1202 Question Bank Biology Form 4

©PAN ASIA PUBLICATIONS

Contents





Must Know iii – viii Chapter 8 Respiratory Systems in
Humans and Animals
Chapter 1 Introduction to Biology and NOTES 73
Laboratory Rules Paper 1 74
NOTES 1 Paper 2 79
Paper 1 2
©PAN ASIA PUBLICATIONS
Paper 2 5 Chapter 9 Nutrition and the Human
Digestive System
Chapter 2 Cell Biology and NOTES 83
Paper 1
84
Organisation
NOTES 7 Paper 2 89
Paper 1 8 Reinforcement and Assessment
Paper 2 13 of Science Process Skills for
Paper 3 (Pratical Test) 93
Chapter 3 Movement of Substances across Chapter 10 Transports in Humans
the Plasma Membrane
NOTES 17 and Animals
Paper 1 18 NOTES 95
Paper 2 25 Paper 1 96
Reinforcement and Assessment Paper 2 101
of Science Process Skills for
Paper 3 (Pratical Test) 30 Chapter 11 Immunity in Humans
NOTES 104
Chapter 4 Chemical Compositions Paper 1 105
in a Cell
Paper 2
109
NOTES 31
Paper 1 32 Chapter 12 Coordination and
Paper 2 38 Response in Humans
Reinforcement and Assessment NOTES 112
of Science Process Skills for Paper 1 113
Paper 3 (Pratical Test) 42 Paper 2 119

Chapter 5 Metabolism and Enzymes Chapter 13 Homeostasis and the Human
NOTES 43 Urinary System
Paper 1 44 NOTES 122
Paper 2 48 Paper 1 123
Reinforcement and Assessment Paper 2 127
of Science Process Skills for
Paper 3 (Pratical Test) 53 Chapter 14 Support and Movement in
Humans and Animals
Chapter 6 Cell Division NOTES 130
NOTES 54 Paper 1 131
Paper 1 55 Paper 2 136
Paper 2 62
Chapter 15 Sexual Reproduction,
Chapter 7 Cellular Respiration Development and Growth in
NOTES 66 Humans and Animals
Paper 1 66 NOTES 139
Paper 2 70 Paper 1 140
Reinforcement and Assessment Paper 2 145
of Science Process Skills for
Paper 3 (Pratical Test) 72 Answers 148

ii




Contents 1202 Biology F4.indd 2 26/05/2022 5:14 PM

MUST


KNOW Important Facts







Chapter 1 Introduction to Biology and Laboratory Chapter 9 Nutrition and the Human Digestive
Rules System
1. Biology is the study of all processes related to life and living. 1. A balanced diet is the diet that contains all seven classes
of food in the right proportions and balanced quantities
2. Safety and rules in the biology laboratory is important to ensure according to one’s needs so that optimal health can be
©PAN ASIA PUBLICATIONS
the safety of the student conducting the experiment, the purity of maintained.
the sample and the cleanliness of the laboratory.
Chapter 10 Transports in Human and Animals
Chapter 2 Cell Biology and Organisation
1. Haemophilia is a hereditary disease caused by the lack or
1. Cell is the basic unit of all living organisms. absence of clotting factors in the blood.

2. Cell → Tissue → Organ → System → Organism 2. A blood clot is a thrombus. If the thrombus is transported by
blood flow, it is called an embolus.
3. Revise the lessons you’ve learned in Form 1.




Important Facts (Chapter 1 & 2) 1 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 9 & 10) 7 @ Pan Asia Publications Sdn. Bhd.


Chapter 4 Chemical Composition in a Cell Chapter 12 Coordination and Response in Humans
1. Water, carbohydrates, proteins, lipids and nucleic acids 1. The nervous system is divided into the central nervous
make up the components in a cell. system and the peripheral nervous system.
2. There are two types of carbohydrates, which are 2. Acetylcholine, noradrenaline, serotonin and dopamine are
monosaccharide and polysaccharide. types of neurotransmitters.
3. Nucleotides are made up of nitrogenous bases, pentose 3. A reflex action is an automatic response that occurs rapidly
sugar and phosphate. and without being controlled by the brain.
4. Drug and alcohol abuse affects the normal functions of the
Chapter 5 Metabolism and Enzyme
nervous system.
1. Anabolism is the process of synthesis of complex molecules
from simple molecules.
2. Catabolism is the process of breaking down complex
compounds into simple molecules.


Important Facts (Chapter 4 & 5) 3 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 12) 9 @ Pan Asia Publications Sdn. Bhd.


Chapter 7 Cellular Respiration Chapter 14 Support and Movement in Humans and
Animals
1. Aerobic respiration is the breakdown of glucose involving
oxygen to produce chemical energy. 1. The three types of skeletons are the hydrostatic skeleton,
exoskeleton and endoskeleton.
2. This occurs in the mitochondrion and cytoplasm.
2. Antagonistic muscles work in the movement of the forearm
3. Fermentation is the process of incomplete decomposition of and leg muscles when we are walking.
glucose in conditions of limited oxygen supply or without
oxygen at all. 3. Earthworms move using the peristalsis movement.
4. This occurs in the cytoplasm. 4. Osteoporosis is very common among older women because
a low oestrogen level can reduce bone density.
5. Human muscle cells will undergo oxygen debt when it
carries out vigorous training.







