ANSWERS
(d) Rate of respiration of yeast (cm/min)
FORM 4
CHAPTER 1: INTRODUCTION TO BIOLOGY
AND LABORATORY RULES
0.9
0.8
Practice 1.1
0.7
1 (a) The study of microorganisms
(b) The study of animals
(c) The study of the environment and its relationship with
organisms
(d) The study of heredity
2 (a) Microbiologist
(b) Zoologist
(c) Ecologist
(d) Geneticist
3 • Biotechnology is a technology that involves the use of living
organisms to develop or make chemicals and products
• Mainly used in agriculture, food science and medicine
• Modern usage includes genetic engineering, cell and tissue
culture technologies
0.6
0.5
0.4
0.3
0.2
0.1
0
1 • Lab coat – To protect skin and clothes from splashes of
chemicals
• Gloves – To protect the hands from hazardous chemicals
and infectious materials
• Goggles – To protect the eyes from hazardous chemicals
2 • Solid wastes (chemical substances, glass, rubber)
• Organic solvents (benzene, acetone)
• Heavy metals (lead, mercury)
• Volatile substances (alcohol, mercury)
(Any three)
3 1. Make the accident area a restricted zone.
2. Ensure the spill does not become any worse by setting the
container upright.
3. Keep the spill from spreading to other areas by using sand.
4. Collect the sand which contain the chemical and dispose
of it safely.
4 Flush with lukewarm running water; turn head side to side
and have water run across both eyes for 15 minutes or use
eye wash station if available.
Aim
Problem statement Hypothesis
Variables (manipulated variable, responding variable and
constant variable)
• Materials and apparatus
• Procedure
• Results
• Discussion
• Conclusion
2 (a) What is the effect of light intensity on the rate of
photosynthesis?
(b) The higher the light intensity, the higher the rate of
photosynthesis.
3 (a) Planning the experiment
(b) Conducting the experiment
(c) Analysing and interpreting data
(d) Writing a report
(c)
SPM FOCUS PRACTICE 1
PAPER 1
1 B
6 C
yeast (cm/min)
4
= 0.4
10
6
= 0.6
10
8
= 0.8
10
8
Concentration
of glucose
solution (%)
Distance
moved by
the coloured
liquid (cm)
Rate of
respiration of
yeast
(cm/min)
10
4
0.4
15
6
0.6
20
8
0.8
20
1 •
•
•
•
1 (a) • Manipulated variable: Concentration of glucose solution
• Responding variable: Distance moved by the coloured
liquid
(b) Distance moved by the
Rate of respiration of
6
15
Practice 1.4
Practice 1.3
4
10
Concentration of glucose (%)
2 • Large and accurate using a sharp pencil
• No shading
• Clear single lines of the drawing, not broken lines
3 (a) Ventral
(b) Dorsal
(c) Superior
(d) Inferior
Practice 1.2
coloured liquid (cm)
5
2 D
7 C
3 C
8 B
4 D
9 D
5 C
PAPER 2
Structured Questions
1 (a) • Constant variable: Surrounding temperature, quantity
of hydrogen carbonate indicator
• Manipulated variable: The content in the specimen tube
• Responding variable: Colour change of hydrogen
carbonate indicator (presence of carbon dioxide)
(b) As a control experiment
(c) Carbon dioxide is emitted during respiration
(d) Yes, because the experiment was conducted systematically
and all the steps required in a scientific investigation
were done.
2 (a) • Manipulated variable: Food sample
• Responding variable: Increase in water temperature,
energy value of food
1
30 MASTERCLASS 2023 SPM BIOLOGY-ANSWERS-(BAHAN QR)-ARUL 1P.indd 1
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(b) Energy value (Jg–1)
(b) Mitochondria
(c) Plasma membrane
(d) Nucleus
(e) Vacuole
(f) Rough endoplasmic reticulum
(g) Cell wall
(h) Golgi apparatus
2 1. Human cheek cells do not have cell walls while the
epidermal cells of an onion have cell walls.
2. Human cheek cells do not have a fixed shape while the
epidermal cells of an onion have a fixed shape.
3. Human cheek cells do not have vacuoles while the
epidermal cells of an onion each has a large vacuole.
3000
2500
2000
1500
1000
500
0
Cashew
nut
Dried
bread
Dried
fish
Food sample
(c) • Cashew nut has the highest energy value.
• Dried bread has the lowest energy value.
Essay Questions
3 (a) (i) Observe the situation. Ask questions regarding
the observation. State an inference to explain the
situation based on variables. Identify the problem by
changing the inference into a question.
(ii) Variables are any factor or condition that can be
controlled, changed and measured. The variable that
is kept the same throughout the experiment is called
the constant variable. The variable that is being
changed in the experiment is called the manipulated
variable. The variable that is being measured and
recorded is called the responding variable.
(iii) A conclusion is made based on the hypothesis.
The conclusion must state whether the hypothesis
is accepted or rejected. If the results support the
hypothesis, then the hypothesis is accepted, if not,
then the hypothesis is rejected.
(iv) The experimental report written must include the
following:
• Aim of investigation
• Problem statement
• Hypothesis
• Variables
• List of apparatus and materials
• Experimental procedure
• Presentation of data/Results
• Analysis and interpreting data
• Conclusion
(b) Disposal of broken glass:
• Wear gloves to pick up the broken pieces of glass
• Wrap sharp waste with papers.
• Place in red, hard plastic sharps container to dispose
Disposal of mercury spillage:
• Remove all jewellery from hands and wrists.
• Keep the spill from spreading by sprinkling sulphur
powder to cover mercury spillage.
• Use a syringe (without a needle) to draw up the
mercury beads.
• Use sticky tape to collect smaller hard-to-see beads.
• Place the mercury and sticky tape in a plastic airtight
container.
• Contact the fire department for disposal.
Practice 2.2
A. Understanding key ideas
1 ✓
2 ✓
3 ✓
6 ✓
7 ✗
8 ✓
Practice 2.3
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
4 ✓
5 ✗
6 ✓
B. Main concepts and facts
1 ribosomes, growth, enzymes
2 structures, oxygen, nucleus, oxygen, diffusion
Practice 2.4
A. Understanding key ideas
1 ✗
2 ✓
3 ✗
4 ✓
5 ✗
B. Main concepts and facts
1 cell specialisation, function, cell organisation, organs,
organ systems
2 epithelial, epidermis
SPM FOCUS PRACTICE 2
PAPER 1
1
6
11
16
21
26
D
A
B
B
D
C
2
7
12
17
22
A
B
B
B
D
3
8
13
18
23
D
B
D
C
B
4
9
14
19
24
B
D
D
A
A
5
10
15
20
25
A
A
B
B
B
PAPER 2
Structured Questions
1 (a) R: Nucleus
S: Rough endoplasmic reticulum
T: Plasma membrane
U: Mitochondrion
(b) (i) Respiration
(ii) Structure U oxidise food (glucose) to release energy
(c) An animal cell, because it does not have a cell wall
2 (a) (i) Chromosomes and nucleolus
(ii) • Controls all the activities of the cell
• Carries genetic information or genes
Practice 2.1
4 ✗
9 ✓
5 ✗
10 ✓
B. Main concepts and facts
1 It enables the protozoa to expel excess water and this
enable the protozoa to survive in its habitat.
2 (a) Respiration
(b) Reproduction
(c) Movement
(d) Osmoregulation (excretion)
(e) Feeding (nutrition)
CHAPTER 2: CELL BIOLOGY AND
ORGANISATION
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
6 ✓
7 ✓
8 ✓
4 ✗
9 ✓
5 ✓
10 ✗
B. Main concepts and facts
1 (a) Chloroplast
2
30 MASTERCLASS 2023 SPM BIOLOGY-ANSWERS-(BAHAN QR)-ARUL 1P.indd 2
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(b) (i) Transportation of proteins, synthesised by ribosomes
to Golgi apparatus
(ii) The cell will not be able to transport the substances
synthesised in the cell
(c) Meristem cell requires a lot of energy to carry out active
cell division to produce new cells.
3 (a) (i) Both organelles have two layers of membrane.
(ii) R: Chloroplast
S: Mitochondrion
(iii) R: Palisade mesophyll cell
S: Sperm cell
(b) • Organelle R carries out photosynthesis. The cell is
able to manufacture food which can be used by the
plant or stored in the plant.
• Organelle S carries out cell respiration to produce the
energy necessary for cell activities.
(c) Has a cell wall/Has a large vacuole
4 (a) (i) Cardiac muscle cell/Muscle cell
(ii) W: Cardiac Muscle cell ➝ X: Cardiac Muscle tissue ➝
Y: Heart ➝ Z: Circulatory system
(b) Cell specialisation
(c) (i) To pump blood to all parts of the body
(ii) Nervous tissue, connective tissue
(d) (i) To transport dissolved respiratory gases, nutrients and
waste products in the blood to and from the body cells
(ii) Structure Z consists of a few organs that work
together to carry out a specific function while
structure Y consists of different types of tissues that
carry out a certain function.
5 (a) Mitochondrion
(b) Muscle cells contract and relax to produce movement.
The energy to contract is supplied by mitochondria
which carry out aerobic respiration to release energy.
(c) There is no aerobic respiration to release energy. Cardiac
muscles of the heart cannot contract to cause pumping
action. Thus, the heart cannot function or will stop beating.
cytoplasm and enclosed in a food vacuole. Enzymes
are secreted into the food vacuole to digest the food
particle. Nutrient from the digested food is then
absorbed into the cytoplasm.
• Movement
Amoeba sp. moves with the help of pseudopodia.
Any part of the cytoplasm can protrude out to form
pseudopodia.
• Respiration
Exchange of gases by diffusion takes place at the
membrane of Amoeba sp.. Oxygen dissolved in water
diffuses through the membrane into the cell. Carbon
dioxide diffuses out from the cell.
• Reproduction
Amoeba sp. as a unicellular organism can reproduce
asexually through a process called binary fission. The
nucleus of Amoeba sp. elongates and divides followed
by the division of the cytoplasm. Two new cells are
formed each time an Amoeba sp. reproduces. The
reproduction of Amoeba sp. by binary fission results
in an increase in the Amoeba’s population.
• In excretion, Amoeba sp. carries out osmoregulation with
the help of its contractile vacuoles to expel excess water
from the cell. This enables it to survive in its habitat.
• It also can grow and responds to changes in the its
environment.
(c) Cells which have many Golgi apparatus and rough
endoplasmic reticulum are mainly secretory cells that
produce enzymes and hormones. An example of secretory
cells is the cells in the salivary glands. These cells secrete
the enzyme amylase. Many rough endoplasmic reticulum
and Golgi apparatus are required in the secretion of amylase.
The ribosomes on the endoplasmic reticulum synthesise
protein which is then transported by the transport vesicles
of the rough endoplasmic reticulum to the Golgi apparatus.
This organelle restructures the protein into a specific protein
like amylase and then packs them into secretory vesicles,
which carry the enzymes to the plasma membrane. The
secretory vesicles then constrict to secrete the enzymes out
of the cell through the plasma membrane.
Essay Questions
6 (a) (i) Root hair cells are epidermal cells of the root. The cell
is long and narrow and extends among the soil particles.
The small size of the root hairs increases the roots’
surface area. This increases the rate of absorption of
water and mineral salts. Hence, root hairs can easily
absorb water and mineral salts from the soil.
(ii) Red blood cells are flat, biconcave discs without
any nucleus. The biconcave disc shape increases
the surface area of the cell for absorption of oxygen.
Oxygen can diffuse at a faster rate into the red
blood cells and combine with haemoglobin to form
oxyhaemoglobin. Red blood cells also contain more
haemoglobin to combine with oxygen because they
do not have any nucleus. The membrane of red
blood cells is elastic. This enables them to change
their shape so that they can squeeze through the tiny
lumen in the blood capillaries.
(b) ‘Cell as a complete unit of life’ refers to a unicellular
organism that can carry out all the life processes as
one cell. Each organism, for example Amoeba sp., is
made up of one cell and yet it can carry out all its life
processes like feeding (nutrition), movement, respiration,
reproduction excretion, growth and response.
• Feeding
Amoeba sp. feeds on decayed organic matter and
bacteria. Amoeba sp. extends its cytoplasm to form
pseudopodia in order to approach its food particle
and surround it. The food particle is taken into the
CHAPTER 3: M
OVEMENT OF SUBSTANCES
ACROSS THE PLASMA
MEMBRANE
Practice 3.1
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
6 ✓
7 ✓
8 ✗
11 ✓
12 ✗
4 ✗
9 ✗
5 ✓
10 ✓
B. Main concepts and facts
1 nutrients, carbon dioxide and waste products
2 carrier, channel, glycoprotein
3
Glycolipid
Channel
protein
Carrier
protein
Glycoprotein
Phospholipid
bilayer
Cholesterol
4 The phospholipid bilayer is made up of two layers of
phospholipid molecules. Each phospholipid molecule
consists of a polar hydrophilic head which attracts water
and a nonpolar hydrophobic tail which repels water
3
30 MASTERCLASS 2023 SPM BIOLOGY-ANSWERS-(BAHAN QR)-ARUL 1P.indd 3
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Practice 3.2
2
A. Understanding key ideas
1 ✓
2 ✓
3 ✓
6 ✓
7 ✗
8 ✓
Cell wall
4 ✗
9 ✗
5 ✗
10 ✗
Plasma
membrane
B. Main concepts and facts
Oxygen
1
Glucose
molecules molecules
3 • Crenation refers to the condition of the red blood cells
which lose its water and shrivels when placed in a
hypertonic solution.
• Plasmolysis refers to the plant cell which lose its water
and becomes flaccid in a hypertonic solution.
4 The concentration of the cell sap is equivalent to the
concentration of the isotonic solution around it.
Plasma
membrane
Energy
Simple Facilitated
diffusion diffusion
across
across
phospholipid carrier
protein
bilayer
Nucleus
Vacuole
Practice 3.4
Active
transport
A. Understanding key ideas
1 ✗
2 ✓
3 ✗
4 ✓
5 ✓
Passive transport
B. Main concepts and facts
1 The excess fertiliser dissolves in the water supplied and
cause the soil water to be hypertonic to the cell sap of root
hair cells. Water molecules from the plant diffuse out of
the root hair cells into the soil water by osmosis. The plant
cell loses its water, wilt and die.
2 • During physical activity, excessive sweating can cause
dehydration due to loss of water and salts such as
sodium and potassium.
• Athletes drink sports drinks which contain water, salts
and glucose to replace the lost water and salts.
• It also maintain the content of the body fluid and
prevent dehydration.
2 Active transport requires energy to pump the substance
across the membrane against its concentration gradient.
The energy is supplied by ATP during respiration. All
living cells carry out respiration and are able to provide
the energy necessary to do this work.
3
Passive transport
Active transport
Does not expend energy
Expends energy
Occurs down the
concentration gradient
Occurs against the
concentration gradient
Occurs through the
phospholipid bilayer or with
the help of protein molecules
to cross the plasma membrane
Needs specific carrier
proteins to transport
substances across the
plasma membrane
4
Similarities
SPM FOCUS PRACTICE 3
PAPER 1
• Both are passive transport
• Both occur down the concentration gradient and do
not expend energy
1
6
11
16
Differences
Simple diffusion
Facilitated diffusion
Does not require the
help of transport
proteins
Requires the help of
transport proteins (carrier
proteins and channel protein)
Occurs across the
phospholipid bilayer
of the membrane
Occurs through the protein
molecules (transport proteins)
Allows small uncharged
molecules (oxygen,
carbon dioxide) and
lipid-soluble substances
to cross the membrane
Allows slightly larger polar
molecules (glucose, amino
acids) and small charged
mineral ions to cross the
membrane
B. Main concepts and facts
1
Solution
Plant cell
4 ✓
Animal cell
Becomes turgid
Haemolysis
(b) Hypertonic
Plasmolysis
Crenation
(c) Isotonic
Maintains its
structure
Maintains its
structure
C
B
C
A
3 A
8 D
13 D
4 B
9 A
14 B
5 A
10 C
15 C
Structured Questions
1 (a) X: Phospholipid bilayer
Y: Carrier protein
(b) (i) X consists of two layers of phospholipid molecules.
Each phospholipid is made up of the polar head
which is hydrophilic and the non-polar tail which is
hydrophobic. The hydrophobic part only allows nonpolar molecules such as oxygen and carbon dioxide
to move across it. The hydrophilic part allows water
molecules to go through it.
(ii) Y helps in the transport of slightly bigger
polar molecules which cannot move across the
phospholipid bilayer of the membrane.
(c) (i) Simple diffusion/Facilitated diffusion
(ii) Dissolved substances or ions required by the cell are
transported across the membrane by carrier proteins
against the concentration gradient and requires
energy.
(d) • Passive transport does not require energy while active
transport needs energy.
• Passive transport occurs down the concentration
gradient while active transport occurs against the
concentration gradient.
5 ✗
(a) Hypotonic
2
7
12
17
PAPER 2
Practice 3.3
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
6 ✗
7 ✓
8 ✓
B
B
A
B
4
30 MASTERCLASS 2023 SPM BIOLOGY-ANSWERS-(BAHAN QR)-ARUL 1P.indd 4
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2 (a)
Percentage of red blood cells bursting (%)
The plant cell maintains its shape and size.
5% sucrose solution is isotonic to the concentration
of the cell sap in the plant cell. Hence, the rate of
movement of water molecules in and out of the plant
cell is the same. The size of the plant cell is maintained.
(ii) In 30% sucrose solution:
100
80
60
40
20
Vacuole shrinks
0
0.30 0.35 0.40 0.45
0.50
–20
Concentration of salt solution (g/100 cm3)
Plasma membrane
pulls away from
the cell wall
(b) 0.39/100 cm3
(c) 0.48/100 cm3 salt solution is isotonic to the concentration
of the red blood cells.
(d) 0.50 g/100 cm3 of salt solution is hypertonic to the red
blood cells. Water molecules diffuse out from the cells
trough osmosis and cause the red blood cells to shrivel.
(e) An onion cell has a cell wall to overcome the hydrostatic
pressure in the cell.
(f) Excess fertiliser will cause the solution around the
roots to become hypertonic to the root hair cells. Water
molecules will diffuse out of the root hair cells by
osmosis. The root hair cells will undergo plasmolysis.
The plant will lose water and wilt. When too much water
is lost, the plant will die from dehydration or dryness.
Water molecules diffuse into cell
Water molecules
Plasma membrane
is firmly pushed
out against the
cell wall
Essay Questions
3 (a) The plasma membrane consists of a phospholipid
bilayer and various types of protein molecules. The
polar head of the lipid layer is hydrophilic while the
nonpolar tail is hydrophobic. This property enables
only certain substances like oxygen, carbon dioxide and
water molecules to go across it. Dissolved substances
like glucose and amino acids as well as charged ions
cannot move across the phospholipid bilayer. These
substances can only move across the membrane with the
help of carrier proteins or channel proteins. Molecules that
are big and complex like sucrose and starch cannot move
across the membrane at all. Hence only certain substances
can move across the membrane, making the membrane
selectively permeable.
(b) An example of simple diffusion in our body is the gas
exchange in the alveolus. The concentration of oxygen
in the alveolus is higher than the concentration of
oxygen in the blood capillary on the surface of the
alveolus. Hence, oxygen molecules diffuse out of the
alveolus into the blood capillary. In the blood capillary,
the concentration of carbon dioxide is higher than in
the alveolus. As a result, carbon dioxide molecules
diffuse out of the blood capillary into the alveolus to be
removed from the body. An example of facilitated
diffusion in our body is the absorption of fructose in villus
of the ileum. The concentration of fructose is higher in the
ileum than in the blood capillaries at the villus. Hence,
fructose molecules diffuse into the blood capillaries of the
villus through the carrier proteins without expending any
energy. This type of diffusion, which needs the help of
carrier proteins, is called facilitated diffusion.
(c) (i) In 5% sucrose solution:
Water
molecules
Cell wall
30% sucrose solution is hypertonic to the
concentration of the plant’s cell sap. As a result,
water molecules diffuse out of the plant cell by
osmosis. The vacuole becomes smaller, the plasma
membrane is pulled away from the cell wall. The
plant cell is plasmolysed and becomes flaccid.
(iii) In 0.1% sucrose solution:
Vacuole
fills with
water
and
expands
Cell becomes
turgid due to
turgor pressure
Water molecules
Water molecules
0.1% sucrose solution is hypotonic to the
concentration of cell sap in the plant cell. As a result,
water molecules diffuse into the cell by osmosis. The
vacuole collects the water and expands, causing a
pressure to be exerted on the cell wall. This pressure
causes the plant cell to expand and become turgid.
CHAPTER 4: CHEMICAL
COMPOSITION IN
A CELL
Practice 4.1
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
4 ✓
5 ✓
B. Main concepts and facts
1 • As a transport medium in and between cells
• As a medium for biochemical reactions
• As a solvent for vitamin B, C and minerals, oxygen and
carbon dioxide
• To provide support in herbaceous plants through
turgidity of cells
2 polar, specific heat capacity, capillary action, cohesive,
adhesive
Practice 4.2
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
4 ✗
5 ✓
B. Main concepts and facts
1 Glucose, fructose, galactose
2 Glucose + Fructose → Sucrose + Water
The process is condensation.
3 Hydrolysis. Polysaccharide is broken down to five
molecules of disaccharide (maltose) and finally broken
down to ten glucose molecules.
Water
molecules
Practice 4.3
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
Vacuole
Water molecules move at the
same rate in both directions
4 ✓
5 ✗
5
30 MASTERCLASS 2023 SPM BIOLOGY-ANSWERS-(BAHAN QR)-ARUL 1P.indd 5
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B. Main concepts and facts
1 amino acid, peptide bond, dipeptide polypeptide
2 Amino acid + Amino acid → Dipeptide + Water
The process is condensation.
3 To build new cells for growth, repair and replace damaged
tissues
Q: Fatty acids
(c) One molecule of glycerol combines with three molecules
of fatty acids to form a triglyceride molecule and water.
(d) Fat/Oil
(e)
Saturated fat
Unsaturated fat
Practice 4.4
A. Understanding key ideas
1 ✗
2 ✗
3 ✓
4 ✗
5 ✓
B. Main concepts and facts
1 lipids, triglyceride, glycerol, fatty acids
2 fatty acid, single, double
3 Glycerol + Fatty acid → Triglyceride (fat) + Water
4 ✓
5 ✗
B. Main concepts and facts
1 deoxyribonucleic acid, ribonucleic acid, genetic, proteins
2
DNA
RNA
Consists of two strands
of polynucleotide
Consists of one strand of
polynucleotide
Pentose sugar in
the nucleotide is
deoxyribose sugar
Pentose sugar in the
nucleotide is ribose
sugar
Nitrogenous base are
adenine, guanine,
cytosine and thymine
Nitrogenous base are
adenine, guanine,
cytosine and uracil
T
G
C
T
A
C
G
SPM FOCUS PRACTICE 4
PAPER 1
1
6
11
16
D
A
C
D
2
7
12
17
C
C
B
C
3
8
13
18
B
C
C
D
4 C
9 A
14 D
Solid at room
temperature
Liquid at room
temperature
Has higher level of
cholesterol
Has lower level of
cholesterol
Essay Questions
4 (a) • Carbohydrate – As a source of energy, to provide
energy for cell activities
• Lipid – As an energy storage in living organisms
• Protein – To build new cells for growth and repair
damaged tissue
• Nucleic acids – Carry genetic information to synthesise
proteins and determine the characteristics of organisms
(b) (i) The three types of polysaccharide:
• Starch – Excess carbohydrate stored in plant cells
• Glycogen – Excess carbohydrate stored in liver
cells and muscles cells
• Cellulose – Builds cell wall in plant cell
(ii) Starch, glycogen and cellulose are each formed from
glucose molecules by the condensation process.
Through a series of condensation reactions, two
glucose molecules are joined to form maltose. More
glucose molecules are added and joined together by
chemical bond to form a polysaccharide molecule.
In starch, the polysaccharide molecule is coiled to
form a spiral shape. In glycogen, it is branched while
in cellulose, it is arranged parallel to one another to
form a network of fibres.
(c) Soluble molecules such as glucose and amino acids
are small and simple molecules. They can be easily
transported in the blood and move across the plasma
membranes to be used by the cells. Insoluble molecules
such as starch and fats are large complex molecules
that cannot be transported and can only be stored in the cell.
5 (a) Nucleic acids contain carbon, hydrogen, oxygen, nitrogen
and phosphorus. The structural unit of nucleic acids is
nucleotide which consists of a pentose sugar, a phosphate
group and a nitrogenous base. There are two types of
nucleic acids: DNA (deoxyribonucleic acid) and RNA
(ribonucleic acid), DNA is a large complex molecule
consisting of two chains of polynucleotide twisted into
a double helix, while RNA is a shorter, single-stranded
molecule of polynucleotide. In DNA, the pentose sugar
in the nucleotide is deoxyribose sugar while in RNA, it
is ribose sugar.
3
A
Contains unsaturated
fatty acids
3 (a) (i) R: Deoxyribonucleic acid (DNA)
S: Ribonucleic acid (RNA)
(ii) R is made up of two polynucleotide chains/
double helix structure while S is made up of one
polynucleotide chain.
(b) Genetic information from R/DNA is copied onto S/RNA.
(c) Polypeptide consists of amino acids while DNA consists
of nucleotide units. Polypeptide is in the form of a single
linear chain while DNA is a double helix strand.
(d) Ribosome reads the genetic code on the RNA. Based on
the sequence of the codon on the RNA, amino acids are
added and joined together to form a polypeptide chain.
Practice 4.5
A. Understanding key ideas
1 ✗
2 ✓
3 ✗
Contains saturated fatty
acids
5 D
10 C
15 D
PAPER 2
Structured Questions
1 (a) P: Peptide bond
Q: Amino acids
(b) Amino acids are joined in a linear sequence by peptide
bond to form a polypeptide chain.
(c) Polypeptide is broken down by an enzyme through
hydrolysis to form dipeptides (products).
(d) • Enzymes
• Hormones
• Antibodies
2 (a) M: Condensation
N: Hydrolysis
(b) P: Glycerol
6
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PAPER 2
(b) • DNA carries genetic information in the form of a
genetic code to direct the cell to synthesise protein.
RNA is involved in the synthesis of protein.
• In the nucleus, mRNA copied genetic information
from DNA. The mRNA leaves the nucleus through
the pore on the nuclear membrane into the cytoplasm
of the cell.
• In the cytoplasm, rRNA on ribosome reads the genetic
code on the mRNA.
• The tRNA brings the specific amino acid based on the
codon on mRNA to the ribosome. The amino acids are
added and joined to form a polypeptide molecule.
• The polypeptide is modified into a specific protein.
• The types of protein in an organism determine the
structure of the organism and control the biochemical
reaction in the cell.
Rate of enzyme reaction
Structured Questions
1 (a) (i) 35°C
(ii) pH 6.5
(b) The rate of reaction increases from low to medium
temperature.
(c) The rate of reaction decreases as the acidity increases.
Salivary
(d)
Practice 5.1
4 ✓
5 ✓
B. Main concepts and facts
1 Metabolism refers to all the biochemical reactions that
occur in the cell.
2 • Anabolism is the process of synthesising a large
complex molecule from small simple molecules while
catabolism is the process of breaking down a large
complex molecule to small simple molecules.
• Anabolism requires energy while catabolism releases
energy.
