in its polypeptide chain.

Importance of proteins in a cell
Proteins are used to build new cells, repair damaged tissues and for the
synthesis of enzymes, hormones, antibodies and haemoglobin. Proteins
also form building blocks such as keratin in the skin, collagen in bones
and myosin in muscle tissues.
The breakdown of proteins or polypeptides by digestive enzymes gives us
the energy to carry out our daily activities. Polypeptides can disintegrate
into amino acids. This amino acid is then used again to build the protein
molecules needed by the body. You will learn in detail about the process
of protein digestion in Chapter 9.

4.3.1 4.3.2 4.3.3 4.3.4

Protein
Protein is a complex compound composed of carbon, hydrogen,
oxygen and nitrogen elements. Most proteins also contain sulphur
and phosphorus. Foods rich in proteins include fish, meat, milk, beans
and eggs. All proteins are composed of one or more polymers known as
polypeptides. Each polypeptide is made up of monomers or small units
known as amino acids. A polypeptide can consist of fifty to thousands
of amino acid molecules. Amino acids are linked together through the
condensation process.
Dipeptides are composed of two amino acid molecules which are linked
together by a peptide bond through the condensation process. In this
process, one water molecule is removed. Further condensation can link
more amino acids to form a polypeptide chain.

1 State the monomers of proteins.

2 Name the reaction that forms
dipeptide.

3 State two importance of proteins.

4 Explain the effect of protein
deficiency in a diet on the hair
and nails of an individual.

4.3Formative Practice

ICT 4.3

Quiz: Test your
understanding of

proteins

Amino acid + amino acid

Amino acid + amino acid

Dipeptide + watercondensation

hydrolysis Dipeptide + water

PHOTOGRAPH 4.5
Foods rich in proteins

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4.4

+

triglycerides waterglycerol fatty acids
hydrolysis

condensation

+ 3H2O

4.4.34.4.24.4.1

PHOTOGRAPH 4.6
Foods rich in lipids

Lipids
Lipids are naturally occurring hydrophobic compounds found in plant and
animal tissues. Like carbohydrates, lipid is made up of carbon, hydrogen and
oxygen elements but with a much higher ratio of hydrogen atoms to oxygen
atoms.
Lipids are insoluble in water but soluble in other organic solvents, for
example, alcohol, ether and chloroform.

Types of lipids
Types of lipids include fats, waxes, phospholipids and steroids.

Fats
Fats and oils are triglycerides. Triglycerides are a type of ester formed from
the condensation of one glycerol molecule with three molecules of fatty
acids. Triglycerides can be hydrolysed again into fatty acids and glycerol
through the reaction of hydrolysis. Glycerols are a type of three carbon
alcohol that contain three hydroxyl groups (–OH).

There are two types of fatty acids, namely saturated fatty acids and
unsaturated fatty acids. The similarities and differences between saturated
fats and unsaturated fats are shown in Tables 4.1 and 4.2.

The presence of fats
can be tested with the
ethanol emulsion test.
The formation of a
white emulsion shows
a positive result of the
presence of lipids.

Biological Lens

TABLE 4.1 Similarities between saturated fats and unsaturated fats

Similarities

Both consist of carbon, hydrogen and oxygen elements.

Both contain glycerol and fatty acids.

Both contain nonpolar molecules.

ICT 4.4

Quiz: Test your
understanding of

lipids

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Wax
Wax contains one molecule of alcohol that combines with another molecule
of fatty acid and is waterproof.

Phospholipid
Phospholipids are a major component of plasma membranes and are made
up of one molecule of glycerol that combines with two molecules of fatty
acid and one group of phosphate.

Steroids
Steroids are lipids that do not contain fatty acids. Examples of steroids are
cholesterol, testosterone, estrogen and progesterone.

Importance of lipids in cells
Fats function as reserved energy for animals. In addition, fats also function
as a liner to protect internal organs and act as a heat insulator for animals.
Waxes are an important component in cuticles that cover the epidermis of
leaves and sebum secreted by our skin. The function of glycolipid is to ensure
the stability of the plasma membrane and to help in the cell identification
process. Cholesterol is important in steroid hormone synthesis.

4.4.5

Saturated fats are
not good for health
because it raises
the cholesterol level
in the blood in the
form of low-density
lipoprotein (LDL).
This increases the
risk of a heart attack.
Saturated fats can
also increase the risk
of diabetes.

Our World of Biology

Saturated fats Unsaturated fats

Fatty acids only have single bonds between
carbon.

Fatty acids have at least one double bond
between carbon.

Do not form chemical bonds with additional
hydrogen atoms because all bonds between
carbon atoms are saturated.

Double bonds can still receive one or more
additional hydrogen atoms because carbon atoms
are unsaturated.

Exist in solid form at room temperature. Exist in liquid form at room temperature.

Source: butter and animal fat Source: olive and fish oil

O

O
C C

H

H

H
–

–
–

=
C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– H–

O

O

C

H –

–

–

=

C

H

C

H

H
C

H

H

–
C

H

H

–
–

– C

H

H
C

H

H

–
C

H

H

–
C

H

H

–
H–C

H

H

C

H

H
–

C

H

H
–

– C

H

H

C

H

H
–

C

H

H
–

C

H

– =

O

O
C C

H

H

H
–

–
–

=
C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– C

H

H

– H–

O

O

C

H –

–

–

=

C

H

C

H

H
C

H

H

–
C

H

H

–
–

– C

H

H
C

H

H

–
C

H

H

–
C

H

H

–
H–C

H

H

C

H

H
–

C

H

H
–

– C

H

H

C

H

H
–

C

H

H
–

C

H

–
=

double
bond

TABLE 4.2 Comparison between saturated fats and unsaturated fats

1 State the elements in lipids.

2 List the types of lipids.

3 Give your opinion on the intake
of synthetic steroids to build
body muscles.

4.4Formative Practice

4.4.478

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4.5

4.5.1 4.5.2 4.5.3

Nucleic Acids
Nucleic acids are one or two polymer chains comprising of nucleotide
monomers. Nucleic acids are formed from the elements of carbon, hydrogen,
oxygen, nitrogen and phosphorus. Each nucleotide consists of a pentose
sugar (5-carbon sugar), a nitrogenous base and a phosphate group that

are combined together through the condensation process
(Figure 4.3). There are two types of pentose sugars, that are,
ribose and deoxyribose. The nitrogenous base consists of
adenine (A) guanine (G), cytosine (C), thymine (T) and
uracil (U).

sugar-phosphate chain

double
helix
structure
of DNA

hydrogen
bonds between
nitrogenous base

T

C

C

T

T

A

G

G

A

A

phosphate group
nitrogenous base

deoxyribose sugar

sugar-phosphate chain

double
helix
structure
of DNA

hydrogen
bonds between
nitrogenous base

T

C

C

T

T

A

G

G

A

A

phosphate group
nitrogenous base

deoxyribose sugar

ICT 4.5

Activity: Making a
DNA model

nucleotide

FIGURE 4.3 Double helix structure of DNA

There are two types of nucleic acids:
• deoxyribonucleic acid (DNA)
• ribonucleic acid (RNA)
Ribonucleic acid (RNA) contains ribose sugar while
deoxyribonucleic acid (DNA) contains deoxyribose
sugar.

Deoxyribonucleic acid (DNA)
DNA consists of two polynucleotide chains that are
intertwined in opposite directions and form the double
helix (Figure 4.3). The nitrogenous base groups on both
polynucleotide chains are matched and bound together
by hydrogen bonds. The nitrogenous bases for DNA are
adenine (A), guanine (G), thymine (T) and cytosine (C).
Adenine will pair with thymine while guanine will pair
with cytosine.

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Ribonucleic acid (RNA)
The RNA structure is a single polynucleotide chain which is shorter
compared to DNA (Figure 4.4).
The nitrogenous bases for RNA are adenine, guanine, cytosine and
uracil. Thymine in DNA is replaced by uracil in RNA.
The three main types of RNA, are messenger RNA (mRNA), ribosomal
RNA (rRNA) and transfer RNA (tRNA). These three RNAs are involved
in the protein synthesis process.

Importance of nucleic acids in a cell
Do you know what determines the characteristics of an organism such
as eye colour or height? DNA is important as a carrier of hereditary
information and a determinant of characteristics in living organisms.
DNA contains genetic codes carried by nitrogenous bases (A, G, C and
T) for the synthesis of polypeptides which form proteins.
The genetic code is written as a series of three bases that determine the
sequence of amino acids in proteins to be synthesised. For example, the
AUG codon (base sequence: adenine, uracil and guanine) on mRNA
is the code for methionine amino acid (Figure 4.5). The three-base
sequence in DNA is transcribed into mRNA codons which are then
translated into the amino acid sequence to form a single polypeptide
chain.  This means that the sequence of nucleotides in DNA determines
the amino acid sequence in the polypeptide chain that builds the
corresponding protein.