Important Facts (Chapter 7) 5 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 14) 11 @ Pan Asia Publications Sdn. Bhd.



Must Know 1202 Biology Form 4.indd 1 24/05/2022 10:38 AM

MUST


KNOW Mnemonics & Important Diagrams







Mnemonic for Biology Biology
comes from Greek word
Cell cycle
logos – means
©PAN ASIA PUBLICATIONS
Interphase Metaphase Telophase Bios – means life Biology
Prophase Anaphase Cytokinesis knowledge
Definition – science of life
by studying the life processes
I Prefer Milk And Tasty Cupcakes and of living organisms
&







Mnemonics (Chapter 6) 13 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 1) 19 @ Pan Asia Publications Sdn. Bhd.


Types of Interference Amoeba
Different leukocyte
Never Let Monkeys Eat Bananas Pseudopodium Amoeba
Projection of Prey Pseudopodia Food vacuole
Neutrophils (60%) Never Contractile cytoplasm that 1 2 3 4
causes
vacuolea
Lymphocytes (30%) Let Osmoregulation movement
Monocyte (6%) Monkeys occurs Feeding of Amoeba - (phagocytosis)
Plasma
Eosinophils (3%) Eat membrane Food
vacuole
Basophils (1%) Bananas Respiration Nucleus Reproduction of Amoeba
(binary fission)
occurs
Cytoplasm
1 2 3 4




Mnemonics (Chapter 11) 15 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 2) 21 @ Pan Asia Publications Sdn. Bhd.


Synapse Chemical Compounds in Cell

Chemical components
synapse

The space between two neurones Organic Non-organic
compounds compounds



Carbohydrates Nucleic Water Mineral
acids salts
Lipid

Protein




Mnemonics (Chapter 12) 17 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 4) 23 @ Pan Asia Publications Sdn. Bhd.



Must Know 1202 Biology Form 4.indd 3 24/05/2022 10:38 AM

MUST


KNOW Important Diagrams






Types of Immunity in Humans Cell Division


Acquired immunity
Immunity developed in life Mitosis Interphase
Meiosis
hormones from ©PAN ASIA PUBLICATIONS
Prophase Prophase I Prophase II
Active immunity Passive immunity
Immunity acquired after Immunity acquired Metaphase Metaphase I Metaphase II
exposed to an infection from someone or Meiosis I Meiosis II
or getting vaccine something Anaphase Anaphase I Anaphase II

Telophase Telophase I Telophase II
Natural Artificial Natural Artificial Cytokinesis Cytokinesis I Cytokinesis II
Antibodies Antibodies Antibodies are Antibodies
are made after are made transmitted acquired
exposed to after from mother from serum
infection vaccination to baby medicine

Important Diagrams (Chapter 11) 32 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 6) 26 @ Pan Asia Publications Sdn. Bhd.


Homeostasis Human Respiratory System


3 Nerve impulse is Control 4 Nerve impulse Nasal cavity
sent by different centre is sent out by
neurone motor
neurone Pharynx
Receptor Effector Larynx
Trachea
Homeostasis Lungs Bronchiole

2 Detected
by Bronchi Alveoli
Beyond or Below 5 Response
1 Change returns
in to balance
variable Diaphragm
Variable (Normal state)

Important Diagrams (Chapter 13) 34 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 7) 28 @ Pan Asia Publications Sdn. Bhd.


Relationship between hormones and thickness Human Blood Circulatory System
of uterus
Pulmonary Lungs Pulmonary
Luteinising hormone (LH)
Level of artery vein
hormones from Follicle stimulating hormone Pulmonary
(FSH)
pituitary gland circulation
Aorta
Level of Progesterone Vena cava
Oestrogen
ovary
Left
Right
Uterus atrium atrium
Uterine
wall
Thickness of Menstruation Right Systemic Left
ventricle
endometrium ventricle
circulation
Days in
2 4 6 8 10 12 14 16 18 20 22 24 26 28 menstrual Body
Menstruation Endometrium increases Secretory phase cycle
in thickness Ovulation
Important Diagrams (Chapter 15) 36 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 8) 30 @ Pan Asia Publications Sdn. Bhd.

Must Know 1202 Biology Form 4.indd 6 24/05/2022 10:38 AM

3 Movement of Substances across the
Chapter Plasma Membrane





NOTES


3.1 Structure of Plasma Membrane
©PAN ASIA PUBLICATIONS
• Hydrophilic head
– Polar head that is attracted to water
Phospholipid 
• Hydrophobic tail
– Non-polar tail that is not attracted to water
Glycolipid
– For cell recognisation
Carbohydrate chain
Outside plasma
Glycoprotein membrane
Globular protein


Hydrophilic head
Phospholipid bilayer
– As a barrier
Hydrophobic tail
to separate
both sides of
Hydrophilic head
membrance
Peripheral protein
Inside plasma
Cholesterol Pore protein Carrier protein membrane
– Makes the plasma – Enables small – As a carrier to transport
membrane more molecules and large molecules such as
stable, stronger ions to pass glucose, amino acids and
and flexible through nucleic acids



3.2 Concept of Movement of Substances Across a Plasma Membrane


Movement of substances across
the plasma membrane





Passive transport Active transport
• Do not require energy or ATP • Requires energy or ATP
• Moves down the concentration gradient • Moves against concentration gradient
• Dynamic equilibrium is achieved • Accumulation or elimination of substances



Hypotonic solution – Solution that has lower
concentration of solutes than the other solution.
Simple Facilitated Osmosis Isotonic solution – Solution that has similar
diffusion diffusion
concentration of solutes with the other solution.
Hypertonic solution – Solution that has higher
concentration of solutes than the other solution.