Practice 5.2
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
6 ✗
7 ✓
8 ✓
4 ✓
9 ✗
5 ✗
10 ✓
B. Main concepts and facts
1 active site, lock, key
2 extracellular, intracellular
3 pH, temperature, substrate
4 doubles, 10°C
5 (a) lock
(b) lock and key hypothesis
(c) substrate
(d) key
(e) active site
(f) enzyme-substrate complex
(g) enzyme
(h) products
0.6
0.5
0.4
0.3
0.2
0.1
A. Understanding key ideas
1 ✗
2 ✓
3 ✗
4 ✓
PAPER 1
C
D
D
B
3
8
13
18
B
B
D
A
Temperature
5 10 15 20 25 30 35 40 45 50 55 60 (°C)
Enzymes are
proteins
SPM FOCUS PRACTICE 5
2
7
12
17
pH
(b) The function of enzyme depends on its shape. At high
temperatures, the polypeptide chains of enzyme are
broken down and cause the specific shape of the enzyme
to change. As a result, the enzyme cannot carry out its
function and is said to be denatured.
(c) Characteristic
Explanation
5 ✓
B. Main concepts and facts
1 enzymes, protease, lipase, amylase, protein, fat, starch
2 protease, tenderise, complex
3 protease, hair, leather
C
A
C
C
1 2 3 4 5 6 7 8 9 10 11 12 13
Essay Questions
3 (a) Rate of enzyme reaction
Practice 5.3
1
6
11
16
amylase Trypsin
enzyme enzyme
(mouth) (duodenum)
(e) The enzyme cannot act on the substrate. The pepsin
enzyme is denatured at a high temperature of 100°C.
The pepsin enzyme also cannot act on protein at pH 7.0.
The pepsin enzyme is most active in an acidic medium.
2 (a) (i) Shirt P: 15°C
Shirt Q: 65°C
Shirt R: 35°C
(ii) Shirt R has the least amount of fat stains remaining.
The temperature of the wash must be at the optimal
temperature of the enzyme (35°C) to break down
most of the fat stains efficiently. Shirt Q has the
largest amount of fat stains remaining. The enzyme
must have been denatured by the high temperature
of the wash at 65°C. Therefore it is unable to break
down the fat stains. Shirt P still has a considerable
amount of stains as the enzyme is inactive at 15 °C.
(b) Add alcohol (an emulsifier) to the stain to break down
the fats into smaller oil globules so as to increase the
surface area for faster hydrolysis by the enzyme/Add in
sodium carbonate to the washing powder to provide an
alkaline medium for optimum enzyme activity.
(c) (i) Lipase
(ii) Protease
(d) The enzymes in the detergent will break down the food
stains and remove it from the clothes.
(e) Enzymes are also used for tenderising meat so that the
meat is soft easier to chew and be broken down.
CHAPTER 5: METABOLISM AND ENZYMES
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
Pepsin
enzyme
(stomach)
4
9
14
19
A
D
C
C
5 A
10 D
15 D
Enzymes are produced by living
cells and have their own specific
shape. The activity of an enzyme
depends on its shape. As proteins,
the shapes of enzymes are easily
changed and this will cause the
enzymes to become inactive.
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(c)
Characteristic
Enzymes
are organic
catalysts
B. Main concepts and facts
1 growth, maintenance, tumour, malignant
2 meiosis, gamete, Down syndrome, 47
Explanation
• Enzymes speed up chemical
reactions
• Catalyse reaction in both directions
• Most enzymes are not changed
or not destroyed at the end of
chemical reactions
Enzymes are
specific in
their actions
Each enzyme only acts on one
specific substrate.
Enzymes are
sensitive to
temperature
• Enzymes are active in the
optimal temperature range
(35 – 40°C)
• At a high temperature (60°C),
enzymes are denatured
• At low temperatures, enzymes
are less active.
Enzymes are
sensitive to pH
The reaction of enzyme is most
active at its optimum pH
SPM FOCUS PRACTICE 6
PAPER 1
1
6
11
16
21
5 ✓
B. Main concepts and facts
1 (a) Karyokinesis – Division of the nucleus to form two
nuclei
(b) Cytokinesis – Division of the cytoplasm to form two
daughter cells
2
Somatic cells
Gametes
All are diploid cells
All are haploid cells
Have paired homologous
chromosomes
Have non–homologous
chromosomes
Have two sets of
chromosomes
Have one set of
chromosome
4 ✓
5 ✓
B. Main concepts and facts
1 two, number of chromosomes, genetic content, 46
2 interphase, S, G2
3 chromosomes, anaphase, sister chromatids
4 tissue culture, clones
Practice 6.3
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
6 ✓
7 ✗
8 ✓
4 ✗
9 ✓
3
8
13
18
23
A
B
B
A
A
4
9
14
19
D
B
A
C
5
10
15
20
C
B
A
A
(c) • The daughter cell from mitosis has 4 chromosomes while
the daughter cell from meiosis has 2 chromosomes.
• The chromosomes in the cell from mitosis do not
undergo crossing over while the chromosomes in the
cell from meiosis undergo crossing over.
(d) The reproductive organs of insect produce haploid gametes
by meiosis. During fertilisation, the male gamete and the
female gamete fuse together to form a zygote which is
diploid. The zygote develops into a new organism which
has a diploid number of chromosomes as the parent.
(e) (i) Increase the number of cells for growth of plants
(ii) Increase the quantity and quality of plants by tissue
culture
4 (a) Chromatid/Chromosome
(b)
Practice 6.2
A. Understanding key ideas
1 ✓
2 ✓
3 ✓
6 ✓
7 ✗
8 ✗
D
D
A
B
D
Structured Questions
1 (a) (i) Interphase
(ii) K : G1 phase
L : S phase
M : G2 phase
(b) In K and M, the cell produces proteins and cytoplasm
organelles for growth and differentiation to carry out
specialised functions. In L, synthesis of DNA occurs and
the chromosomes replicate.
(c) Mitosis
2 (a) X: Metaphase
Y: Prophase
Z : Anaphase
(b) Chromosomes condense, become thick and short, nuclear
membrane breaks down, nucleolus disappears, spindle
fibres begin to form.
(c) X : Chromosomes line up at equator of cell
Z : Sister chromatids separate and move to opposite
poles of the cell
3 (a) 4
(b)
Practice 6.1
4 ✗
2
7
12
17
22
PAPER 2
CHAPTER 6: CELL DIVISION
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
C
B
C
C
D
5 ✗
(c) (i) End of process Y
B. Main concepts and facts
1 gametes, haploid, diploid
2 synapsis, prophase I, crossing over
3 paired homologous chromosomes, sister chromatids
4 paired homologous chromosomes, sister chromatids
5 characteristics, inherited, diploid number
(ii) Causes variation
(d) (i)
Practice 6.4
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
4 ✓
5 ✓
Prophase (mitosis)
Prophase I (meiosis)
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(ii) Sperm and ovum
(iii) Three chromosomes (haploid number of chromosomes)
(e) A tumour is formed
• During prophase, the chromosomes become short and
thick. Each chromosome consists of two chromatids
joined together at the centromere.
• Nuclear membrane and nucleolus disintegrate. Spindle
fibres begin to form.
• During metaphase, the chromosomes with its
chromatids line up at the equator of the cell.
• During anaphase, the centromere splits, each
chromatid separates and moves to opposite poles.
• During telophase, the chromatids reach the respective
poles and become chromosomes. Nuclear membrane
and nucleolus reform in each group of chromosomes
to form two daughter nuclei. Spindle fibres disperse.
• Cell plate is formed at the equator of the cell and
grows outwards to merge with the plasma membrane
and formed cell wall.
• Two daughter cells identical to the parent cell are
formed.
(b) (i) If the cell divides out of control by mitosis, then the cell
will divide very fast to produce many cells and form a
tumour. If the tumour is located at its original place
and does not invade to other cells, then the tumour is
benign. If the cells invade other tissues and destroy
them, then the tumour is malignant, that is, cancer.
(ii) • Substances which can cause uncontrolled mitosis
to occur are ultraviolet rays, X-rays, radioactive
rays and carcinogenic substances such as asbestos,
benzene and cigarette smoke.
• Uncontrolled mitosis can be avoided by practising
a balanced diet, avoiding exposure to carcinogenic
substances and having a healthy social life.
Essay Questions
5 (a) • During meiosis I, the parent cell with four
chromosomes will produce haploid gametes with
two chromosomes. Meiosis I consists of four stages:
prophase I, metaphase I, anaphase I and telophase I.
• During prophase I, homologous chromosomes pair up
through the process called synapsis. Each chromosome
has two chromatids.
• During metaphase I, the paired homologous
chromosomes line up at the equator of the cell.
• During anaphase I, the paired homologous
chromosomes separate and move to opposite poles.
• During telophase I, two nuclei are formed at opposite
poles. Each nucleus has two chromosomes.
• Cytokinesis occurs to divide the cell into two daughter
cells, each having two chromosomes. Hence the
diploid number of chromosomes (4) in the parent cell
has been reduced to two chromosomes in the haploid
daughter cell.
(b) • After meiosis I, two haploid daughter cells (n = 2) are
produced. Each daughter cell enters meiosis II which is
similar to mitosis. It consists of four stages: prophase
II, metaphase II, anaphase II and telophase II.
• In prophase II, the chromosomes become thick
and short. Each chromosome consists of two sister
chromatids attached at the centromere. The nuclear
membrane and nucleolus disintegrate. Spindle fibres
are formed.
• In metaphase II, each chromosome with two
chromatids line up at the equator of the cell.
• In anaphase II, the centromere splits. Each chromatid
separates and moves to opposite poles.
• The chromatids become new chromosomes when
reach the pole.
• In telophase II, two nuclei are formed at opposite
poles.
• Nuclear membrane and nucleolus reform.
• Cytokinesis occurs to divide the parent cell into two
daughter cells. Each daughter cell is haploid with two
chromosomes, similar to the parent haploid daughter
cell. This is similar to mitosis where the parent cell
produces two daughter cells which are identical to it.
(c) (i) Crossing over occurs in prophase I. In prophase
I, the non-sister chromatids of two homologous
chromosomes criss-cross. The chromatids break off
and join again to another chromatid. This results in the
exchange of genetic materials called crossing over. The
four gametes produced have different genetic
materials.
(ii) Crossing over results in exchange of genetic
materials, producing gametes of different genetic
content. Hence crossing over results in genetic
variation in the offspring.
6 (a) • The type of cell division that causes growth at the
shoots is called mitosis. Before mitosis, the cell is in
interphase.
• During this stage, the cell prepares itself for cell
division. Proteins and cytoplasmic organelles are
produced and the chromosomes replicate.
• The chromosomes appear as thread-like structure
called chromatin. Cell division by mitosis starts with
prophase, metaphase, anaphase and telophase.
CHAPTER 7: C
ELLULAR RESPIRATION
Practice 7.1
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
4 ✓
5 ✓
B. Main concepts and facts
1 • Cell division for growth
• Formation of gametes for reproduction
• Contraction of muscles for movement
2 Aerobic respiration is the process of glucose breakdown
in the presence of oxygen and occurs in mitochondria of
the cell. Anaerobic respiration is the process of glucose
breakdown in the absence or limited supply of oxygen and
occurs in the cytoplasm of the cell.
Practice 7.2
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
4 ✗
5 ✓
B. Main concepts and facts
1 Aerobic respiration is the process of breaking down
glucose in the presence of oxygen to release energy.
2 Glucose + Oxygen → Carbon dioxide + Water + Energy
3 • Glycolysis – The breakdown of glucose to pyruvate
which occurs in the cytoplasm
• Oxydation of pyruvate which occurs in the mitochondria
Practice 7.3
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
4 ✓
5 ✓
B. Main concepts and facts
1 Aerobic respiration is the breakdown of glucose in the
presence of oxygen while fermentation is the breakdown
of glucose in the absence of oxygen or in situations with
limited supply of oxygen.
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2 Differences
Aerobic
respiration
• The volume of air or gas in the syringe decreases.
• This decreases the pressure of air or gas in the syringe
and causes the higher atmospheric pressure to push
the oil droplet to the right or towards the seeds.
(ii) • The movement of the oil droplet slows down or
stops.
• The rate of respiration decreases because oxygen
has been used up.
Fermentation
Occurs in the
presence of
oxygen
Occurs in the absence of oxygen
or in situations with limited
supply of oxygen
Releases large
amounts energy
(38 ATP or
2 898 kJ)
Releases less energy
• Two ATP molecules or
150 kJ through lactic acid
fermentation in muscle cells
• Two ATP molecules or 210 kJ
through alcohol fermentation
of yeast
Produces carbon
dioxide, water
and energy
Lactic acid fermentation produces
lactic acid and energy while
alcohol fermentation produces
ethanol, carbon dioxide and energy
Glucose is
completely
broken down
Glucose is not completely broken
down
Occurs in the
mitochondria
and cytoplasm
Occurs in the cytoplasm only
Essay Questions
3 (a) Cellular respiration in tissue X:
• Tissue X carries out aerobic respiration.
• The glucose is completely broken down to release
carbon dioxide, water and energy.
• The quantity of energy produced is higher, about 38
molecules of ATP.
Cellular respiration in cell Y:
• Cell Y carries out anaerobic respiration.
• Glucose is not completely broken down to produce
carbon dioxide, ethanol and energy.
• The quantity of energy produced is lower, about 2
molecules of ATP.
(b) • In organ R, the rate of respiration and breathing rate
of individual P is higher than in Q.
• More oxygen is needed for the oxidation of glucose
during cellular respiration to produce more energy to
carry out vigorous activity.
• In organ S, the rate of heartbeat of individual P is
higher than in individual Q.
• This is to ensure that the blood flows to the tissues in
individual P faster and more oxygen can be supplied
to his tissues.
• In organ T, the quantity of urine produced in individual
P is less and more concentrated than in individual Q.
• This is because more water is lost during the vigorous
exercise through sweating. More water is absorbed from
the kidney tubules into the blood capillaries in individual
P to maintain the osmotic pressure of the blood.
(c) (i) • An increased pulse rate due to an increased
heartbeat rate during vigorous exercise enables
more glucose and oxygen to be transported more
quickly to muscles.
• This causes cell respiration to occur rapidly to
produce sufficient energy for the vigorous muscle
activity.
• Blood circulation increases in order to remove
carbon dioxide formed during rapid cell respiration.
Carbon dioxide is transported in the blood plasma to
the lungs and then, expelled out during exhalation.
(ii) • The pulse rate takes some time to return to normal
after a vigorous exercise because a large amount of
oxygen is still required by the muscle cells.
• The oxygen is needed to break down the lactic
acid which is accumulated in the muscles from
anaerobic respiration. When all the lactic acid is
broken down, the oxygen debt is said to be paid
and the pulse rate returns to normal.
• The heart rate is also high during the recovery
period so that blood can be circulated rapidly.
• This enables the high concentration of carbon
dioxide formed during cell respiration to be
transported and removed from the lungs and more
oxygen is transported from the lungs and supplied
to muscle cells.
(d) (i) • During alcohol fermentation in yeast, glucose is
broken down in the absence of oxygen to release
carbon dioxide, ethanol and energy.
3 alcohol fermentation, ethanol, carbon dioxide, energy
SPM FOCUS PRACTICE 7
PAPER 1
1 D
6 B
11 C
2 A
7 B
12 A
3 C
8 B
13 B
4 D
9 A
14 B
5 D
10 D
PAPER 2
Structured Questions
1 (a) (i) After exercise:
1 breath = 3.5 dm³
20 breaths = 20 × 3.5 = 70 dm³ per minute
(ii) • After exercise, during the recovery stage, aerobic
respiration occurs.
• The breathing rate is fast and deep (panting) and
the heartbeat rate increases to increase the supply
and transport of oxygen to the muscle cells.
• The increased supply of oxygen is used to oxidise
the lactic acid that has accumulated in the muscle
cells during the vigorous exercise.
• When all the lactic acid has been oxidised, the
oxygen debt that has been incurred is paid off.
• The muscles are no longer fatigued and at rest.
The breathing rate and heartbeat rate slowly
return back to normal.
(b) • During the vigorous activity, muscle cells need more
energy to contract repeatedly and rapidly. Thus the
muscle cells need more oxygen for aerobic respiration.
• But the supply of oxygen is not enough for aerobic
respiration. As such, the muscle cells carry out
anaerobic respiration to produce the required energy.
• Lactic acid is produced and accumulates in muscle cells,
which causes muscle fatigue or muscle cramp.
2 (a) Glucose + Oxygen ➝ Carbon dioxide + Water + Energy
(b) Temperature, mass of soda lime, volume of air in the
syringe, volume / size of syringe, mass of seeds
(Any three)
(c) (i) • The oil droplet moves to the right or towards seeds.
• The seeds absorb oxygen and give out carbon
dioxide which is absorbed by the soda lime.
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2 • Insects – Breathing mechanism involves the contraction
and relaxation of muscles in the abdomen
• Fish – Breathing mechanism involves the movement of
the operculum and the floor of the mouth
• Amphibians (frogs) – Breathing mechanism involves
the movement of the floor of the buccopharyngeal
cavity and the elasticity of the lungs
• Humans – Breathing mechanism involves the action of
intercostal muscles and diaphragm muscles
• When yeast is mixed with flour and sugar, alcohol
fermentation occurs to release carbon dioxide.
• When the bread dough is heated, the carbon dioxide
trapped in the dough expands and rises up to escape
from the dough. This causes the bread dough to rise
and makes the baked bread light and soft.
(ii) • Yeast is used in brewing to make wine and beer.
• During alcohol fermentation in yeast, glucose is
broken down in the absence of oxygen to produce
ethanol.
• This ethanol is used to make wine and beer.
Practice 8.3
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
6 ✓
7 ✓
8 ✗
CHAPTER 8: RESPIRATORY SYSTEMS IN
HUMANS AND ANIMALS
Practice 8.1
4 ✓
5 ✓
Alveolus
O2
B. Main concepts and facts
Organism
Respiratory
surface
Human
Alveolus
Insect
Fish
Frog
Tracheole
Gill filament
Skin, lining
of walls
inside of the
mouth cavity,
lungs
Characteristics of
adaptation
Blood capillary
• Large surface area
• Thin and moist wall
Oxygen is
transported
in the form of
oxyhaemoglobin
O2
Body cells
2 haemoglobin, oxyhaemoglobin, respiration,
carbaminohaemoglobin, bicarbonate ions
• Large surface area
• Thin and moist wall
• Network of blood
capillaries
3
• Large surface area
• Thin and moist
surface or lining of
walls
• Network of blood
capillaries
Components
of air
Inhaled air
alveolus,
Exhaled air
Oxygen
21%
16%
Carbon dioxide
0.03%
4%
Nitrogen
79%
79%
Water vapour
Varies depending Saturated with
on humidity of air water vapour
Temperature
(heat)
Varies depending
on atmospheric
temperature
Body temperature
(37°C)
Practice 8.4
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
Practice 8.2
A. Understanding key ideas
1 ✓
2 ✓
3 ✓
Red blood cell
Carbon dioxide CO2
is transported
in the form of
carbonic acid,
carbaminohaemoglobin
and bicarbonate
ions
CO2
• Large total surface
area
• Thin and moist wall
of alveolus
• Network of blood
capillaries
2 Gaseous exchange occurs directly between the tracheoles
and muscle cells. This enables oxygen to diffuse directly
into the muscle cells to be used for respiration to release
energy for muscle contraction. It does not have to depend
on blood capillaries to transport oxygen to muscle cells.
As a result, oxygen is readily available for the muscle
activities of the insects.
4 ✗
5 ✓
B. Main concepts and facts
respiratory, alveoli, total surface area, oxygen
4 ✗
5 ✗
SPM FOCUS PRACTICE 8
B. Main concepts and facts
1
5 ✗
B. Main concepts and facts
1
A. Understanding key ideas
1 ✗
2 ✓
3 ✗
1
4 ✓
7
Air moves into the lungs
4
Diaphragm flattens
2
The ribcage moves upwards and outwards
6
Pressure in the thoracic cavity decreases
5
The volume of the thoracic cavity increases
3
Diaphragm muscles contract
1
The external intercostal muscles contract, internal
intercostal muscles relax
PAPER 1
1 B
6 A
11 C
2 B
7 D
3 C
8 A
4 B
9 B
5 A
10 A
PAPER 2
Structured Questions
1 (a) P: Moist surface, thin alveolus wall, a network of blood
capillaries
R: Thin wall, fluid at its end
(b) (i) Structure P has a network of blood capillaries while
structure R does not have any blood capillaries.
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(c)
(d)
2 (a)
(b)
(c)
(d)
(ii) In R, gaseous exchange occurs directly between
tracheoles and muscular cells. Hence blood
capillaries are not necessary to transport oxygen.
• Dissolved oxygen in the water diffuses into the gill
filaments due to the high concentration of oxygen in
the water compared to the concentration of oxygen in
the blood vessels in the gill filament.
• In the blood vessels of the gill filament, the
concentration of carbon dioxide is higher than in the
water. Hence carbon dioxide diffuses out of the blood
into the water.
• Thin epithelium layer (one cell thick) to facilitate
movement of gases across it.
• A layer of moisture on the surface of the epithelium
layer for the gas to dissolve and then diffuse across
the membrane.
• A large surface area to increase the rate of diffusion
of respiratory gases.
P: Internal intercostal muscle
Q: Rib
P contracts and causes the ribs to move downwards and
inwards. This reduces the volume of the thoracic cavity
and increases the air pressure in the lungs causing the air
to flow out of the lungs.
• The breathing rate becomes slow.
• Changes in the volume of the thoracic cavity can only
depend on the action of the intercostal muscles to
move the ribs.
• Hence, the changes in the volume of the thoracic cavity
is reduced and this causes the difference in the air
pressure in the atmosphere and in the lungs to be small.
As a result, air will flow in and out of the lungs slowly.
• It enables the oxygen in the alveoli to diffuse into the
blood capillaries and combine with the haemoglobin
in the red blood cells to form oxyhaemoglobin. In the
form of oxyhaemoglobin, oxygen can be transported
to the body cells.
• It enables carbon dioxide from the cells to diffuse out from
the blood capillaries around the alveoli into the lungs and
to be removed from the lungs during exhalation.
(b)
(c)
4 (a)
Essay Questions
3 (a) • During inhalation, the external intercostal muscles of the
ribs contract and the internal intercostal muscles relax.
• This causes the ribcage to move upwards and outwards.
• Simultaneously, the diaphragm muscles contract
causing the diaphragm to flatten. Both these actions
increase the volume of chest cavity and reduce the
internal pressure on the lungs.
• Air from the outside, where the atmospheric pressure
is higher, is then forced into the lungs through the
nasal cavity.
(b)
• Air that enters the nasal cavity flows through the
pharynx into the larynx and enters the trachea.
• The trachea is strengthened by C-shaped cartilages
which prevent the tube from collapsing during breathing.
• The air passes down the trachea into bronchi and then
into the bronchioles which end in alveoli.
• Oxygen diffuses across the epithelium of the alveolus
and then diffuses into the blood capillaries around the
alveolus. In the blood capillaries, the oxygen diffuses
into the red blood cells.
• Oxygen combines with haemoglobin in the red blood
cells to form oxyhaemoglobin.
• Oxyhaemoglobin is transported away from the lungs
to the body cells.
• Numerous and small alveoli increase the surface area
for diffusion and gaseous exchange.
• Thin (one-cell thick) epithelium of the alveolus allows
diffusion to occur easily across it.
• The layer of moisture at the alveolus epithelium allows
the gases to dissolve in the moisture and then diffuse
across the thin epithelium of the blood capillary into
the blood plasma.
• The alveolus is surrounded by a network of blood capillary
to increase the rate of diffusion and transport of oxygen to
the body cells and carbon dioxide into the alveolus.
• Smoking dries up the epithelium layer of the air passage
such as trachea, bronchus, bronchiole and alveolus.
• This hinders the process of gaseous exchange of the
alveolus.
• Cigarette smoke contains acidic gases such as oxides
of nitrogen which dissolve in the moisture and corrode
the membrane of the alveolus, damaging the surface
for gaseous exchange.
• Tar deposits in the lungs and thickens the wall of the
alveolus and decreases the rate of gaseous exchange.
• Carbon monoxide in cigarette smoke combines
more readily with haemoglobin than oxygen to form
carboxyhaemoglobin and reduces the amount of
oxygen absorbed by red blood cells.
• This causes a reduction of oxygen supply to the body cells.
Different organisms have different respiratory structures
for breathing. Humans breathe through lungs. Insects
have a tracheal system for breathing. Fish breathe
through gills. Frogs breathe through the skin and lungs.
Similarities:
• Humans and animals have respiratory systems to carry
out their breathing mechanisms.
• All have breathing mechanisms that involve changes
in volume and changes in pressure to enable air to
flow in and out of the respiratory organ.
• All involve the contraction and relaxation of muscles
to change the volume of the respiratory cavity.
Differences:
Aspect
Insects
Fish
Amphibians
Humans
Respiratory aperture
Spiracles
Mouth and
operculum
Nostrils
Nostrils
Structure where
change in volume
and pressure occurs
Trachea
Buccal cavity
Mouth cavity
Thoracic cavity
Breathing
mechanism
Action of abdominal muscles
which causes rhythmic
movements of the body
Action of the
floor of the mouth
and the operculum
Action of the muscles
in the mouth cavity and
elasticity of the lungs
Action of intercostal
muscles and
diaphragm muscle
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2 • Glucose, amino acids, minerals, vitamins B and C are
absorbed into the blood capillaries of the villus.
• Glycerol, fatty acids, vitamins A, D, E and K are
absorbed into lacteal of villus.
(c) • Air enters the lungs through inhalation.
• Action of intercostal muscles:
– The external intercostal muscles contract and the
internal intercostal muscles relax.
– Ribcage moves upwards and outwards.
– Volume of the thoracic cavity increases.
• Action of diaphragm muscle:
– The diaphragm muscle contracts.
– The diaphragm flattens.
– The volume of thoracic cavity increases.
– Both the actions of the intercostal muscles and
diaphragm muscle increases the overall volume of
the thoracic cavity.
– The air pressure in the thoracic cavity decreases, to
be lower than the atmospheric pressure.
– Air in the atmosphere flows into the lungs.
Practice 9.4
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
4 ✓
5 ✓
6 ✓
Practice 9.1
B. Main concepts and facts
1 assimilation, liver, liver, hepatic portal vein, deamination,
detoxification
2 • Excess glucose is converted to glycogen and stored in
liver and muscles.
• Excess amino acids are converted to urea in the liver
and excreted in the urine.
• Excess minerals such as iron is stored in the liver.
• Excess vitamins B and C are removed in the urine.
• Excess lipids are converted to adipose tissue and stored
below the skin.
• Vitamins A and D are stored in liver.
A. Understanding key ideas
1 ✗
2 ✓
3 ✗
Practice 9.5
CHAPTER 9: NUTRITION AND THE HUMAN
DIGESTIVE SYSTEM
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
B. Main concepts and facts
1 Mouth, oesophagus, stomach, duodenum, ileum, large
intestine
2 Mouth - salivary glands, Stomach- gastric glands, Small
intestine - intestinal glands
5 ✗
B. Main concepts and facts
1 peristalsis, contraction, rectum, defaecation
2 • Absorption of water and vitamin
• Synthesis of vitamins K and B12
Practice 9.2
A. Understanding key ideas
1 ✗
2 ✗
3 ✓
6 ✓
7 ✗
8 ✓
4 ✗
Practice 9.6
4 ✗
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
5 ✓
4 ✓
5 ✗
B. Main concepts and facts
1 Energy value is the quantity of energy released when one
gram of food is completely oxidised.