FIGURE 4.4 Structure of RNA

FIGURE 4.5 Genetic code on mRNA is translated into protein

H2N

N
N

N

N
Adenina

H

H

NH2

H-N
N

N

N
O

Guanina

H

NH2

O

N
H

H
N

Sitosina

O

H3C O

N-HH
H-N

Thimina

H2N

N
N

N

N
Adenina

H

H

NH2

H-N
N

N

N
O

Guanina

H

NH2

O

N
H

H
N

Sitosina

O

H O

N-HH
N

Urasil

DNA RNA

sugar and
phosphate
backbone

nitrogenous base

4.5.3 4.5.4 4.5.5

ICT 4.6

Quiz: Test your
understanding of

nucleic acids

Formation of chromosomes from DNA and
proteins
Chromosomes are formed from DNA polynucleotide chains that are wound
around a protein called histone. Histones do not carry genetic information.
DNA molecules combine with histone proteins to form nucleosomes.
Nucleosomes are intertwined to form the chromosome structure.

DNA

mRNA

transcription

translation

codons

Polypeptides

A U CG G A U A C

methionine arginine tyrosine

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nitrogenous base pairs
adenine (A)
thymine (T)
guanine (G)
cytosine (C)

gene

nucleus

DNA

FIGURE 4.6 Formation of chromosome from DNA and proteins

AR

Water

Characteristics:
water polarity,
specific heat
capacity of
water, adhesive
force and
cohesive force
of water

• Each nucleotide
consists of
pentose sugar,
nitrogeneous
base and
phosphat group

• Type:
 - DNA
 - RNA

• Type: fats, wax,
phospholipids and
steroids

• Comparison
between
saturated fats and
unsaturated fats

Type:
• monosaccharide:

glucose, fructose
and galactose

• disaccharide:
maltose, sucrose
and lactose

• polysaccharide:
starch, glycogen
and cellulose

Amino acid

Dipeptides

Polypeptides

Carbohydrates Proteins Lipids Nucleic Acids

CHEMICAL COMPOSTION IN A CELL

NUCLEOSOME

HISTONE

CELL

CHROMOSOME

Summary

1 State two types of nucleic acids.

2 State the components in a
nucleotide.

3 Explain why the RNA structure is
shorter than the DNA structure.

4 Explain the possibility that may occur
if a cell does not have a nucleic acid.

4.5Formative Practice

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1 Wax is a type of lipid. Wax can be found in the leaf cuticle, fruits and seeds. Explain the
function of wax on a fruit skin.

2 Figure 1 shows a nucleotide.

(a) Name the components P, Q and R.

(b) Complete Figure 2 to show a complete DNA molecule.

3 State the differences between

(a) the structure of a DNA molecule and a polypeptide molecule

(b) the structure of DNA and RNA

Have you mastered the following important concepts?
• Properties of water and its importance in cells
• Types of carbohydrates and its importance in cells
• Elements of proteins and its importance in cells
• Types of lipids and its importance in cells
• Structure of nucleic acids and its importance in cells
• Formation of chromosomes

Summative Practice 4

Self Reflection

P

R Q

FIGURE 1 FIGURE 2

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4 (a)   How does water help in the respiration and digestion process?

(b) What are the characteristics of aquatic life that allows them to live throughout winter?

5 Figure 3 shows the reaction between the formation and the decomposition of lipids.

(a) (i)   Name the parts labelled K, L and M.

(ii)  State processes P and Q.

(b) Fatty acids are divided into saturated
fatty acids and unsaturated fatty acids.
State the four differences between
saturated fats and unsaturated fats.

Essay Questions

6 Figure 4 shows the production of carbohydrates through the process of photosynthesis.
Carbohydrate is a type of macromolecule.

(a) Explain the meaning of organic compound macromolecules.

(b) Explain the formation of polysaccharide.

(c) Discuss the importance of organic compounds in
cells.

Enrichment

7 Ariff practises a diet low in carbohydrates. Suggest
what needs to be done to reduce the starch in potatoes
during food preparation.

8 The results of biological studies have been widely
applied in various industries to enhance quality and
productivity and to overcome problems. Among the
studies being used by environmentalists are biodegradable plastics and environmentally
friendly batteries. A group of scientists in Malaysia have succeeded in inventing
environmentally friendly batteries using seaweed pulp. In your opinion, how did research on
chemical elements in seaweed help scientists to invent environmentally friendly products?

water

sunlightoxygen

carbon
dioxide

FIGURE 4

FIGURE 3

K L

+

P

Q
M

+  H2O

Complete answers are
available by scanning the
QR code provided

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• What types of metabolism occur in a cell?
• What are the characteristics of the enzyme and its mechanism of action?
• What are the factors that influence the reaction of enzymes?
• What is the application of enzymes on everyday life?

Do you KNOW...

What is the purpose of
the enzyme extracted
from fungi?

Metabolism
and Enzymes5

CHAPTER

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5.1 Metabolism

5.1.1 Define metabolism.
5.1.2 State the types of metabolism in a cell:
  • anabolism
  • catabolism

5.2 Enzymes

5.2.1 Define enzymes.
5.2.2 Reason out the necessity of enzymes

in metabolism.
5.2.3 Describe the naming of enzymes with

the addition of –ase to their substrates.
5.2.4 Characterise the general properties of

enzymes.
5.2.5 Communicate about the involvement

of specific organelles in the production
of:

• intracellular enzymes
  • extracellular enzymes
5.2.6 Explain the mechanism of enzyme

action using the ‘lock and key’
hypothesis.

5.2.7 Interpret energy diagrams to explain
the mechanism of enzyme action.

5.2.8 Correlate the mechanism of enzyme
action with the change in the following
factors:

• temperature
  • pH
  • substrate concentration
  • enzyme concentration
5.2.9 Design and conduct experiments to

study the effects of temperature and
pH on the activities of amylase and
pepsin.

5.3 Application of Enzymes in Daily Life

5.3.1 Explain by using examples the
application of enzymes in daily life.

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Metabolism
Metabolism refers to all chemical reactions that occur in a living
organism. The processes in metabolism involve the conversion of food
into energy in the form of ATP, and the formation of carbohydrate,
protein, lipid and nucleic acid.

Types of metabolism in a cell
There are two types of metabolism, which are catabolism and anabolism.
Catabolism is the process of breaking down complex substances into
simple substances. This reaction releases energy. For example, the
breakdown of glucose during cellular respiration to generate energy.
Generally, catabolic reactions are as follows:

A                             B + C

(substrate) (products)

Anabolism is the process of synthesising complex molecules from
simple molecules. This reaction uses or absorbs energy. For example,
the formation of glucose during photosynthesis.
Generally, anabolic reactions are as follows:

A + B                             C

(substrate) (product)

5.1.15.2.15.1.1 5.1.25.2.25.1.2

Enzyme
In a cell, biochemical reactions occur at a high rate to protect the living
processes. Biochemical reactions can occur quickly in cells due to the
presence of enzymes that help speed up reactions.
An enzyme is an organic catalyst that is mostly made up of proteins
and is produced by living cell organisms. However, not all enzymes
are synthesised from proteins. The substances needed for an enzyme
reaction are called substrates. Substrates will bind with enzymes at a
specific site known as the active site and form an enzyme-substrate
complex (Figure 5.1).

5.2

FIGURE 5.1 Formation of an enzyme-substrate complex

5.1

The Alcanivorax
borkumensis bacteria
produces hydroxylase
enzymes to break down
an oil spill in the sea and
change it to a non-toxic
substance.

Biological Lens products

substrate

active site

enzyme

enzyme

enzyme-substrate
complex

H2O

products

substrate

active site

enzyme

enzyme

enzyme-substrate
complex

H2O

products

substrate

active site

enzyme

enzyme

enzyme-substrate
complex

H2O

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5.2.45.2.3

Enzyme nomenclature
In the 1960s, The International Union of Biochemistry and Molecular
Biology (IUBMB) introduced the enzyme nomenclature based on the
substrate or reaction it catalyses. The name of the enzyme is derived by
adding ‘-ase’ to the name of the substrate it catalyses.
An example of the ‘-ase’ added to substrate is the lactase enzyme, which
catalyses the hydrolysis of lactose.

However, there are also a few enzymes that do not follow this naming
system, especially enzymes that were discovered before the systematic
naming system was introduced. Some examples are: trypsin, pepsin, and
renin.

The reaction of enzymes
are specific. Only
substrate that fit the
form completely with an
enzyme in an active site
can combine.

Enzymes are biological
catalysts that speed
up biochemical
reactions.

General characteristics of enzymes
 Enzymes are only

required in small
quantities and are
reusable.

Enzyme activity can be
slowed down or stopped
by enzyme inhibitors. An
example of an inhibitor is
heavy metal such as lead
or mercury.

The structure of
enzymes remain
unchanged and are
not destroyed after a
reaction.

Most of the reactions
catalysed by enzymes
are reversible.

Some enzymes need
cofactors to work more
efficiently. An example of
a cofactor is vitamin B,
and magnesium ion.

Enzymes act rapidly.

GENERAL
CHARACTERISTICS

OF ENZYMES

Collect information
on the enzyme
nomenclature
conventionally
and based on The
International Union
of Biochemistry and
Molecular Biology
(IUBMB). Present it in
class.

Activity Zone

Lactose + water                       Glucose + galactose
Lactase

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Intracellular and extracellular enzymes
Enzymes that are synthesised in a cell for its own use are called
intracellular enzymes. For example, the hexokinase enzyme that is used
in the glycolysis process during cellular respiration.
On the other hand, enzymes that are secreted outside the cell are known
as extracellular enzymes. For example, the trypsin enzyme is produced
by the pancreatic cells and secreted into the duodenum to break down
polypeptides.
How are extracellular enzymes produced? The production of enzymes
involves a few specific components of cells (Figure 5.2).