17




C03 1202 Biology Form 4.indd 17 24/05/2022 4:33 PM

10
Chapter Transport in Humans and Animals





NOTES



10.1 Types of Circulatory System

Lung capillaries Lung capillaries
Heart
Gill capillaries
1 ventricle
Heart 2 atria 2 atria
1 atrium 1 ventricle 2 ventricle
Heart Heart
Body capillaries
Body capillaries Body capillaries
Haemocoel
Insect Fish Amphibian Human

10.3 Mechanism of heartbeat



Blood group compatibility
SA
node SA node generates Donor Recipient
electrical impulses The impulse travels
along the AV bundle
and the bundle O O
Stimulus spreads across branches to the
the atrial surface and Purkinje fibres to
AV ©PAN ASIA PUBLICATIONS
reaches the AV node
node whole ventricle walls A A
Both atria contract Both ventricles B B
AV simultaneously and contract
bundle
blood is forced into simultaneously and
the ventricles blood is pumped out
Moderator Bundle Purkinje AB AB
band branches fibres of the heart



Clotting factor
(Thrombokinase) Calcium and vitamin K

Prothrombin Thrombin
Fibrinogen (soluble) Fibrin (Insoluble)

Broken Activated Fibrin Blood
vessel platelet strands clot










Damaged blood vessel Formation of platelet plug Development of clot
Injury to vessel lining triggers Vasoconstriction limits blood flow Fibrin strands adhere to the
the release of clotting factors and platelets form a sticky plug plug to form an insoluble clot



95




C010 1202 Biology Form 4.indd 95 20/04/2022 3:15 PM

PAPER 1


Each question has four different answers A, B, C and D. For each question, choose one answer only.

3.1 Structure of Plasma Membrane 5. Diagram 2 shows the fluid mosaic model of plasma
membrane.
1. The term ‘fluid’ in the fluid mosaic model refers to W
A the lipid molecule shows a bit of movement to X Y
avoid becoming packed.
©PAN ASIA PUBLICATIONS
B the main compartment of the lipid bilayer that
contains fluid in mosaic formation.
C the ability of the lipid to attract water.


2. The movement of particles through the plasma
membrane will be regulated by plasma membrane.
Diagram 2
Which of the following does not explain the What is Y?
importance of the function of the plasma membrane? A Pore protein
HOTS Applying B Carrier protein
A To provide nutrients for metabolism. C Glycoprotein
B To eliminate toxic waste products.
C To secrete useful substances. 6. Diagram 3 shows the structure of plasma membrane.
D To maintain the shape of plant cells.
R
S
3. Diagram 1 shows plasma membrane which consists  Q
of molecules arranged in a double layer. T
U
P
P

Q
Diagram 3
Which structures are hydrophobic?
A P and Q C R and S
B Q and S D T and U

Diagram 1 7. Diagram 4 shows the structure of plasma membrane.
The parts labelled as P and Q are R
A both hydrophobic. S  U
B both hydrophilic.
C hydrophilic and hydrophobic respectively. T
D hydrophobic and hydrophilic respectively.
Diagram 4
4. Which structure attracts water?
A R C T
• X is a component of plasma membrane. B S D U
• X is the precursor to synthesize steroid.
8. Structure P makes the plasma membrane become
What is X? HOTS Applying more flexible by strengthening and stabilizing the
A Triglyceride plasma membrane. What is P? HOTS Applying
B Protein
A Glycolipid
C Glycoprotein
B Phospholipid
D Cholesterol
C Cholesterol
D Glycoprotein
18
Question 2: Question 8:
SOS TIP Question 4: SOS TIP
Consider the functions of the plasma membrane.
P is a large, lipid molecule.
Steroid needs a substance with two five-carbon rings and two six-carbon
rings.