2 cardiovascular, carbohydrates, obesity, diabetes mellitus
B. Main concepts and facts
1 (a) Mouth
(b) Stomach
(c) Duodenum
2
Part of the
Practice 9.7
Enzyme involved
alimentary canal
Mouth
Salivary amylase
Stomach
Pepsin, rennin
Duodenum
Lipase, amylase, trypsin
Ileum
Maltase, sucrase, lactase,
erepsin, lipase
A. Understanding key ideas
1 ✗
2 ✓
3 ✗
6 ✓
7 ✓
4 ✓
5 ✓
B. Main concepts and facts
1 (a) Obesity
(b) Anorexia nervosa
(c) Gastritis
(d) Bulimia nervosa
2 Constipation/Colon cancer/Haemorrhoids
Practice 9.3
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
4 ✗
5 ✓
SPM FOCUS PRACTICE 9
PAPER 1
B. Main concepts and facts
1
1 A
6 D
11 D
Epithelial cell
2 A
7 A
12 A
3 C
8 B
13 B
4 A
9 A
14 B
5 C
10 B
15 A
PAPER 2
Structured Questions
1 (a) P: Stomach
Q: Pancreas
R: Liver
S: Large intestine
(b) Q: Secretes pancreatic juice which contains enzymes for
digestion of food
S: Absorbs water
(c) (i) Protein
Lacteal
Blood capillary
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(ii) Pepsin
Peptone and polypeptides
(iii) • Digestion: Intestinal glands on the wall of ileum
secrete a few enzymes to complete the digestion
process.
• Absorption: The wall of ileum has many
projections called villi to absorb the products of
digestion.
2 (a)
(ii) Pancreas secretes pancreatic juice which consists
of enzyme amylase, lipase and trypsin. Enzyme
amylase hydrolyses starch in the rice to maltose.
Enzyme lipase hydrolyses fat into glycerol and fatty
acids. Enzyme trypsin acts on protein in the chicken
and hydrolyses it to peptides.
(iii) Ileum carries out two functions: digestion and
absorption. Intestinal juice secreted by the glands in
the ileum contains many enzymes for digestion. The
enzyme maltase hydrolyses maltose into glucose.
The enzyme peptidase/erepsin hydrolyses peptides
into amino acids. The enzyme lipase hydrolyses fats
into glycerol and fatty acids. The end products such
as glucose and amino acids are absorbed into the
blood capillaries of the villus in the ileum. Glycerol
and fatty acids are absorbed into the lacteal of the
villus.
(b) • Glucose is transported to the liver by hepatic portal
vein. In the liver, much of the glucose is converted
to glycogen and is then stored in the liver cells and
muscle cells. Glucose is also carried to the heart and
then distributed to all parts of the body. Glucose is
oxidised during cell respiration to release energy for
cells activities.
• Amino acids are transported to the liver by the hepatic
portal vein and then to the heart, to be distributed to
all parts of the body. Amino acid is used for growth, to
build new protoplasm, to repair damaged tissues and to
build enzymes and hormones. Excess amino acids are
converted to urea in the liver and excreted in the urine.
• Glycerol and fatty acids are absorbed by the lacteal
of the villus and transported out of the ileum by the
thoracic duct to the subclavian vein. Then glycerol
and fatty acids are transported in the blood to all parts
of the body.
• Lipids are used to build plasma membrane. If glucose
is not enough, then lipids will be oxidised to provide
energy. Excess lipids are stored as adipose tissue
below the skin, around the organs and in the abdomen.
6 (a) (i) Balanced diet because it has all the classes of food.
Milk provides protein, lipid, vitamins and minerals.
Bread provides carbohydrate and fibres. Margarine
contains lipid, protein, vitamins and minerals.
(ii) Bread contains carbohydrate. The bread is first
chewed into small particles by the teeth in the mouth.
Salivary amylase then acts on the starch in the bread
and hydrolyses it to maltose.
Salivary
Starch + Water
➝ Maltose
amylase
In the stomach, the pH is not suitable for the digestion
of starch. The reaction of salivary enzyme on starch
stops because it is too acidic. In the duodenum, any
starch that is not digested yet is hydrolysed to maltose
by the enzyme amylase secreted by the pancreas.
In the ileum, the enzyme maltase hydrolyses maltose
to glucose.
Maltase
Maltose + Water
➝ Glucose
Epithelium
Lacteal
Microvilli
Blood capillary
(b) • Has a very thin epithelium that is only one-cell thick
• Has many microvilli to provide a large surface area
• Has a lot of blood capillaries and lacteal
(c) Circulatory system in the structure absorbs simple
sugars, amino acids, minerals and vitamins B and C.
Lymphatic system in the structure absorbs fat droplets
that is fatty acids and glycerol, as well as vitamins A,
D, E and K.
3 (a) • As a source of energy
• For storage/stored as adipose tissue
• Shock absorber
• Insulator/Reduce heat loss
(Any three)
(b) Gall bladder. Stores bile
(c) Fat is emulsified by bile. Fat is broken down into small fat
droplets to increase surface area for enzyme lipase to act.
(d) • Fatty acids and glycerol diffuse into the lacteal/lymph
capillary of the villus of the ileum
• Carried out of ileum by lymphatic vessel/thoracic duct
to the left subclavian vein
• The products in thoracic duct enter into left subclavian
vein.
• Carried to the heart/enter into circulatory system
(pump) to all body cells
4 (a)
Location
Enzyme
Substrate
Product(s)
of enzyme
production
Salivary
amylase
Starch
Maltose
Salivary
glands/Mouth
Lipase
Fats
Fatty
acids and
glycerol
Pancreas
Pepsin
Protein
Polypeptide
and peptone
Stomach
Erepsin
Peptide
Amino
acids
Small intestine
(ileum)
Maltase
Maltose
Glucose
Small intestine
(ileum)
(b) (i) Assimilation
(ii) • Enzymes/Antibodies/Fibrinogen (Plasma protein)
• Insulin
Glucose is the end product of the digestion of starch.
(b) (i) Not suitable because children need more protein for
growth. The potato chips have very little protein
compared with carbohydrate and lipid. It also lacks
in minerals. Minerals such as calcium is important
for building strong bones and teeth. There is also
no vitamins which are needed to maintain a healthy
body.
Essay Questions
5 (a) The food taken by the student consists of the following
classes of food: carbohydrate, protein and lipid.
(i) Liver secretes bile to emulsify the fats from the fried
chicken into small droplets to increase the surface
area for the reaction of enzyme.
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2 The numbness is due to poor circulation of the blood at the
legs. The skeletal muscles at the legs relax. The blood in
the veins at the legs cannot flow back to the heart because
there is no force exerted on the veins by the skeletal
muscles to cause the valves to open and enable the blood
to flow back to the heart.
(ii) • Growth of body, bones and teeth is not at normal
rate due to lack of protein
• Frequent or easily contract diseases because lack
of vitamins
• Too much lipid can cause obesity or cardiovascular
diseases
• Lack of fibre can cause constipation
(iii) • Increase the amount of vitamins and minerals
• Increase the amount of protein
• Decrease the amount of lipid
Practice 10.4
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
CHAPTER 10: TRANSPORT IN HUMANS
AND ANIMALS
B. Main concepts and facts
1 Vitamin K, calcium ions
2 Haemophilia, thrombosis
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
6 ✓
7 ✗
8 ✓
Practice 10.5
Practice 10.1
4 ✓
5 ✓
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
B. Main concepts and facts
Jugular vein
Head and neck
Subclavical artery
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
Lungs
Pulmonary
vein
Aorta
Left atrium
Left ventride
Right ventricle
Hepatic artery
Hepetic
portal vein
Mesentery
artery
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
6 ✓
7 ✗
Stomach and intestines
Renal vein
Renal artery
Femoral artery
Hind limbs
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
4 ✗
5 ✓
5 ✓
5 ✓
SPM FOCUS PRACTICE 10
PAPER 1
1 D
6 A
11 B
Practice 10.3
4 ✓
4 ✓
B. Main concepts and facts
If the lymphatic vessel is blocked, 10% of the tissue fluid
cannot diffuse into the lymph capillary. It accumulates in the
spaces between the body cells and causes swelling of that part
of the body, resulting in oedema.
B. Main concepts and facts
1 carbon dioxide, oxygen, haemoglobin
2 nutrients, waste products, plasma proteins
3 • Artery has a small lumen while vein has a large lumen.
• Artery has thick muscular wall while vein has thin less
muscular wall.
• Artery transports blood away from the heart while vein
transports blood towards the heart.
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
4 ✗
Practice 10.8
Practice 10.2
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
5 ✓
B. Main concepts and facts
1 skeletal muscles, lymphatic vessels
2 (a) blood capillaries (b) lymphatic system
Kidneys
Femoral vein
4 ✓
Practice 10.7
Liver
Hepatic vein
5 ✓
B. Main concepts and facts
Similarities:
• Both atherosclerosis and arteriosclerosis are conditions due
to blockage of the arteries.
• Both conditions restrict the flow of blood.
• Both cause the lumen of the artery to become small.
Difference:
• Atherosclerosis is caused by deposits of saturated fats
and cholesterol while arteriosclerosis is due to deposits of
calcium and plaque on the inner wall of arteries.
Vena cava
Right
atrium
4 ✗
Practice 10.6
Forelimbs
Pulmonary
artery
5 ✓
B. Main concepts and facts
1 Blood group A and blood group AB
2 Group B
Carotid artery
Subclavical vein
4 ✗
5 ✓
2 B
7 C
12 C
3 A
8 B
13 B
4 C
9 A
14 C
5 A
10 B
15 A
PAPER 2
Structured Questions
1 (a) Cell X: Lymphocyte
Cell Y: Red blood cell (erythrocyte)
(b) Cell X: Produces specific antibodies to react with
particular antigen
Cell Y: Transports oxygen and carbon dioxide.
B. Main concepts and facts
1 • The sound of the heart beat is “lub-dub”.
• “Lub” is the first sound occurs when the tricuspid and
bicuspid valves close during systole.
• “Dub” is the second sound occurs when the semilunar
valves close during diastole.
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(c) Platlets (Z) can release enzyme (thrombokinase) that can
convert prothrombin into thrombin needed to clot blood.
2 (a) (i) Humans/Other mammals
(ii) Fish
(b) Similarity: Both are closed circulatory system.
Difference: X is a double circulatory system while Y is
a single circulatory system.
(c) A: Pulmonary circulation B: Systemic circulation
(d) X: Lungs
Y: Gills
3 (a) Aorta and pulmonary veins
(b) P
(c) Contraction of the left ventricle muscles wall generate a
high pressure to close the bicuspid valve and open the
semilunar/R valve at the aorta. This causes the blood to
flow out of the heart to all the body tissue through the aorta.
(d) Coronary artery
(e) (i) Left atrium, left ventricle, pulmonary vein, aorta
(ii) Right atrium, right ventricle, pulmonary artery, vena
cava
4 (a) (i) Tissue fluid
(ii) High hydrostatic pressure in the blood capillary (E)
forces some of the blood plasma with its dissolved
substances to diffuse out through the wall of E into
the space between the body cells to form fluid D.
(iii) 90% of fluid D diffuses back into the blood capillaries on
the venule side where the pressure is low. The remaining
10% of fluid D diffuses into the lymph capillary to
form lymph. The lymph flows in the lymphatic system
and is finally returned to the circulatory system by the
thoracic duct via the left subclavian vein and by the right
lymphatic duct via the right subclavian vein.
(b) (i) Blood capillary
(ii) Thin wall, only one cell thick. Exchange of
substances occurs between E and the body cells.
• No valves as the blood is still under moderate pressure
(b) • The condition of the artery shows that the patient has
developed atherosclerosis or arteriosclerosis.
• The lumen of the artery becomes narrow because of
deposition of plaque (saturated fats and cholesterol)
at the inner wall of the artery. If calcium is also
deposited on the plaque, the artery becomes hardened
and inelastic.
• Further development of plaque will make the lumen
narrower and restrict further the flow of blood.
• This can trigger the blood clotting process to occur in
the artery and formation of a blood clot (thrombus)
resulting in thrombosis.
• Blockage of the coronary artery at the heart due to
plaque or blood clot will lead to heart attack and
eventually death.
• If this blockage occurs at the blood vessel in the brain,
it can cause stroke.
6 (a) (i)
P
Q
Essay Questions
5 (a) • Blood vessel P is vein. Function of blood vessel P is
to carry blood from organs to the heart.
• The wall of vein is thin as it carries blood at a low
pressure and blood flow is slow. The size of the lumen
is large to reduce resistance to blood flow.
• It has valves to prevent backflow of the blood.
• It has less elastic wall, less fibrous tissues to ensure
that blood flows smoothly.
• Blood vessel Q is artery. The function of artery is to
carry blood away from the heart to all organs in the body.
• Has thick muscular wall to withstand the high pressure
of blood.
• Has elastic wall with fibrous tissues to enable it to stretch to allow blood to flow in surges. This enables
the heartbeat to be felt as pulse. It can also recoil to
maintain blood pressure.
• Has a small lumen to help maintain blood pressure.
• Does not have valves as high pressure of blood can
prevent backflow of blood.
• Blood vessel R is capillary. The function of R is for
exchange of substances with the body cells.
• Has very thin wall, one-cell thick and permeable.
Gases, nutrients and waste products can diffuse easily
and quickly across it.
• No muscle or elastic fibres. This allows diffusion of
substances between capillary and surrounding cells.
• Has a very small lumen, one red blood cell wide. Blood
cells pass through slowly to allow diffusion of substances.
• Has pores on the wall, so white blood cells can
squeeze through the pores on capillary wall.
The system consists of
a heart which functions
as a pump to circulate
the blood in the body.
The system does not
have any organ that
functions as a pump to
circulate the lymph in
the body.
Among the blood
vessels, only veins
have valves.
All lymphatic vessels
have valves.
The blood capillaries
are continuous from
the arteriole side to the
venule side.
The lymph capillaries
have closed ends, not
continuous.
The blood cells that
are present in the
blood are red blood
cells, white blood cells
and platelets.
The blood cells that are
present in the lymph are
only white blood cells
such as lymphocyte
and phagocyte.
The blood contains
low concentrations
of fatty acids and
glycerol.
The lymph contains
high concentrations
of fatty acids and
glycerol.
The blood flow is aided
by the pumping of the
heart and contraction
of skeletal muscles
around the veins.
The lymph flow
is aided by the
contraction of skeletal
muscles around the
lymphatic vessels.
(ii) • Q/lymphatic system is joined to the circulatory
system via the left and right subclavian vein.
• The two main lymphatic vessels are thoracic duct
and right lymphatic duct. The thoracic duct is joined
to the circulatory system via the left subclavian
vein while the right lymphatic duct is joined to the
circulatory system via the right subclavian vein.
• The lymphatic system collects the tissue fluid and
returns it to the circulatory system.
• The thoracic duct collects two third of the tissue
fluid in the body and returns it to the circulatory
system.
• The right lymphatic duct collects one third of
tissue fluid in the body and returns it to the
circulatory system.
(b) Similarities:
• Both circulatory systems consist of heart, blood
vessels and blood.
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SPM FOCUS PRACTICE 11
• Both circulatory systems transport substances required
by the cells and remove waste products.
• The heart of both circulatory systems has valves.
Differences:
Human
PAPER 1
1 B
6 D
11 C
Fish
Has a double closed
circulatory system
Has a single closed
circulatory system
The blood flows through the
heart twice in one complete
circulation
The blood flows
through the heart
once in one
complete circulation
The heart has four cavities
(two atria and two
ventricles)
The heart has two
cavities (one atrium
and one ventricle)
The right side of the heart
carries deoxygenated blood.
The left side of the heart
carries oxygenated blood.
Deoxygenated blood
in the heart (both
atrium and ventricle)
Has septum
No septum
Exchange of gases occurs
at lungs
Exchange of gases
occurs at gills
Aorta carries oxygenated
blood to all body tissues
Aorta carries
deoxygenated blood
to the gills
5 ✗
B. Main concepts and facts
1 (a) The first line of defence, hair in the nose, the entry of
microorganisms
(b) The second line of defence, phagocytosis, neutrophil
(c) The third line of defence, antibodies
2 second, third, pathogens
Practice 11.2
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
4 ✓
5 ✗
B. Main concepts and facts
1 specific, antitoxin, toxin, agglutinin, phagocytes
2 lysis
Practice 11.3
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
4 ✓
5 ✓
B. Main concepts and facts
1 antiserum, immunity, temporary/short–lived
2 vaccine, lymphocytes, Memory cells, long lasting
3 (a) production of antibodies
(b) receiving antibodies
(c) After recovering from a disease or infection
(d) Injection of vaccine
(e) Through placenta and breast milk
(f) Injection of serum containing antibodies
Practice 11.4
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
4 ✓
4 C
9 B
5 B
10 B
Structured Questions
1 (a) Individual M: Vaccine
Individual N: Antiserum containing antibodies
(b) Individual M: Artificial active immunity
Individual N: Artificial passive immunity
(c) • Immunity acquired by M is long-lasting while
immunity acquired by N is for short-term (temporary).
• Immunity acquired by M is to give protection against
the disease. Immunity acquired by N is to give
immediate treatment to recover from a disease.
(d) • Individual M: To increase the production of antibody
quickly so that the level of antibody is above the
level of immunity thus providing effective immunity
against the disease
• Individual N: To provide enough antibodies so that the
level of antibody is above the level of immunity and
the antibodies can immediately destroy the pathogens
(e) Artificial passive immunity
2 (a) Cell P: Lymphocyte
Cell R: Phagocyte
(b) Antibody
(c) • Produces antibodies to react with a particular
pathogen/antigen/bacteria by binding to the pathogen
• Label or mark the pathogen/antigen so that it is easier
to be destroyed by phagocytosis
(d) • Cell R carries out phagocytosis
• Antibodies bind to a bacteria and mark it
• Cell R approaches and surrounds the bacterium
• Cell R engulfs the bacterium and forms a phagocytic
vesicle/phagosome
• Enzymes are secreted into the phagosome to digest/
break down the bacterium and destroy it
3 (a) • K has acquired natural active immunity towards the
disease.
• In K, lymphocytes produce antibodies to destroy the
pathogen thus help K recovers from the disease.
• Some lymphocytes remain in the plasma and form
memory cells, which can produce antibodies immediately
if K is attacked by the same pathogen again.
(b) • M has acquired artificial active immunity
• M is injected with a suspension of vaccine which
contains dead or weakened measles virus.
• The presence of antigen in blood stimulates lymphocytes
to produce specific antibodies against the pathogen.
Antibodies and memory cells are formed to produce
antibodies anytime the same pathogen enters the body.
(c)
Active immunity
Passive immunity
Practice 11.1
4 ✓
3 B
8 B
PAPER 2
CHAPTER 11: IMMUNITY IN HUMANS
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
2 B
7 B
Produce antibodies
Receive antibodies
Give long-term immunity
Give short-term immunity
Have memory cells to
produce antibodies
No memory cells are
produced
Essay Questions
4 (a) (i) Second line of defence
• Involves the process of phagocytosis when
pathogens succeed in entering the body
• Both the circulatory system and lymphatic system
have leucocytes which act as phagocytes to
destroy the pathogens in the body.
5 ✗
B. Main concepts and facts
Acquired Immuno Deficiency Syndrome, pneumonia, Human
Immunodeficiency Virus (HIV), lymphocytes, phagocytes
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Practice 12.2
• The phagocytes approach the pathogens such as
bacteria which enters the body. The phagocytes then
surround and engulf; ingest and digest the pathogens,
hence preventing the infection of diseases.
• In the lymphatic nodes, phagocytes are present to
destroy any pathogens in the lymph.
• At infected wounds, phagocytes squeeze through
the wall of blood capillary and destroy the
pathogens by phagocytosis.
(ii) Third line of defence
• Involves the production of antibodies by
lymphocytes
• Involves the specific action of antibody on antigen
in the body
• Lymphocytes in the blood and lymphatic nodes
are stimulated by antigens such as pathogens in
the body to produce antibodies to act against the
specific antigen.
• The level of antibody must be above the level of
immunity in order to destroy the pathogen and
give immunity.
(b) Immunity against chickenpox
The immunity acquired against chickenpox is natural
active immunity. The immunity is acquired after a
person has recovered from the disease. When a person
is infected with chickenpox, lymphocytes are stimulated
to produce antibodies against the specific antigen. After
recovery from the disease, the antibodies still remain
in the body and is above the level of immunity. This
will give immunity against the disease when the same
pathogen invades the body again in the future.
Immunity against hepatitis B
The immunity acquired against hepatitis B is artificial
active immunity. A vaccine containing hepatitis B virus
that is dead or weakened is injected into the body. The
presence of antigens in the body will stimulate the
lymphocytes to produce antibodies against the specific
pathogen. In the first injection, only a small amount of
antibodies are produced. The quantity of antibody has
not reached the immunity level and cannot act against
the pathogen effectively. A second injection is given
so that the antibodies are produced at a faster rate and
the quantity is increased to above the level of immunity
quickly. The antibodies remain in the body and give
immunity up to a certain period of time only.
Immunity against an infected cut
The immunity acquired is artificial passive immunity. An
antiserum containing antibodies is injected into the body
to act against the pathogen immediately. Immediate action
has to be taken to prevent the infection of the wound.
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
6 ✓
7 ✗
B. Main concepts and facts
1 W: Brain
Y: Cranial nerve
2 P: Spinal nerve
R: Ventral root
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
B. Main concepts and facts
1
Sensory neurone
Long dendron, short axon
X: Spinal cord
Z: Spinal nerve
Q: Ganglion of dorsal root
S: Grey matter
4 ✓
5 ✓
Motor neurone
Short dendron, long axon
Cell body at the side of the Cell body at the end of the
neurone (not at the end)
neurone
Begins with receptor
Ends with effector
2 The synaptic knob has synaptic vesicles which contain
neurotransmitters that diffuse across the synaptic gap to
the dendrite of another neurone in one direction. Dendrites
do not have neurotransmitter.
Practice 12.4
A. Understanding key ideas
1 ✗
2 ✓
3 ✗
4 ✓
5 ✓
B. Main concepts and facts
(a) K: relay neurone
L: motor neurone
M: sensory neurone
N: muscle/effector
(b) (i) stimulus
(ii) sensory receptor
(iii) nerve impulse
(iv) sensory
(v) relay neurone, grey matter
(vi) motor
(vii) effector
(viii) contract
Practice 12.5
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
4 ✓
5 ✓
B. Main concepts and facts
memory, intellectual/mental, tremors, weakness
Practice 12.6
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
6 ✗
7 ✓
8 ✗
Practice 12.1
4 ✗
5 ✓
Practice 12.3
CHAPTER 12: COORDINATION AND
RESPONSE IN HUMANS
A. Understanding key ideas
1 ✓
2 ✓
3 ✓
4 ✓
4 ✓
5 ✗
B. Main concepts and facts
1 Adrenaline, glucose, glycogen, glucose, energy
2 (a) Antidiuretic hormone
(b) Thyroid-stimulating hormone
(c) Thyroxine
(d) Ovaries and testes
(e) Development of eggs in ovaries and the formation of
sperms in testies
5 ✓
B. Main concepts and facts
1 stimulus, response, sensory neurone, receptor, motor
neurone, effector(s)
2 (a) Touch and pressure
(b) Light
(c) Body temperature
(d) Chemical substances
(e) Blood pressure
(f) Pain
Practice 12.7
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
4 ✗
5 ✓
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B. Main concepts and facts
(a) Diabetes insipidus
(b) Dwarfism
(c) Gigantism
(d) Hypothyroidism
(e) Diabetes mellitus
(f) Hyperthyroidism
4 (a) (i) Thyroid-stimulating hormone
(ii) Stimulates the thyroid gland to secrete thyroxine.
(b) Follicle-stimulating hormone. Pituitary gland.
(c) (i) Insulin— converts excess glucose to glycogen.
Glucagon—converts glycogen to glucose.
(ii) Diabetes insipidus due to lack of antidiuretic hormone
Diabetes mellitus due to lack of insulin
(d) Because it secretes hormones that control the secretion
of hormones by other endocrine glands.
SPM FOCUS PRACTICE 12
PAPER 1
1
6
11
16
2
7
12
17
D
A
D
C
C
C
C
D
3
8
13
18
C
D
B
C
4
9
14
19
D
C
D
B
Essay Questions
5 (a) (i) The sensory neurone is located between the receptor
and the central nervous system (brain or spinal cord).
This neurone is found in the cranial nerves and spinal
nerves of the peripheral nervous system. Sensory
neurone carries nerve impulses from the receptor
to the central nervous system. The dendrite of the
sensory neurone is at the receptor where stimuli can
be detected and nerve impulses are generated. The
nerve impulses are carried along the neurone via the
cranial nerves direct to the brain or via the spinal nerve
and dorsal root into the spinal cord where it ends. The
information in the spinal cord is then carried to the
brain or to the effector by other neurones.
(ii) Synapse is the narrow gap between the axon terminal
of a neurone with the dendrite of another neurone.
Synapse can be found in the grey matter of the
brain and spinal cord where neurones are joined.
Synapse enables nerve impulses to be transmitted
from one neurone to another. Nerve impulses are
transmitted by neurotransmitters across the synapse.
The neurotransmitters diffuse across the synapse and
trigger nerve impulses to be generated in the dendrite
of another neurone. The nerve impulses are then
carried away by this neurone.
(b) (i) Differences between coordination by nervous system
and endocrine system are as follows:
5 B
10 A
15 A
PAPER 2
Structured Questions
1 (a) Neurone G: Motor neurone
Neurone H: Sensory neurone
(b) Fibre X carries nerve impulses into the cell body of the
neurone. Fibre Y carries nerve impulses away from the
cell body of the neurone.
(c)
Synapse
2 (a) Electrical impulse/signals
(b) (i) Nerve impulses that reach the synaptic knob will
stimulate structure U to release neurotransmitters,
which then diffuse across space T to the dendrite of
another neurone. Nerve impulses are triggered and
carried away by the neurone.
(ii) To enable information to be transmitted from one
neurone to the next neurone.
(c) Neurotransmitters are produced only at the synaptic knob
of the axon terminal in a neurone.
(d) (i) Acetylcholine/Noradrenaline
(ii) The transmission of nerve impulses across the
synapse by neurotransmitters requires energy. The
energy is supplied by mitochondria.
3 (a)
Relay neurone
Nervous system
Endocrine system
Involves nerve impulses
Involves hormones
Impulses are transmitted
by neurone
Hormones are
transported in the
bloodstream
K
Biceps
muscle
Hot object
Responses are
Responses are slow,
immediate, localised,
widespread, involve
involve specific effectors more effectors
Sensory neurone
Motor neurone
(b) Sensory nerve endings (sensory receptors) of the skin detect
the stimulus (hot object) and trigger the nerve impulses
which are carried by the sensory neurone to the spinal cord.
The nerve impulses are then transmitted across the synapse
to the relay neurone in the spinal cord. From there, the
nerve impulses are transmitted directly to the motor neurone
across the synapse. The motor neurone carries the impulses
to the effector (biceps muscle) which contract and causes
the hand to withdraw from the hot object.
(c) • The reflex action is an automatic involuntary action in
response to an external stimulus. It does not involve
the brain. The decision to lower the arm is a voluntary
action which involves the brain.
