Q & A STPM 2022 Biology

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Semester 1 2 3 Q & A STPM Q



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FEATURES TITLES IN THIS SERIES: Biology
• Based on the latest syllabus ■ Pengajian Am Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
• Arranged according to subtopics ■ Bahasa Melayu
• Tips to guide students in answering ■ Sejarah
questions effectively ■ Geografi
• Common Errors to show common ■ Ekonomi
mistakes frequently made by students ■ Pengajian Perniagaan STPM
while answering questions ■ Mathematics (T)
• Model Paper which follows the latest ■ Physics
STPM assessment format to prepare ■ Chemistry
students for the actual exam ■ Biology . .
Pengajian Am Ekonomi Physics Semester 1 . 2 . 3 Semester 1 2 3
Bahasa Melayu Pengajian Perniagaan Chemistry
Sejarah Mathematics (T) Biology
Biology
Geograf


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ISBN: 978-967-2720-51-5



PELANGI N. Nair

CONTENTS
CONTENTS


STPM Scheme of Assessment
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Semester 1

CHAPTER Biological Molecules
1 1.1 Water 1
1.2 Carbohydrates
1
1.3 Lipids 1
1.4 Proteins 11
1.5 Nucleic Acids 11


CHAPTER Structure of Cells and Organelles
2 2.1 Prokaryotic and Eukaryotic Cells 22
2.2 Cellular Components
22
2.3 Specialised Cells 39


CHAPTER Membrane Structure and Transport
3 3.1 Fluid-Mosaic Model 47
3.2 Movement of Substances across Membrane
54

CHAPTER Enzymes
4 4.1 Catalysis and Activation Energy 65
65
4.2 Mechanism of Action and Kinetics
4.3 Cofactors 77
4.4 Inhibitors 77
4.5 Classification of Enzymes 84
4.6 Enzyme Technology 84


CHAPTER Cellular Respiration
5 5.1 The Need for Energy in Living Organisms 88
5.2 Aerobic Respiration
88
5.3 Anaerobic Respiration 104


iv




Contents.indd 4 11/2/21 9:15 AM

CHAPTER Photosynthesis
6 6.1 Autotrophs 111
6.2 Light-dependent Reactions
114
6.3 Light-independent Reactions 125
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6.4 Limiting Factors 139

STPM Model Paper 1 964/1 142
Semester 2

CHAPTER Gaseous Exchange
7 7.1 Gaseous Exchange in Humans 150
7.2 Breathing Cycle
164
7.3 Gaseous Exchange in Plants 168


CHAPTER Transport in Animals and Plants
8 8.1 Transport System in Mammals 170
8.2 Transport System in Plants
181

CHAPTER Control and Regulation
9 9.1 Nervous System 189
203
9.2 Hormones

CHAPTER Reproduction, Development and Growth
10 10.1 Sexual Reproduction in Humans 213
224
10.2 Sexual Reproduction in Flowering Plants
10.3 Seed Germination 227
10.4 Growth Curves and Patterns of Growth 231

CHAPTER Homeostasis

11 11.1 Importance of Homeostasis 240
245
11.2 Liver
11.3 Osmoregulation in Mammals 249
11.4 Osmoregulation in Plants 259



v




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CHAPTER Immunity
12 12.1 Immune System 261
12.2 Development of Immunity
265
12.3 Concept of Self and Non-self 271
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12.4 Immune Disorder 271

CHAPTER Infectious Diseases

13 13.1 Infectious Disease 278
278
13.2 Dengue
13.3 Cholera 278
13.4 Tuberculosis 278
13.5 Malaria 278

STPM Model Paper 2 964/2 291

Semester 3

CHAPTER Taxonomy and Biodiversity
14 14.1 Taxonomy 293
14.2 Diversity of Organisms
302
14.3 Biodiversity in Malaysia 320
14.4 Threats to Biodivesity 320
14.5 Conservation of Biodiversity 320