C03 1202 Biology Form 4.indd 18 24/05/2022 4:33 PM

9. When phospholipid molecules are placed in water, the 14. A concentration gradient exists when HOTS Applying
force that promotes formation of the bilayer is
A a semi-permeable membrane is present.
HOTS Applying B the volume of solution is unequal.
A the hydrogen bond that exists between C a membrane carries charges.
phospholipid and water molecules. D the solute concentrations on the two sides of a
B the electrostatic force that exists between membrane are different.
phospholipid and water molecules.
C the hydrophobic interaction between the fatty
acid tail and water molecules. 15. Which of the following diagrams shows osmosis
D the hydrophilic interaction between charged taking place in a root hair of a plant? HOTS Analysing
©PAN ASIA PUBLICATIONS
phospholipid molecules and water molecules. Keys:
10. The relative impermeability of membranes to water- Solute molecule
soluble molecules is mainly due to the HOTS Analysing
Water molecule
A presence of phospholipids in the lipid bilayer
B presence of cholesterol on the membranes Cell wall and plasm membrane
C presence of large proteins that extend through
both sides of the membranes
A C
D non-polar nature of water molecules
Soil Root hair Soil Root hair
3.2 Concept of Movement of
Substances Across a Plasma
Membrane
B D
11. Which of the following substances can be transported
by simple diffusion? Soil Root hair Soil Root hair
I Fatty acids III Carbon dioxide
II Glucose IV Amino acids
A I and II C II and IV
B I and III D III and IV
16. Diagram 5 shows a U-tube with two types of solutions
12. Among the types of membrane below, which
and a membrane between them.
membrane performs the fastest rate of diffusion?
HOTS Analysing
A C
20% 0.5%
sucrose sucrose
solution solution
P Q
Membrane with Membrane with a
microvilli rounded layer of cells
Semi-permeable membrane
B D
Diagram 5
At the end of the experiment, HOTS Analysing
I the concentration of P is higher than Q.
II the water level in P increases while the water
Membrane with Membrane with a level in Q decreases.
a villus flat layer of cells
III the water level in Q increases while the water
level in P decreases.
13. Which of the following processes involves the IV the osmotic pressure in P is equal to Q.
movement of substances across membranes with the A I and II
help of carrier proteins but without the requirement of B I and III
energy? C II and IV
A Active transport C Simple diffusion D III and IV
B Osmosis D Facilitated diffusion
19
Question 9: Question 12:
SOS TIP Question 10: Question 16: SOS TIP
Consider the component in the phospholipid molecule.
Consider from the perspective of surface area.
Consider the contents on both sides.
Think about the structure of the plasma membrane.





C03 1202 Biology Form 4.indd 19 24/05/2022 4:33 PM

17. Diagram 6 shows an experiment with osmosis 20. How are simple diffusion and facilitated diffusion
process. similar? HOTS Applying
I Both processes involve the movement of
substances from a region of higher solute
concentration to a region of lower solute
concentration.
II Both processes involve the movement of
substances from a region of lower solute
concentration to a region of higher solute
X Y concentration.
©PAN ASIA PUBLICATIONS
III Both processes do not need energy.
Diagram 6 IV Both processes occur with the help of transport
What would be the correct solutions for X and Y that proteins.
would cause the water level (indicated by the arrow) A I and II C II and IV
in the capillary tube to rise the highest after an hour? B I and III D III and IV
HOTS Analysing
21. Diagram 8 shows the active transport of sodium ions
X Y across a plasma membrane.
A 25 % sucrose solution Water Outside of cell
B 35 % sucrose solution Water
Inside of cell Na +
C 25 % sucrose solution 15 % sucrose solution

18. Diagram 7 shows the movement of solute molecules Na + Na +
ATP
through plasma membrane.

Diagram 8
Which of the following is true? HOTS Applying
Movement of solute molecules
A Sodium ions move from a region of higher
solute concentration to a region of lower solute
concentration.
B Dynamic equilibrium of sodium ions will be
Higher concentration Lower concentration achieved at both sides of the cell.
of molecules of molecules C Energy from the ATP (adenosine triphosphate)
Diagram 7 molecule causes the shape of the carrier protein
to change.
What type of movement of molecules are shown in
the information given in Diagram 7? HOTS Analysing D The concentration of sodium ions on the outside
I Osmosis of water through membrane. of the cell is lower than on the inside of the cell.
II Active transport through active proteins.
22. Diagram 9 is a graph that shows the concentration of
III Simple diffusion through phospholipid.
ions in pond water and in the cell sap of algae that
IV Facilitated diffusion through carrier proteins.
live in the pond.
A I and II C II and IV
B I and III D III and IV Concentration of ion Keys:
Pond
water
19. Which of the following processes will take place
when the root hair cells are deprived of oxygen? Cell sap of
Nitella sp.
HOTS Analysing
A Diffusion and osmosis
B Diffusion only
C Active transport only Na + K + Ca Mg 2+ Cl – Mineral ions
2+
D Active transport and osmosis
Diagram 9
20
Question 17: Question 21:
SOS TIP Question 19: SOS TIP
Take note of the presence of ATP.
Relate to the difference of concentration.
Relate oxygen to the cellular respiration.






C03 1202 Biology Form 4.indd 20 24/05/2022 4:33 PM

PAPER 2


Section A

Answer all questions.

1. Diagram 1 shows two different types of movement of substances across the plasma membrane.

Glucose
Outside Potassium ion
the cell
X

Y

Inside
the cell
ATP

Diagram 1

(a) Based on Diagram 1, what are X and Y? [2 marks]





(b) State one characteristic of structure Y. [1 mark]



(c) Certain molecules can pass through structure Y by simple diffusion. State one characteristic of these molecules.
[1 mark]
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(d) Glucose molecules and potassium ions are transported across the plasma membrane through the processes shown
in Diagram 1.
(i) Explain how glucose molecules are transported across the plasma membrane. HOTS Applying [2 marks]








(ii) Describe how potassium ions are transported into the cell. HOTS Applying [2 marks]
















25
Question 1:
SOS TIP (ii) Take note of the charge on the potassium ions. SOS TIP
(d) (i) Consider the characteristic of the plasma membrane.








C03 1202 Biology Form 4.indd 25 24/05/2022 4:33 PM

2. Diagram 2.1 shows two regions, X and Y which are separated by a semi-permeable membrane.
Semi-permeable
membrane
Water molecule

Sucrose
molecule
Region X Region Y
Diagram 2.1
©PAN ASIA PUBLICATIONS
(a) (i) A dynamic equilibrium is achieved between region X and Y after 30 minutes.
Complete Diagram 2.2 by showing: HOTS Analysing
• The number of sucrose molecules and water molecules in both regions after 30 minutes.
• The level of the solution in both regions after 30 minutes.