• The reflex action is an immediate and fast response to
an external stimulus while the decision to lower the
arm is a controlled action which can be slow or fast.
(d) Some of the sense of touch is lost or no ability to sense
stimuli but the arm can still move.
Responses are rapid and
short-term
Responses are slow
and long-lasting
Involve voluntary and
involuntary actions
Involve involuntary
actions only
(ii) When a person witnessed a robbery, the receptors
in the eyes are stimulated and nerve impulses
are generated. The nerve impulses are carried by
the sensory neurone to the brain. From the brain,
the nerve impulses are transmitted through the
motor neurone to effectors such as adrenal glands.
Adrenaline is produced and transported by the
bloodstream to target organs. Adrenaline is carried to
the heart where it increases the heart rate. Adrenaline
is also transported to the lungs where it increases the
rate of breathing. It raises the blood glucose level by
converting glycogen stored in the liver and muscles
to glucose which is required for the increased
metabolic rate to release more energy to prepare the
body to fight or flight.
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6 (a) (i) • When the tendon below the kneecap is tapped, the
quadriceps muscles at the thigh will be stretched.
The stretching of the quadriceps muscle is detected
by the stretch receptor at the quadriceps muscle.
• Nerve impulse is triggered and is carried by the
sensory neurone (N) via the spinal nerve and
dorsal root to the grey matter of the spinal cord.
• Nerve impulse is transmitted across the synapse to
the motor neurone (M).
• The motor neurone carries the nerve impulse out of
the spinal cord via the ventral root and spinal nerve
to the effector which is the quadriceps muscle.
• The quadriceps muscles contract and causes the
leg to be kicked up, which is called knee jerk.
(ii) • It is a voluntary action.
• When the student hears the telephone ringing,
the sensory receptor in the ear is stimulated and
triggered to release nerve impulses.
• The nerve impulse is carried by the sensory
neurone to the brain.
• The brain interprets and integrates the
information and sends out the nerve impulses via
the motor neurone to the effector which is the
biceps muscle.
• The biceps muscle contracts, to bend the arm and
pick up of the phone.
(b) Similarities:
• Both actions are controlled by the central nervous
system.
• Both actions involve stimulus, receptors, and effectors.
• Both involve skeletal muscles as the effector.
Differences:
Action of (a)(i)
Action of (a)(ii)
It is an involuntary action
called reflex action
It is a voluntary action
Controlled by the spinal
cord
Controlled by the
cerebrum of the brain
Occurs automatically
without any conscious
control
Action occurs
according to the will,
with conscious control
Occurs rapidly and
spontaneously
Occurs according to
own pace.
2 (a)
(b)
(c)
(d)
(e)
(f)
Practice 13.2
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
6 ✓
7 ✗
5 ✓
pressure,
Practice 13.3
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
4 ✗
5 ✓
B. Main concepts and facts
• Diets:
– Eating a diet that is high in protein, sodium (salt) and
sugar may increase the risk of kidney stones.
– Especially a high-sodium diet. Too much salt in the
diet increases the amount of calcium in the kidneys and
causes stones.
• Dehydration – Not drinking enough water each day can
result in concentrated urine and the minerals in urine to
crystallise and form kidney stones.
SPM FOCUS PRACTICE 13
PAPER 1
1 B
6 A
11 C
Practice 13.1
4 ✓
4 ✗
B. Main concepts and facts
1 ultrafiltration,
reabsorption,
hydrostatic
Bowman’s capsule, kidney tubule/renal tubule
2 (a) (i) Bowman’s capsule
(ii) Proximal convoluted tubule
(iii) Loop of Henle
(iv) Glomerulus
(v) Distal convoluted tubule
(vi) Collecting duct
(b) (i) ultrafiltration
(ii) proximal convoluted tubule
(iii) water, salts
(iv) antidiuretic hormone
(v) aldosterone
CHAPTER 13: HOMEOSTASIS AND THE
HUMAN URINARY SYSTEM
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
• Thyroid gland, adrenal glands not activated to
secrete hormones. Rate of metabolism is slow. No
heat is produced.
Insulin converts excess glucose to glycogen
Glucose used by cells in respiration increases
Glucose is converted to lipids (fat)
Glucagon converts glycogen to glucose
Less glucose used in respiration
Lipids are converted to glucose
2 C
7 A
12 B
3 A
8 B
13 C
4 B
9 B
14 C
5 A
10 C
PAPER 2
5 ✓
Structured Questions
1 (a)
X
B. Main concepts and facts
1 (a) Thermoreceptors
in
the
hypothalamus,
Thermoreceptors in the skin
(b) • Smooth muscles at arteriole relax. Vasodilation
occurs. More blood flows to the blood capillary.
Excess heat is lost by conduction and radiation.
• Sweat glands are active. Sweating occurs.
Evaporation of sweat causes excess heat to be lost
from the body.
• Hair erector muscles relax. Hair lies flat. Thin layer
of air is trapped between the hair. Excess heat is
easily lost from the body by convection.
• Skeletal muscles not activated to carry out
involuntary action. No shivering occurs. No heat
is produced.
B
W
A
Y
(b) (i) Glomerulus
(ii) • Y contains more water.
• Y contains glucose and amino acid while W does
not.
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(c) (i) Protein
(ii) The size of protein is too large to diffuse through the
walls of glomerulus.
(iii) Reabsorption of water along the renal tubule reduces
the volume of water in the tubule and hence increases
the concentration of urea. Excess urea in the blood
capillaries is also secreted into the filtrate at the distal
convoluted tubule to increase the concentration of urea.
2 (a) (i) Skin
(ii) • Regulating body temperature
• As an excretory organ / Eliminating urea
• As a protective layer
(b) • P contains groups of different tissues.
• Erector muscle contains a group of muscle cells which
perform a specific function.
(c) • Integumentary system
• Excretory system
More water is reabsorbed into the blood capillary
by osmosis. The osmotic pressure of the blood is
lowered to the normal level. The urine produced is
concentrated and in a small amount.
4 (a) Ultrafiltration:
• Blood flows from afferent arteriole to efferent arteriole
through glomerulus resulting in a high hydrostatic
pressure in the glomerulus.
• This causes some of the blood plasma with its dissolved
substances to diffuse out from the glomerulus into
Bowman’s capsule.
• The filtrate in Bowman’s capsule is called glomerular
filtrate which contains mainly water, urea, glucose,
amino acids and mineral salts.
Reabsorption:
• Glomerulus filtrate flows into the proximal convulated
tubule.
• At the proximal convoluted tubule, all glucose and
amino acids are reabsorbed from the tubule into blood
capillary by active transport.
• About 75% of the water in the filtrate is reabsorbed
by osmosis.
• Some mineral salts are reabsorbed by active transport.
• Urea is not reabsorbed.
• At the loop of Henle, about 15% of water in the
filtrate is reabsorbed by osmosis.
• Some mineral salts are reabsorbed by active transport.
• Urea is not reabsorbed.
• At the distal convoluted tubule, the amount of water
and mineral salts reabsorbed depends on the osmotic
pressure of the blood.
Secretion:
• Nitrogenous waste products such as urea and uric
acids are secreted out of the blood capillary into the
distal convulated tubule and collecting duct by active
transport.
• Drugs and alcohol are secreted out of blood capillaries
into the tubule by simple diffusion.
• The final content of the filtrate is urine which flows
into the collecting duct and into the pelvis of kidney.
(b) • During the race, more water is lost through sweating.
• Osmotic pressure in the blood increases.
• The high blood osmotic pressure is detected by
osmoreceptor in the hypothalamus.
• Pituitary glands is stimulated to secrete antidiuretic
hormone which increases the permeability of the walls
of the distal convoluted tubule and collecting duct
towards the reabsorption of water.
• More water is reabsorbed from the tubule into the
blood capillaries.
• Adrenal glands are not stimulated to secrete
aldosterone.
• Less mineral salt is reabsorbed.
• This decreases the osmotic pressure of the blood and
returns it to the normal level.
• The urine produced is a small amount and concentrated.
Essay Questions
3 (a) In a cold environment of 5°C, the body reacts through the
negative feedback system in homeostasis to regulate the
body heat and maintain a constant body temperature. The
mechanisms to maintain a constant body temperature are
as follows:
• The arterioles in the skin constrict, causing the
diameter to become smaller. Less blood flows through
the blood capillaries. The blood capillaries are also
further away from the body surface. Hence, heat
loss by conduction and radiation is reduced during
vasoconstriction and the heat is conserved in the body.
• The hair erector muscles contract causing the hairs to
stand erect. A thick layer of air is trapped in between
the hairs. The trapped air is a good insulator of heat.
Heat loss is reduced and the trapped air warms up the
body in the cold environment.
• The sweat glands are inactive. Sweating is reduced.
Hence, latent heat is kept in the body.
• The skeletal muscles contract and relax to produce
shivering. This increases the respiration rate in muscle
cells and more heat is produced to warm the body.
• The adrenal glands are stimulated to secrete adrenaline
which causes an increase in the metabolic rate. More
heat is produced to increase the body temperature to
the normal level.
• The thyroid gland is stimulated to secrete thyroxine
which increases the metabolic rate. As a result, more
heat is produced to raise the body temperature.
(b) (i) Effects of drinking seawater on the body fluids and
cells:
Seawater contains a lot of mineral salt. When
seawater is consumed, the mineral salt is absorbed
into the bloodstream and increases the composition
of salt in the blood. As a result, the osmotic pressure
of the blood increases. When the osmotic pressure
of the blood is higher than the osmotic pressure in
the cells, water is lost from the cells by osmosis.
The cells contract and become dehydrated. In high
osmotic pressure of the blood, erythrocytes will also
lose water by osmosis. The erythrocytes contract and
undergo crenation and unable to transport oxygen.
(ii) When a person is thirsty, the osmotic pressure of the
blood increases. Osmoreceptors in the hypothalamus
are stimulated and nerve impulses are transmitted
to the pituitary gland to stimulate it to release
antidiuretic hormone (ADH) into the blood. ADH
is transported to the kidneys by the blood. ADH
increases the permeability of the walls of the distal
convoluted tubule and collecting duct towards water.
CHAPTER 14: S
UPPORT AND MOVEMENT
IN HUMANS AND ANIMALS
Practice 14.1
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
4 ✓
5 ✗
B. Main concepts and facts
• To provide locomotion
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• To provide shape and support
• To protect internal organs
(b)
Practice 14.2
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
4 ✓
20
21
22
23
24
4 ✓
5 ✗
4 ✗
5 ✓
B. Main concepts and facts
1 triceps, biceps, ulna
2 antagonistic, myotome, tail
Practice 14.4
B. Main concepts and facts
1 Osteoporosis, porous, brittle
2 Arthritis, joints
3 Osteoarthritis, cartilage
4 Gout, uric acid
SPM FOCUS PRACTICE 14
PAPER 1
1 D
6 A
11 C
2 A
7 C
12 D
3 B
8 D
13 C
4 C
9 A
14 C
5 A
10 B
15 B
PAPER 2
Structured Questions
1 (a) Supports the body, allows movement
(b) P – Axial skeleton, Q – Appendicular skeleton
(c) C – Rib, D – Vertebra/Vertebral column
(d) Elbow – hinge joint
Function: allow movement in one plane
Knee – ball and socket joint
Function: allow movement in all planes
2 (a)
Triceps
Y
(c) Thoracic vertebra has two facets, one on the transverse
process and the other at the side of centrum, to articulate
with the rib. This allows the rib to move upwards and
downwards during breathing.
(d) The vertebral column consists of many vertebrae
which are joined together through the facets to form
a flexible vertebral column. This allows the backbone
or vertebral column to bend and to move from side to
side. The cartilage disc between the vertebrae helps to
absorb shock and reduce friction. The vertebrae have
projections to provide surface for attachment of muscles
for movement and to provide support for the body.
4 (a) Bone in Photograph 1.2 is porous or brittle with low bone
mass compared to the bone in Photograph 1.1.
(b) Milk contains calcium which is necessary for the
formation of strong bones.
(c) Rickets, osteomalacia, arthritis (any two).
(d) Having a balanced diet, having a good posture, using
proper attire for daily activities (any one).
5 (a) The fluid in the body acts as a hydrostatic skeleton to
extend the body forward by changing the hydrostatic
pressure in the body and direction of the flow of body
fluid.
(b) Circular muscle and longitudinal muscle
(c) The circular muscles and the longitudinal muscles on the
body wall contract and relax antagonistically. When the
circular muscles contract and the longitudinal muscles
relax the anterior part of body elongates. The hydrostatic
pressure is transferred from the anterior to the posterior.
This causes the body fluid to flow to the posterior end.
When the longitudinal muscles contract, the anterior part
is shortened to pull the posterior end forward.
(d) By using the chaetae to anchor in the posterior part of
the body
(e) Leech and caterpillar
Practice 14.3
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
X
5 ✗
B. Main concepts and facts
1 (a) Appendicular skeleton
(b) Skull
(c) Ribcage
(d) Pelvic girdle
(e) Lower limb
2 (a) P: Biceps muscles
Q: Tendon
R: Cartilage
S: Synovial membrane
T: Ligament
U: Synovial fluid
(b) (i) Biceps muscles/muscles
(ii) Tendon
(iii) Ligament
(iv) Synovial fluid, synovial membrane
(v) Cartilage
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
1
2
3
4
5
6
7
Essay Questions
6 (a) Cervical vertebrae – 7 at the neck, thoracic vertebrae –
12 at the thorax, vertebra lumbar – 5 at the waist, sacrum
– 5 sacral vertebrae fused together at the hip, coccyx – 4
caudal vertebrae fused together at the tail.
(b) (i) Hinge joint.
Cartilage – to absorb shocks and reduce friction
between the bones at the joint.
Ligaments – join bone to bone at the joint.
Synovial membrane – secretes synovial fluid.
Synovial fluid – acts as a lubricant to reduce friction
at the joint.
Biceps
Ligament
Tendon
(b) X: Humerus Y: Radius Z: Ulna
(c) Refer 2(a)
3 (a) X: Cervical vertebra
Y: Lumbar vertebra
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Practice 15.2
(ii) • Biceps muscle contracts while triceps muscle relaxes.
• A pulling force is produced by the contraction of
biceps muscle.
• This force is transmitted through the tendon to the
radius.
• The radius is pulled upwards and causes the arm
to bend at the elbow.
• The forearm moves upwards.
(c) When standing, the head, body and limbs should be in a
vertical line, with the vertebral column nearly parallel to
the vertical axis. When lifting heavy objects, squat with
the arms extended out straight to the object but with the
head and body straight. As the object is being lifted up, the
thorax and abdomen should remain straight and upright.
Humerus Coracoid
7 (a)
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
B. Main concepts and facts
1
2n
Spermatogonium
Meiosis I
n
n
Secondary
spermatocyte
n
Spermatid
Meiosis II
n
n
n
Differentiation
Sperm
Sternum
Pectoralis major
2 P: Primary follicle
Q: Secondary follicle
R: Graafian follicle
S: Secondary oocyte
T: Corpus luteum
X: Ovulation
Birds fly either by flapping their wings due to the
antagonistic action of breast muscles: pectoralis major
and pectoralis minor. When the bird is moving its wings
downwards, the pectoralis major, which is large and
strong contracts. The air resistance produced as a result
of moving the wings down provide an upthrust on the
wings. The upthrust is transmitted from the wings to the
coracoid until the sternum. As a result, the whole body is
lifted upwards. When the pectoralis minor contracts, the
pectoralis major relaxes. Air resistance is low. The wings
are pulled upwards to return to the starting position.
(b) • Fish has an endoskeleton for the attachment of muscles.
• The muscles in fish are in segmental blocks called
myotomes on both sides of the flexible backbone
• Fish moves forward due to the contraction and
relaxation of myotomes on either side of the body.
• When the myotomes on the right contract, the
myotomes on the left relax.
• The body bends and swept the tail to the right.
• Alternate contractions of right and left myotomes
cause the body and the tail to sweep from side to side.
• This produces a forward thrust that propels the fish
forward in a straight path.
• A lateral thrust is produced in the opposite direction,
so it is cancelled off.
• Fish has fins to balance the body in the water and to
control the direction of movement.
Practice 15.3
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
6 ✗
7 ✓
8 ✓
B. Main concepts and facts
1
Hormone
Site of secretion
CHAPTER 15: SEXUAL REPRODUCTION,
DEVELOPMENT AND
GROWTH IN HUMANS AND
ANIMALS
4 ✗
4 ✗
9 ✓
5 ✓
10 ✗
Function
Folliclestimulating
hormone
Pituitary gland
Stimulates the
development of
follicle in the
ovary
Oestrogen
Follicle cells in
the ovary and
tissues of the
ovary
Repairs and
thickens the
endometrium of
the uterus
Luteinising
hormone (LH)
Pituitary gland
Stimulates
ovulation
Progesterone
Corpus luteum
Thickens and
maintains
thickness of the
endometrium
2 (a)
Practice 15.1
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
5 ✓
Growth and
development
Primary
spermatocyte
2n
Wings
move downwards
Upward
thrust
4 ✗
Thickness of
endometrium
5 ✓
B. Main concepts and facts
Fertilisation
Male reproductive system
Female reproductive system
Seminal vesicle
Prostate gland
Scrotum
Testis
Penis
Sperm duct
Urethra
Uterus
Ovary
Cervix
Vagina
Fallopian tube
0
5 14
28
14
28 14
28
First
Second
Third
month
month
month
Time
(day)
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B. Main concepts and facts
1 sigmoid growth curve, lag phase, exponential phase, slow
growth phase, stationary phase, senescence phase, death phase
2
Adult
(b)
Hormonal level
Length of the body (mm)
Progesterone
Oestrogen
0
5 14
28
14
28
First
Second
month
month
Time
(day)
14
28
Third
month
Instar
4
Instar
3
Instar
2
Instar
1
Egg
1st ecdysis
Practice 15.4
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
4 ✗
Time interval for each stage (day)
5 ✓
SPM FOCUS PRACTICE 15
B. Main concepts and facts
1 Fertilisation → Zygote → Morula → Blastocyst → Embryo
→ Foetus → Baby
2 uterine, exchange, Oxygen, nutrients, carbon dioxide, urea
3 Umbilical vein, foetus, Umbilical artery, the foetus, the
placenta
PAPER 1
1
6
11
16
4 ✗
B. Main concepts and facts
1 one ovum, one sperm, one, completely
2 two ova, two, two, placenta
3
• Both involve the fertilisation of sperm with ovum
• Both involve mitosis in the development of the zygote
Two ova are fertilised by two
different sperms
Both foetus share one
placenta
Each foetus has its own
placenta
Both twins look alike
Both have some similar and
some different characteristics
The zygote divides after
fertilisation
The zygote does not divide
after fertilisation
Both twins are of the same
sex
Both twins may or may not
be of the same sex
Practice 15.6
A. Understanding key ideas
1 ✗
2 ✗
3 ✓
4 ✓
5 C
10 B
15 B
200
175
150
125
100
75
50
25
Adult
5
4
3
2
0 stage
1
1st ecdysis
(b) (i) A series of steps
(ii) This is the egg stage, where the egg has not hatched
and the length is the length of the egg.
(c) (i) Nymph
(ii) Ecdysis
(iii) During ecdysis, air is sucked in to expand the body
and to break the hard exoskeleton. Before the new
exoskeleton hardens, the insect grows rapidly to
increase its size.
Practice 15.7
4 ✗
D
B
B
A
25 50 75 100 125 150175 200 225250 275300 325 350
Time interval for each stage (day)
5 ✗
B. Main concepts and facts
1 Artificial insemination and sperm bank
2 • In man – blocked sperm ducts and low sperm count
• In woman – blocked Fallopian tubes and no ovulation
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
4
9
14
19
A
A
D
A
(b) • R rebuilds and thickens the uterine wall.
• S maintains the thickness of the uterine wall.
(c) (i) Day 12th –16th
(ii) The embryo will be implanted onto the uterine wall.
These uterine tissues and the embryonic tissues will
form the placenta that allows the exchange of gases,
nutrients and waste products.
2 (a)
Length of the body (mm)
One ovum is fertilised by one
sperm
3
8
13
18
28 1 2
26 27
3
4
25
5
24
6
23
7
22
8
21
9
20
Ovulation
10
19
11
18
12
17
16 15 14 13
Similarities
Fraternal twins
A
B
C
A
Structured Questions
1 (a) (i) P: Follicle-stimulating hormone
Q: Luteinizing hormone
R: Oestrogen
S: Progesterone
(ii)
Menstruation period
5 ✓
Identical twins
2
7
12
17
A
C
A
B
PAPER 2
Practice 15.5
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
Instar
5
5 ✓
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3 (a) X: 4
Y: 2
(b) Cell X undergoes meiosis to produce cell Y which is
haploid. Cell Y receives half the number of chromosomes
from cell X.
(c) Spermatogenesis
4 (a) P: Secondary follicle
Q: Graafian follicle
R: Secondary oocyte
S: Corpus luteum
(b) N: Meiosis I
O: Meiosis II
Size of gamete produced
(secondary oocyte) is
large
Sperm produced has
head, middle piece and
tail structure
Secondary oocyte
produced is round in
shape
Sperm produced is
mobile
Secondary oocyte
produced is not mobile
The process starts before
birth but stops for
several years at prophase
I until at puberty
CHAPTER 1: O
RGANISATION OF PLANT
TISSUES AND GROWTH
Practice 1.1
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
4 ✓
5 ✓
B. Main concepts and facts
1 collenchyma, photosynthesis, store, collenchyma tissue,
sclerenchyma tissue
2 (a) Divides by mitosis to produce new cells
(b) Transports water and minerals from roots to all parts
of the plant
(c) Protects underlying cells
(d) Provides support to the plant
(e) Transports dissolved organic substances from leaves
to all parts of the plant
(f) Carries out photosynthesis and stores food
Practice 1.2
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
4 ✓
5 ✗
6 ✓
B. Main concepts and facts
1 tip of shoots, tip of roots, apical meristem, height
2 lateral meristem, vascular cambium, cork cambium,
circumference, secondary xylem tissues, secondary phloem
tissues, secondary cortex
3 The plant undergoes secondary growing
Secondary
xylem
Differences
Size of gamete (sperm)
produced is small
Meiosis I and II
occur continuously in
spermatogenesis
FORM 5
• Both process take place in the reproductive organ
• Both process involve meiosis
• Both process produce new cells which have half
the numbers of chromosomes of the parent cell
Oogenesis
Only one gamete
produced in oogenesis is
involved in fertilisation
(c) • After fertilisation, a zygote is formed.
• As a zygote develops, it moves down the Fallopian
tubes due to the peristalsis process of the wall of
Fallopian tubes.
• The zygote undergoes repeated mitosis to form
two-celled embryo, four-celled embryo, eight-celled
embryo, sixteen-celled embryo until a morula is formed.
• The morula at the end of the Fallopian tubes moves
down into the uterus.
• The morula then transforms into a fluid-filled cell
called blastocyst in the uterus.
• The blastocyst is implanted in the endometrium and
develops into an embryo.
Essay Questions
5 (a) Oogenesis is the process of secondary oocyte formation
in the ovary. It begins in the ovary of a female foetus.
The primordial germ cells divide repeatedly through mitosis
to form diploid oogonia (2n). Each oogonium grows and
develops into a primary oocyte (2n). The primary oocyte
undergoes meiosis I and completes meiosis I at puberty,
to form two haploid cells; a secondary oocyte and polar
body. During ovulation, the secondary oocyte is released
from the ovary. When fertilisation occurs, the secondary
oocyte completes meiosis II to form an ovum (n) and
a polar body (n). The first polar body also undergoes
meiosis II to form another two haploid polar bodies. All
three polar bodies will eventually degenerate.
(b) (i) After ovulation, if fertilisation does not occur, the
thickened endometrium of the uterus will breakdown.
The excess blood and the endometrium tissues together
with the unfertilised egg will leave the uterus through
the vagina. This process is called menstruation and it
usually lasts for three to seven days.
(ii) After ovulation, if fertilisation occurs, the
endometrium continues to thicken due to the
increasing progesterone level produced by the corpus
luteum. The zygote continues to develop until a
blastocyst is formed and implanted into the thickened
endometrium. The blastocyst then develops into an
embryoand also formed the placenta.
6 (a) Spermatogenesis is the formation of sperms in the testes
by meiosis. It occurs in the seminiferous tubule of the
testes. It begins with the primordial germ cells (2n) in
the germinal epithelial of the wall of seminiferous tubule.
Primordial germ cells divide by mitosis repeatedly to
form many spermatogonia (2n). Spermatogonia then
grow to form primary spermatocytes (2n). The primary
spermatocyte undergoes meiosis I to form two haploid
secondary spermatocytes (n). The secondary spermatocyte
undergoes meiosis II to form haploid spermatids (n). The
spermatids differentiate to form four haploid sperms.
(b)
Similarities
Spermatogenesis
All gametes produced
in spermatogenesis are
involved in fertilisation
Primary
xylem
Primary
phloem
Secondary
phloem
Cambium ring
Practice 1.3
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
4 ✓
5 ✗
B. Main concepts and facts
1 (a) Plants that can live from year to year
(b) Plants that can live for one year
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(c)
2 (a)
(b)
(c)
Plants that can live for two years
Perennial plants
Biennial plants
Annual plants
and the height of the plant also increases. Thus, growth
occurs and the process is irreversible.
(b) Secondary growth in the eudicot stem
The vascular cambium in the vascular bundle divides
laterally to form a ring of cambium. The vascular
cambium cells in the vascular bundle divide to
produce new cells. The cells in the inner part of the
vascular cambium differentiate to form secondary
xylem, while the cells in the outer part of the vascular
cambium form the secondary phloem. As a result, the
primary xylem is pushed towards the pith while the
primary phloem is pushed towards the epidermis. The
cambium cells in the ring divide to form the secondary
parenchyma cells. The cells of the cork cambium
below the epidermis divide to form an outer layer of cork
cells and an inner secondary cortex. The addition of all the
secondary tissues causes the circumference of the stem to
increase and the epidermis to be stretched and split.
(c) (i) Three differences between primary and secondary
growth
SPM FOCUS PRACTICE 1
PAPER 1
1 A
6 C
11 D
2 C
7 B
12 D
3 B
8 C
13 C
4 B
9 C
5 A
10 B
PAPER 2
Structured Questions
1 (a) Tissue W is the apical meristem.
It divides actively through mitosis to produce new cells
to increase the length of the shoot.
(b) (i) Type of growth: Primary growth
(ii) Significant change: Shoot increases in height
(c) (i) Lateral meristem consists of the vascular cambium
and the cork cambium
(ii) Secondary growth in stem produces more xylem and
phloem tissues to provide additional transport to the
plant. The xylem tissues with its walls thickened
with lignin also provide additional support to sustain
the extra weight of the plant. Increase in diameter of
stem provides support and stability to the plant.
2 (a) Zone B = Zone of cell differentiation
Zone C = Zone of cell elongation
Zone D = Zone of cell division
(b) Mitosis occurs at zone D. The apical meristem divides
to produce new cells.
(c) (i) Primary growth
(ii) It causes an increase in the length of the roots to
enable the roots to grow deep into the ground to
absorb water and mineral salts.
3 (a) In stage X, stored food in the seed is used to provide
energy for the germination of seed. This is indicated by
the small decrease in mass followed by an increase. when
the seed has germinated and growth rate increases. In
stage Y, the young plants have green leaves which can
carry out photosynthesis. The dry mass increases and the
growth rate is high.