5.2.5

FIGURE 5.2 Production of extracellular enzymes

Once it reaches the end of the
rough endoplasmic reticulum,
the membrane buds off to form
transport vesicles.

Proteins synthesised by
ribosome enter the lumen of the
rough endoplasmic reticulum and
are transported through it.

Ribosome is the site of
protein synthesis.

The transport vesicle
that contains proteins
moves towards the Golgi
apparatus and then fuses
with it.

In the Golgi apparatus, the
proteins are modified into
enzymes and secreted in
secretory vesicles that are
formed from the tip of the
Golgi apparatus.

Secretory vesicles will
move towards the plasma
membrane and fuse with it to
secrete extracellular enzymes.

‘Glycolysis’ originates
from the words
‘glucose’ and ‘lysis’
(break down) and
is the breakdown
of glucose by the
hexokinase enzyme
to produce energy
and pyruvate.

Biological Lens

1

4 5 6

2 3

nucleus pore
nuclear membrane

rough endoplasmic
reticulum

secretory vesicle

protein secreted
as enzymes

ribosome

proteinsmooth
endoplasmic reticulum

Golgi apparatus

transport vesicle

plasma membrane

1

2

4

3

6

5

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Mechanism of enzyme action
The ‘lock and key’ hypothesis
Most enzymes are complex proteins made up of polypeptide chains that
are folded into three-dimensional structures. This three-dimensional
structure has an active site with a specific configuration that complements
a specific substrate molecule. The binding of a molecule substrate on
an active site of enzymes is specific like a ‘lock and key’ combination
(Figure 5.3). The enzyme is represented by a ‘lock’ and the substrate is
represented by a ‘key’.

5.2.6    5.2.7
5.2.6 5.2.7

AR

enzyme reaction

products

reactants

activation energy
with enzyme

activation energy
without enzyme

en
er

gy
 in

sy
ste

m

FIGURE 5.4 The presence of enzymes will
reduce the activation energy of a reaction

FIGURE 5.3 The ‘lock and key’ hypothesis

Most reactions inside the cell require high activation energy. Activation
energy is the energy needed to break the bond in the substrate molecule
before reaction can occur. Enzymes function by lowering the activation
energy (Figure 5.4). By doing so, the rate of biochemical reactions in the
cell is accelerated.

Enzyme
substrate complex

enzyme
enzyme

productssubstrate

Enzyme
substrate complex

enzyme
enzyme

productssubstrate

Enzyme
substrate complex

enzyme
enzyme

productssubstrate

The specific substrate
approaches the
enzyme.

A reaction takes place and a product is
created. The product then leaves the active
site after the reaction is complete.

1

2

3

Substrate combines with the
active site to form enzyme-
substrate complex.

enzyme

Indicator:

reaction without enzyme

reaction with enzyme

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Effects of pH

1 2 3 4 5 6 7 8 9 10
pH

optimum pH
for pepsin

optimum pH
for trypsin

ra
te

of
re

ac
tio

n

0

optimum pH for
salivary amylase
and most enzymes
in animal cells

FIGURE 5.6 The effect of pH on the enzyme activities of
pepsin, salivary amylase and trypsin

5.2.8

Effects of temperature
Figure 5.5 shows the effects of temperature on the rate of
biochemical reaction controlled by enzymes.

FIGURE 5.5 The effects of temperature
on the reaction rate of an enzyme

• The enzyme activity is influenced by the
pH of the surrounding solution. Generally,
all enzymes react most effectively at their
optimal pH.

• Most enzymes are most active in the range
of between pH 6 and 8. For example, the
salivary amylase works at pH 6.8.

• However, there are some exceptions.
For example, the pepsin enzymes in the
stomach act at the optimal pH range of
between 1.5 and 2.5. The trypsin enzyme
in the duodenum, on the other hand, only
works well in an alkaline medium, at a pH
of around 8.5 (Figure 5.6).

The mechanism of enzyme action and factor changes
There are many factors that affect the mechanism of enzyme action. The chemical bond
in enzymes can be easily transformed as a result of chemical changes and physical conditions.
Among these factors are temperature, pH, enzyme concentration and substrate concentration.

1

2

3

• At a low temperature, the rate of reaction catalysed by
enzymes is low.

• When the temperature rises, the kinetic energy of the
substrate molecules and enzymes also increases. This
increases the frequency of effective collision between
the substrate molecules and enzyme molecules.

• The rate of reaction between enzymes and substrate
molecules increases.

• With every rise in temperature of 10°C, the rate of
reaction controlled by the enzymes will double until it
reaches the optimal temperature.

At optimal temperature, the enzyme reaction is at its
maximum. The optimal temperature for enzyme
reactions in the human body is around 37°C.

• After reaching the optimal temperature, any further increase in temperature will reduce the
enzyme activity rapidly until it stops at 60°C.

• At this temperature, the enzymes become denatured as the chemical bonds in the enzyme
molecules break at extreme temperatures.

• The enzymes are unable to retain the three-dimensional form. The active site of enzymes
changes. The substrate does not complement the active site of enzymes.

1

2

3

rate of reaction

optimum temperature

37°C

temperature (°C)
10 20 30 40 50 60

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5If the concentration of enzymes is fixed while the
concentration of substrate is increased, the reaction
rate controlled by enzymes will also increase, leading
to an increase in products created (Figure 5.8).

The effect of substrate concentration

5.2.8

FIGURE 5.8 The effect of substrate concentration
on the rate of enzyme reaction

The effect of enzyme concentration

substrate concentration

rate of reaction

maximum level
10

0.5

0

FIGURE 5.9  Effects of enzyme
concentration on the enzyme reaction rate

• The change in pH value changes the charge
(ion H+) of the active site of enzymes and the
substrate surface. As a result, the enzyme-
substrate complex cannot be formed.

• When the pH of the environment returns to
the optimum level, the charge on the active
site will be restored. The enzyme will return
to function as normal.

• The extreme change in the pH value
will break the structural chemistry bond
and change the active site of enzymes.
Denatured enzyme (Figure 5.7).

1

2

Molecule of a normal enzyme

Molecule of a denatured enzyme

1

1

2

2

3

• When substrate concentration increases, the
opportunity for an effective collision between
the substrate and enzyme molecules also
increases.

• The reaction rate continues to increase until
it reaches the maximum level. The rate of
reaction is constant.

When the concentration of enzyme
increases, the rate of enzyme reaction
will increase because of the presence
of more active sites that are ready for
catalytic action.

• At the maximum level, the concentration of
enzymes becomes a limiting factor. The
reaction rate can only increase when the
concentration in enzymes increases.

• After reaching the maximum level, all active sites
of enzymes are saturated with substrate and are
involved in the catalytic reaction.

If the concentration of an enzyme in one
reaction is doubled, the amount of
substrate converted to products per
unit of time is also doubled with the
condition that there is an excess supply
of substrate.

At the maximum rate, the concentration
of the substrate becomes the limiting
factor. The rate of reaction can only be
increased by adding more substrate.

rate of reaction

enzyme
concentration

maximum level

3

2

1

91

FIGURE 5.7  Molecule of a denatured enzyme

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5.2.9

Activities2.1Activity 5.1 Experiment
Studying the effect of temperature on the
amylase enzyme activity

Problem statement
What is the effect of temperature on the reaction rate of an amylase enzyme?

Hypothesis
An increase in temperature increases the reaction rate of an amylase enzyme up to an optimum
temperature. The reaction rate of an enzyme decreases after the optimum temperature.

Variables
Manipulated: Temperature
Responding: Reaction rate of an amylase enzyme
Fixed: The concentration of amylase and starch suspension (substrate) and the pH of the reaction
medium

Materials
1% starch suspension, 0.5% amylase enzyme solution, iodine solution, ice and filtered water

Apparatus
Beaker, test tube, syringe, dropper, glass rod, white grooved tile, thermometer, Bunsen burner, tripod
stand, wire gauze, test tube rack, measuring cylinder and stopwatch

Procedure
 1 Using a syringe, put 5 ml of 1% starch suspension into each test tube labelled A1, B1, C1, D1

and E1.

2  Using another syringe, put 2 ml amylase enzyme solution in each test tube labelled A2, B2, C2,
D2 and E2.

3 Test tubes A1 and A2, B1 and B2, C1 and C2, D1 and D2, E1 and E2, are placed respectively
in 5 separate water baths at fixed temperatures of 20°C, 30°C, 40°C, 50°C and 60°C.

4 Incubate all test tubes for 5 minutes.

5  Meanwhile, prepare a dry, white grooved tile and put a drop of iodine solution into the tile.

6 After incubating for 5 minutes, pour the starch suspension in test tube A1 into test tube A2.
Stir the mixture with a glass rod. Start the stopwatch immediately.

7 Use the dropper to extract a drop of the mixture from test tube A2 and drop it immediately into
the first groove on the tile containing the iodine solution (The first groove is considered as
zero minute).

8 Repeat the iodine test every 30 minutes. Rinse the dropper
with the water from the beaker after each sampling. Record
the time taken to complete the starch hydrolysis, that is, the
time when the mixture remains brownish yellow in colour
when tested with the iodine solution.