Region X Region Y
Diagram 2.2
[2 marks]
(ii) Name the process involved in the result in 2(a) (i). [1 mark]

(b) (i) Diagram 2.3 shows the condition of okra after being immersed in tap water.


Tap water After being immersed for
three hours
Okra is cut halfway
longitudinally Final observation
Initial observation
Diagram 2.3
Explain the observation based on Diagram 2.3. HOTS Evaluating [2 marks]





(ii) Diagram 2.4 shows a fish immersed in a concentrated salt solution.


After being immersed
for three hours
Concentrated
salt solution Salted fish
Initial observation
Diagram 2.4
Based on Diagram 2.4, explain why salted fish can be stored for four months. HOTS Creating [2 marks]



26
Question 2:
SOS TIP (b) (i) Consider the movement of water molecules. SOS TIP
(a) (i) Think about the movement of the water and solute particles.
(ii) Relate to the process that takes place.







C03 1202 Biology Form 4.indd 26 24/05/2022 4:33 PM

3. A small amount of potassium manganate (VII) is put into a beaker of water as shown in Diagram 3.1.

Potassium manganate(VII) Purple solution of
crystal potassium manganate(VII)









At the beginning At the end
©PAN ASIA PUBLICATIONS
of experiment of experiment
Diagram 3.1
(a) (i) How does the process in Diagram 3.1 occur? HOTS Applying [2 marks]








(ii) Several examples of movement of substances in daily life are shown in Table 3.1.
In Table 3.1, tick (3) on the correct examples of the process as shown in Diagram 3.1.
Table 3.1

A drop of red ink is added into a glass of water, causing the water to turn red.

Gaseous exchange between the alveoli and blood cells.

Mineral ions move from soil water into the root hair cells.

[2 marks]
(b) Diagram 3.2 shows the steps in preparing homemade crispy sweet potato chips.









1) Cut the sweet 2) Soak the sliced 3) Rinse a few times 4) Fry the sliced sweet
potatoes into sweet potatoes and dry the sliced potatoes in hot cooking
thin slices. into concentrated sweet potatoes. oil.
salt solution.

Diagram 3.2
In your point of view, explain which step is important to produce crispy sweet potato chips. HOTS Creating
[3 marks]










27
Question 3:
SOS TIP SOS TIP
(a) (i) Consider the process of the movement of solute particles.








C03 1202 Biology Form 4.indd 27 24/05/2022 4:33 PM

SECTION B
5. (a) Explain why plasma membrane is known as a semi-permeable membrane. [5 marks]

(b) Diffusion and osmosis are always applied in food industries.
Describe how these concepts help in food production. HOTS Analysing [5 marks]

(c) Describe how
(i) water is absorbed by plant roots.
(ii) oxygen is absorbed by Paramecium sp.
©PAN ASIA PUBLICATIONS
(iii) iodine is absorbed by an alga (Nitella sp.) which lives in the sea. [10 marks]
SECTION C

6. (a) Give the meaning of the terms
(i) haemolysis
(ii) plasmolysis [4 marks]

(b) There are a variety of conditions that may lead to haemolysis of erythrocytes in the human body. Certain
medications, infections, toxins, poison or a very low solute concentration may contribute to the premature death
of erythrocytes due to haemolysis.
Diagram 5.1 is a graph that shows the percentage haemolysis of erythrocytes when the erythrocytes are immersed
in sodium chloride solution with different concentrations.
Percentage of erythrocytes undergo haemolysis (%)

100
80
60

40

20
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.0
Concentration of sodium chloride solution (%)
Diagram 5.1
Based on Diagram 5.1, explain how different concentrations of sodium chloride solution affect the condition of
erythrocytes. HOTS Analysing [8 marks]
(c) Diagram 5.2 shows a plant in one condition.













Diagram 5.2
Based on Diagram 5.2, explain what happens to the plant and two possible reasons that cause the plant to have
this condition. HOTS Evaluating [8 marks]

29
Question 5:
SOS TIP SOS TIP
(b) Consider the water content in the treated food.








C03 1202 Biology Form 4.indd 29 24/05/2022 4:33 PM

SECTION C
7. (a) State six properties of water. [6 marks]
(b) Briefly describe the significance of water to living organisms. [6 marks]
(c)
There are several components in cells that forms organic and inorganic compounds. Organic compounds such
as carbohydrates, proteins, lipids and nucleic acids are normally found in cells and each of the compound
plays an important role in the cells.