(b) Annual plants
(c) The growth curve is S-shaped known as sigmoid curve.
It starts with germination of seed. The dry mass decreases
because food stored is used for germination.
Growth rate then rapidly increases as green leaves
develop and the plant starts to carry out photosynthesis.
This causes the dry mass to increase.
The plant reaches its maturity stage and is at its maximum
size. Dry mass is constant and growth rate is zero.
Finally ageing process occurs. Dry mass decreases
because there is shedding of leaves, flowers and seed
and less photosynthesis occurs.
Primary growth
Secondary growth
Involves the apical
meristem
Involves the lateral
meristem
Occurs at the shoot
tip and root tip of
plants
Occurs at the stem
Increases the length
of shoot and root of
plants
Increases the diameter/
circumference of stem
and root
(ii) Benefits to plants that undergo secondary growth
compared to those that do not undergo secondary
growth
Essay Questions
4 (a) The meaning of growth
At the shoot tip in zone 1 and zone 2, the meristem cells
divide by mitosis to produce new cells. This causes the
number of cells to increase at the shoot tip. At zone
3, the cells elongate. New vacuoles are formed and the
vacuoles are enlarged. This causes the cells to increase
in size. At zone 4, the cells differentiate and undergo
cell specialisation to become special cells to carry out a
specific function. The length of the shoot tip increases
Plants with
secondary growth
Plants without
secondary growth
The plants have a
longer life span. Thus,
they are able to bear
fruits and produce
many offspring.
The plants have a
shorter life span.
They usually die
after bearing fruits.
Thus, they have few
offspring.
The plants are better
adapted for survival
because they are taller
and bigger. They
have better access
to get maximum
sunlight. They have
more xylem and
phloem to give
additional strength
and support to the
plants and additional
transportation of
water, minerals and
food. They have cork
tissues to protect the
internal tissues.
The plants are less
adapted for survival
because they are
shorter and smaller.
They have less access
to obtain sunlight.
They have less xylem
and phloem, hence
cannot give additional
strength and support.
They have basic
transportation of
water, minerals and
food. They do not
have a cork layer to
protect the internal
tissues.
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The plants are
woody plants of high
economic values as
they can be used as
timber.
The plants are
herbaceous plants or
non woody plants,
not beneficial in the
timber industry.
Has strong and hard
wood, can last longer
and widely used in
the wood industry.
The wood is not hard,
cannot last long and
is not used in the
wood industry.
The plants can live
for a long time, thus
needs no replanting
The plants live for
only a short time,
thus needs replanting.
• In dim light, both photosynthesis and respiration
occur. The amount of carbon dioxide released during
respiration is equivalent to the amount of carbon dioxide
used up during photosynthesis. This is because the rate
of photosynthesis is equal to the rate of respiration.
• In bright light, the absorption of carbon dioxide is more
than the release of carbon dioxide. This is because
the rate of photosynthesis is faster than the rate of
respiration. Hence more carbon dioxide is needed.
SPM FOCUS PRACTICE 2
PAPER 1
1 A
6 B
11 B
CHAPTER 2: LEAF STRUCTURE AND
FUNCTION
4 ✓
5 ✗
4 ✓
5 ✓
Practice 2.2
B. Main concepts and facts
close, photosynthesis, decreases, guard cells, epidermal cells,
osmotic pressure, guard cells, epidermal cells, flaccid
Practice 2.3
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
4 ✗
5 ✓
4 ✓
9 ✓
5 ✓
10 ✓
B. Main concepts and facts
1 loss, water, water vapour, stomata
2 (a) increases
(b) increases
(c) increases
(d) decreases
Practice 2.4
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
6 ✗
7 ✗
8 ✓
B. Main concepts and facts
1 (a) Chlorophyll
(b) Hydroxyl ions
(c) Hydrogen ions
(d) Water
(e) Oxygen
(f) Hydrogen atom
(g) Glucose
2 During the light-dependent reaction, photolysis of water occurs.
The water is split into hydroxyl ions and hydrogen ions. The
hydroxyl ions are neutralised to form oxygen and water.
The hydrogen ions are neutralised to form hydrogen atom.
During the light-independent reaction, reduction of carbon
dioxide occurs. The hydrogen atom from the light-dependent
reaction is used to reduce carbon dioxide to form glucose.
Practice 2.5
A. Understanding key ideas
1 ✗
2 ✗
3 ✓
4 ✓
4 A
9 C
14 A
5 A
10 B
15 B
Structured Questions
1 (a) (i) Chloroplast
(ii) Palisade mesophyll cells, spongy mesophyll cells,
guard cells
(b) Organelle R carries out photosynthesis. The cells in the
leaves that have organelle R are able to synthesise food
which can be used by the plant or stored in the plant.
(c) (i) Granum/Thylakoid
(ii) Y contains chlorophyll to absorb energy from the
sunlight for photosynthesis.
(d) The light-dependent reaction occurs in Y.
Photolysis of water occurs. The energy from the sunlight
is absorbed by the chlorophyll to split the water molecules
and form hydroxyl ions and hydrogen ions. Water and
oxygen are produced in the light-dependent reaction. 2 (a) Oxygen
(b) • Increase the concentration of carbon dioxide in the water
by dissolving some sodium bicarbonate in the water.
• Increase the intensity of the light.
• Increase the temperature of the water.
(c) To allow the water weed to adjust to the new conditions
and the chemical reaction in photosynthesis to be in an
equilibrium state. Therefore, the rate of oxygen produced
is more stable, producing more accurate results.
(d) The distance for the water meniscus in the glass tube
to move over a fixed time can be measured more
accurately in millimetres using a ruler as the glass tube is
narrow. The rate of the production of oxygen is directly
proportional to the rate of photosynthesis. The rate of
photosynthesis can be calculated as follows:
Rate of photosynthesis = distance moved by the meniscus
in the glass tube (mm) time taken (sec)
This method allows the rate of photosynthesis to be
measured with greater accuracy and also enables results
to be obtained more quickly.
(e) No gas is released. Photosynthesis cannot occur because
at 75°C, the enzymes for photosynthesis are denatured.
3 (a) (i) Carbon dioxide and oxygen
(ii) Gaseous exchange occurs through the stomata in the
leaves and lenticels in the stem by simple diffusion.
In the leaves, oxygen diffuses through the stomata
into the intercellular air spaces. Oxygen dissolves in
the layer of moisture on the surface of the cells and
diffuses into the cells for aerobic respiration.
At the same time, carbon dioxide produced
during respiration diffuses out of the cells into the
intracellular air spaces and then diffuses out into the
atmosphere through the stomata.
B. Main concepts and facts
1 veins, phloem tissue
2 water, mineral salts, organic, leaves
A. Understanding key ideas
1 ✗
2 ✓
3 ✗
3 D
8 B
13 B
PAPER 2
Practice 2.1
A. Understanding key ideas
1 ✗
2 ✓
3 ✗
2 B
7 D
12 A
5 ✓
B. Main concepts and facts
1 oxygen, carbon dioxide, simple diffusion
2 • In the dark, that is, low light intensity, only respiration
occurs. Carbon dioxide is released to the atmosphere.
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(b) The factors that affect the rate of transpiration are:
• Temperature
High temperature increases the rate of evaporation
of water from the mesophyll cells and consequently
increases the water loss through stomata. Hence
increases rate of transpiration.
• Light intensity
In high intensity of light, the stomata opens and
water is easily lost from the leaf. Hence, the rate of
transpiration is higher.
• Relative air humidity
In conditions with high relative air humidity, less
water is evaporated and lost through the stomata.
Hence, the rate of transpiration is low.
• Air movement
In moving air or windy conditions, the rate of
transpiration is high while in still air, the rate of
transpiration is low.
(Any two)
(c) Water is lost through the stomata during transpiration.
Stoma opens or closes to control the rate of transpiration.
Stoma opens when the guard cells are turgid to increase
rate of transpiration and closes when guard cells are
flaccid to decrease rate of transpiration.
the reduction of carbon dioxide by hydrogen atom to
produce glucose
• The overall photosynthesis rate will be greatly
increased
CHAPTER 3: N
UTRITION IN PLANTS
Practice 3.1
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
6 ✗
7 ✗
8 ✓
4 ✓
5 ✓
4 ✓
5 ✓
B. Main concepts and facts
1 carbon dioxide, water
2 magnesium, nitrogen, iron
Practice 3.2
A. Understanding key ideas
1 ✗
2 ✓
3 ✗
B. Main concepts and facts
root hairs, surface area, osmosis, active transport
Practice 3.3
A. Understanding key ideas
1 ✗
2 ✗
3 ✓
Essay Questions
4 (a) (i) Photosynthesis is the process where glucose is
synthesised in the leaves from water and carbon
dioxide in the presence of chlorophyll and sunlight.
(ii) • Photolysis of water is the light-dependent reaction
of photosynthesis
• Light energy is absorbed by chlorophyll to break
down water molecules to produce hydrogen ions
and hydroxyl ions.
• Each hydroxyl ion is neutralised to form a
hydroxyl group. Four hydroxyl groups combine
together to form water and oxygen.
• Each hydrogen ion is neutralised to form a
hydrogen atom.
• Light energy is converted to chemical energy.
• Chemical energy is used to break down water
molecules to produce hydrogen ions and hydroxyl
ions.
• A lot of hydroxyl ions will combine with each
other to produce water and oxygen.
• Reduction of carbon dioxide is the lightindependent reaction of photosynthesis.
• Hydrogen atom reduces carbon dioxide in a series
of reactions catalysed by enzymes to produce
glucose.
• Overall equation:
Water
sunlight
Glucose +
+
Carbon dioxide chlorophyll Oxygen + Water
4 ✗
5 ✓
B. Main concepts and facts
Carnivorous plant
Parasitic plant
Epiphytic plant
Obtains nutrients
by trapping and
digesting its prey
Obtains all its
nutrients from
other plants by
using haustorium
Obtains nutrients
from its
surrounding
Can carry out
photosynthesis
Cannot carry out
photosynthesis
Can carry out
photosynthesis
Does not have host
plant
Has host plant
Has host plant
SPM FOCUS PRACTICE 3
PAPER 1
1 D
6 B
2 D
7 D
3 B
8 D
4 C
9 B
5 D
PAPER 2
Structured Questions
1 (a) To determine the type of nutrients involved in plant growth
(b) (i) To destroy the microorganisms on the apparatus and
in the culture solution
(ii) To prevent the growth of algae in the culture solution
(iii) To supply the roots of the seedling with sufficient
oxygen for respiration
(c) To replace the nutrients that have been absorbed by the
seedling
(d) (i) Area with sufficient sunlight
(ii) To enable the seedling to carry out photosynthesis
2 (a) (i) To absorb water and mineral salts from the soil
(ii) It is numerous and small in size to provide a large
surface area to increase the rate of absorption of
water and mineral salts.
(b) (i) X: Carnivorous plant
Y: Parasitic plant
Z: Epiphytic plant
(ii) X and Z
(iii) It has a snap trap which can catch and trap prey.
Enzymes are secreted to digest its prey into simple
molecules that can be absorbed.
(b) • To increase crop yield: need to increase the rate of
photosynthesis
• This will increase the production of glucose and crop
production.
• External factors that affect photosynthesis are light
intensity and concentration of carbon dioxide
• Light intensity can be controlled by placing light
sources at the farm especially at night
• So that photolysis of water can occur continuously
• Concentration of carbon dioxide can be increased by
adding sources of carbon dioxide
• This will increase the supply of carbon dioxide for
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SPM FOCUS PRACTICE 4
Essay Questions
3 (a) Parasitic plant. The plant (Rafflesia sp.) has a long
modified root called haustorium that is extended into
a living host. The haustorium can penetrate into the
stem and fuse with the vascular tissues in the stem. The
haustorium can absorb water and mineral salts from
the xylem tissues of the host. It also absorbs dissolved
organic substances from the phloem tissues in the stem.
In this way, it obtains nutrients from its living host and
causes harm to the host in the process.
(b)
Carnivorous
plant
Parasitic plant
Epiphytic
plant
Obtains nutrients
by trapping and
digesting its prey
Obtains all its
nutrients from
other plants by
using haustorium
Obtains
nutrients
from its
surrounding
Can carry out
photosynthesis
Cannot carry out
photosynthesis
Can carry out
photosynthesis
Does not have
host plant
Has host plant
Has host plant
PAPER 1
1 A
6 A
11 B
4 ✓
5 ✗
4 ✓
5 ✓
B. Main concepts and facts
xylem vessel, tracheid, hollow, lignin
Practice 4.2
A. Understanding key ideas
1 ✗
2 ✓
3 ✗
B. Main concepts and facts
(a) Cortex cells in root
(b) osmosis
(c) root pressure
(d) capillary action
(e) transpiration pull
(f) Spongy mesophyll cells
(g) osmosis
(h) transpiration
(i) transpirational pull
Practice 4.3
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
5 A
10 A
Essay Questions
3 (a) • Tissue X is xylem tissue which consists of xylem
vessel and tracheid to transport water and mineral
salts. Tissue Y is phloem tissue which consists of
sieve tube and companion cell to transport dissolved
organic substances.
• Adaptation of tissue X for transport:
– Xylem vessel is a fine, long and hollow continuous
tube (like capillary tube), from the roots through
the stem to the leaves.
– This enables the xylem vessel to transport a
continuous column of water and mineral salts from
the roots to the leaves.
– The cell wall of xylem vessel is thickened with
lignin to provide mechanical support to the xylem
vessel and the plant.
– Tracheid is a long, narrow, hollow tube with
tapering closed ends.
– The cell wall of tracheid is strengthened with lignin
and has small openings called pits.
– The narrow tube enables water and minerals to flow
from one tracheid to another tracheid through the pits.
• Adaptation of tissue Y for transport:
– Sieve tubes are joined to form a long continuous
tube from the root to the leaves. Each sieve tube
has streamlike cytoplasm which can flow from one
sieve tube to another through the sieve plate.
– Dissolved organic substances in the form of sucrose
and amino acids are carried in the cytoplasm through
the sieve tubes from the leaves to all parts of the plant.
– Companion cell which is always beside a sieve tube
has many mitochondria.
Practice 4.1
5 ✗
4 C
9 B
Structured Questions
1 (a) Organ
(b) P: Epidermal tissue
Q: Vascular tissue
R: Ground tissue
(c) Xylem and phloem
2 (a) (i) K: Sieve tube
L: Companion cell
(ii) • Sieve tube/K are joined to form a long continuous
tube from the root to the leaves. Each sieve tube
has streamlike cytoplasm which can flow from
one sieve tube to another through the sieve plate.
• Dissolved organic substances in the form of
sucrose and amino acids are carried in the
cytoplasm through the sieve tubes from the leaves
to all parts of the plant by active transport.
• Companion cell which is always beside a sieve
tube has many mitochondria. Companion cell
supplies energy to the sieve tube for the transport
of these substances by active transport.
(b) (i) S: Swollen
T: Shrivelled/Reduced in diameter/shrink
(ii) • The organic substances accumulate above the
ring of bark as the phloem tissue is removed and
causes S to swell or to increase in diameter.
• The organic substances cannot be transported
through the ring of bark to T and causes T to shrivel.
CHAPTER 4: TRANSPORT IN PLANTS
4 ✓
3 C
8 C
13 B
PAPER 2
(c) (i) Active transport
(ii) The concentration of P is lower in the soil water than
in the root hair cell. P diffuses from the soil into the
root hair cell against the concentration gradient. The
movement of P into the root hair cell requires energy.
(iii) There are numerous root hair cells which are small in
size. This increases the surface area for absorption of
mineral salts from the soil.
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
6 ✓
7 ✗
2 A
7 D
12 B
B. Main concepts and facts
(a) Source: Tuber
Sink: Shoot
(b) Sucrose is transported by the sieve tube of the phloem
tissue from the cells in the tuber to the shoot for respiration
and growth.
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(b) •
•
•
•
•
•
•
•
Practice 5.3
– The transport of dissolved organic substances by
the sieve tube is by active transport which requires
energy. Companion cell supplies energy to the
sieve tube for the transport of these substances.
Water is absorbed from the soil by the root hairs.
Water from the dilute soil solution with high water
potential diffuses into the more concentrated cell sap of
the root hair cells (low water potential) by osmosis.
This causes the root hair cells to have a lower osmotic
pressure compared with the adjacent cells of the root
cortex.
As a result, water diffuses into the adjacent cortex
cells by osmosis and finally across the root cortex to
xylem tissue of the root.
As water diffuses across the cortex cells in the root,
the root cells become turgid.
This generates a root pressure which causes the water
to move into the xylem tissues in the root and up to
the stem. The water is then transported by the xylem
tissues in the stem up to the leaves by two types of
forces: capillary action and transpirational pull.
The cohesion and adhesion force of water in the xylem
tissue of the stem produce capillary action which
causes water to flow up the stem.
Transpirational pull is a suction force produced at the
leaves as a result of transpiration which pulls up a
continuous column of water from the roots toall the
leaves on the plant.
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
4 ✓
5 ✓
B. Main concepts and facts
1. As a weed killer
The hormone auxin is used as a herbicide (weed killer) in
high concentration to kill eudicot weeds while monocots
are unharmed.
2. In synchronised fruiting
• Hormones such as auxin and ethylene are sprayed onto
fruits, to make the fruits develop and ripen at the same
rate.
• This controls fruiting and allows growers to harvest
economically with machinery.
3. To produce seedless fruits (parthenocarpy)
• Auxins are sprayed on ovaries of flowers
• This stimulate the ovaries to develop into seedless fruits
without fertilisation taking place.
SPM FOCUS PRACTICE 5
PAPER 1
1 D
6 B
11 D
2 C
7 D
12 A
3 C
8 D
13 A
4 C
9 A
14 C
5 B
10 C
15 C
PAPER 2
Structured Questions
1 (a)
CHAPTER 5: RESPONSE IN PLANTS
Practice 5.1
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
Unilateral
light
4 ✗
5 ✓
1
B. Main concepts and facts
Tropism
Nastic response
Responds to directional
stimuli
Responds to nondirectional stimuli
Response is irreversible
Response is reversible
It is a growth response
It is not a growth response
Response depends on
direction of stimulus
Response does not depend
on direction of stimulus
Occurs at the shoots and
roots
Occurs at any part of the
plant
Stimuli are light, gravity,
water, touch, chemical
substances
Stimuli are light, day and
night, contact, temperature,
vibration
Practice 5.2
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
6 ✓
7 ✗
4 ✓
2
3
4
5
(b) (i) Seedling 1
(ii) The seedling grows and bends towards the light.
(c) In seedling 2, the coleoptile growth is retarded. In
seedling 3, the coleoptile grows straight and upwards
a little. In seedling 4, the coleoptile grows upwards
and bends towards the light. In maize seedling 5, the
coleoptile growth is retarded.
2 (a) Tropism is a growth response of the shoot tip and root
tip towards or away from an external stimulus
(b) Positive phototropism
(c) The seedling is able to grow towards light to enable the
leaves to obtain maximum light for photosynthesis.
(d) • Auxin is produced in the apical meristem of the tip of
seedling F. Auxin diffuses down into the zone of cell
elongation.
• More auxin accumulates in the shaded region.
The concentration of auxin in the shaded region is
higher than in the lighted region. A high concentration
of auxin in the shoot tip stimulates cell elongation.
• Hence, the cells in the shaded region with a higher
concentration of auxin elongate faster than the cells
in the lighted region. This causes the shoot to bend
and grow towards the unidirectional light.
5 ✗
B. Main concepts and facts
1 positive geotropism, water, mineral salts, photosynthesis,
turgor pressure, negative geotropism, light, photosynthesis
2
Essay Questions
3 (a) • Thigmotropism – Response to touch
• Geotropism – Response to gravity
• Phototropism – Response to light
• Hydrotropism – Response to water
• Chemotrophism – Response to chemical substance
Unilateral
light
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30 MASTERCLASS 2023 SPM BIOLOGY-ANSWERS-(BAHAN QR)-ARUL 1P.indd 30
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B. Main concepts and facts
1
(b) (i) When a shoot is exposed to light from one direction,
the shoot grows and bends towards the light showing
positive phototropism. The main stages in the
response are as follows:
• Auxin is produced in the shoot tip and moves into
the zone of cell elongation.
• At the zone of cell elongation, more auxin moves
to the shaded side, away from the light.
• More auxin accumulates in the shaded side
resulting in a higher concentration of auxins in the
shaded region than in the region exposed to light.
• The cells in the shaded region of the shoot
elongate more than the cells in the other region.
• As a result, the shoot grows and bends towards
the direction of the light showing positive
phototropism.
(ii) Auxin is produced at the root tip and moves to
the lower side of root. More auxin accumulates
on the lower side of the root, resulting in a higher
concentration of auxin in the lower side. A high
concentration of auxin in the root inhibits elongation
of cells. Hence the cells on the lower side of the root
grow slower than the cells on the upper side. As a
result, the root grows and bends downwards towards
the pull of gravity showing positive geotropism.
(iii) Commercial values of auxin in agriculture:
• Promote growth of plants
• Produce seedless fruits in parthenocarpy
• As a weedkiller when in high concentration
(Any two)
(c) The benefits of tropism of plants:
• Positive phototropism of the shoots enables the shoots
to grow towards sunlight and absorb as much sunlight
as possible for photosynthesis.
• Positive geotropism of the roots will ensure the roots
grow downwards into the soil due to the pull of
gravity. This enables the roots to absorb water and
mineral ions as well as provide anchorage for the
plant.
Meiosis
Pollen mother
cell (2n)
Tube
nucleus
Matured pollen
grains
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
It is the male reproductive
organ of a flower
It is the female reproductive
organ of a flower
Consists of filament and
anther
Consists of stigma, style
and ovary
Produces pollen grains
Produces embryo sac
The anther releases pollen
grains to the environment
The stigma receives pollen
grains
Polar nuclei
Ovum (egg cell)
Synergid cells
3 ovum or egg cell, embryo, two polar nuclei, triploid
endosperm nucleus, endosperm
Practice 6.4
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
6 ✓
7 ✓
8 ✓
4 ✓
4 ✗
5 ✗
4 ✓
5 ✗
B. Main concepts and facts
1 (a) fruit
(b) seed
(c) embryo
(d) endosperm
(e) seed coat/testa
2 (a) M
(b) K
(c) N
(d) L
Practice 6.5
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
B. Main concepts and facts
• Seeds can store food for the embryo to grow into a new
plant
• Seeds can be dispersed to new location to ensure them to
colonise a larger areas.
• Seeds can remain dormant in unfavourable condition for
long time to ensure their survival.
Practice 6.2
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
5 ✓
Male nuclei
B. Main concepts and facts
Carpel
4 ✓
B. Main concepts and facts
1 Stigma → Style → Ovary wall → Micropyle → Embryo
sac
2
Antipodal cells
5 ✓
Stamen
Four
microspore (n)
Practice 6.3
Practice 6.1
4 ✓
Generative
nucleus
2 R, T, P, S and Q
CHAPTER 6: SEXUAL REPRODUCTION IN
FLOWERING PLANTS
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
Tetrad of
haploid cells
5 ✗
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SPM FOCUS PRACTICE 6
• The other three haploid cells in the embryo sac
become the antipodal cells.
PAPER 1
1
6
11
16
B
C
B
D
2 C
7 B
12 B
3 B
8 D
13 C
4 A
9 B
14 D
5 B
10 A
15 B
Nucellus
cell
Ovule
meiosis
develops
PAPER 2
Young embryo sac
Haploid cells
Nucleus
degenerate
undergoes
mitosis Four haploid cells
Four nuclei three
Three antipodal cells
times
at each
Polar
end
nuclei
Two polar
nuclei
Egg
Egg cell/ovum
cell
Eight nuclei in
Micropyle
Two synergid cells
the embryo sac
Embryo sac
Mature ovule
Structured Questions
1 (a) (i) Germination
(ii) Pollen tube
(iii)
Pollen grain
Male nuclei
(b) (i)
Tube nucleus
(b) After pollination, the pollen grain on the stigma germinates
and forms a pollen tube. Enzymes are secreted to facilitate
the growth of the pollen tube through the style and ovary
wall. In the pollen grain, the tube nucleus moves to the
end of the pollen tube and the generative nucleus is behind
it. The generative nucleus then divides to form two male
nuclei. The tube nucleus guides the growth of the pollen
tube towards the matured ovule and then into the embryo
sac through the micropyle. When the pollen tube reaches
the embryo sac, the tip of the pollen tube will burst. The
tube nucleus will degenerate and the two male nuclei are
released into the embryo sac of the ovule. One of the
male nucleus will fuse with the ovum to form a diploid
zygote and the first fertilisation occurs. The other male
nucleus will fuse with the two polar nuclei to form a
triploid endosperm nucleus. This is the second fertilisation.
Hence, in flowering plants, fertilisation occurs twice and is
therefore called double fertilisation.
(ii) Double fertilisation
(c) One of the male nucleus in the pollen tube fuses
with the ovum to form diploid zygote. Another male
nucleus fuses with the two polar nuclei to form triploid
nucleus.
2 (a) W: Pollen tube
X: Embryo sac
(b) (i)
Polar nuclei
Diploid nucleus
Megaspore mother cell
CHAPTER 7: A
DAPTATIONS OF PLANTS IN
DIFFERENT HABITATS
Egg cell
Practice 7.1
A. Understanding key ideas
1 ✗
2 ✗
3 ✓
(ii) • Double fertilisation occurs
• One Y structure fertilises an egg cell to form a
diploid zygote.
• One more Z structure fuses with two polar nuclei
to form a triploid endosperm nucleus.
(c) (i) • Keep V in a dry place.
• Moisture will initiates germination.
(ii) Spraying a sugary solution onto V.
B. Main concepts and facts
1 Xerophytes
3 Hydrophytes
4 ✓
5 ✓
2 Halophytes
4 Mesophytes
SPM FOCUS PRACTICE 7
PAPER 1
1 C
6 C
Essay Questions
3 (a) • The nucellus cell in the ovule develops to become the
megaspore mother cell.
• The megaspore mother cell divides through meiosis to
form four haploid cells.
• Three of these four haploid cells will degenerate.
• The remaining cell develops and enlarges to form an
embryo sac.
• The nucleus in the embryo sac divides through mitosis
three times to form eight haploid nuclei; with four
nuclei at each end of the embryo sac.
• One nucleus from each end of the embryo sac moves
to the centre and form two polar nuclei.
• One of the three haploid cells near the micropyle
end becomes the ovum. The other two haploid cells
become the synergid cells.
2 B
7 A
3 D
8 B
4 C
9 A
5 D
PAPER 2
Structured Questions
1 (a)
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(b) (i) Photosynthesis
(ii) Diffusion. Oxygen diffuses out from the leaf
through the stomata into the atmosphere down a
concentration gradient.
(c) • Hairs prevent movement of water vapour.
• Stomata are sunken to trap water vapour.
• Thick waxy cuticle prevents water loss from outer
surface.
• Rolled-up leaf traps air inside and reduces water loss.
(Any three)
(d) Sand dunes do not have much water as water drains
away very fast. Plants here face difficulties replacing any
significant water loss and the ability to reduce water loss
is important for survival.
• There are hydathodes on the leaves to secrete excess
salt from the plants and to control the osmotic pressure
of the cells in the plant
Xerophytes:
Example: Cactus
Adaptive features:
• Stem has many chloroplasts to carry out photosynthesis.
• Leaves in the form of spines or thorns to reduce water
loss by transpiration.