9 Keep all test tubes immersed in the respective water baths
throughout the experiment. Repeat steps 5 to 8 for each pair
of test tubes B1/B2, C1/C2, D1/D2 and E1/E2.

5 ml starch
suspension

2 ml
amylase
solution

water bath at
respective
temperatures

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5.2.9

Activities2.1Activity 5.2
Experiment

Problem statement
What is the optimum pH of a pepsin enzyme reaction?

Hypothesis
pH 2 is the most optimum for the pepsin enzyme reaction.

Variables
Manipulated: pH medium reaction
Responding: The clarity or turbidity of the mixture reaction
Fixed: Albumen concentration, pepsin solution concentration and temperature of reaction medium

Materials
Albumen suspension (egg white), 1% pepsin solution, 0.1 M hydrochloric acid, 0.1 M sodium
hydroxide solution, water bath at the temperature of 37°C, pH paper and distilled water

Apparatus
Beaker, dropper, thermometer, test tube, 5 ml syringe (without needle), stopwatch, Bunsen burner,
tripod stand, wire gauze, thermometer and test tube rack

10 Record your results in an appropriate table. Plot your graph to show the reaction rate of the
enzyme (minute-1) against temperature (°C).

Results

Time for starch hydrolysis

to complete (minute)

Rate of reaction

(minute-1)

20

30

40

50

60

Temperature
(°C)

Discussion
 1 Why do you need to incubate the test tubes in the respective water baths for 5 minutes at the

beginning of the experiment?

2 What is the result of amylase action on starch?

3 What is the function of the iodine solution?

4 Based on the drawn graph, explain the effect of temperature on the enzyme activity.

Conclusion
Is the hypothesis accepted? Suggest a suitable conclusion for this experiment.

Studying the effect of pH on the pepsin
enzyme activity

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5.2.9

Procedure
 1  Prepare three test tubes and label as P, Q and R.

2 Put 5 ml albumen suspension in each test tube
using a syringe.

3 Add the following solution into each test tube:

•  P: 1 ml 0.1 M hydrochloric acid + 1 ml 1%
pepsin solution

•  Q: 1 ml distilled water + 1 ml 1% pepsin solution
•  R: 1 ml 0.1 M sodium hydroxide solution + 1 ml 1%

pepsin solution
 4 Dip a piece of pH paper into each test tube and record the pH

of the mixture.

5 Put all the test tubes in a water bath with a fixed temperature
of 37°C for 20 minutes.

6 Observe the state of the mixture at the beginning of the experiment and after 20 minutes.

7 Record the results in a table.

Results

Test tube pH
Clarity or turbidity

0 minute After 20 minutes

P

Q

R

Discussion
 1 Why is it necessary to incubate the test tube in a water bath with a fixed temperature of 37°C?

2 What is the effect of pH on pepsin action on albumen?

3 Discuss the result obtained for test tubes P, Q and R.

Conclusion
Is the hypothesis accepted? Suggest a suitable conclusion for this experiment.

1 What will happen if ribosomes in a pancreas cell do not function?

2 Why must our body temperature be maintained at 37°C?

3  What is the optimum pH for a pepsin enzyme action?

4  Y is the enzyme that catalyses the hydrolysis of sucrose into
glucose and fructose. What is enzyme Y?

5.1Formative Practice

thermometer

test tube

P Q R

water bath
37°C

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5.3.1

Application of Enzymes in
Daily Life
Enzymes have long been widely used in the commercial sector and for
everyday use. The enzymes used are extracted from natural resources
such as bacteria or are produced synthetically.
Immobilized enzymes are enzymes that combine with inert and
insoluble substances to increase the resistance of enzymes towards
change in factors such as pH and temperature. With this method,
the enzyme molecules will remain in the same position throughout
the catalytic reaction and then be separated easily from its product.
This technology is known as immobilized enzyme technology. This
technology is used in various industrial applications (Photograph 5.1).

5.3

5.2Formative Practice

Digestive enzymes
are used in the
medical sector

Pectinase and
cellulase enzymes are
used in juice production

Lactose
enzymes
are used in
lactose-free
milk

Protease enzyme
separates the fish skin

Amylase, lipase,
protease and cellulase
enzymes in bio detergent

Trypsin enzyme extracts
fur from animal hide to make
leather products

1 How are enzymes produced?

2 How does immobilized enzyme technology help to accelerate
the enzyme reaction?

3 Give examples of industries that use enzymes in the
manufacturing of products.

PHOTOGRAPH 5.1 Enzyme
immobilization technology is used
in various industrial application

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METABOLISM AND ENZYMES

• Anabolism
• Catabolism

• Enzyme nomenclature: adding -ase to the
substrate

• General characteristics of enzymes
• Intracellular and extracellular enzymes
• Enzyme action mechanism—‘lock and key’

hypothesis
• Factors that influence the enzyme action
 – Temperature
 – pH
 – Substrate concentration
 – Enzyme concentration

Metabolism Enzyme

• Digestive
enzymes

• Lactose-free
milk and fruit
juice

• Bio detergent
• Leather product

(separate fur
from animal
hide)

Application of
Enzymes in Daily

Life

Have you mastered the following important concepts?
• Types of metabolism
• General characteristics of enzymes
• Mechanism of enzyme action
• Factors that influence the mechanism of enzyme action
• Application of enzymes in daily life

Summary

Self Reflection

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Summative Practice 5
 1 Some chefs sometimes wrap meat in papaya leaves and the meat is marinated for 5 hours

before it is cooked. What is the purpose of wrapping with papaya leaves?

2 Why are apples that have been boiled after they are peeled, do not change colour to
brown?

3 (a) Enzymes are used in industries and everyday life. Explain the use of enzymes to extract
agar-agar from seaweed.

(b) State one function of lipase in the food industry.

4 (a) State two characteristics of enzymes.

(b) Explain why only certain substrate can combine with enzymes.

(c)  (i)    What is the hypothesis that is used to explain the mechanism of enzyme action? In
this hypothesis, what represents the structure of enzymes and the structure of the
substrate?

(ii)   Which characteristics of enzymes can explain this hypothesis?

Essay Questions

5 (a) If you are a food entrepreneur, suggest an enzyme that you can use to process meat
and fish. State the function of this enzyme.

(b)  Discuss how the characteristics of the enzymes can influence its action.

Enrichment

6 The enzymes that exist in the bacteria strain which live in hot spring areas can be extracted
and added to laundry detergent. Suggest why enzymes from these bacteria are suitable to
be used as laundry detergent.

7 Why does cyanide poisoning cause immediate death?

8 Fresh fruits can be processed to produce juice. Fruits are crushed and squeezed before
the juice is extracted. Plant cells contain strong cellulose walls. However, if enzymes
that contain pectinase enzymes are used, more juice can be extracted. Based on this
information, suggest one laboratory experiment that can extract more fruit juice than the
pressing method.

Complete answers are
available by scanning the
QR code provided

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• How does growth happen?
• How does an organism produce new cells?
• How is genetic variation created?

Do You KNOW…

Can identical
organisms be
produced?

6
CHAPTER

Cell Division

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6.1 Cell Division

6.1.1 Describe:
  •  karyokinesis (nuclear

division)
  •  cytokinesis

(cytoplasmic division)
6.1.2 Describe the terms

haploid, diploid,
chromatin, homologous
chromosomes, paternal
chromosome and
maternal chromosome.

6.2 Cell Cycle and Mitosis

6.2.1 Describe the phases in a
cell cycle:

•  interphase
  •  G1 phase

•  S phase
  •  G2 phase
  •  M phase
  •  mitosis
  •  cytokinesis
6.2.2 Arrange the stages of

mitosis in the correct
order.

6.2.3 Communicate about the
cell structure of each
stage of mitosis and
cytokinesis by using
labelled diagrams.

6.2.4 Compare and contrast
mitosis and cytokinesis in
animal and plant cells

6.2.5 Discuss the necessity of
mitosis in:

•  development of
  embryo
 •  growth of organisms
 •  healing of wounds on
  the skin
 •  regeneration
 •  asexual reproduction

6.3 Meiosis

6.3.1 State the meaning of
meiosis.

6.3.2 Identify types of cells that
undergo meiosis.

6.3.3 State the necessity of
meiosis in:

•  the formation
    of gametes
    (gametogenesis).

•  producing genetic
    variation

•  maintaining diploid
    chromosomal numbers
    from one generation to
    another.

6.3.4 Explain the stages of
meiosis in the correct
order:

•  meiosis I
•  meiosis Il

6.3.5 Draw and label the cell
structure in each stage of
meiosis I, meiosis II and
cytokinesis.

6.3.6 Compare and contrast
meiosis and mitosis.

6.4 Issues of Cell Division on
Human Health

6.4.1 Explain the effects of
abnormal mitosis on
human health:

•  tumour
•  cancer

6.4.2 Evaluate the effects
of abnormal meiosis
on Down syndrome
individuals.

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Cell Division
Cells in our body always grow, divide and die. As such, the dead cells
must be replaced with new cells. Cells in the body produce new cells
through the cell division process. Cell division involves two stages, that
is karyokinesis and cytokinesis.
• Karyokinesis involves the division of the nucleus.
• Cytokinesis involves the division of the cytoplasm.
The organism’s body cells are divided into somatic cells and reproductive
cells or gametes.