Explain the importance of carbohydrates and proteins in a cell. HOTS Evaluating [8 marks]
8. Briefly explain the molecular structures, properties and location to find the following substances.
©PAN ASIA PUBLICATIONS
(a) Triglycerides [6 marks]
(b) Phospholipid [8 marks]
(c) Steroid [6 marks]

Reinforcement and Assessment of Science Process Skills
for Paper 3 (Practical Test)


You are required to carry out an experiment to determine the presence of reducing sugar and non-reducing sugar.
Procedure: Glucose solution
1. 1 ml of Benedict’s solution and 1 ml of glucose solution are +
Benedict’s solution
put into the same boiling tube.
Benedict’s
2. The solution is heated in a water bath for 5 minutes. solution Water bath
3. The changes that occurred in the contents of the boiling tube
are recorded.
4. Steps 1 to 3 are repeated on glucose solution, fructose
solution, maltose solution, lactose solution and sucrose
solution.
Diagram 1
5. The result of the experiment is shown in the following table.
Solution Type of sugar Initial colour of solution Colour change of solution
Distilled water
Glucose
Fructose
Maltose
Lactose
Sucrose
(a) Complete the table above. [3 marks]
(b) Based on the results in the table, state two inferences. [2 marks]
(c) State the variables in this experiment.
(i) Manipulated variable (ii) Responding variable (iii) Fixed variable [3 marks]
(d) Describe how you
(i) manipulate the variable
(ii) observe the responding variable
(iii) set the controlled variable [3 marks]
(e) Why is distilled water used in this experiment? [1 mark]
(f) State the operational definition for reducing sugar. [1 mark]
(g) The mixture of Benedict’s solution and sugar is heated using a water bath. Explain why a water bath is used.
[1 mark]
(h) State the conclusion for this experiment. [1 mark]

Please download sample data from the QR code at the back cover of this book in order to answer the questions.
42
Question 7:
SOS TIP SOS TIP
(c) Consider the functions of carbohydrates and proteins in the cell.








C04 1202 Biology Form 4.indd 42 27/05/2022 9:55 AM

Reinforcement and Assessment of Science Process Skills
for Paper 3 (Practical Test)


SAMPLE DATA

Chapter 3 Chapter 5


Note: If you are unable to conduct the experiments Note: If you are unable to conduct the experiments
©PAN ASIA PUBLICATIONS
described in Page 30, you can answer this question described in Page 53, you can answer this question
using the sample data obtained from the observation using the sample data obtained from the observation
in Table 1 shown below. in Table 3 shown below.
Table 1 Table 3

Initial length Final length Time taken for the
Solution of potato of potato Test Temperature hydrolysis of starch to
strip (mm) strip (mm) tube (°C) be complete (minutes)

30 ml of distilled 50 50
water A2 0 Not completed after
10 minutes
30 ml of 0.5 M 50 46
sucrose solution B2 28 2

30 ml of 1.0 M
sucrose solution 50 41 C2 37 1
D2 45 3

Chapter 4 Not completed after
E2 60 10 minutes
Note: If you are unable to conduct the experiments
described in Page 42, you can answer this question
using the sample data obtained from the observation
in Table 2 shown below. Chapter 7
Table 2 Note: If you are unable to conduct the experiments

Initial colour Colour change described in Page 72, you can answer this question
Solution
of solution of solution using the sample data obtained from the observation
in Table 4 shown below.
Distilled water Blue Remain blue
Table 4
Brick-red
Glucose Blue
precipitate Reagent bottle Change in limewater
Brick-red
Fructose Blue Presence of cockroach Turns cloudy
precipitate
Brick-red Without cockroach Unchanged
Maltose Blue
precipitate
Brick-red
Lactose Blue
precipitate
Sucrose Blue Remain blue








ii




Sample Data 1202 Biology F4.indd 2 25/05/2022 11:48 AM

Answers Complete Answers (Paper 1)

https://bit.ly/3JtHzOl



CHAPTER 1 experiment.
In the fifth step, the researcher starts to carry out investigation
by handling the apparatus, materials and specimens used
Paper 1 correctly.
Next, the researcher can collect and present data in the
1. C 2. A 3. D 4. A 5. D experiment accurately.
6. B 7. A 8. D 9. A 10. B After this step, the researcher needs to analyse data and
11. A 12. A 13. C 14. C 15. B interpret data by analysing the relationship between the
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manipulated variable and the responding variable. [2]
16. D 17. B 18. A 19. C 20. B Finally, the researcher can make conclusion and write a report
21. D 22. C 23. D 24. A 25. A about the experiment.
26. B 27. C 28. C 29. B 30. D
SECTION C
PAPER 2 4. (a) Procedure:
1. Prepare six sets of the apparatus shown in Diagram 1 in
the question. The weight and the size of the plant, and the
SECTION A number of leaves should be the same. [2]
1. (a) (i) Presence of yeast in the dough 2. Place each set 10 cm from the light source.
(ii) Size of dough 3. The water temperature in each beaker is maintained at
(iii) Initial size of dough 10°C, 20°C, 30°C, 35°C, 40°C and 50°C respectively.
(b) The dough with yeast will increase in size after baking. 4. For each set, the number of bubbles released per minute is
Chapter 1 – Chapter 2
(c) counted.
5. Record the results in a table. Draw a graph for number of
Type of dough Size of dough bubbles released per minute against temperature. [2]
(b) Manipulated variable: Temperature
Dough P (with yeast) Increase in size Responding variable: Number of bubbles released per minute
Fixed variable: Carbon dioxide concentration and light
Dough Q (without yeast) Size remains the same intensity
(d) The presence of yeast causes the dough to rise and increase in (c) Personal protective equipment that can be worn during a
size. surgical experiment is a mask, safety googles, gloves and
laboratory coat.
2. (a) (i) Type of fertilisation These equipment are worn to prevent from direct contact with
(ii) The alleles representing the hereditary traits spurted blood or other bodily fluids.
(iii) Nutrients provided to the plants (d) Determine the problem statement, make a hypothesis, plan an
(b) A cross of tall pea plant with dwarf pea plant results in tall investigation, determine and control variables, perform the
progeny. experiment, collect, analyse, then interpret the data, make an
(c) The data is analyzed using statistics (probability). conclusion and write a report.
(d) Results show that the dominant trait would mask the recessive
trait in a heterozygous cross but when the heterozygous
plants are allowed to self-fertilise, the recessive trait can be CHAPTER 2
recovered.