• Have shallow roots to absorb water from surface of
the ground during lightest rainfall and deep roots to
penetrate deep into the ground to absorb water
CHAPTER 8: B
IODIVERSITY
Practice 8.1
Essay Questions
2 (a) • Mesophytes are plants that grow in an environment
with a moderate supply of water.
• Hydrophytes are plants that grow in or near water.
• Halophytes are plants that grow in soil or water of
high salinity.
• Xerophytes are plants that grow in arid conditions.
(b) Mesophytes:
Examples: Rose plant, sunflower, palm tree
(Any land plant growing with average supply of water)
Adaptive features:
• Well-developed root system to absorb water by
osmosis and mineral salts by active transport
• The leaves are usually broad, flat and thin to provide a
large surface area for maximum absorption of sunlight
for photosynthesis.
• Contains xylem tissues which are thickened with
lignin to provide support for the plant.
Hydrophytes:
Examples: Hydrilla sp., Elodea sp. (submerged water
plants)
Adaptive features:
• The leaves and stems have many air spaces to enable
the plants to float upright in the water.
• Do not have roots and no xylem tissues
• The whole plant is covered by a thin and permeable
cuticle or no cuticle to allow diffusion of water,
mineral and gases (carbon dioxide and oxygen)
throughout the whole surface of the plant.
• The plant has thin, small leaves and fine stems to
reduce resistance to water current.
Example: Water lily (floating plant)
Adaptive features:
• Stem and leaves have aerenchyma tissues which
consist of air sacs and many air spaces to enable the
plant to float on the surface of water.
• No stomata on lower epidermis of leaves to prevent
water from entering into air spaces.
• Leaves have very thin cuticles as they do not need to
conserve water.
Halophytes:
Example: Mangrove trees
Adaptive features:
• The root systems cover large surface areas to provide
support in the soft muddy soil.
• Have breathing roots for gaseous exchange in
waterlogged soil.
• The roots have tissues with higher osmotic pressure
of the cell sap compared to the surrounding water to
prevent plasmolysis and to enable the root cells to
absorb water by osmosis.
A. Understanding key ideas
1 ✗
2 ✓
3 ✗
4 ✓
5 ✓
6 ✓
B. Main concepts and facts
1 Domain → Kingdom → Phylum → Class → Order →
Family → Genus → Species
2 Class, phylum, kingdom, domain
3 (a) Protista
(b) Archaebacteria
(c) Fungi
(d) Animalia
(e) Plantae
Practice 8.2
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
4 ✗
5 ✓
B. Main concepts and facts
Fungi
Animalia
Plantae
Protista
Practice 8.3
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
4 ✓
5 ✗
B. Main concepts and facts
1
Aspect
Viruses
2
6 ✓
Bacteria
Covered by
Protein coat
Cell wall
Cell membrane
No
Yes
Cytoplasm
No
Yes
Genetic
material
DNA or RNA –
only a few genes
DNA – enough
for several
hundred genes
Living or nonliving
Non–living
unless in host
Living
Microorganism
(a) Rhizobium sp.
(bacteria)
Role
Convert nitrogen to nitrate
in the root nodules of
leguminous plants
Azotobacter sp.
and Clostridium
sp. (bacteria)
Convert nitrogen to nitrate
in the soil
Blue-green algae
(Nostoc sp.)
Convert nitrogen to nitrate
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(b) Bacteria and
fungi that are
saprophytes
• Species diversity:
– Refers to the variety and abundance of species
living in an area
– The higher the variety and abundance of plants
and animal species in an area, the greater the
biodiversity.
• Genetic diversity:
– Refers to the variety of genes within a single
species
– This leads to a diversity of genetic characteristics,
reflecting the biodiversity within the species.
(b) (i) Organism X is a bacterium. The characteristics are as
follows:
• Unicellular organism
• Has cell wall made up of a complex mixture of
protein, sugar and lipids. Some have capsules
outside the cell wall.
• No true nucleus because no nuclear membrane
• Genetic material is in the form of a single circular DNA found in the cytoplasm
• Reproduces asexually by binary fission
• Has flagellum for movement
Organism Y is a virus. The characteristics are as follows:
• Genetic material is a single strand of nucleic acid:
ribonucleic acid (RNA) or deoxyribonucleic acid
(DNA)
• Has a protein coat to protect the nucleic acid
• Has no cytoplasm, nucleus or plasma membrane
• Shows no features of living things such as
respiration, nutrition or excretion
• Only shows signs of life in living cells of hosts
when it reproduces.
• Non-cellular organism that exists in various forms
Decompose dead plants
and animals to ammonium
compounds
Microorganism
Role
(c) Nitrosomonas sp.
(bacteria)
Oxidise ammonium
compounds to nitrite
Nitrobacter sp.
(bacteria)
Oxidise nitrite to nitrate
SPM FOCUS PRACTICE 8
PAPER 1
1
6
11
16
21
2
7
12
17
22
C
A
C
D
C
B
B
C
B
B
3
8
13
18
23
C
A
A
A
D
4
9
14
19
24
B
B
B
D
D
5
10
15
20
25
A
A
A
D
B
PAPER 2
Structured Questions
1 (a) (i) Prokaryotes: C and D.
Eukaryotes: A, B, E and F
(ii) Prokaryotes do not have a true nucleus because there
is no nuclear membrane while eukaryotes have a true
nucleus surrounded by a nuclear membrane.
(b) • In Kingdom A, the organisms have chlorophyll while
in Kingdom F, do not have chlorophyll.
• In Kingdom A, the cell wall of the organism is made
up of cellulose while in Kingdom F, it is made up of
chitin.
(c)
E/Animalia
F/Fungi
(ii)
A/Plantae
B/Protista
C/D
Eubacteria
C/D
Archaebacteria
Common ancestor
2 (a) (i) Rhizobium sp.
(ii) The microorganism is a symbiont. It lives in the root
nodules of leguminous plants. It converts nitrogen in
the soil to nitrates which are absorbed by plants.
(b) The microorganisms involved in process Q are
saprophytic bacteria and fungi which are decomposers.
These bacteria and fungi break down the proteins in
dead and decaying plants and animals into ammonium
compounds.
(c) Process R: Nitrosomonas sp.
Process S: Nitrobacter sp.
Virus
Bacterium
The outer layer is a
protein coat
The outer layer is
cell wall
Does not have plasma
membrane
Has plasma
membrane
Does not have cytoplasm
Has cytoplasm
Genetic material is DNA
or RNA
Genetic material is
DNA
Is non-living unless in a
living cell
Is a living cell
CHAPTER 9: E
COSYSTEM
Practice 9.1
A. Understandingkey ideas
1 ✓
2 ✓
3 ✓
6 ✓
7 ✗
8 ✓
4 ✓
9 ✓
5 ✗
10 ✗
B. Main concepts and facts
1 different, common basic needs
2 Avicennia sp., Sonneratia sp., cable/widespread, mud
3 (a) Commensalism is an interaction in which only one
organism benefits while the other organism neither
benefits nor is harmed. In mutualism, both organisms
living together benefit.
(b) In saprophytism, the host is the dead and decayed
organism while in parasitism, the host must be a living
organism.
(c) Epizoit is an animal that attaches to another animal
to obtain leftover food, transport and protection.
Epiphyte is a plant that attaches on another plant to
obtain support and maximum sunlight.
Essay Questions
3 (a) (i) Biodiversity refers to the large variety of plants,
animals and other organisms on Earth or in a
particular habitat.
(ii) • Ecosystem diversity:
– Refers to the variety of ecosystems in an area
or on Earth
– Each ecosystem consists of all the living
organisms that interact with one another and
with the environment.
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(d) Ectoparasite is a parasite that lives on the external
body surface of its host and obtain its food from
the host. Endoparasite is a parasite that lives in the
body of the host and obtain its food by absorbing the
nutrients directly from the intestine of the host.
4 (a) Heterotrophic nutrition
(b) chemoautotrophic nutrition
(c) photoautotrophic nutrition
(d) holozoic nutrition
(e) saprotrophic nutrition
(f) parasitic nutrition
3 (a)
(b)
(c)
Practice 9.2
A. Understandingkey ideas
1 ✓
2 ✗
3 ✗
6 ✓
7 ✗
8 ✗
4 ✓
9 ✓
5 ✓
10 ✗
4 (a)
B. Main concepts and facts
1 (a) Capture-mark-release-recapture technique
(b) Quadrat sampling technique
2 Hypothesis:
1. The population growth of Lemna sp. is fastest in a
neutral pH or nearly neutral pH.
2. High light intensity is most suitable for the population
growth of Lemna sp.
(b)
SPM FOCUS PRACTICE 9
PAPER 1
1
6
11
16
21
C
B
C
B
D
2
7
12
17
22
C
A
C
B
A
3
8
13
18
23
B
D
A
B
C
4
9
14
19
24
B
B
C
C
C
5
10
15
20
25
C
C
C
C
C
Essay Questions
5 (a) (i) • Biotic components are living organismsin an
ecosystem which interact with one another.
– Example: plants and animals
• Abiotic components are non-living component
such as the physical environment that can affect
an organism in an ecosystem.
– Example: pH value, temperature, light intensity,
microclimate, topography, air humidity
(ii) Abiotic components that affect the organisms in a
pond:
• pH of the pond
– Most organisms survive in a neutral or nearly
neutral surrounding
– Aquatic organisms can die if there are changes
in the water pH
• Temperature of surrounding
– High temperature will make the enzymes
denature
– Aquatic organisms live in a low temperature
surrounding
– To maintain the moisture of their skin
– Example: frog that depends on its moist skin
for respiration process
• Light intensity that penetrates into the pond
– Aquatic plants need light for photosynthesis
– If the light intensity that can penetrate into the
pond is too low, the photosynthesis rate is also
low
– This will cause the oxygen and food in the
pond to also decrease
– It will also affect the number of organisms in
the pond
(b) • The number of rats increase and decrease in a dynamic
equilibrium with predators.
• When the number of rats increases, the number of
owls and snakes also increase.
PAPER 2
Structured Questions
1 (a) R: Secondary consumer
Q: Primary consumer
(b) (i) • Contractions of muscle for movement
• Maintain a constant body temperature/for growth/
reproduction
(ii) L: Respiration
M: Excretion
(c) (i) Chemical energy
(ii) The sun’s energy enters the food chain through the
producers at trophic level P. Light energy from the
sun is converted to chemical energy by photosynthesis
in the producers (green plants). This chemical energy
is transferred to herbivores (trophic level Q) which
feed directly on the producers, and subsequently,
transferred to successive organisms in the food chain.
2 (a)
Zone I
Zone II
• The high salinity sea water in the mangrove
swamp. The seeds are protected from dehydration
through this reproduction.
(i) Process P ➝ Q: Colonisation
(ii) Process R ➝ S: Succession
Q: Avicennia sp. and Sonneratia sp.
R: Rhizophora sp.
S: Bruguiera sp.
• The cable root system of the mangrove trees in zone
Q traps more mud and silt as well as organic matter
from decaying plant parts.
• As the mud and silt accumulate, the bank is slowly
raised and contains less water. Zone Q is now more
suitable for another mangrove tree species. The soil
becomes more compact and the shore level is raised
with time.
X: Commensalism
Y: Saprophytism
Z: Mutualism
(i) P: Epiphyte
Q: Saprophyte
(ii) The presence of aerial roots which absorb moisture
from the air. /has roots that can absorbed water in the
crevices of the tree bark
(iii) • Organism Q is a decomposer.
• It breaks down dead decayed organic matter and
animal waste into simpler substances that can be
absorbed and used by plants.
Zone III
(b) Avicennia sp. or Sonneratia sp.
(c) (i) Pneumatophore
(ii) Gaseous exchange through the lenticels
(d) (i) Viviparity seeds
(ii) • Lack of oxygen in the waterlogged soil. The seeds
can still obtain oxygen directly from the atmosphere.
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• This is because the prey for the predator is increased.
• After some time, the higher number of predator will
cause more rats to be eaten.
• The number of rats will start to decrease.
• When the number of rats decreases, this will cause
lack of food for predators.
• Some predators will move to another place to search
for food.
• This will cause the number of predators to decrease.
• When the number of predators decreases, the
condition is favourable for the rats to reproduce and
their number increases.
• When the rat population increases again, the number
of predators at that place also increases, hence causing
the rat population to decrease. The population of
the rats and its predators will continue to fluctuate
together.
6 (a) • Pioneer species that are suitable to grow in the extreme
conditions of mangrove swamps are Avicennia sp. and
Sonneratia sp.
• The mangrove trees have pneumatophores roots for
breathing in waterlogged soil.
• The roots also trap sediments and dead organic
material.
• As more sediment accumulates, the bank is slowly
raised and contains less water.
• This condition becomes favourable for Rhizophora sp.
• Rhizophora sp. has prop roots and viviparity seeds.
• This condition allows more Rhizophora sp. grow at
the area and replace the pioneer species.
• The Rhizophora sp. roots trap more sediment and dead
organic substances.
• The banks are raised up even higher. The soil becomes
more compact, more fertile and less saline.The soil that
is harder and drier now is not suitable for Rhizophora
sp. but more suitable for Bruguiera sp.
• Bruguiera sp. have buttress roots.
• The buttress roots is a bigger root system compared
to prop roots.
• Bruguiera sp. will replace Rhizophora sp.
• The roots will trap more sediments and organic
substances.
• The soil becomes harder and dry land is formed.The
dry land is not suitable for Bruguiera sp. and the
mangrove trees are gradually replaced by land plants.
(b) (i) Energy flows in one direction in the food chain
and it is either utilised or lost into the surroundings
and it cannot be recycled. Green plants can carry
out photosynthesis to store the energy from the
Sun as chemical energy in the organic compounds
formed. When green plants are eaten as food, this
energy is taken into the consumers and used for
growth, metabolic and biological activities. Energy
passes along the food chain from one trophic level
to another. However, there is a decrease in the total
amount of energy as it is passed along the food chain
to the consumers. The energy is progressively lost at
each trophic level as heat from cell respiration. Only
about 10% of the energy is obtained by the consumer
at each trophic level. The remainder 90% is lost at
each trophic level, mainly as:
• heat from respiration, which is lost to the
surrounding by radiation
• heat in urine and other excretion and faeces
(ii) Energy is lost as it is transferred from one trophic
level to the next. Hence, a longer food chain would
result in a smaller amount of energy passed on to
successive trophic levels. Less energy is available
to higher trophic consumers. Therefore feeding crop
plants to animals results in adding another trophic
level to the food chain. More energy would be lost
and the higher consumers such as man would get less
amount of energy.
(iii) The longer the food chain, the more energy will
be lost. Hence the shorter the food chain, the
more energy will be available for higher trophic
consumers. Food chains with two or more trophic
levels have greater efficiency as they provide more
energy to consumers in the higher trophic levels. In a
longer food chain, less energy is available for higher
consumers. Hence it is better for humans to be the
last organism in a short food chain rather than the last
organism in a long food chain.
CHAPTER 10: ENVIRONMENTAL
SUSTAINABILITY
Practice 10.1
A. Understanding key ideas
1 ✓
2 ✗
3 ✗
6 ✓
7 ✗
8 ✓
4 ✓
5 ✓
B. Main concepts and facts
1 (a) sound
(b) acid rain
(c) thermal pollution
(d) oxygen, organic matter
(e) polluted, low, decreases
2 • High concentrations of nitrates and phosphates in the
fertilisers (which leach from farmland) and sewage
cause alga bloom in the pond.
• Algae grow rapidly, covering the surface of pond and
preventing sunlight from reaching the plants in water.
• The aquatic plants cannot carry out photosynthesis and
cannot carry out photosynthesis and die. Some old
algae died. Decomposition by aerobic bacteria occurs
which uses the remaining dissolved oxygen in the water.
• The concentration of dissolved oxygen in the water is
reduced.
• Aquatic animals in the water die due to lack of oxygen.
Practice 10.2
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
4 ✓
5 ✗
B. Main concepts and facts
1 (a) Preservation of ecosystem refers to the efforts to
maintain an intact ecosystem in its original state or
prevent it from being damaged.
(b) Conservation of ecosystem refers to the sustainable
use and maintenance of the ecosystem through the
management of natural resources for the present and
future generation.
(c) Restoration of ecosystem refers to the recovery of
ecosystems that have been damaged, degraded or
destroyed back to a stable, healthy and sustainable state.
2 (a) In-situ conservation (on-site conservation)
i. In-situ conservation refers to the conservation of
plant and animal species in their natural environment.
ii. Specifically, protecting an endangered plant or
animal species in its natural habitat
iii. It is done either by protecting the habitat itself,
or by defending the species from predators.
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30 MASTERCLASS 2023 SPM BIOLOGY-ANSWERS-(BAHAN QR)-ARUL 1P.indd 36
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iv. It involves the conservation of total ecosystems
in protected areas such as national parks (Taman
Negara), marine parks and wildlife sanctuaries.
v. Necessity of in-situ conservation:
• Protects the flora and fauna in natural habitats
without human interference
• Enables the life cycles of organisms to occur
in a natural way
• Provides the necessary green coverage and its
benefits to the environment
• Protects the total ecosystem
• Protects the interests of the indigenous people
(b) Ex-situ conservation (off-site conservation)
i. Ex-situ conservation refers to the conservation of
plants and animals outside their habitats.
ii. This is important for some endangered species.
iii. It conserves the breeding populations of plants
and animals and to provide for its reintroductions.
iv. Examples of ex-situ conservation include operations
in zoos, botanical gardens and tissue culture centre
v. Necessity of ex-situ preservation:
• Prevents the decline of the population of plant
and animal species
• Ensures endangered species on the verge of
extinction continues to survive
• Enables threatened species to be bred in captivity
and then released into the natural habitats
• Observation of wild animals, which is
otherwise not possible
• Enables research and scientific work to be
conducted on different species
(b) (i) Pollution of the river causes an increase in the
population of aerobic bacteria which carry out
decomposition and depletes the oxygen in the
water. This results in insufficient dissolved oxygen
available for the fishes, leading to the death of the
fishes.
(ii) The discharge of untreated sewage into the river
(c) • The fishes that survived the harmful effects of the
pollution require time to grow, mature and reproduce
to replace the number of fishes that have died.
• Other toxic inorganic chemicals may be present in the
sewage that polluted the river, causing the fishes to die
and killing the eggs.
(d) (i) Relative unit
BOD value
Number of fish
Number of bacteria
P
4 ✗
5 ✓
B. Main concepts and facts
1 Efficient use of water, use of renewable energy, waste
reduction and recycling, efficient use of energy and
conservation, domestic and toxic waste management, practise
biological control and efficient eco–friendly transport
2 Food security is defined as having an assurence that all
people at all times have availability of food, sufficient
access to food and safe food utilisation. The components
of food security are availability, ccessibility, utilisation and
stability of food.
Practice 10.4
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
4 ✓
5 ✗
B. Main concepts and facts
1 The four pillars are energy, environment, economy and
social.
2 Awareness, Knowledge, Attitudes, Skills, Participation
SPM FOCUS PRACTICE 10
A
A
D
A
2
7
12
17
B
B
C
B
3
8
13
18
B
D
C
D
4
9
14
19
B
C
C
A
5
10
15
20
20
30
40 50
Distance
Essay Questions
3 (a) • Organic and inorganic fertilisers that contain
nitrates and phosphates flow into the river, causing
eutrophication. The algae grow rapidly and increase
in population, preventing sunlight from reaching the
aquatic plants in the river. As a result, the aquatic
plants die. The number of bacteria increases to
decompose the dead plants, thus, using much oxygen
and reducing the concentration of oxygen in the river.
Many aquatic animals die. The BOD value increases
and the water is polluted.
SPM Focus Practice 10
PAPER 1
1
6
11
16
10
downstream (km)
(ii) The higher the BOD value, the higher the level of
pollution, the higher the number of bacteria and the
lower the number of fish.
2 (a) (i) • Reduced photosynthesis due to lesser trees leads
to decreased removal of carbon dioxide from
atmosphere or increased carbon dioxide in the
atmosphere
• Increased burning or decomposition because
more trees are converted into fuel or more rotting
of dead trees
(ii) • Soil erosion
• Increased leaching by rain because soil not
covered (bare) or exposed to rain and wind leads
to infertile soil
(iii) • Disrupted food chains or altered balance in food
web
• Destruction of potential resources
• Loss of genetic pool material
• Loss of biodiversity
• Destruction or loss of habitats
(Any two)
(b) (i) Methane/nitrogen oxide/CFC
(ii) • Sunlight enters Earth’s atmosphere as solar radiation.
• Some solar radiation is reflected back to space.
• Most of the radiation is absorbed by Earth as heat.
• Part of the heat is reflected back into space as
infrared radiation.
• Most heat is absorbed and trapped by increased
carbon dioxide levels in the atmosphere. This
increases the temperature of the atmosphere
causing the greenhouse effect.
Practice 10.3
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
Concentration of
dissolved oxygen
D
C
A
B
PAPER 2
Structured Questions
1 (a) 10 km
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30 MASTERCLASS 2023 SPM BIOLOGY-ANSWERS-(BAHAN QR)-ARUL 1P.indd 37
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• Agrochemicals such as pesticides, herbicides and
insecticides used on plants are non-biodegradable and
especially toxic and harmful, as can be passed on in
food chains. These agrochemicals, accumulate in the
body and cause harmful effects. This causes activites
also water pollution when the pollutants leach into
nearby rivers. These toxic substances are harmful to
aquatic organism.
(b) • Increase in the number and size of aquatic animals
Sewage contains nitrates and phosphates which are
nutrients to aquatic plants. When a small amount
of sewage is added to the pond, the nitrates and
phosphates are readily absorbed by aquatic plants
such as algae. As a result, the algae grow rapidly
and its population increases. Algae is a producer and
a source of food for the aquatic animals. The large
population of algae provides a rich source of food
4 (a)
for the aquatic animals, hence promoting the growth
and reproduction of aquatic animals. As a result, the
aquatic animals increase in number and size.
• Aquatic animals die
When sewage is added in large amounts, the excess
nitrates and phosphates causes the algae to grow at
such a rapid rate that it covers the surface of the pond.
The layer of algae prevents sunlight from penetrating
into the pond to reach the aquatic plants at the base of
the pond. Thus, the aquatic plants in the pond soon die
because there is no sunlight for photosynthesis. The
process of decomposition occurs and the oxygen supply
in the water will be used up by bacteria that carry out
the decomposition process. The large amount of algae
will also use up the oxygen supply in the water. This
causes the oxygen supply in the water to be greatly
reduced and causing the aquatic animals to die.
Type of pollution and source of pollutants
Effect of pollutants
1. Air pollution
Air pollution is caused by the emission of
smoke and fumes from chimneys of factory
site A to the surroundings areas such as
town B and forest C.
(a) Combustion of fossil fuels causes
factory to produce acidic gases such as
sulphur dioxide and nitrogen oxide
(b) Incomplete combustion of fossil fuels
in factory produces pollutants such as
carbon monoxide, soot, greenhouse
gases and carbon dioxide.
• Sulphur dioxide and nitrogen oxide are carried into the air by
the prevailing wind and dissolves in the air’s water vapour to form
droplets of acid in the air. When it rains in forest C, the acid dissolves
in the rain and comes down as acid rain. The acid rainwater causes
the soil and river water to become so acidic that, the plants and trees
in forest C become badly damaged and the fish in the river are killed.
• The acidic gaseous pollutants and smoke are also carried by the
prevailing winds to town B. These pollutants irritate the eyes,
damage respiratory tract and lungs of town B’s population when
they breathe in the air. These pollutants also aggravate diseases such
as asthma and bronchitis. The acid rain in town B also corrodes
metal structures in the buildings.
• Soot in smoke carried by winds to town B, forest C, farm D and
plantation E blackens buildings, covers the stomata of leaves, prevents
the absorption of carbon dioxide and reduces the rate of photosynthesis.
• Carbon monoxide is a toxic gas which combines more readily with
haemoglobin compared with oxygen, thus reducing the capacity of
the blood to transport oxygen.
2. Water pollution
Water pollution is caused by the discharge
of chemical wastes into the river along with
pesticides and excess fertilisers from farms.
• Chemical wastes and pesticides are non-biodegradable, toxic and
harmful as they can be passed on in food chains and accumulate
in the human body, causing destruction of cells and development
of cancer.
Type of pollution and source of pollutants
Effect of pollutants
(a) Chemical wastes from factory site A are
discharged into river through waste pipes
(b) Run-off from excessive use of pesticides
to control pests in plantation E
(c) Excessive use of artificial fertilisers
in farm D results in the leaching of
fertilisers into the nearby river.
• Artificial fertilisers used in farm D contain nitrates and phosphates
which cause eutrophication. The nutrients absorbed by algae cause
the algae to grow rapidly and increase the population of algae.
The large population of algae uses up the oxygen supply in the
water and also covers the surface of the pond, causing the aquatic
plants to die. Decomposition that occurs will also use up the
oxygen supply in the water. The dissolved oxygen in the water is
greatly reduced, causing the aquatic animals to die.
(b) (i) Recycling of paper reduces the usage of natural
resources such as wood pulp, water and energy from
fossil fuels. This results in less air and water pollution,
reduces solid waste and the amount of timber used,
hence slowing down the rate of deforestation. This
helps to control the harmful effects of deforestation
such as:
• Destruction of natural habitats of plants and
animals causing the extinction of many plants and
animals.
• Soil erosion which causes the land to become
infertile and clogs up rivers leading to floods.
• Increase in the concentration of carbon dioxide due
to the decrease in photosynthesis which causing
an increase in the temperature of the environment.
This contributes to the greenhouse effect.
(ii) Pesticides are used to control pests and increase
agricultural productivity. However, the widespread
and excessive use of pesticides have caused a lot of
harmful effects such as:
• Pesticides can be poisonous to other organisms
like fish and birds. Insecticides can also destroy
useful insects in soil.
• Pesticides are stable and non-biodegradable.
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30 MASTERCLASS 2023 SPM BIOLOGY-ANSWERS-(BAHAN QR)-ARUL 1P.indd 38
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These pesticides can be passed through the food
chain and gradually accumulate in the body reaching
a dangerous level that can cause sickness or death.
• Pesticides may be washed into rivers and lakes
and accumulate to significant amounts that can
poison aquatic life.