6.1
What happens when
cells cannot undergo
cell division?

Brainstorm!

• Body cells apart from gametes.
• Somatic cells are produced

through the mitosis process.
• It contains a diploid number

of chromosomes, that is,
each cell contains two sets of
chromosomes or 2n. In human
somatic cells, 2n = 46.

• Gametes are reproductive cells.
• Gametes are produced through

the meiosis process.
• It contains a haploid number

of chromosomes, that is,
each cell contains one set of
chromosomes or n. In human
gametes, n = 23.

SOMATIC CELL GAMETE

FIGURE 6.1 Haploid and diploid chromosomes

6.1.26.1.1

Importance of
cell division

(Accessed on 21
August 2019)

ICT 6.1

In diploid cells, one set of chromosomes originate from the male parent
or paternal chromosomes and another set is from the female parent
or maternal chromosomes. Both paternal and maternal chromosomes
have the same structural characteristics. This pair of chromosomes
are called homologous chromosomes (Figure 6.1). Chromatin is a
chromosome that looks like a long thread.

Haploid (n) A copy of each set of chromosome Diploid (2n) Two copies of each chromosome

Three chromosomes without pairs Three pairs of homologous chromosome
(one set of paternal chromosomes, one set of
maternal chromosomes)

Haploid (n) A copy of each set of chromosome Diploid (2n) Two copies of each chromosome

Three chromosomes without pairs Three pairs of homologous chromosome
(one set of paternal chromosomes, one set of
maternal chromosomes)

Haploid (n): A copy of each chromosome

Three chromosomes without pairs Three pairs of homologous chromosomes
(one set of paternal chromosomes, one set of
maternal chromosomes)

Diploid (2n): Two copies of each chromosome

ORGANISM CELL

1 Give the definition of the following terms:

(a) karyokinesis   (c) chromatin

(b) cytokinesis     (d) homologous chromosomes

2 Predict what will happen if the cells in the
reproductive organs of humans are unable to
produce haploid cells.

6.1Formative Practice

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G1 PHASE

Cell Cycle and Mitosis
What is a cell cycle?
The cell cycle refers to the sequence of events that involves DNA
multiplication and cell division to produce two daughter cells. The cell
cycle consists of interphase and M phase. Interphase is the longest
phase in the cell cycle. This phase is made up of the G1, S and G2  phase.

6.2

INTERPHASE

G1

G2

S

M
it

os
is

C
yt

ok
in

es
is

S PHASE

M PHASE

FIGURE 6.2 Cell cycle

6.2.1

M PHASE

G2 PHASE

1

2

Cells grow. Cell components
such as mitochondrion and
endoplasmic reticulum are
produced at this stage. Proteins
used in the cell cycle are also
synthesised during this time. At
this stage, the nucleus looks big
and the chromosome is in the
form of chromatin.

DNA synthesis occurs in the S phase.
The DNA in the nucleus is replicated.
Each chromosome multiplies into two
identical chromosomes known as sister
chromatids. Both chromatids contain
the same copy of the DNA molecule.
Both chromatids are joined at the
centromeres.

3 The cells will continue to grow and remain active
metabolically during the G2 phase. Cells gather energy
and make final arrangements to enter the next stage
of cell division. After the interphase stage, the cell will
enter the M phase.

4 M phase is made up of mitosis
and cytokinesis. Mitosis
involves prophase, metaphase,
anaphase and telophase.

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Mitosis
Mitosis is defined as the division of the nucleus of parent cell into two
nuclei (Photograph 6.1). Each nucleus contains the same number of
chromosomes and  genetic content with the nucleus of parent cell.

PHOTOGRAPH 6.1 Mitosis at the tip of the plant root

PROPHASE

• In the nucleus, chromatin starts to shorten and
thicken to form a chromosome structure that
can be seen through a light microscope.

• The chromosome is seen to be made up of two
identical threads called sister chromatids.

• Both sister chromatids are joined at the
centromere.

• The nucleus membrane
disintegrates, the nucleolus
disappears, the centriole moves
to the opposite poles and the
spindle fibres start to form.

6.2.36.2.2

Activity: Design three
dimensional models of

the mitotic stages

ICT 6.2 anaphase
telophase

prophase

metaphase

centriole

nucleus membrane
disintegrates

centromere

chromosomes

centromere

spindle fibres

The failure of mitotic
division in somatic
cells will not be
inherited by the next
generation.

Our World of Biology

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METAPHASE

ANAPHASE

TELOPHASE

• Centrioles are at the opposite
poles of the cell.

• The spindle fibres maintain
the chromosomes at the
equatorial plane.

• The chromosomes
become aligned in a
single row on the
equatorial plane.

• Metaphase
ends when the
centromere
begins to divide.

• The centromere divides
into two and the sister
chromatids separate.

• Spindle fibres shorten,
contract and the
sister chromatids are
attracted to the opposite
pole cells.

• Anaphase ends when the
chromatid arrives at the
pole of the cell.

• When the chromatids are at the
opposite poles, they are now called
the daughter chromosome.

• Each pole contains one set
of complete and identical
chromosomes.

• Chromosomes are shaped again
as fine chromatin threads.

• Nucleoli are formed again.
• Spindle fibres disappear.
• A new nucleus

membrane is
formed.

• The telophase
stage is followed
by cytokinesis.

6.2.36.2.2

FIGURE 6.3 Mitosis

sister chromatids

pole of the cells

daughter
chromosomes

spindle
fibres

nuclear
membrane

nuclear
membrane

daughter
cells

centriole

sister
chromatids

equatorial plane
centromere

spindle fibres

chromosomes

chromosomes

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The differences between mitosis and
cytokinesis in animal cells and plant cells
Plant cells do not contain centrioles. However, plant cells can still form
spindle fibres during mitosis.
Cytokinesis is different between animal cells and plant cells. Cytokinesis
is the division of cytoplasm that happens immediately after the nucleus
is formed, that is, at the end of telophase. Cytokinesis occurs in animal
cells when the plasma membrane constricts in the middle of the cell
between the two nuclei (Figure 6.4). Microfilaments at the point of
constriction will contract, causing the cell to constrict until it splits to
form two daughter cells.
Cytokinesis in plant cells also begins when the formed vesicles combine
to form cell plates at the centre of the cell (Figure 6.5). The cell plates are
surrounded by a new plasma membrane and a new cell wall substance
is formed among the spaces of the cell plates. The cell plates expand
outwards until they combine with the plasma membranes. At the end of
cytokinesis, cellulose fibres are produced by the cells to strengthen the
new cell walls. Two daughter cells are formed. Each cell has a diploid
condition.

FIGURE 6.5 Cytokinesis in plant cells

FIGURE 6.4 Cytokinesis in
animal cells

PHOTOGRAPH 6.2 The necessity of mitosis
for living organisms

6.2.56.2.4

cleavage furrow
microfilaments

contractile ring of
microfilaments

daughter cell

daughter cell

vesicles form cell
plates

parent cell wall
new cell wallcell plate

The necessity of mitosis
Mitosis is important for the following life processes.

For embryo development
and organism growth, mitosis
ensures that rapid cell growth
occurs.

When the body is injured, mitosis will
produce new cells to replace cells that are
dead or damaged.

Through the mitosis process, the
lizard is able to grow a new tail
(regeneration) if the tail breaks.

Mitosis aids organisms such as
hydra to produce new individuals
through the formation of new buds.

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PHOTOGRAPH 6.3 The
application of mitosis in
the fields of medicine and
agriculture

6.3.1 6.3.2 6.3.3

Meiosis
Meiosis is the process of cell division that occurs in reproductive organs to
produce gametes that contain half the number of chromosomes (haploid)
of the parent cells (diploid). Meiosis occurs in the testis (male) and ovary
(female) for animals and humans.

The need for meiosis
Meiosis forms gametes through the process of gametogenesis and ensures
that the diploid chromosome number of organisms that carry out sex
reproduction is always maintained from one generation to the next. Meiosis
also produces genetic variation in the same species. Meiosis is divided into
two stages of cell division, that is meiosis I and meiosis II (Figure 6.6).
a. Meiosis I comprises of prophase I, metaphase I, anaphase I

and telophase I.

6.3

PHOTOGRAPH 6.4
Homologous chromosome

In agriculture, the technique of culturing plant
tissues is used to produce young plants through the
culturing of parent cells without going through the
fertilisation process.

b. Meiosis II comprises of prophase II, metaphase II, anaphase II
and telophase II.

1 State the application of
mitosis in the field of
agriculture.

2 Explain the process that
occurs during the S phase.

3 Predict what will happen
if the spindle fibres fail to
develop.

4 Explain the necessity of
mitosis for life processes.

6.2Formative Practice

The culturing technique uses stem cells from
animals which are then cultured in laboratories
to produce meat.

Stem cell therapy uses stem
cells from bone marrows to treat
damaged cartilage.

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• Chromatin shortens, thickens and forms visible chromosomes. The
pairing of homologous chromosomes (synapsis) forms bivalent (or known
as a tetrad, that is four chromatids for each homologous chromosome).

• The crossing over process that is an exchange of genetic material
between non-identical chromatids takes place. Crossing over produces a
combination of genes that are new in chromosomes. The point where
the chromatids cross over is called chiasma. At the end of prophase I, the
nucleus membrane and nucleoli will start to disappear. Both centrioles will
move towards the opposite pole cells. Spindle fibres are formed among the
centrioles.