SECTION B PAPER 1
1. A 2. C 3. D 4. A 5. B
3. (a) Biology is the organized study of life and living things, and the 6. B 7. D 8. C 9. A 10. B
interactions with their natural and physical environment. [2]
The study or knowledge of Biology is important because it 11. D 12. A 13. C 14. C 15. A
helps us to better understand the interactions in life. 16. D 17. D 18. B 19. A 20. A
Other than that, it is applied in many fields such as 21. C 22. B 23. D 24. D 25. A
biotechnology, genetic engineering and food technology.
Furthermore, it is the core subject in an array of careers such 26. A 27. B 28. C 29. C 30. A
as medicine, nursing and research. 31. D 32. B 33. A 34. A 35. C
(b) The five different fields of study are ecology, botany, zoology, 36. D 37. D 38. B 39. C 40. A
microbiology and biotechnology. [5]
(c) The scientific investigation method is the process of gathering 41. D 42. B 43. A 44. C 45. A
facts based on an observable event or phenomenon.
In biological research, the first step is to identify the PAPER 2
problem.
Next, the researcher needs to construct a hypothesis which is
a tentative explanation for an observed event. SECTION A
In the third step, the researcher needs to identify and control
variables in the experiment by determining the manipulated 1. (a) (i) Nucleus
variable, the responding variable and the fixed variable. (ii) It modifies proteins from the rough endoplasmic
After that, plan the investigation which involves collecting reticulum to form glycoproteins.
relevant information or the scientific background of an (b) It contains grana with chlorophyll to capture light energy and
experiment, preparing the apparatus and the materials convert the light into chemical energy (glucose).
needed, and determining the procedures in conducting the (c) R will shrink and the cell is plasmolysed. The water diffuses

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Answer 1202 Biology Form 4.indd 148 24/05/2022 10:43 AM

(b)
Reinforcement and Assessment of Science Process Skills
for Paper 3 R
+
(Practical Test) Substrate
(a)
Enzyme + Substrate
Type of Initial colour Colour change
Solution
sugar of solution of solution
Distilled No sugar Blue Remain blue
water present
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Glucose Reducing Blue Brick-red
sugar precipitate Enzyme – substrate complex

Fructose Reducing Blue Brick-red
sugar precipitate
+
Maltose Reducing Blue Brick-red
sugar precipitate Product
Enzyme + Product
Lactose Reducing Blue Brick-red
sugar precipitate 2. (a) (i) Lactose
(ii) The enzyme is highly specific because it has a specific
shape of active site, which means only the shape of
Chapter 4 – Chapter 5
Sucrose Non-reducing Blue Remain blue substrate X is complimentary to the shape of the active
sugar site and fits into it.
(iii) Lactase enzyme is not used up or destroyed after the
(b) Reducing sugar is present in the solution. reaction. It can be reused to combine with other substrate
Non-reducing sugar is present in the solution. molecules for the reaction.
(c) (i) Type of sugar present (b) Molecule P is an inhibitor. It can mimic substrate X, thus it can
(ii) Colour change at the end of experiment compete with substrate X to combine with the lactase enzyme
(iii) Volume of solution at the active site to slow down or stop the enzymatic reaction.
(d) (i) Add the same volume of the different solutions into Benedict’s
solution. 3. (a) Protease and lipase enzymes can hydrolyse the protein and
(ii) Observe the colour change on the mixture. lipid into their monomers. Hence, the baby can easily consume
(iii) Replace the sugar solution with distilled water. and digest the food.
(e) To act as a control set for the experiment. (b) Enzyme pectinase is applied to the apple juice to break down
(f) Reducing sugar is the sugar that undergoes a colour change from pectin in the apple juice to make the apple juice looks clear.
blue to brick-red precipitate in the presence of Benedict’s solution. Other than that, pectinase also increases the yield of apple
(g) It is easier to maintain constant temperature when using a water juices.
bath. (c) Enzymes can catalyse biochemical reaction as it can lower the
(h) Glucose, fructose, maltose and lactose are reducing sugars but activation energy of the biochemical reaction to speed up the
sucrose is a non-reducing sugar. reaction.
4. (a) S1: There are more enzymes compared to substrates and the
CHAPTER 5 rate of reaction of enzyme is low.
S2: The substrate concentration increases but there are still
more enzymes than substrates.
PAPER 1
S3: The amount of enzymes is equal to the amount of
1. A 2. B 3. A 4. C 5. D substrates. The rate of reaction approaches maximum capacity.
6. D 7. C 8. A 9. C 10. A S4: There are more substrates than enzymes and the rate of
reaction of enzyme remains at its maximum.
11. B 12. C 13. D 14. A 15. A (b) All active sites of enzymes have been bound with substrates.
16. D 17. B 18. A 19. C 20. B Hence, the increase of substrate concentration does not cause
21. B 22. C 23. D 24. A 25. A the increase of the rate of reaction.
26. B 27. C 28. C 29. A 30. A (c) Increase the enzyme concentration. The V max increases
proportionately to the enzyme concentration.
Rate of reaction
PAPER 2
SECTION A 2 V max 2x enzyme
concentration
1. (a) (i) Organelle X (nucleus) contains genetic information in
the deoxyribonucleic acid that can be transcribed into 2 V 1x enzyme
mRNA, which then carries it out to the cytoplasm. max concentration
(ii) Protein that is formed at the ribosome enters the rough
endoplasmic reticulum and is transported into the Golgi
apparatus. The protein will be modified into extracellular
enzyme in the Golgi apparatus and then packed into the Substrate
Golgi apparatus for excretion. concentration
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Answer 1202 Biology Form 4.indd 154 24/05/2022 10:43 AM