CHAPTER Ecology
15 15.1 Levels of Ecological Organisation 326
331
15.2 Biogeochemical Cycles
15.3 Energy Flow 337
15.4 Population Ecology 350
15.5 Carrying Capacity 350
15.6 Quantitative Ecology 362










vi




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CHAPTER Selection and Speciation
16 16.1 Natural and Artificial Selection 371
388
16.2 Speciation
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CHAPTER Inheritance and Genetic Control
17 17.1 Types of Genetic Crosses and Breeding System 398
398
17.2 Non-Mendelian Inheritance
17.3 Genetic Mapping 425
17.4 Population Genetics 425
17.5 DNA Replication 447
17.6 Gene Expression 447
17.7 Regulation of Gene Expression 469
17.8 Mutation 469


CHAPTER Gene Technology
18 18.1 Recombinant DNA Technology 497




CHAPTER Biotechnology
19 19.1 Roles of Biotechnology 520
520
19.2 Applications of Biotechnology
STPM Model Paper 3 964/3 528

Answers 534



















vii




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viii




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Chapter

3 Membrane Structure


and Transport


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3.1 Fluid-Mosaic Model Semester


Section A Multiple-choice Questions
1
Question 1
Which of the following component of the plasma membrane controls its
fluidity and flexibility?
A Hydrophilic head of the phospholipid molecules
B Oligosaccharide chains attached to the peripheral proteins
C Cholesterol molecules
D Intrinsic proteins

Answer : C
One of the components of the plasma membrane is cholesterol which are
inserted among phospholipid molecules. These molecules increase the
flexibility and stability of membranes. Without it, the membranes break up.


Question 2
Which of the following properties of lecithin are important in the formation
of the structure of a cell membrane?
I Lecithin can be hydrolysed
II Two alcohol groups bind with one fatty acid respectively
III The tips of two hydrocarbon chains are non-polar
IV The tips of the molecule with the phosphate and nitrogen groups are polar
A I and III
B I, III and IV
C II, III and IV
D I, II, III and IV

Answer : B
The most common phospholipid is lecithin which has a positively charged
choline group bound to the phosphate group. It is an important component
of the cell membrane.





47




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Biology STPM Chapter 3 Membrane Structure and Transport

Question 3
Which of the following factors would tend to increase membrane fluidity?
A A greater proportion of unsaturated phospholipids
B A lower temperature
C A relatively high protein content in the membrane
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D A high membrane protential
Answer : A
Semester
1 The unsaturated fatty acid in the phospholipid molecules have double bonds
which form kinks. This will prevent close packing of the phospholipid
molecules, thereby, increasing fluidity.


Question 4
Which of the following parts of a phospholipid molecule contributes most to
the thickness of a cell surface membrane?
A Phosphate group
B Hydrophilic head
C Glycerol
D Hydrocarbon chain

Answer : D
The phospholipid molecule consists of two fatty acid long chain of hydrocarbon
and a phosphate hydrophilic head.

Phosphate Choline
hydrophilic head
Glycerol
residue

Hydrocarbon
chains


Fatty acids
Phospholipid molecule
Question 5
A cell lacking in oligosaccharides on the external surface of its plasma
membrane would likely be inefficient in
A cell recognition.
B transporting ions against an electrochemical gradient.
C maintaining fluidity of the phospholipid layer.
D establishing a diffusion barrier to charged molecules.



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Biology STPM Chapter 3 Membrane Structure and Transport
Answer : A
Some proteins and phospholipids conjugate with oligosaccharides to form
glycoproteins and glycolipids on the outer surface. These are important for
cell recognition.
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Question 6 Semester
Which of the following statements is not true about the cell membrane?
A Membrane allows phagocytosis and pinocytosis to occur.
B Membrane components consist of hydrophilic and hydrophobic molecules. 1
C Fatty acid in the membrane only consists of unsaturated fatty acids.
D The transportation of soluble substances through membrane occurs by
diffusion or active transport.


Answer : C
The phospholipid molecules in the membrane have both saturated and
unsaturated fatty acids. However, unsaturated fatty acids have double bonds
that form kinks that keep the molecules from packing together, increasing
membrane fluidity.