PROPHASE I

• The homologous
chromosomes are arranged
at the equatorial plane.

• One chromosome
from each pair of the
homologous chromosome
is tied to the spindle fibres
from one pole cell and its
homologous is tied to the
spindle fibres from the
opposite pole cell.

• The sister chromatids are
still tied together because
the centromere has not
separated.

METAPHASE I

• The spindle fibres
contract and cause
each homologous
chromosome to separate
from its homologous
pair and be pulled to the
opposite poles.

• Each chromosome is
still made up of a pair of
sister chromatids tied to
a centromere and move
as one unit.

ANAPHASE I
• The chromosomes arrive at

the opposite pole cells.

• Each polar cell contains
a number of haploid
chromosomes that are
made up of one set of
chromosomes only.

• The spindle fibres will then
disappear.

• Nucleoli will reappear and
the nuclear membrane is
formed.

TELOPHASE I

AR

• Telophase I is succeeded by
the cytokinesis process that
produces two daughter cells.

• Both daughter cells
produced are in the haploid
condition.

• The interphase for meiosis I
is usually short and the DNA
does not replicate.

6.3.56.3.4

FIGURE 6.6 Meiosis

centriole

plasma
membrane
constricts

chiasma
non-identical
chromatid

spindle
fibrebivalent/

tetrad

equatorial
plane

homologous chromosome
separated and pulled to
opposite poles

homologous chromosome
arranged at the equatorial
plane centriole

sister chromatids
are still tied to the
centromere

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• Chromosomes are arranged at
random on the equatorial plane
for each daughter cell.

• Each chromatid is tied to
the spindle fibres at the
centromere.

• Metaphase II ends when the
centromere separates.

METAPHASE II

• Chromosomes arrive at the pole of the cell.
• Spindle fibres disappear. The nuclear membrane and

the nucleoli are reconstructed.
• The number of chromosome for each daughter cell

is half the number of parent chromosomes.

• Telophase II ends with the process of cytokinesis
that produces four daughter cells that are haploid.

• Each haploid cell contains half the number of
parent cell chromosomes. The genetic content is
also different from the diploid parent cell. The haploid
cells develop into gametes.

TELOPHASE II

6.3.4 6.3.5

• The nucleoli and the nuclear
membrane disappear.

• Each chromosome is made
up of sister chromatids that
are joined at the centromere.

• The spindle fibres start to
form in both daughter cells.

PROPHASE II
• The sister chromatid centromere starts

to separate.
• The sister chromatid pair separates

and moves towards the opposite poles
led by the centromere.

• Each chromatid at this stage is known as
a chromosome.

ANAPHASE II

four haploid daughter
cellssister chromatids separate

nuclear membrane

two haploid
daughter cells

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Comparison and contrast between meiosis and mitosis
You have learned about two types of cell divisions, that is the mitosis and
meiosis. What is the main event that differentiates mitosis and meiosis
and between meiosis I and meiosis II? Compare and contrast the two
types of cell division.

Issues of Cell Division on
Human Health
The cell cycle is controlled by a special control system at each G1, S, G2 and
M phase to ensure proper division of the cells. However, uncontrolled
cell division sometimes can lead to the formation of tumours.
Tumour is divided into two types which are benign tumour and
malignant tumour. A benign tumour is not dangerous and can be
removed surgically. A malignant tumour is also called cancer. Cancer
is caused by several factors such as radiation (x-ray, gamma rays
and ultraviolet rays), chemical substances (such as tar in tobacco),
carcinogens (such as formaldehyde and benzene), genetic factors, and
also bacteria and viruses. This will cause the cells to divide continuously
and develop into a tumour. The cancer cells will spread and destroy
normal cells around them. This condition will affect the functions of the
tissues around them. Cancer that is not identified at the early stage can
cause damage to the organs and finally death (Figure 6.7).

6.4

6.3.6 6.4.1

The tumour grows
from a single cell.

A new tumour
develops on other
organs.

Cancer cells compete to get
nutrients from other tissues
around them.

The cancer cells spread
through the lymph vessels and
blood vessels to other parts of
the body.

FIGURE 6.7 The development of breast cancer

Video: Cancer

ICT 6.3

Build a thinking tool
to compare and
contrast:
(a) meiosis I and

meiosis II
(b) meiosis and

mitosis

Activity Zone

1 State the most obvious
difference between meiosis I
and meiosis II.

2 Explain how meiosis I can
reduce the number of
chromosomes in the
daughter cell.

6.3Formative Practice

cancer
cell

lymph
vessel

blood
vessel

grandular
tissues

a tumour

(Accessed on 21 August 2019)

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Any abnormality during the division of meiosis can also
cause genetic diseases such as Down syndrome. This happens
because the spindle fibres fail to function during anaphase I
or anaphase II. As a result, the chromosome fails to separate
(nondisjunction). Gametes will have an abnormal number
of chromosomes (22 or 24 chromosomes). If fertilisation
between a normal gamete (23 chromosomes) and an abnormal
chromosome (24 chromosomes) occurs, the zygote will carry
47 chromosomes which is an abnormal condition (Figure 6.8).

PHOTOGRAPH 6.6
The complete
chromosome set of an
individual with Down
syndrome

PHOTOGRAPH 6.5
A child with Down syndrome
displays certain characteristics
such as stunted body growth and
mental retardation

6.4.2

An individual with Down syndrome has 47 chromosomes, which is an extra
chromosome at the 21st set. This condition is known as trisomy 21. This
syndrome can cause mental retardation, slanted eyes and a slightly protruding
tongue.

Diploid chromosome
number, 2n +1= 47

( Three copies of
chromosome 21 )

The diploid
chromosome number

in humans
2n = 46

The haploid
number

n = 23

The haploid
number

n = 23

MEIOSIS

FERTILISATION

FIGURE 6.8 Formation of trisomy 21

In a normal meiosis division,
the chromosomes are divided
evenly among the gametes.

If the homologous chromosome or
sister chromatids fail to separate, the
distribution of parent chromosomes
during meiosis will be uneven.

The diploid
chromosome number

in humans
2n = 46

The haploid
number

n = 24

The haploid
number

n = 22

MEIOSIS

nondisjunction
during
anaphase

1 Explain why radiotherapy is used to control
or stop the growth of cancer cells.

2 Nondisjunction conditions in humans can
cause genetic diseases such as Down
syndrome. State the number of chromosomes
and the characteristics that are found in an
individual with Down syndrome.

6.4Formative Practice

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Have you mastered the following important concepts?
• Definitions of karyokinesis, cytokinesis, haploid, diploid, chromatin, homologous

chromosomes, paternal and maternal chromosomes
• Cell cycle
• Stages of mitosis
• The differences between mitosis and cytokinesis between animal cells and plant cells
• Stages of meiosis
• Differences and similarities between meiosis and mitosis
• The need for mitosis and meiosis
• The effects of mitosis and meiosis abnormality towards human health

Self Reflection

Karyokinesis
(nuclear division)

Cytokinesis
(cytoplasmic
division)

Cell Division Cell Cycle and Mitosis Meiosis Issues of Cell Division
on Human Health

CELL DIVISION

The effects
of meiosis
abnormality:
Down syndrome

The effects
of mitosis
abnormality:
tumour, cancer

M phase
(mitosis and
cytokinesis)

Meiosis I and
meiosis II

The similarities
and differences
between
meiosis and
mitosis

Interphase
(G1, S and
G2  phase)

Summary

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1 Name the sequences in the mitosis process.

2 What is the function of the centriole in the division of animal cells?

3 State one difference between mitosis metaphase and meiosis metaphase I.

4 (a) Explain the importance of cell division that happens at the tip of a plant root.

(b) A farmer wants to plant a large number of quality breed mango trees in a short time for
commercial purposes. State and explain the techniques that can be used by the farmer.

5 Figure 1 shows a cell at stage M in a cell cycle. Draw
both cells that will be formed if the P chromosome
does not separate.

Essay Questions

6 Figure 2 shows the complete set of
chromosomes of an individual.

(a) State the genetic disorder this individual
has.

(b) Explain how this individual is born with this
genetic disorder.

7 Cancer cells are formed after normal cells are exposed to factor Y.
  (a) Explain the formation of cancer cells.
  (b) State two examples of factor Y that causes the formation of cancer cells.
  (c) State two ways to avoid the development of cancer cells.

Enrichment

8 The development of plant tissue culture has allowed scientists to improve the quality and
quantity of a crop. Scientists in Malaysia have succeeded in patenting a product that
can be sprayed on orchid plants to overcome infections caused by a virus. This branch
of biotechnology is called RNA interference technology. In your opinion, can the spray
technology be used for all organisms as protection against infections?

Summative Practice 6

FIGURE 1

P chromosome

Answer

P chromosome

FIGURE 2

Complete answers are
available by scanning the
QR code provided

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• Why is energy required for the metabolic process?
• What is the main substrate in the production of energy?
• What are the types of respiration?• What are the processes that occur in aerobic respiration  and fermentation?

Do you KNOW...

How is tempe
processed?