Answers






CHAPTER 1 5. B Centrioles produce spindle fibres to help the movement of
chromosomes during cell division.
6. B Matured erythrocytes do not contain nucleus because it gives
Paper 1
more space for haemoglobin.
1. C Biology is the science of life and living things. 7. D –
2. A Biology is a branch of scientific knowledge and encompasses 8. C Pancreatic cells contain high amount of rough endoplasmic
manipulative techniques in science. reticulum to produce digestive enzymes.
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3. D – 9. A Lysosome contains digestive enzymes which also named as
4. A Biology curriculum enables students to understand living things hydrolytic enzymes to break down the macromolecules and
and sharpens students’ thinking. worn-out organelles.
5. D Biology is the science that covers only living things, and the 10. B Organelle P is rough endoplasmic reticulum which function to
relationship between all living things and the environment. transport protein which is synthesised by ribosome.
6. B Gregor Johann Mendel is the Father of Modern Genetics. 11. D Endurance training increases the number and size of
7. A Gynaecology is a medical practice that addresses the health of mitochondria.
the female reproductive system. 12. A –
8. D Kitchen apron is usually for cooking in the kitchen. 13. C –
9. A Face mask can help to protect the entire face of the researcher. 14. C –
10. B Materials with a pH value between 5 and 9 will not damage and 15. A –
corrode the laboratory sink. 16. D The nucleus has chromosomes that contain ribonucleic acid,
11. A Category A is sharp wastes. Sharp waste is easy to break the while mitochondria and chloroplasts contain little amount of
thin containers. Therefore, it needs to be inserted into specially deoxyribonucleic acids.
prepared containers. 17. D Rough endoplasmic reticulum transports protein made by
12. A Microorganisms can only be destroyed at a temperature of 121°C ribosomes, and smooth endoplasmic reticulum synthesises and
and a pressure of 15 psi for 20 minutes. transports lipids.
13. C Only harmless specimens can be dumped in the bin. 18. B Cytoplasm can help to fill the whole cell and to maintain the cell
14. C – shape.
15. B – 19. A Nucleolus is inside the nucleus and is used to synthesise DNA.
16. D A good biological drawing requires a label. 20. A Muscle cell needs large number of mitochondria to produce Chapter 1 – Chapter 2
17. B – energy for movement.
18. A – 21. C A is a nerve cell, B is a white blood cell and D is a sperm cell
that are found in a multicellular organism. C is a Paramecium
19. C – which is a unicellular organism.
20. B – 22. B In freshwater, Amoeba sp. is hypertonic towards water. The water
21. D Magnification factor of drawing = 10 × 10 = 10 from the surrounding undergoes osmosis into Amoeba sp. So,
10 Amoeba sp. requires contractile vacuoles to remove excess water
22. C The control set is a normal experimental set and does not carry from its body.
out changes to the material used in the experiment. 23. D P is cilium which helps to trap food and assist movement in
23. D Conclusions are made based on the situation observed. water.
24. A – 24. D Pseudopodia can be used for eating during phagocytosis as the
25. A – pseudopodia engulf the food.
26. B The steps of gathering relevant information, determining the 25. A As seawater is hypertonic towards the cytoplasm of Amoeba sp.,
apparatus and materials needed and identifying the variables the water will undergo osmosis to move out of the Amoeba sp.
involved are steps in planning the investigation. making it shrink.
27. C – 26. A Phagocytosis is the process involved in eating for Amoeba sp.
28. C – 27. B Both of them are made up of only one cell.
29. B – 28. C –
30. D – 29. C This is an adipose tissue. The adipose tissue helps to store fats.
30. A –
CHAPTER 2 31. D Sperm cells can be found in the male reproductive system and
red blood cells can be found in the circulatory system.
PAPER 1 32. B Mitochondrion is required in large amounts to provide enough
1. A Mitochondrion has an oval shape, and two layers of membrane, energy to cardiac muscle tissue to function.
which is smooth on the outside and folded on the inside. 33. A Trachea contains epithelial tissue with cilia to filter the dust
2. C P is the Golgi apparatus that processes, modifies and packages particles in inhaled air.
proteins but Q is the smooth endoplasmic reticulum that 34. A Heart contains cardiac muscle tissues, but skeletal muscle tissues
synthesizes lipid. are attached to bones.
3. D Ribosome is the site to synthesise protein in the cell. 35. C –
4. A P is cell wall of the plant cell. It allows substances to move 36. D Cartilage is a soft connective tissue to provide protection and
through it freely. support for the body.

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Answer QR1202 Biology Form 4.indd 1 26/04/2022 5:25 PM

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