Question 7
The table below shows the components of a cell membrane and its functions.
Components of a cell Functions
membrane
P Lecithin (i) Maintains the fluid characteristic of the
membrane.
Q Oligosaccharide (ii) As an indicator for cell recognition.
(iii) Acts as a receptor to a certain molecule
R Cholesterol such as hormone.
(iv) Gives the semipermeable and selective
S Protein characteristics to the membrane.

Which of the following is the correct match for the components of a cell
membrane and their functions?
P Q R S
A (iv) (ii) (i) (iii)
B (ii) (iii) (i) (iv)
C (iv) (i) (iii) (ii)
D (iv) (iii) (ii) (i)




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Biology STPM Chapter 3 Membrane Structure and Transport
Answer : A
The membrane consists of a phospholipid (lecithin) bilayer as well as extrinsic
proteins on the outer and inner surfaces of the membrane and intrinsic
proteins partially embedded in the membrane or penetrate through the
membrane.
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Some of the proteins and phospholipids conjugate with oligosaccharides to
form glycoproteins and glycolipids on the outer surface.
Cholesterol molecules are found between phospholipid molecules.
Semester
1 Question 8
Lipoprotein is present in the cell membrane. This determines the characteristic
of
A the movement of water molecules through xylem.
B the selective permeability towards molecules or ions.
C protein synthesis.
D the diffusion of water molecules and ions.

Answer : B
The membrane consists of phospholipid bilayer and proteins. Hence,
membranes are selectively permeable and regulate the movement of substances
in and out of the cell.


Section B Structured Questions

Question 9
The diagram below shows a section of a cell surface membrane.

T R
S
Q




P





(a) Name the structures labelled P, Q, R, S and T. (5 marks)
(b) Explain how proteins are held in the membrane. (1 mark)
(c) Give two functions of proteins in the membrane. (2 marks)




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Biology STPM Chapter 3 Membrane Structure and Transport
Answer :
(a) P : Phospholipid bilayer
Q : Intrinsic protein
R : Extrinsic protein
S : Channel or pore protein
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T : Glycolipid Semester
(b) – Proteins are amphoteric.
– They have both basic and acidic proteins as they are zwitterions.
– Hence, their hydrophilic (polar) regions are in the aqueous parts, 1
either facing in or out of the membrane.
– Proteins also contain regions of hydrophobic amino acids which
enable them to interact with hydrophobic tails in the phospholipid
bilayer.

(c) – Provides structural support to the cell membrane.
– Allows transport across membranes by acting as pumps, carriers and
channel proteins.
– Are receptor sites for enzymes, hormones or antibodies.
– As antigens.
– Act as enzymes.
– Helps to link up neighbouring cells.





• The pore formed by the proteins encloses the channel which is the space
that allow substances to flow through the membrane.
• Glycolipids involve attachment of carbohydrate to the hydrophilic heads
of the phospholipid bilayer.
• Carbohydrates attached to the proteins form glycoproteins which act as
antigens.
• The proteins in the membrane have hydrophobic amino acid regions that
interact with the fatty acid ‘tails’ to repel water.
• The functions of the proteins include the functions of channel protein,
carrier proteins and proteins as receptor molecules for chemical signalling
between cells.









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Biology STPM Chapter 3 Membrane Structure and Transport

Question 10
The diagram below shows the fluid-mosaic model of the cell surface
membrane.

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Channel
Carbohydrate
chains

Semester
1





(a) Explain why the cell surface membrane is described as a fluid-mosaic.
(2 marks)
(b) State the function of the channel labelled in the diagram. (1 mark)
(c) State three features of a substance which influence its ability to pass
through a cell membrane. (3 marks)
(d) Many substances are said to be actively transported across membranes.
Explain what is meant by ‘active transport’. (2 marks)


Answer :
(a) – It is due to the scattered mosaic arrangement of proteins that float
in the phopholipid bilayer.
– The phospholipid bilayer is fluid as the lipid and protein molecules
can move laterally and rotate on their axis.
(b) – To allow selective transport of polar molecules and ions across the
membrane by facilitated diffusion.
(c) (i) Size – Small molecules pass through the cell membrane faster than
large molecules.
(ii) Charge – Polar molecules are repelled by the non-polar
hydrophobic lipid membrane.
(iii) Solubility – Water-soluble molecules diffuse slower across the
membrane than fat-soluble molecules.
(d) Active transport is the transport of substances across cell membranes
against a concentration gradient, that is from a region of low
concentration to a region of high concentration and requires energy in
the form of ATP.