Cellular
Respiration7

CHAPTER

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7.1 Production of energy through cellular
respiration

7.1.1 Justify the necessity of energy in
metabolic processes.

7.1.2 Identify the main substrate used in
energy production.

7.1.3 List the types of cellular respiration:
  • aerobic respiration
  • anaerobic respiration
  • fermentation

7.2 Aerobic respiration

7.2.1 Conceptualise energy production
from glucose during aerobic
respiration in cells.

7.2.2 Write a word equation for aerobic
respiration in cells.

7.2.3 Conduct an experiment to study
aerobic respiration.

7.3 Fermentation

7.3.1 State the factors that cause
fermentation to occur in cells.

7.3.2 Explain by using examples of energy
production from glucose during
fermentation in:
• human muscle cells

• Lactobacillus
  • yeast
  • plants such as paddy
7.3.3 Write and explain word equations for:
  • lactic acid fermentation
  • alcohol fermentation
7.3.4 Conduct an experiment to study

fermentation in yeast.
7.3.5 Compare and contrast aerobic

respiration and fermentation.

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Production of energy
through cellular respiration
In Chapter 5, you have learned about two types of metabolic reaction,
which are anabolism and catabolism. Both of these reactions involve
energy.
• The catabolism process releases energy.
• The anabolism process uses energy.

Without energy, the anabolic processes such as protein formation which
is the basic muscle substance will not occur.

The main substrate in energy production
Cellular respiration is carried out to generate the energy needed by all
living cells. Cellular respiration is the oxidation process of organic
molecules through several stages to release energy. The main substrate
for cellular respiration is glucose. Chemical energy found in glucose is
released to produce energy required by cells. In humans and animals,
glucose is obtained through the digestion of carbohydrates from the
food eaten.
In green plants, light energy can be trapped by chlorophyll for the
photosynthesis process to produce glucose.

Types of cellular respiration
There are two types of cellular respiration, which are aerobic and
anaerobic respiration. Aerobic respiration occurs in the presence
of oxygen. Anaerobic respiration occurs in the absence of oxygen.
Fermentation is an alternative pathway of obtaining energy besides
cellular respiration. In fermentation, the breakdown of glucose is
incomplete in conditions of limited oxygen or without oxygen. This
chapter focuses only on aerobic respiration and fermentation.

7.1

7.1.27.1.1 7.1.3

1 Give five examples of the necessity of
energy in a metabolic process.

2 State the main substrate in the production
of energy.

3 State the meaning of cellular respiration and
the types of cellular respiration.

4 Explain how humans, animals and plants
acquire glucose to produce energy.

Formative Practice 7.1

Conduct a group
discussion about
the energy
requirements in the
metabolic process.

Activity Zone

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7.2.1 7.2.2

ICT 7.1

Video: Aerobic respiration
(Accessed on 21 August 2019)

The number of
mitochondrion
in the muscle
cells of an athlete
increases after
intensive training.
Explain how this
contributes to
the achievement
of the athlete as
compared with
those who do not
undergo intensive
training.

Brainstorm!

Aerobic respiration is the breakdown of glucose involving oxygen to
produce chemical energy. Oxygen is used to oxidise glucose to produce
carbon dioxide, water and energy.
The aerobic respiration process begins with the glycolysis process.
Glycolysis means the breakdown of glucose by enzymes. This process
occurs in the cytoplasm. One glucose molecule is broken down into two
pyruvate molecules.
The following process occurs in the mitochondrion. Pyruvate produced
from glycolysis is then oxidised through a series of reactions to produce
carbon dioxide, water and energy. A large amount of this energy is used
to produce  adenosine triphosphate (ATP) molecules.

Aerobic Respiration7.2

The aerobic respiration is simplified as follows.
• ATP molecules are produced when a group of non-organic phosphate is

added to adenosine diphosphate (ADP).

ADP + phosphate  ATP
energy

• ATP molecules have weak phosphate links.

• When the phosphate links on ATP molecules are broken, the energy
released is supplied to cells to help us carry out our daily activities.

ATP  ADP + phosphate
energy

The complete process of glucose oxidation is simplified as follows:

Word equation:

Glucose  +  oxygen    Carbon dioxide + water + energy
(2898 kJ)

Glucose  Pyruvate

(Occurs in the cytoplasm)

Glycolysis

Carbon dioxide + water + energy

(Occurs in the mitochondrion)

Oxidation of Pyruvate

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7.2.3

Activities 2.1 ExperimentTo study aerobic respirationActivity 7.1
Problem statement
Do living organisms carry out aerobic respiration?

Hypothesis
Living organisms use oxygen and release carbon dioxide during aerobic
respiration.

Variable
Manipulated: Presence of living organisms
Responding: Increase in the level of coloured liquid
Fixed: Initial level of coloured liquid

Materials
Water, coloured liquid, soda lime, living organism (cockroach) and petroleum jelly

Apparatus
Boiling tubes, screw clip, wire gauze, 250 ml beaker, capillary tube, ruler, rubber tube and water bath

screw clip

capillary tube capillary tube

boiling tube A boiling tube B

wire gauze

level of
coloured
liquid

level of coloured
liquid

soda lime water bath to maintain
temperature

cockroach

rubber tube

Apparatus set-up to study aerobic respiration process

Procedure
 1 Prepare the apparatus as shown in the figure above.

2 Prepare two boiling tubes labelled A and B.

3 Fill both boiling tubes with 10 g soda lime.

4 Put the wire gauze in the middle of boiling tube A.

5 Put a cockroach on the wire gauze in boiling tube A while the boiling tube B is left empty.

6 Wipe all connections of the apparatus with petroleum jelly.

7 Close the screw clip and mark the height of the initial level of the coloured liquid in the capillary
tube for both boiling tubes.

8 Leave the apparatus for an hour.

9 Measure and record the final height of the coloured liquid in both capillary tubes after an hour with
a ruler.

10 Record your observations in the following table.

Wipe all connectors
with petroleum jelly
to ensure that the
apparatus prepared
is airtight.

Take Note!

The apparatus
set-up is called a
respirometer. It is
used to measure the
rate of respiration
of an organism by
estimating the rate of
oxygen used.

Biological Lens

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7.3.1

Fermentation is the incomplete breakdown of
glucose in conditions of limited oxygen or
without oxygen. Fermentation is different
from aerobic respiration in its metabolic
pathway after the glycolysis stage.
After glycolysis, the pyruvate
produced will undergo either
alcohol fermentation or lactic
acid fermentation.

Fermentation7.3

Results

Boiling tube Initial level (cm) Final level (cm)
Difference in levels

(cm)

A

B

Discussion
 1 What is the purpose of preparing boiling tube B?

2 What is the function of soda lime in the boiling tube?

3 Is there a change in the level of coloured liquid in capillary tube A? Explain your answer.

Conclusion
Is the hypothesis accepted? Suggest a suitable conclusion.

1 State the meaning of aerobic respiration.

2 Suggest another substrate apart from
glucose that can be used by cells for cellular
respiration.

3 State the word equation for aerobic
respiration.

4 Describe the processes involved in aerobic
respiration to produce energy.

Formative Practice 7.2

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• Ethanol is used in the making of
beer and wine.

• The released carbon dioxide makes
bread dough rise.

• Paddy plants that grow in waterlogged areas with less
oxygen are able to carry out alcohol fermentation.

• Ethanol produced in the tissues during the
fermentation process is toxic to most plants but
the cells of paddy plants have a higher
tolerance for ethanol compared to
other species.

• Paddy plants produce plenty
of alcohol dehydrogenase
enzymes that can break down
ethanol molecules into non-toxic
carbon dioxide.

YEAST PLANTS

The incomplete breakdown of glucose to ethanol, carbon dioxide and energy.

Glucose  Ethanol + carbon dioxide + energy (210 kJ)

ALCOHOL FERMENTATION
ICT 7.2

Video: Anaerobic respiration
(Accessed on 21 August 2019)

The incomplete breakdown of glucose in limited or no
oxygen conditions.

FERMENTATION

Activities 2.1 Project
Produce and market food products
produced through fermentation

Activity 7.2

Procedure
 1 Your teacher will divide your class into a few groups.

2 Each group will choose one food product that is produced through the fermentation process and
market that product in school. Examples of products are tapai, yoghurt or bread.

3 Each group needs to prepare a proposal before starting the project. The proposal must contain:

• introduction of the project including the objectives
  • execution cost
  • production and marketing plan
  • expected outcome

4 If necessary, get advice from your teacher or parents to ensure the smooth delivery of the project.

5 Conduct the project as planned.

6 At the end of the project, each group must prepare a complete report.

7.3.2 7.3.3

Alcohol fermentation.

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• This process is carried out by the muscle cells during vigorous training.

• During vigorous training, the rate of oxygen used exceeds the oxygen supplied by the blood
circulatory system.

• The muscle is in an oxygen-deficiency state and is said to undergo oxygen debt.

• During this process, glucose cannot break down completely. For each glucose molecule that is
broken down, only two ATP molecules or 150 kJ energy will be produced.

• The produced lactic acid accumulates until it reaches a level of concentration that can cause fatigue
and muscle cramps.

The breakdown of glucose into lactic acid and energy.

Glucose  Lactic acid + energy

LACTIC ACID FERMENTATION

Some bacteria
can only survive
in anaerobic
conditions. Predict
what can happen to
this type of bacteria
when oxygen is
supplied.