Phenotype ratio:
9 coloured, smooth seed: 3 coloured, rough seed: 3 non–coloured,
smooth seed: 1 non–coloured, rough seed
2
Independent
assortment
during
metaphase I
CHAPTER 11: INHERITANCE
Practice 11.1
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
6 ✗
7 ✓
8 ✓
4 ✓
9 ✗
Parent cell
with two
B
pairs of
homologous H
chromosomes
5 ✗
10 ✗
b
h
B. Main concepts and facts
×
Father
Parental genotype:
Gametes:
Mother
Aa
aa
A
a
Gametes
produced
B
b
B
H
h
H
b
B
b
H
h
H
b
and
h
B
and
h
Practice 11.3
a
A. Understanding key ideas
1 ✓
2 ✓
3 ✗
a
4 ✓
5 ✗
B. Main concepts and facts
Offspring
genotype:
Aa
Aa
aa
aa
Offspring
phenotype:
Curly
hair
Curly
hair
Straight
hair
Straight
hair
Phenotypic ratio: 1 (curly hair) : 1 (straight hair)
Phenotype of the 50% of the offspring will have curly hair
and 50% will have straight hair
offspring:
S
Heterozygous
Q
Gene
P
Homologous chromosomes
U
Gene loci
R
Homologous dominant allele pair
V
Alleles
T
Homologous recessive allele pair
Practice 11.4
Practice 11.2
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
6 ✓
7 ✗
8 ✗
4 ✗
B. Main concepts and facts
Coloured,
1 Parental
phenotype:
smooth seed
(Heterozygous)
Parental
genotype:
YyLl
×
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
6 ✗
7 ✗
8 ✗
5 ✓
Coloured,
smooth seed
(Heterozygous)
Gametes:
YL
Yl
yL
yl
YL
YYLL
coloured,
smooth
seed
YYLl
coloured,
smooth
seed
YyLL
coloured,
smooth seed
YyLl
coloured,
smooth seed
Yl
YYLl
coloured,
smooth
seed
YYll
coloured,
rough
seed
YyLl
coloured,
smooth seed
Yyll
coloured,
rough seed
yL
YyLL
coloured,
smooth
seed
YyLl
coloured,
smooth
seed
yyLL
non-coloured,
smooth seed
yyLl
non-coloured,
smooth seed
yl
YyLl
coloured,
smooth
seed
Yyll
coloured,
rough
seed
yyLl
non-coloured,
smooth seed
Yyll
non-coloured,
rough seed
A
B
O
O
Offspring
Genotype: AO
AO
BO
BO
Phenotype: Blood group A
Blood group B
2 22 + X, 44
Male normal
Female
3 Parents
×
Phenotype:
blood clotting
haemophilia
Genotype:
XHYXhXh
YL Yl yL yl
Gamete
5 ✓
10 ✓
B. Main concepts and facts
1 Not justified because in this inheritance, the child can only
have either blood group A or B. It is not possible to have
any child with the blood group AB.
Parents
Male
×
Female
Genotype:
AB
OO
YyLl
Gametes:
YL Yl yL yl
F1 generation:
4 ✗
9 ✓
Gametes:
XH
Fertilisation
Offspring
Genotype: XH Xh
Y
Xh
Xh
XH Xh
Xh Y
Xh Y
Phenotype:
Female, normal
Male,
blood clotting
haemophilia
(carriers)
All the sons will have haemophilia. All the daughters will
have normal blood clotting trait but they are also carriers.
Hence, the recessive allele for haemophilia can be inherited
by the next generation.
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30 MASTERCLASS 2023 SPM BIOLOGY-ANSWERS-(BAHAN QR)-ARUL 1P.indd 39
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SPM FOCUS PRACTICE 11
The possible blood groups of the baby from couple Y
are A and O. Based on the possible blood groups of the
babies from both couples, it can be concluded that the
baby X was mistakenly exchanged with baby Y.
(b) (i) Parents
×
Father
Mother
PAPER 1
1
6
11
16
2
7
12
17
B
A
B
A
3
8
13
18
D
C
C
C
4
9
14
19
C
D
C
D
5
10
15
20
C
C
B
D
C
B
D
D
phenotype: Colour blind
Structured Questions
1 (a) (i) R: dominant allele for red
Parents
Phenotype:
Gamete:
Plant S
×
White-flowered
plant
Genotype:
F1 generation
Genotype:
Phenotype:
R
r
Rr
rr
r
Rr
rr
(ii) 1 red-flowered plant :1 white-flowered plant
(b) (i) Monohybrid inheritance
(ii) Mendel’s First Law/Law of Segregation
2 (a) (i) Down syndrome
(ii) There is an extra 21st chromosome
(b) 45 autosomes
(c) The karyotype consists of 23 pairs of chromosomes.
Each pair is made up of two homologous chromosomes.
The 23rd pair consists of two X chromosomes.
(d) The sex of a baby is determined by the type of sex
chromosome present in the sperm. During fertilisation, if
the sperm with an X chromosome fuses with the ovum, a
female baby is born. If the sperm with a Y chromosome
fuses with the ovum, then a male baby is born.
3 (a) Couple X
Husband
Gamete:
A
B
Children
genotype:
AB
AB
Phenotype:
Blood group AB
Parents
Genotype:
Gametes:
Husband
BB
B
B
BB
BB
Fertilisation
A
Fertilisation
Children
genotype:
Phenotype:
Blood group B
Wife
×
AB
BO
B
B
O
AB
AO
BB
BO
Blood
group AB
Blood
group A
Blood
group B
Blood
group B
Parents
phenotype:
Genotype:
The possible blood groups of the baby from couple X are
A, B and AB.
Couple Y
Parents
genotype:
Husband
Gamete:
A
AA
XBY
Xb Y
X
Normal Colour
colour blind
vision
×
Plant Q
Plant R
(Red-flowered)
(White-flowered)
RR
Gamete:
R
F1 generation
genotype:
Rr
rr
R
r
r
Rr
Rr
Wife
×
A
Xb
Y
X
Essay Questions
4 (a) (i) Phenotype is the observable characteristic of an
organism.
Genotype is the genetic composition of an organism
that cannot be seen.
(ii) Homozygote refers to an organism that has
two similar alleles, for example, TT or tt while
heterozygote refers to an organism that has two
different alleles, for example, Tt.
(iii) Dominant allele will display its dominant trait when
at least one dominant allele is present while recessive
allele will only display the recessive trait when both
alleles are recessive.
(b) Mendel’s First Law:
During the formation of gametes, only one allele from each
pair of alleles for a particular trait is present in a gamete.
Mendel’s Second Law:
During the formation of gametes, two or more pairs of
alleles segregate independently of each other./During
gamete formation, each allele from a pair of alleles can
combine randomly with any allele from another pair of
allele.
(c) (i) R: dominant allele for red
Plant Q: Red-flowered plant
Plant R: White-flowered plant
Wife
×
AB
B
Female
Male
(ii) 1/2 or 50%
(iii) 0%
(iv) A woman has two X chromosomes and will only
display colour blindness when both the alleles linked
to the X chromosomes are recessive. A man has
one X chromosome and one Y chromosome. The Y
chromosome does not carry colour blindness gene.
Hence, even if only one recessive allele is linked to
his X chromosome, he will be colour blind.
RedWhiteRedWhiteflowered flowered flowered flowered
plant
plant
plant
plant
Parents
genotype:
XBXb
Children
XbXb
genotype: XbXB
Phenotype: Carrier Colour
blind
Rr
r
b
Fertilisation
Plant P
Red-flowered
plant
rr
Gamete:
XbY
Genotype:
PAPER 2
Normal colour
vision (carrier)
OO
O
O
Rr
Pink-flowered
Phenotype:
Fertilisation
Self-pollination
Children
genotype: AO
Phenotype:
AO
AO
AO
Parents
phenotype:
Genotype:
Blood group A
Parents
genotype:
Husband
Gametes:
A
AO
O
Wife
×
Gamete:
OO
O
40
OO
Rr
R
r
Pink-flowered
plant
Rr
R
r
Rr
rr
O
Fertilisation
Children
AO
genotype: AO
Phenotype: Blood group A
Pink-flowered
×
plant
OO
Blood group O
30 MASTERCLASS 2023 SPM BIOLOGY-ANSWERS-(BAHAN QR)-ARUL 1P.indd 40
F2 generation
genotype:
Phenotype:
RR
Rr
RedPink-flowered Whiteflowered
flowered
12/01/2023 5:20 PM
Genotypic ratio: 1RR : 2Rr : 1rr
Phenotypic ratio: 1 Red-flowered : 2 Pink-flowered:
1 White-flowered
(ii) L: Dominant allele for long horns
W: Dominant allele for white face
Parents
Genotype:
LLWW
Gamete:
LW
Progeny
Genotype:
Phenotype:
Male
Therefore, the probability of having a boy or a girl is
1/2 or 50% because the chances of either an X sperm
or Y sperm fertilising the ovum are equal.
6 (a) T: dominant allele for tall
Parents
Female
×
LlWw
LW
Lw
lW
Genotype:
lw
Meiosis
Gamete:
LLWW
LLWw
LlWW
LlWw
Long
Long
Long
Long
horns,
horns,
horns,
horns,
white face white face white face white face
Phenotypic ratio: All have long horns and white
faces
The back cross involves crossing each progeny
with the homozygous recessive goat for both traits.
If the progeny is homozygous dominant for both
traits, all progeny produced will be heterozygous
for each trait. If the progeny is heterozygous for
each trait, then half of the offspring produced will
have dominant phenotype and another half will have
recessive phenotype for that trait.
5 (a) (i) Curly hair is controlled by a dominant allele (R),
hence the trait will be displayed whenever there is
a dominant allele present, in either homozygous
dominant (RR) or heterozygous (Rr).
(ii) The genotype for straight hair is homozygous
recessive where both alleles are recessive. Hence,
when both parents have straight hairs, only the
recessive allele will be inherited to their children.
Thus, all children will have straight hair.
(iii) Parents
Father
Mother
phenotype:
Genotype:
Gamete:
Offspring
Genotype:
Phenotype:
Phenotypic
ratio:
×
Curly hair
R
RR
Rr
Curly hair
R
r
Rr
rr
A
Phenotype:
Phenotypic
ratio:
Offspring
AB
AO
Genotype:
Blood
Phenotype: Blood
group AB group A
O
BO
OO
Parents
Phenotype:
Genotype:
Gamete:
Offspring
genotype:
Phenotype:
Blood
Blood
group B group O
(ii) Sex is determined by the sex chromosomes:
X chromosome and Y chromosome. A boy has the sex
chromosomes XY and a girl has XX chromosomes.
There are two types of sperms: sperm with the X
chromosome and sperm with the Y chromosome.
Meanwhile, there is only one type of ovum with the
X chromosome.
Parents
genotype:
44 + XY
Gamete:
22 + X 22 + Y
Father
Offspring
genotype: 44 + XX 44 + XX
Phenotype: Female (girl)
Phenotypic 1 girl : 1 boy
ratio:
×
×
Heterozygous
recessive
Smooth stem plant
Hh
H
hh
h
h
hh
hh
h
Hh
Hh
Hairy stem
Smooth stem
1 Hairy stem : 1 Smooth stem
Ahmad
Blood group A
Wife
× Blood group B
AO
BO
A
O
AB
AO
Blood
Blood
group AB group A
B
BO
Blood
group B
O
OO
Blood
group O
Practice 12.1
A. Understanding key ideas
1 ✗
2 ✗
3 ✗
6 ✓
7 ✓
8 ✗
4 ✗
9 ✓
5 ✓
10 ✓
B. Main concepts and facts
1 Continuous variation:
(a) Height (b) Skin colour
(c) gradual
(d) normal distribution curve
44 + XX
22 + X 22 + X
44 + XY 44 + XY
30 MASTERCLASS 2023 SPM BIOLOGY-ANSWERS-(BAHAN QR)-ARUL 1P.indd 41
Heterozygous
Hairy
stem plant
CHAPTER 12: VARIATION
Mother
Male (boy)
Tt
Therefore, the probability of the offspring to get either
hairy stem or smooth stem is 50%.
(c) Siti is the biological daughter of Ahmad. The genotype of
Ahmad’s blood group is either AO or AA. The genotype
of his wife’s blood group is either BO or BB. Since the
blood group of Siti is O, the genotype of Ahmad must
be AO and his wife is BO. From the genetic diagram
below, the possibility of the children’s blood group from
the parents with genotype AO and BO are AB, A, B
and O.
Mother
Blood group B
B
Tt
Tt
F1 generation
Genotype:
BO
O
t
Fertilisation
3 curly hair : 1 straight hair
×
t
All tall plants
Gamete:
Straight hair
Parents
Father
phenotype: Blood group A
Genotype:
AO
T
Phenotype:
Genotype:
Meiosis
Rr
r
T
tt
Fertilisation
F1 generation
genotype:
Tt
Parent 2
Phenotype:
Curly hair
Rr
TT
Dwarf plant
(homozygous
recessive)
(b) Let H be the dominant allele for hairy stem and h the
recessive allele for smooth stem. The cross between
the purebreed, hairy stem plants (HH) and purebreed,
smooth stem plants (hh) would produce offspring with
heterozygous hairy stem plants (Hh). To produce offspring
with hairy stems and smooth stems in a ratio 1: 1, the
possible genotypes of the plants are heterozygous hairy
stem (Hh) and homozygous recessive smooth stem (hh).
(b) (i) The parents must be heterozygous for the blood
group A and B, so that the child will inherit the
recessive allele (O) from both parents and thus
possesses blood group O.
Gamete:
Tall plant
(homozygous ×
dominant)
41
12/01/2023 5:20 PM
(e) genetic factor
(f) environmental factor
Discontinuous variation:
(a) Blood group
(b) Ability to roll tongue
(c) distinct
(d) discrete distribution
(e) genetic factor only
2 (a) Zygote with different genetic content is formed
(b) Gametes with different combination of chromosomes
are formed
(c) Gametes formed have permanent change in the genetic
content
(d) There is exchange of genetic materials
(e)
(f)
(g)
(h)
(i)
SPM FOCUS PRACTICE 12
PAPER 1
1
6
11
16
21
26
Practice 12.2
4
9
14
19
24
B
C
B
B
B
D
B
C
D
D
5
10
15
20
25
A
C
D
A
D
Structured Questions
1 (a), (b) & (c) Normal distribution curve
3 Type of earlobe Discontinuous variation
Can be inherited
4 Skin colour
Continuous variation
Cannot be
inherited
5 Colour
blindness
Discontinuous variation
Can be inherited
6 Ability to taste
PTC
Discontinuous variation
Can be inherited
7 Body weight
Continuous variation
Cannot be
inherited
8 Fingerprint
pattern
Discontinuous variation
Can be inherited
9 Present of
dimple
Discontinuous variation
Can be inherited
10 Albinism
Discontinuous variation
Can be inherited
11 Haemophilia
non-sister
chromatids of
Discontinuous variation
Can be inherited
12 Eye colour
Discontinuous variation
Can be inherited
1
2
2
1
Length of shell (mm)
(d) (i) Continuous variation
(ii) There are no distinct differences in continuous
variation and there are intermediate characteristics.
2 (a) (i) Figure 1.2
(ii) Figure 1.1: Discontinuous variation
Figure 1.2: Continuous variation
(b) Discontinuous variation is determined by one single gene.
It is often caused by genetic factor or the spontaneous
mutation of genes and chromosomes.
Continuous variation iscontrolled by several genes and
influenced by environmental factors. Hence it is caused
by genetic factors and environmental factors.
(c) (i) Tongue rolling, eye colour, blood group, fingerprint
patterns, type of earlobes
(ii) Body weight, skin colour, height
Essay Questions
3 (a) Continuous variation has characteristics with no distinct
differences, hence are not easily distinguished and
exhibits many intermediate characteristics. It is controlled
by several genes and is influenced by the environment
factors. For example, body weight and height.
Discontinuous variation has contrasting characteristics
with distinct differences and no intermediate traits. It
is caused by one single gene and is not influenced by
environmental factors. For example, the ABO blood
group and tongue rolling.
(b) (i) R is inversion. A segment of the chromosome breaks
between gene AB and gene CD, rotates 180°, then
rejoins. As a result, genes B and C are in reverse
order. The structure of chromosome is changed and
a mutant chromosome is formed. S is translocation.
A section of one chromosome breaks between B
and C and gets attached to another chromosome
with genes V and W. There is transfer of genes and
mutant chromosomes are produced. T is duplication.
The section of chromosome with genes C and D is
Practice 12.3
4 ✗
9 ✗
2
56 – 60
Cannot be
inherited
4
51 – 55
Continuous variation
4
46 – 50
2 Height
7
5
41 – 45
Can be inherited
6
21 – 25
Discontinuous variation
8
8
36 – 40
Number of snails
1 Ability to roll
tongue
10
31 – 35
5 ✗
Can or cannot
be inherited
Type of variation
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
6 ✓
7 ✗
8 ✓
3
8
13
18
23
B
A
D
D
A
C
PAPER 2
4 ✗
B. Main concepts and facts
Characteristic
2
7
12
17
22
27
B
A
B
C
C
C
26 – 30
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
Down syndrome
Turner syndrome
X
47
sex
5 ✗
10 ✓
B. Main concepts and facts
1 (a) Structure
(b) Number
(c) Deletion
(d) duplicated/doubled/copied
(e) inversion
(f) another
2 (a) Polydactylism
(b) haemoglobin
(c) recessive
(d) melanin pigment
42
30 MASTERCLASS 2023 SPM BIOLOGY-ANSWERS-(BAHAN QR)-ARUL 1P.indd 42
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CHAPTER 13: GENETIC TECHNOLOGY
duplicated. The mutant chromosome has doubled
the genes C and D. U is deletion. The chromosome
breaks between genes A and B and also between C
and D. The fragment broken off is lost. Hence the
mutant chromosome has lost some genes.
(ii) Three characteristics: fur colour, tail length, ear size.
(ii) Parent P
Parent Q
Offspring
Black fur
White fur
Black fur, white fur
Short tail
Long tail
Short tail, long tail
Big ears
Small ears
Big ears, small ears
Practice 13.1
A. Understanding key ideas
1 ✗
2 ✓
3 ✓
4 ✗
5 ✓
B. Main concepts and facts
(a) Production of crop with additional vitamins
(b) Production of crop with herbicide tolerance
(c) Production of pink coloured flesh and extra sweetness
(d) Production of crops with altered fatty acids composition
(e) Production of human insulin
(f) Production of crops with insects resistance
Practice 13.2
Based on the differences in characteristics, the
offsprings inherited some characteristics from
parent P and some from parent Q. The inheritance
of the characteristics is due to the genetic factors in
variation. During the formation of gametes in the
gonads at prophase I of meiosis, crossing over occurs
between non-sister chromatids of the homologous
chromosome. There is exchange of genetic materials
which results in variation. At metaphase I, the
independent assortment of homologous chromosomes
results in gametes that are haploid and with different
combinations of maternal and paternal chromosomes.
Hence a variety of gametes are formed which results in
variation among individuals of the offspring. Random
fertilisation of a sperm and an ovum results in a
variety of diploid zygote. This also causes variation
between the individuals of the offspring.
4 (a) Fingerprint pattern is a discontinuous variation. Each
pattern is distinctively different. There is no intermediate
pattern of fingerprint. Each pupil has one pattern of
fingerprint. The distribution of fingerprint patterns
is discrete. It is caused by genetic factors and can be
inherited. Height is a continuous variation. All the pupils
show a wide range of height. The difference in height
among the pupils is not distinct but is gradual. There are
intermediate characteristics for height. The distribution
of height is a normal distribution curve. This variation
is caused by environmental and genetic factors. The
environmental factors interact with the genetic factors
to cause the variation. Nonetheless, this variation cannot
be inherited.
(b) Environmental factors
Environmental factors that can cause variation are diet,
fertility of soil, pH, humidity, sunlight and nutrients.
Non-environmental (genetic) factors
• Independent assortment of homologous chromosomes
during metaphase I of meiosis results in various
combinations of chromosomes in gametes. Genetic
recombination occurs during crossing over in prophase I
of meiosis to form gametes with different genetic
composition.
• Changes in genetic composition will cause changes in
the genetic information, resulting in variation.
• Random fertilisation of gametes produces zygotes
of different genetic composition. Mutation causes
changes in the structure of genes and chromosomes,
resulting in mutant genes and chromosomes that will
cause genetic disorders.
(c) A thumbprint is a discontinuous variation and a
photograph is a continuous variation. By having both
the thumbprint and photograph on an identity card, an
individual’s identity can be ascertained by checking both
the photograph and the thumbprint.
A. Understanding key ideas
1 ✓
2 ✗
3 ✓
4 ✓
5 ✗
B. Main concepts and facts
1 (a) To identify individuals
(b) To determine the father in a paternity dispute
(c) To solve criminal cases through forensic science
(d) To identify genetic diseases
2 (a) To produce Bt cotton/Bt brinjal/Bt maize
(b) To cure or prevent genetic disease
(c) Use of bacteria that can breakdown hydrocarbon in
water or soil
(d) To make wine beer and antibiotic
(e) To identify criminal identity genetic disease, settle
paternity and maternity dispute
SPM FOCUS PRACTICE 13
PAPER 1
1
6
11
16
D
C
A
B
2
7
12
17
D
A
B
C
3
8
13
18
B
A
D
A
4
9
14
19
C
B
B
C
5 A
10 C
15 C
PAPER 2
Structured Questions
1 (a) Yeast
(b) (i) Flour/starch/sugar/glucose/sucrose/carbohydrate
(ii) Respiration
(iii) Carbon dioxide
(c) • In step 3, the temperature 35°C is the optimum or
best temperature for the respiration of yeast. Carbon
dioxide is released in the dough. It expands and causes
the dough to rise.
• In step 5, 200°C is a high temperature. Organism A
or yeast is killed and enzymes are denatured.
• The high temperature is needed to cook the dough. All
the carbon dioxide trapped, expand and are forced out
of the dough to make the bread rise more. The high
temperature also causes the ethanol in the dough to
evaporate.
(d) Biofuels, wine, beer, penicillin, antibiotic
2 (a) Bacterium Bacillus thuringiensis
(b) The gene for pest resistance or gene produces toxins is
isolated from the bacterium Bacillus thuringiensis and
inserted into the DNA of cotton plant cells.
(c) • Can kill caterpillars or insects or pests without using
pesticide
• Higher yields or increased crop production
• Reduced losses due to damage caused by pests
• Higher quality cotton
• Less insecticide used so there is less pollution/less cost/
less risk to workers’ health (from using insecticides)/
less chance of pests becoming insecticide-resistant
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3 (a)
KERTAS 2
Number
of stage
Description of the stage
The plasmids are removed from the
bacterial cell
5
A chromosome is removed from a
healthy human cell
2
Plasmids are returned to the bacterial cell
8
Restriction endonuclease enzyme is used
3/6
Bacterial cells are allowed to reproduce
in a fermenter
9
Bahagian A
1 (a) (i) Kepekatan/Tekanan separa oksigen adalah lebih
tinggi di dalam alveolus berbanding dengan di dalam
kapilari darah.
There is a higher concentration/partial pressure of
oxygen in the alveolus than in the blood capillary.
(ii) Luas permukaan yang besar/dinding setebal satu sel/
permukaan yang lembap/terdapat jaringan kapilari
darah
Large surface area/one-cell thick wall/moist surface/
presence of network of blood capillaries
(b) (i) • Kecerunan kepekatan oksigen antara udara dalam
atmosfera dengan di dalam kapilari darah pada
alveolus adalah kecil.
The concentration gradient of oxygen between
the air in the atmosphere and the blood capillary
around alveolus is small.
• Kadar resapan oksigen dari atmosfera ke dalam
alveolus adalah rendah. Kurang oksigen meresap
ke dalam alveolus.
Rate of diffusion of oxygen from the atmosphere
into the alveolus is low. Less uptake of oxygen
into the alveolus/less oxygen diffuse into the
alveolus.
(ii) • Apabila terdapat lebih banyak sel darah merah,
terdapat lebih banyak haemoglobin untuk
mengangkut oksigen ke sel-sel otot.
When there is more red blood cells, there will be
more haemoglobin to transport more oxygen to
the muscle cells.
• Hal ini akan meningkatkan kadar respirasi selsel otot. Lebih banyak tenaga dibebaskan untuk
meningkatkan prestasi.
This increases the rate of respiration in the
muscles cells. More energy is released to allow
better performance.
2 (a) (i) D dan/and H
(ii) B, D, H dan/and I
(b) (i) Pembentukan hempedu/Penyimpanan glikogen/
Pembentukan urea (deaminasi)/Penyahtoksinan
Formation of bile/Storage of glycogen/Formation of
urea (deamination)/Detoxification
(ii) Penghasilan jus pankreas/Penghasilan enzim
pencernaan/Penghasilan insulin dan glukagon
Production of pancreatic juice/Production of
digestive enzymes/Production of insulin and
glucagon
(c) • Pencernaan lemak akan terjejas/menjadi perlahan
Digestion of fats will be affected/slow
• Hempedu tidak dapat diangkut ke duodenum untuk
mengemulsikan lemak.
No bile can be transported to the duodenum to
emulsify the fats.
3 (a) (i) Xilem/Xylem
(ii) Floem/Phloem
(b) Ketiga-tiga kepala anak panah akan menghala ke bawah.
All three arrow heads point downwards.
(c) (i) • Menjalankan kejuruteraan genetik
Carry out genetic engineering
• Gen daripada bakteria yang menghasilkan racun
serangga dimasukkan ke dalam DNA tanaman.
Tanaman dengan gen ini tahan serangga.
Gene from bacteria that can produce insecticide
is inserted into the DNA of crop. The crop with
this gene is insect-resistant.
(b) Insulin is used to treat diabetes mellitus patients. The GM
insulin is produced from bacteria which is neutral. Hence
it is acceptable to people of diverse religious beliefs.
4 (a) • Half of the bands of the child’s DNA profile match
that of the mother’s DNA profile and the other half
match that of the father’s DNA profile.
• This is because the child is the result of fertilisation
of the mother’s egg and the father’s sperm.
• The child received the mother’s DNA from the egg
and the father’s DNA from the sperm
(b) Suspect 3. The DNA profile has the most DNA bands
that match the bands in the blood sample found at the
crime scene.
(c) • Zookeepers can check the DNA profiles of animals
which are to be bred together.
• They can look for the greatest difference in DNA
profiles and only breed those with the greatest
differences in DNA profiles. This will ensure the
greatest genetic variation in the offspring
(d) (i) • To identify genetic disorders
• To know whether there is any possibility to develop
hereditary diseases, cancer or other genetic diseases.
• To understand the type of genetic diseases
• To help to give the best treatment
(ii) • Advantages: To prevent embryos that are positive
for hereditary disease from being implanted into
the mother's uterus. The embryos implanted
should not carry any genes for hereditary disease
or medical condition from any of its parents
• Disadvantages: Embryos that carry genes for
hereditary disease are usually destroyed. This
raises ethical concerns and issues as embryos
are considered as life. Further, some may be tempted
to use this technique to create “super babies”.
(e) The steps in genetic engineering include:
1. Identify genes that code for valuable characteristics.
2. Isolate those genes from the DNA of the animal or
plant that has this characteristic.
3. Transfer the genes to the DNA of another organism
that can grow rapidly (bacteria).
4. This organism now carries the genes in its DNA that
can produce the valuable characteristic.
5. This modified DNA is known as recombinant DNA.
SPM MODEL TEST
KERTAS 1
1
6
11
16
21
26
31
36
C
C
B
A
D
A
B
C
2
7
12
17
22
27
32
37
B
B
A
D
A
D
D
D
3
8
13
18
23
28
33
38
C
A
A
B
C
A
C
C
4
9
14
19
24
29
34
39
B
C
B
D
A
A
C
A
5
10
15
20
25
30
35
40
D
A
A
D
B
B
D
D
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• Gen ini menghasilkan toksin untuk membunuh
serangga perosak (larva serangga) yang memakan
tanaman.
The gene can produce toxins to kill the pests
(insect larvae) which eat the crops.
(ii) Serangga yang membantu pendebungaan; Serangga
seperti lebah menghasilkan madu
Insects help in pollination; Insect such as bees
produce honey
4 (a) (i) Sebagai pengguna ketiga dan keempat
As the tertiary consumer and quarternary consumer
(ii) Fitoplankton ➝ Zooplankton ➝ Ikan kod Artik
➝ Burung artik
Phytoplankton ➝ Zooplankton ➝ Arctic cod
➝ Arctic bird
(iii) 7
(b) (i) Spesies yang mempunyai populasi yang kecil,
kurang peluang untuk membiak dan mungkin pupus
The species with a small population which has less
chances to reproduce and may become extinct.