52




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Biology STPM Chapter 3 Membrane Structure and Transport

Section C Essay Questions

Question 11
(a) With the aid of a large labelled diagram, describe the structure of a cell
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membrane. [8]
(b) Explain the function of the various types of membrane proteins. [7]
(15 marks) Semester

Answer : 1
(a) – The model for cell membrane is the fluid-mosaic model by Singer
and Nicholson in 1972.
– The cell membrane is a dynamic structure that consists of a
phospholipid bilayer with protein molecules that float on or within
the membrane.
– The non-polar long hydrocarbon tails are hydrophobic and face
inwards and are attracted to one another by hydrophobic interactions
and van der Waals forces.
– The polar phospholipid heads are hydrophilic, face outwards and are
attracted to the water in the cell and the outer aqueous surrounding.
– The protein molecules are scattered between phospholipid molecules
forming a mosaic pattern.
– Extinsic proteins are on the outer and inner surfaces of the membrane.
– Intrinsic proteins are proteins that are partially embedded or
penetrate the membrane (integral or transmembrane protein).
– The phospholipid membrane is fluid because the phospholipid
molecules and some protein molecules can move laterally or change
places.
– Unsaturated fatty acids have double bonds which form kinks. These
prevent close packing of the phospholipid molecules, thus increase
the membrane fluidity.
– Short, branched carbohydrate chains bind to proteins to form
glycoproteins and those that bind to lipids form glycolipids. These
are important for cell recognition.
– There are cholesterol molecules between some phospholipid
molecules.
– These molecules help to stabilise the membrane structure and
regulate membrane fluidity.
– The membrane is supported by interacellular protein filaments at the
inner surface, which act as a cytoskeleton.







53




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Biology STPM Chapter 3 Membrane Structure and Transport
(b) – Some membrane proteins act as enzymes that function to catalyse
chemical reactions in the cell.
– Some proteins are receptors for specific substances such as hormones.
For example, hormone insulin molecules bind to the receptor on the
cell membrane of the liver cell and promotes the conversion of excess
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glucose to glycogen.
– Protein molecules form channels and carriers for passage of ions and
polar molecules to enter or leave the cell.
– Some protein molecules act as carrier proteins that are able to actively
Semester
1 transport molecules across the cell membrane. The process involves
the use of energy obtained from ATP.
– Cell adhesion proteins act to bind adjacent cells together to form
tissues that carry out specific functions.
– Glycoproteins with branching oligosaccharides are involved in cell-
cell recognition and act as antigen.
– Some protein molecules provide attachment sites for the cytoskeleton
filaments.



3.2 Movement of Substances across Membrane


Section A Multiple-choice Questions

Question 1
Which of the following methods of transport across the plasma membrane
uses ATP and is able to transport across a substance against concentration
gradient?
A Active transport
B Active transport and receptor-mediated pinocytosis
C Facilitated diffusion and active transport
D Facilitated diffusion, active transport and receptor-medicated pinocytosis


Answer : B
The active transport is the transport of molecules or ions across the membrane
against the concentration. The receptor-mediated endocytosis is a form of
endocytosis that makes use of receptors attached to the cell membrane to
engulf molecule. Both of these processes require energy.







54




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Biology STPM Chapter 3 Membrane Structure and Transport

Question 2
Four groups of potato cells, each with a water potential [ψ] of –300 kPa are
placed in four solutions P, Q, R and S of different water potentials.
In which of the four solutions will there be a net movement of water molecules
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into the potato cells?
Solution Water potential, ψ (kPa) Semester
A P 0
B Q –300 1
C R –500
D S –700


Answer : A
Pure water has the highest water potential, (ψ=0). Water will diffuse from a
region of higher water potential (less negative or zero value) to a region of
lower water potential (more negative value) until the concentration of water
molecules in both areas are equal.