Brainstorm!

HUMAN MUSCLE CELLS

oxygen debt repaid

oxygen intake
during exercise

exercise

lack of oxygen (oxygen debt)

recovery

ox
yg

en
 in

tak
e

at the beginning
of exercise

at the end
of exercise

Time (minutes)

at the end
of recovery

FIGURE 7.1  Lack of oxygen in muscles and oxygen debt repaid

7.3.2 7.3.3

• The bacteria Lactobacillus carries out milk fermentation to
produce yoghurt.

• Lactobacillus acts on the lactose (milk sugar) and turns it into
lactic acid.

• The lactic acid will then coagulate casein (milk protein) to
produce yoghurt.

• Lactic acid is the source of a sour taste in yoghurt.

LACTOBACILLUS

• Once the vigorous activity stops,
the intake of excess oxygen
will oxidise the lactic acid into
carbon dioxide, water and energy.
When all the lactic acid has been
expelled, the oxygen debt is

repaid.

• Figure 7.1 shows a lack
of oxygen in muscles
and oxygen debt is
repaid.

Lactobacillus bacteria

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7.3.4

Activities 2.1 ExperimentTo study the process of yeast fermentationActivity 7.3
Problem statement
What are the products of yeast fermentation?

Hypothesis
Yeast fermentation produces energy, carbon dioxide and ethanol.

Variables
Manipulated: Presence of yeast
Responding: Changes in temperature, lime water and ethanol smell
Fixed: The volume of boiled glucose solution and the anaerobic condition

Materials
5% yeast suspension, 5% boiled glucose solution, lime water and paraffin oil

Apparatus
Boiling tube, test tube, thermometer, measuring cylinder, delivery
tube and cork

Procedure
 1 Fill 2 boiling tubes with 15 ml of 5% glucose solution that has

been boiled and left to cool.

2 Label the boiling tubes as A and B.

3 Put 5 ml 5% yeast suspension into boiling tube A.

4 Add paraffin oil into both of the boiling tubes.

5 Close both boiling tubes with the cork that has a hole and
a delivery tube. Prepare 2 test tubes with 2 ml of lime water
respectively. Dip the end of each delivery tube into each test
tube that contains lime water.

6 Leave the apparatus for 1 hour.

7 Measure and record the initial and final temperature using a
thermometer.

8 Record your observations in the table below.

Results

Boiling
tube

Temperature (°C)
Change in lime water

Smell of
solutionBeginning of

experiment
End of

experiment

A

B

Discussion
 1 How is the anaerobic condition maintained to ensure that the fermentation process is complete?

2 What is the function of preparing boiling tube B?

Ensure that the end
of the delivery tube is
soaked in lime water.

Take Note!

thermometer

boiling
tube A

boiling
tube B

delivery
tube

-1
0

0
10

20
30

40
50

60
70

80
90

10
0

11
0

-1
01010

5% boiled glucose
solution + yeast
suspension

paraffin
oil

lime
water

thermometer

delivery
tube

-1
0

0
10

20
30

40
50

60
70

80
90

10
0

11
0

-1
010100

5% boiled
glucose
solution lime

water

paraffin
oil

thermometer

boiling
tube A

boiling
tube B

delivery
tube

-1
0

0
10

20
30

40
50

60
70

80
90

10
0

11
0

-1
01010

5% boiled glucose
solution + yeast
suspension

paraffin
oil

lime
water

thermometer

delivery
tube

-1
0

0
10

20
30

40
50

60
70

80
90

10
0

11
0

-1
010100

5% boiled
glucose
solution lime

water

paraffin
oil

Apparatus set-up to study the
yeast fermentation process

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Comparison between aerobic respiration and
fermentation
There are a few similarities and differences between fermentation and
aerobic respiration (Figure 7.2 and Table 7.1).

FIGURE 7.2 Similarities between aerobic respiration and fermentation

7.3.5

Why is aerobic
respiration more
suitable for yeast
compared with
fermentation?

Brainstorm!

3 What is the purpose of boiling the glucose solution earlier?

4 How do the results show that fermentation has taken place in boiling tube A?

Conclusion
Is the hypothesis accepted? Suggest a suitable conclusion.

The breakdown
process of glucose
and its conversion
to chemical energy

Produces chemical energy
in the form of ATP

The process begins with glycolysis when
glucose is converted to pyruvate

Occurs in yeast,
bacteria, animals
and plants

The process
begins in the
cytoplasm

DIFFERENCES

Aerobic Respiration Fermentation

The breakdown process of glucose is completed
in the presence of oxygen.

The breakdown process of glucose is incomplete
without oxygen or in limited oxygen conditions.

Occurs in cytoplasm and mitochondrion. Occurs in cytoplasm.

Produces water. Does not produce water.

Glucose is oxidised completely into carbon
dioxide and water.

Glucose is not oxidised completely into ethanol
and carbon dioxide or lactic acid.

One molecule of glucose generates 2898 kJ of
energy

One molecule of glucose generates 210 kJ
(alcoholic fermentation) or 150 kJ (lactic acid
fermentation) of energy

TABLE 7.1  Differences between aerobic respiration and fermentation

SIMILARITIES BETWEEN AEROBIC RESPIRATION AND
FERMENTATION

1 State where the process of fermentation
usually occurs.

2 Give three examples of microorganisms and
food produced by the fermentation process.

3 While helping your father to cut the grass at
the farm, you come across a snake. Terrified,
you run away from the snake. Explain the

cellular respiration that takes place in the
muscle cells of your legs.

4 State the differences between aerobic
respiration and fermentation.

Formative Practice 7.3

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The main substrate
in energy production
is glucose

Fermentation

Production of energy
through cellular respiration

CELLULAR RESPIRATION

Have you mastered the following important concepts?
• The necessity of energy in metabolic processes
• The main substrate in the production of energy
• Types of cellular respiration
• Energy production from glucose during aerobic respiration in cells
• Word equation for aerobic respiration in cells
• Factors that cause fermentation to occur in cells
• Example of energy production from glucose during fermentation
• Lactic acid fermentation and alcohol fermentation
• Yeast fermentation process
• Differences between aerobic respiration and fermentation

Summary

Self Reflection

The incomplete breakdown of
glucose in limited oxygen or no
oxygen

• Alcohol fermentation
• Lactic acid fermentation

The breakdown of glucose in the
presence of oxygen to produce
chemical energy

Occurs in cytoplasm
Glucose  Pyruvate

Occurs in mitochondrion (pyruvate oxidation)
  Carbon dioxide + water + energy

Aerobic Respiration Anaerobic Respiration

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1 What are the uses of alcohol fermentation products?

2 Why do muscles carry out cellular respiration that produces lactic acid during vigorous
training?

3 Why does cellular respiration in muscles that produce lactic acid supply less energy
compared to aerobic respiration?

4 Explain why an individual usually feels tired faster compared with an athlete, when both of
them are running together.

5 A 100-metre sprinter usually holds his breath while running compared with a long-distance
runner. After running, the sprinter needs seven litres of oxygen to remove the lactic acid in
his muscle cells. Explain this difference between the sprinter and the long-distance runner.

6 Photograph 1 shows the activities by two individuals, P and Q.

Q

P

(a) (i) Based on Photograph 1, identify the respiration that occurs in the muscles of
individuals P and Q.

(ii) State the products of respiration in P and Q.

(b) During the 100-metre sprint on Sports Day, a pupil experienced muscle cramps and
had to stop running. Explain why muscle cramps happen.

(c) Paddy plants grown in waterlogged areas have tolerance to ethanol compared
with other plants.

(i) State the type of fermentation that occurs in paddy plant cells.
   (ii) Write the word equation for the fermentation process that occurs in the paddy

plant cells.
   (iii) Suggest another cell that can carry out the fermentation process as in

question c(ii).

Summative Practice 7

PHOTOGRAPH 1

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Essay Questions

7 (a) Explain why energy is required in metabolic processes.

(b) Compare aerobic respiration with fermentation.

(c) Microorganisms such as yeast and bacteria usually play an important role in the
fermentation process to produce food. Explain why yoghurt can spoil if it is not kept in
the refrigerator.

Enrichment

8 A person who is not used to exercising will experience muscle cramps when doing
vigorous exercise because of the accumulation of lactic acid in the cells. However, for
high-performance athletes, such problems do not occur because their bodies have a high
tolerance for lactic acid. In your opinion, how do high performance athletes overcome the
problem of lactic acid accumulation? Give your reasoning.

9 Studies have shown that intake of sodium bicarbonate or baking powder (baking soda) can
increase muscle efficiency during intense activities that involve muscle fermentation. Give
your justification.

10 While conducting an experiment using yeast, Mei Ling found that if grape juice is kept with
yeast in a covered container, the yeast will slowly break down the glucose in the grapes.
However if the container does not contain any oxygen, the yeast will break down the
glucose at a faster rate, and the alcohol content in the container will rise very fast. At the
end of the experiment, Mei Ling found that the breakdown rate of glucose becomes slow
again even though there are some grapes that have not been oxidised. Explain Mei Ling’s
observation.

11 Susan tried to make bread using dry yeast bought from a shop. When she mixed the yeast
with plain flour, she found that her bread did not rise after half an hour. Explain how you can
help Susan solve her problem.

Complete answers are
available by scanning the
QR code provided

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