(ii) Populasi ikan paus pembunuh meningkat; beruang
kutub makan kurang anjing laut; populasi anjing laut
meningkat; lebih banyak makanan untuk ikan paus;
beruang kutub makan kurang ikan kod Artik; lebih
banyak makanan untuk anjing laut berjalur/anjing
laut pelabuhan
Killer whale population rises; polar bears eat less
seals; seal population rises; more food for killer
whales; polar bears eat less Arctic cod; more food
for ringed / harbour seals
(c)
Helang/Eagle
menghasilkan antibodi untuk bertindak ke atas
antigen/patogen tersebut.
Fluid F is the lymph which has many leucocytes such
as phagocytes and lymphocytes. The phagocytes
destroy pathogens by phagocytosis, while the
lymphocytes destroy pathogens by producing antibodies to react with antigens/pathogens.
6 (a) (i) X: Tisu tulang/Bone tissue
(ii) Y: Tisu otot/Muscle tissue
(b) • X/Tisu tulang keras dan kuat untuk memberi sokongan
dan perlindungan.
X/Bone tissues are hard and strong to provide support
and protection.
• Y/Tisu otot adalah serat elastik yang dapat mengecut
dan mengendur untuk menghasilkan pergerakan.
Y/Muscle tissues are elastic fibres that can contract
and relax to produce movement.
(c) • Sendi engsel/Hinge joint
• Bulatkan sendi pada siku, lutut dan jari
Circle the joint at the elbow, knee and phalanges
(d) Membengkokkan tangan/Bending of the arm
• Otot bisep mengecut manakala otot trisep mengendur.
Biceps muscles contract while triceps muscles relax.
• Daya tarikan dihasilkan oleh pengecutan otot bisep
dan dipindahkan ke tulang radius melalui tendon.
A pulling force is produced by the contraction of biceps
muscles and transmitted to radius through tendon.
• Tulang radius ditarik ke atas dan menyebabkan tangan
membengkok.
The radius is pulled upwards and causes the bending
of the arm.
7 (a) (i) • Apabila V/debunga jatuh pada stigma bunga,
suatu cecair bergula akan dihasilkan daripada
stigma yang akan merangsang V/debunga untuk
bercambah dan membentuk tiub debunga.
When V/pollen grain falls on the stigma of a
flower, sugary fluid is produced from the stigma
which stimulates V/pollen grain to germinate and
form W/pollen tube.
• W/Tiub debunga mengandungi nukleus tiub dan
dua nukleus jantan.
The W/pollen tube contains a tube nucleus and
two male nuclei.
(ii) Tanpa W/tiub debunga, dua nukleus jantan/gamet
jantan tidak dapat dipindahkan ke pundi embrio/Y
untuk persenyawaan.
Without W/pollen tube, the two male nuclei/ gametes
cannot be transported to the embryo sac/Y for
fertilisation.
(b) (i)
Ular/Snake
Burung/Bird
Kutu/Tick
Rusa/Deer
Rumput/Grass
5 (a) (i) D: Bendalir tisu/Tissue fluid
(ii) Sel darah merah/platlet dan protein plasma/fibrinogen
Red blood cell/ platelets and plasma protein/ fibrinogen
(b) Tekanan hidrostatik yang tinggi di dalam kapilari (pada
sebelah arteriol) menyebabkan sebahagian plasma darah
meresap diserap keluar dari kapilari ke ruang antara sel
badan. Bendalir yang mengisi ruangan ini adalah bendalir D.
The high hydrostatic pressure in the capillary (on the
arteriole side) forces some of the blood plasma to diffuse
out from the capillary into the space between the body
cells. The fluid that fills these spaces is fluid D.
(c) (i) Bendalir E terlibat dalam proses pembekuan darah.
Fibrin terbentuk pada luka semasa proses pembekuan
darah, oleh itu, dapat menghalang bakteria/patogen
daripada memasuki badan melalui luka.
Fluid E is involved in blood clotting. Fibrin is formed
on the wound during the blood clotting process thus
preventing bacteria/pathogen from entering the body
through the wound.
Atau/Or
Fagosit seperti neutrofil terdapat dalam bendalir E.
Jika bakteria/patogen memasuki badan, fagosit akan
menghapuskan bakteria tersebut melalui fagositosis.
Phagocytes such as neutrophils are present in
fluid E. If bacteria/pathogens enter the body, the
phagocytes will destroy them by phagocytosis.
(ii) Bendalir F adalah limfa yang mengandungi
banyak leukosit seperti fagosit dan limfosit.
Fagosit memusnahkan patogen melalui fagositosis,
manakala limfosit memusnahkan patogen dengan
Nukleus kutub
Polar nuclei
Sel telur
Egg cell
(ii) Z adalah dua gamet jantan. Satu gamet jantan/Z akan
memasuki pundi embrio dan bercantum dengan sel
telur untuk membentuk zigot diploid. Gamet jantan/Z
yang satu lagi bercantum dengan dua nukleus kutub
di dalam pundi embrio untuk membentuk nukleus
endosperma triploid/zigot triploid.
Z are two male nuclei. One male gamete/Z enters the
embryo sac and fuses with the egg cell to form a diploid zygote. The other male nucleus/Z fuses with the
two polar nuclei in the embryo sac to form triploid
endosperm nucleus/triploid zygote.
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dalam udara panas atau di dalam ketuhar. Kaedah
pengeringan menyingkirkan air daripada makanan. Ini
dapat mengelakkan mikroorganisma daripada bertumbuh
kerana mikroorganisma tidak dapat hidup tanpa air. Selain
itu, ikan dapat direndam dalam larutan garam pekat untuk
menyingkirkan air daripada sel melalui osmosis. Proses
ini dikenali sebagai pengontangan. Larutan garam pekat
adalah hipertonik terhadap sel ikan. Molekul air meresap
keluar daripada sel ikan dan tersejat daripada permukaan
ikan. Sel kehilangan air dan seterusnya mikroorganisma
tidak dapat hidup dan membiak.
The fish can either be dried under the sunlight, in hot
air or in the oven. Drying method removes water from
the surface of food. This prevents microorganisms from
growing because microorganisms cannot live without
water. Other than that, the fish can also be soaked in
concentrated salt to remove the water from the cells
through osmosis. This process is called dehydration. The
concentrated salt solution is hypertonic to the cytoplasm of
fish cells. Water molecules diffuse out of the fish cells and
evaporate from the surface of the fish. The cells lose water
and therefore, microorganisms cannot live and reproduce.
Betik/Papaya
Penjerukan adalah kaedah untuk mengawet betik. Betik
dipotong kepada kepingan-kepingan nipis dan direndam
dalam larutan cuka atau gula yang hipertonik terhadap
sel betik. Molekul air meresap keluar daripada sel betik,
ke dalam larutan tersebut. Air dihilangkan daripada sel
betik melalui osmosis. Tanpa air, mikroorganisma tidak
boleh tumbuh dan membiak. Maka betik boleh tahan
lama.
Pickling method is used to preserve papaya. The papaya
is cut into thin slices and then soaked in a vinegar or
sugar solution, which is hypertonic to the papaya cells.
Water molecules diffuse out of the papaya cells, into the
solution. The cells lose water through osmosis.Without
water, microorganism cannot grow and reproduce.
Therefore, the papaya can last longer.
(c) Larutan sukrosa 5% adalah isotonik terhadap sap sel bagi
sel bawang. Tiada kecerunan kepekatan antara sap sel dan
larutan sukrosa. Oleh itu, kadar resapan molekul air ke
dalam dan keluar sel adalah sama. Akibatnya, struktur sel
atau saiz sel kekal sama. Sel itu kemudiannya direndam
dalam larutan sukrosa 35%, yang hipertonik terhadap
sel. Terdapat kecerunan kepekatan antara sap sel dengan
larutan sukrosa di sekeliling sel. Molekul air meresap
keluar dari vakuol dan sitoplasma sel bawang ke dalam
larutan. Isi padu air di dalam vakuol dan sitoplasma
berkurang. Vakuol menjadi lebih kecil dan membran
plasma tertarik daripada dinding sel. Sel bawang
mengalami plasmolisis. Sel tersebut kemudiannya
direndam di dalam air suling yang hipotonik terhadap
sap sel bawang. Terdapat kecerunan kepekatan antara
sap sel bawang dengan larutan di sekelilingnya. Molekul
air meresap ke dalam sel bawang melalui osmosis.
Pergerakan air ke dalam sel menyebabkan vakuol
membesar. Tekanan yang dikenakan terhadap dinding
sel menyebabkan sel mengembang dan menjadi segah
semula.
There is no concentration gradient between the cell sap
and the 5% sucrose solution which is isotonic to the
onion cell sap. Hence, the rate of diffusion of water
molecules in and out of the cells is the same. As a result,
the structure of the cells or the size remains unchanged.
The cell is then immersed in the 35% sucrose solution,
which is hypertonic to the cell sap. A concentration
(c) Menjadi buah/Becomes a fruit
8 (a) Mutasi gen ialah perubahan pada urutan bes DNA/
perubahan struktur gen
Gene mutation refers to a change in the base sequence
of DNA/change in the structure in the gene
(b) XBY
(c) (i) Individu 13
/Person 13
(ii) • Individu 4 dan 5 masing-masing menerima satu
kromosom X daripada induk. Alel resesif Xb bagi
buta warna adalah daripada bapa dan alel normal
XB daripada ibu.
Person 4 and 5 each receives one X chromosome
from each parent. The recessive allele Xb for
colour blindness is from the father and the
normal allele XB is from the mother.
• Semua anak perempuan adalah heterozigot/XBXb.
All female offspring are heterozygous/XBXb.
(d) (i) Gamet/Gametes: XB,Y, Xb , Xb
Genotip anak/Offspring genotypes: XBXb, XBXb,
XbY, XbY
Fenotip anak: Perempuan pembawa, perempuan
pembawa, lelaki buta warna, lelaki buta warna
Offspring phenotypes: Carrier female, carrier
female, colour blind male, colour blind male
(ii) 50% atau/or 1/2
Bahagian B
9 (a) Proses ini adalah resapan ringkas. Pada awal
eksperimen, kepekatan kuprum sulfat pada dasar bikar
adalah lebih tinggi daripada air suling. Maka, terdapat
kecerunan kepekatan antara kuprum sulfat dengan
air suling. Molekul-molekul kuprum sulfat meresap
dari kawasan berkepekatan kuprum sulfat tinggi ke
kawasan berkepekatan kuprum sulfat rendah. Pada akhir
eksperimen, semua molekul kuprum sulfat meresap ke
dalam air suling dan disebarkan dengan sekata dalam air
untuk membentuk larutan biru kuprum sulfat.
The process is simple diffusion. At the beginning of the
experiment, the concentration of copper sulphate at the
base of the beaker is higher than the distilled water.
Thus, there is a concentration gradient between the
copper sulphate and the distilled water. Molecules of the
copper sulphate will diffuse from the region of a higher
concentration to the region of a lower concentration of
copper sulphate. At the end of the experiment, all the
copper sulphate molecules have diffused into the distilled
water and are evenly distributed in the water to form a
blue solution of copper sulphate.
(b) Susu segar/Fresh milk
Pempasteuran merupakan satu kaedah untuk mengawet
susu. Dalam kaedah ini, susu segar dipanaskan ke
suhu 63oC selama 30 minit, diikuti dengan penyejukan
serta-merta dan cepat. Susu juga dapat diawet dengan
memanaskan susu pada suhu 72oC selama 15 saat
dan diikuti dengan penyejukan serta-merta. Kaedahkaedah ini akan memusnahkan mikroorganisma serta
mengekalkan nutrien dan rasa semula jadi susu.
Pasteurisation is a method to preserve milk. In this
method, the fresh milk is heated to 63°C for 30 minutes,
followed by immediate and rapid cooling. Milk can also
be preserved by heating the milk at 72°C for 15 seconds
and followed by rapid cooling. These methods will
destroy microorganisms while maintaining the nutrients
and the natural flavour of milk.
Ikan/Fish
Ikan dapat dikeringkan di bawah cahaya matahari,
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gradient exists between the cell sap and the surrounding
sucrose solution. Water molecules diffuse out of the
vacuole and the cytoplasm of the onion cell into the
solution. The volume of water in the vacuole and the
cytoplasm becomes less. The vacuole becomes smaller
and the plasma membrane is pulled away from the cell
wall. The onion cell is plasmolysed. The cell is then
immersed in distilled water, which is hypotonic to the
onion cell sap. A concentration gradient exists between
the onion cell sap and the surrounding solution. Water
molecules diffuse into the onion cell through osmosis.
A net inflow of water into the cell causes the vacuole
to become larger. The pressure exerted against the cell
wall causes the cell to expand and becomes turgid again.
10 (a) (i) • Semasa hari sejuk, tumbuhan segah dan tegak.
During a cold day, the plant is firm and upright.
• Sel tumbuhan segah kerana mengandungi air yang
mencukupi.
The plant cells are turgid because they contain
enough water.
• Keadaan ini membenarkan daun tumbuhan
terbuka untuk meningkatkan luas permukaan bagi
penyerapan cahaya semasa fotosintesis.
This condition enables the leaves to open to
increase the surface area for the absorption of
sunlight for photosynthesis.
• Semasa hari panas, tumbuhan layu.
During a hot day, the plant wilts.
• Sel tumbuhan dalam daun menjadi flasid kerana
penyejatan meningkat dan kadar kehilangan air
tinggi.
The cells in the leaves become flaccid as
evaporation increases and the rate of water loss
is high.
• Tumbuhan dan daun layu untuk mengurangkan
luas permukaan yang terdedah pada cahaya
matahari dan mengurangkan kehilangan air
daripada daun.
The plant wilts with the leaves drooping to reduce
the surface area exposed to sunlight and reduce
loss of water from the leaves.
• Stoma juga tertutup untuk mengurangkan kadar
transpirasi/menghalang kehilangan air melalui
transpirasi.
Stomata also close to reduce rate of transpiration/
prevent loss of water by transpiration
(ii) • Pada keamatan cahaya tinggi, kadar transpirasi
adalah tinggi.
At high light intensity, the rate of transpiration is
high.
• Stoma terbuka/The stomata are open.
• Fotosintesis berlaku dalam sel pengawal.
Photosynthesis occurs in the guard cells.
• Sukrosa dihasilkan/Sucrose is produced.
• Tekanan osmosis dalam sel pengawal meningkat.
The osmotic pressure in guard cells increases.
• Ion kalium dipam keluar daripada sel epidermis di
sekeliling ke dalam sel pengawal.
Potassium ions are pumped out from adjacent
epidermal cells into guard cells.
• Hal ini meningkatkan tekanan osmosis dalam sel
pengawal.
This increases the osmotic pressure in guard cells.
• Tekanan osmosis keseluruhan dalam sel pengawal
lebih tinggi daripada sap sel pada sel epidermis di
sekelilingnya.
•
•
•
•
(b) (i) •
•
•
•
•
(ii) •
•
•
•
•
The overall osmotic pressure in the guard cells is
higher than the cell sap of surrounding epidermal
cells.
Air meresap daripada sel epidermis ke dalam sel
pengawal melalui osmosis.
Water diffuses from epidermal cells into guard
cells by osmosis.
Sel pengawal mengembang dan menjadi segah.
The guard cells expand and become turgid.
Dinding dalam sel pengawal lebih tebal daripada
dinding luar. Maka, sel pengawal melengkung
keluar menyebabkan stoma terbuka.
The inner wall of guard cells is thicker than
the outer wall. Hence the guard cells curved
outwards causing the stoma to open.
Keamatan cahaya paling tinggi pada waktu
tengahari apabila stoma terbuka dan kadar
transpirasi adalah paling tinggi/pada tahap
maksimum.
Light intensity is highest around noon when
stomata are open and the rate of transpiration is
the highest/at its maximum.
Apabila otot fleksor di kaki belakang mengecut,
otot ektensor mengendur, dan kaki belakang
dilipat dalam bentuk Z.
When the flexor muscle on the hind leg contracts,
the extensor muscle relaxes and the hind leg is
folded into a Z shape.
Apabila belalang bersedia untuk melompat,
otot ekstensor mengecut manakala otot fleksor
mengendur.
When the grasshopper is prepared to jump/leap,
the extensor muscle contracts while flexor muscle
relaxes.
Kaki belakang diluruskan.
The hind legs are straightened.
Daya ke bawah dan ke belakang dihasilkan.
A downward and backward force is produced.
Hal ini menghasilkan daya ke atas dan ke hadapan
yang menolak belalang ke udara/untuk melompat.
This produces an upward and forward force
which propels the grasshopper up into the air
/to jump.
Lengkung pertumbuhan belalang adalah tidak
selanjar, berbentuk tangga. Telur menetas
menghasilkan nimfa yang pertama.
The growth curve of grasshopper is not
continuous, in the form of a series of steps. Egg
hatches to form the first nymph.
Nimfa serupa dengan belalang dewasa tetapi
mempunyai kurang segmen pada abdomen,
serta sayap dan organ pembiakan yang belum
berkembang sepenuhnya.
Nymph is identical to the adult grasshopper
except it has less segments in the abdomen and
wings as well as its reproductive organs are not
fully developed.
Setiap peringkat nimfa dikenali sebagai instar.
Each stage of the nymph is called instar.
Nimfa yang pertama merupakan peringkat instar
yang pertama.
The first nymph is the first instar stage.
Sewaktu peringkat instar, kadar pertumbuhan
adalah sifar, kerana pertumbuhan dihalang oleh
rangka luar yang keras. Namun begitu, banyak
aktiviti berlaku semasa peringkat instar:
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• Adrenalina meningkatkan tahap gula dalam darah/
adrenalina menyebabkan pertukaran glikogen
kepada glukosa
The adrenaline increases the level of blood sugar/
causes the conversion of glycogen to glucose
• Kadar degupan jantung dan kadar pernafasan
meningkat
The rate of heartbeat and breathing rate are
increased
• Lebih banyak oksigen dipam dengan cepat ke
otot sel untuk respirasi bagi menghasilkan lebih
banyak tenaga untuk menghadapi situasi tersebut.
More oxygen is pumped rapidly to the muscle
cells for respiration to produce more energy to
overcome the situation
• Aliran darah ke organ-organ meningkat
Blood flow to other organs increases
• Untuk meningkatkan kadar metabolisme dan
menghasilkan lebih banyak tenaga
To increase metabolic rate and produce more
energy
(ii) • Auksin dihasilkan di hujung pucuk/koleoptil
Auxin is produced at the tip of shoot/coleoptile
• Auksin meresap ke dalam zon pemanjangan sel
Auxin diffuses into the zone of cell elongation
• Lebih banyak auksin berkumpul di kawasan yang
jauh daripada sumber cahaya, menyebabkan
taburan auksin yang tidak sekata
More auxin accumulates in the region that is
away from the light source, causing an uneven
auxin distribution
• Kepekatan auksin lebih tinggi dalam kawasan
yang gelap
The concentration of auxin is higher in the dark
region
• Kepekatan auksin yang tinggi pada pucuk
merangsang pemanjangan sel
The high concentration of auxin in the shoot
stimulates cells elongation
• Kawasan yang lebih gelap dengan kepekatan
auksin tinggi, memanjang lebih cepat berbanding
dengan kawasan yang cerah
The darker region with higher concentration of
auxin elongates faster than the brighter region
• Pucuk membengkok dan tumbuh ke arah cahaya
matahari, menunjukkan fototropisme positif.
The shoot bends and grows towards the sunlight,
showing positive phototropism.
(b) (i) Ujian air kencing yang positif menunjukkan
kehadiran gula penurun. Ini bermakna air kencing
individu Y mengandungi glukosa. Individu Y
menghidap diabetes melitus.
Positive urine test shows that reducing sugar is
present. This means that the urine of individual Y
contains glucose. Individual Y has diabetes mellitus.
(ii) Kandungan gula yang tinggi dalam darah adalah
disebabkan oleh pankreas yang rosak. Akibatnya,
insulin tidak dihasilkan dan glukosa berlebihan tidak
dapat ditukar kepada glikogen. Glukosa berlebihan
dalam tubul ginjal tidak dapat diserap semula ke
dalam kapilari darah di sekelilingnya. Maka, glukosa
disingkirkan dalam air kencing, yang memberikan
keputusan ujian kencing positif.
The high level of sugar in the blood is due to a
damaged pancreas. As a result, insulin is not
produced and excess glucose cannot be converted to
During the instar stage, growth rate is zero,
because the growth is hindered by the hard
exoskeleton. But many activities occur in the
instar stage:
– Mitosis berlaku untuk menambahkan bilangan
sel dan meningkatkan jisim.
Mitosis occurs to increase the number of cells
and mass.
– Organ menjadi lebih kompleks.
Organs become more complex.
– Rangka baharu yang lembut dibina di bawah
rangka luar yang keras.
A new soft skeleton is built below the old hard
exoskeleton.
• Di akhir peringkat instar, ekdisis bermula.
At the end of the instar stage, ecdysis begins.
• Ekdisis merupakan peringkat menegak dalam
lengkung pertumbuhan serangga.
Ecdysis is the vertical stage in the growth curve
of insects.
• Semasa eksidisis, pertumbuhan adalah mendadak
dalam satu tempoh yang singkat. Rangka luar
dipecahkan untuk membolehkan pertumbuhan
berlaku.
During ecdysis, the growth is very rapid in a
short period of time. The outer skeleton is shed
(moulting) to allow growth to occur.
• Semasa
ekdisis,
udara
disedut
untuk
mengembangkan badan dan memecahkan rangka
luar lama yang keras.
During ecdysis, air is sucked in to expand the
body and to break the old hard exoskeleton.
• Sebelum rangka baru mengeras, pertumbuhan
mendadak berlaku.
Before the new exoskeleton hardens, rapid growth
occurs.
• Panjang dan saiz badan meningkat, organ
berkembang dan menjadi lebih kompleks.
The body length and size increases, the organs
develop and become more complex.
• Ekdisis tamat apabila rangka luar baharu
mengeras.
Ecdysis stops when the new exoskeleton harden.
• Nimfa memasuki peringkat instar seterusnya.
The nymph enters the next instar stage.
• Proses pertumbuhan berulang melalui setiap
instar dan ekdisis sehingga belalang mencapai
saiz maksimum dan menjadi dewasa.
The process of growth is repeated through
each instar and ecdysis until the grasshopper
achieves its maximum size and becomes an
adult.
Bahagian C
11 (a) (i) • Rangsangan bahaya dikesan oleh mata/organ
deria
The stimulus of danger is detected by the eye/
sensory organ
• Maklumat disampaikan ke otak dan diterjemah
sebagai ketakutan
The information is transmitted to the brain and
interpreted as fear
• Perasaan takut merangsang kelenjar adrenal untuk
merembeskan lebih banyak adrenalina
The feeling of fear stimulates the adrenal gland
to secrete more adrenaline
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After 10 minutes, record the shape of Visking tubing, the
colour of solution in the Visking tubing and the colour of
solution in the beaker.
(b) (i) Kekal coklat kekuningan/warna iodin
Remains yellowish brown/colour of iodine
(ii) Tiada kehadiran kanji. Kanji tidak dapat meresap keluar
daripada tiub Visking.
Starch is not present. Starch cannot diffuse out from the
Visking tubing.
(c) (i)
Sebelum
Selepas
glycogen. The excess glucose in the kidney tubules is
not reabsorb into the surrounding blood capillaries.
Hence, it is excreted in the urine, giving the test a
positive result.
(iii) Individu Y dinasihatkan untuk mengurangkan
pengambilan makanan yang kaya dengan karbohidrat.
Ini kerana, karbohidrat menghasilkan glukosa apabila
dicernakan dan akhirnya meningkatkan tahap glukosa
darah.
Individual Y is advised to reduce her/his intake of food
rich in carbohydrates. This is because carbohydrate
produces glucose when digested and eventually
increases the blood glucose level.
(iv) Selain itu, individu Y perlu disuntik dengan insulin
untuk menukarkan glukosa berlebihan kepada
glikogen. . Maka dengan itu dapat menurunkan tahap
glukosa dalam darah kembali ke julat normal.
Besides that, individual Y has to be injected with
insulin in order to convert the excess glucose to
glycogen and thus reducing the glucose level in the
blood to its normal range.
Before
Tiub Visking
lembut
Visking tubing
is soft
Ampaian
kanji jernih
Starch
suspension
is clear
After
Tiub Visking lebih
besar dan keras
Visking tubing
bigger and firm
Ampaian kanji
berwarna biru
gelap
Starch
suspension
is blue black
(ii) • Molekul iodin di dalam bikar boleh meresap ke dalam
tiub Visking dan menyebabkan larutan kanji bertukar
warna daripada jernih ke biru gelap.
Iodine molecules in the beaker can diffuse into the
Visking tubing and cause the starch solution to change
colour from clear to blue black.
• Molekul air di dalam bikar dapat meresap ke dalam
tiub Visking menyebabkan tiub Visking mengembang
dan segah.
Water molecules in the beaker can diffuse into the
Visking tubing, causing the Visking tubing to become
bigger and turgid/firm.
(d) Molekul air dan iodin adalah kecil manakala molekul kanji
adalah besar.
Water and iodine molecules are small whereas starch
molecules are large.
(e) (i) Larutan di dalam tiub Visking kekal jernih atau coklat
kekuningan, larutan di dalam bikar kekal coklat
kekuningan.
The solution in the Visking tubing remains clear or
yellowish brown, the solution in the beaker remains
yellowish brown.
(ii) Tiada kanji di dalam tiub Visking atau di dalam bikar.
Kanji telah dihidrolisiskan oleh amilase kepada maltosa.
Starch is not present in the Visking tubing or in the
beaker. The starch has been hydrolysed by amylase into
maltose.
(f) Membran separa telap adalah membrane yang membenarkan
molekul kecil seperti molekul iodin dan molekul air
merentasinya tetapi bukan molekul besar seperti kanji.
A semi-permeable membrane is a membrane which allows
smaller molecules such as iodine and water to pass through
it but not large molecules such as starch molecules.
KERTAS 3
(a) 1. Letakkan tiub Visking di bawah air paip yang mengalir
untuk melembutkannya.
Place Visking tubing below running tap water to soften it.
2. Ikat satu hujung tiub Visking dengan ketat menggunakan
benang untuk mengelakkan kebocoran.
Tie one end of the Visking tubing with thread tightly to
prevent leakage.
3. Isi tiub Visking dengan 20 ml ampaian kanji.
Fill the Visking tubing with 20 ml starch suspension.
4. Ikat satu lagi hujung tiub Visking dengan ketat
menggunakan benang.
Tie the other end of Visking tubing tightly with thread.
5. Bilas permukaan luar tiub Visking untuk menyingkirkan
sebarang kanji.
Rinse the outer surface of the Visking tubing to remove
any starch.
6. Rekodkan bentuk tiub Visking dan warna larutan di
dalamnya.
Record the shape of Visking tubing and the colour of
solution in it.
7. Isi bikar dengan 300 ml air suling dan 10 ml larutan
iodin.
Fill the beaker with 300 ml of distilled water and 10 ml
of iodine solution.
8. Rekodkan warna larutan tersebut.
Record the colour of the solution.
9. Rendam tiub Visking ke dalam bikar. Mulakan jam
randik.
Immerse the Visking tubing in the beaker. Start the
stopwatch.
10. Selepas 10 minit, rekod bentuk tiub Visking, warna
larutan di dalam tiub Visking dan di dalam bikar.
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