Question 3
Which of the following curves best shows the relationship between the
concentration outside the membrane and the movement rate through the
membrane by simple diffusion and facilitated diffusion respectively?
Movement rate

d
a c
b





Concentration
outside membrane
Simple diffusion Facilitated diffusion
A d b
B d c
C c a
D c d



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Biology STPM Chapter 3 Membrane Structure and Transport
Answer : B
Facilitated diffusion shows saturation at a higher solute concentration when
all the binding sites on the transport proteins are used up. In simple diffusion,
a maximum rate of diffusion is usually not achieved.
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Question 4
The diagram below shows the water potentials of three neighbouring plant
cells.
Semester
1
X Y
0 kPa -1500 kPa



Z
-4500 kPa


In which direction will water move?
A From cell X to cells Y and Z.
B From cell Y to cells X and Z.
C From cell Z to cell X.
D From cell Z to cell Y.


Answer : A
Water moves from a region of high water potential to a region of lower (more
negative) water potential, down a water potential gradient.


Question 5
A cell takes in dissolved materials by forming tiny vesicles around fluid droplets
trapped by folds of the plasma membrane. This process is
A active transport.
B receptor-mediated endocytosis.
C pinocytosis.
D facilitated diffusion.

Answer : C
In pinocytosis, the material taken up by the cell is liquid. Tiny droplets are
trapped by the folds of the membrane. It is then pinched off into the cytosol
as tiny vesicles.




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Biology STPM Chapter 3 Membrane Structure and Transport

Question 6
An example of a process that involves active transport is
A an influx of sodium ions into a nerve axon during the conduction of a
nerve impulse.
B the movement of sodium ions from glomerular filltrate into blood plasma
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during reabsorption in the nephron tubule.
C the movement of potassium ions from blood plasma into the lumen of a Semester
Bowman’s capsule during ultrafiltration.
D a shift of chloride ions across the membrane of a red blood cell during
carbon dioxide transport. 1

Answer : B
The movement of sodium ions into the axon of the nerve is through open
sodium channels. Ultrafiltration occurs in the glomerulus and Bowman’s
capsule. The chloride shift involves diffusion.


Question 7
A plant cell with a solute potential of –0.65 MPa maintains a constant volume
when immersed in a solution with a solute potential of –0.3 MPa. Which of
the following conclusion can be made about the cell?
A The pressure potential of the cell is +0.65 MPa.
B The water potential of the cell is –0.65 MPa.
C The pressure potential of the cell is +0.3 MPa.
D The pressure potential of the cell is +0.35 MPa.

Answer : D
The water potential of a plant is the sum of its solute potential and pressure
potential, ψ ψ ψ . P
S
2

Question 8
Plasma membrane is selectively permeable to certain substances. Which of the
following substances is / are not permeable to the cell surface membrane?
I Hydrophobic molecules
II Polar and tiny molecules
III Polar and large molecules
IV Ions
A IV only
B I and II only
C III and IV only
D I, II and III only




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Biology STPM Chapter 3 Membrane Structure and Transport
Answer : C
The phospholipid bilayer is not permeable to charged ions such as Na , K ,
+
+
Cl , HCO and hydrophilic molecules such as glucose and macromolecules.
–
–
3
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Question 9
The diagram below shows the movement of three substances P, Q and R into
a cell through the cell surface membrane.
P Q R
Semester
1




Which of the following are P, Q and R?
P Q R
A Ethanol Oxygen Amino acid
B Carbon dioxide Acetone Amino acid
C Acetone Glucose Oxygen
D HCO – Oxygen Glucose
3

Answer : D
The phospholipid bilayer is permeable to very small, uncharged polar
molecules like oxygen and carbon dioxide, and hydrophobic molecules such
–
+
as steroids. Hydrophilic substances such as Na , HCO , ions, water molecules
3
diffuse across the membranes with the help of transport proteins.
Question 10
Which of the following does not involve facilitated diffusion?
A Movement of sodium ions across the membrane of axon during action
potential.
B Transportation of glucose into the cell.
C Diffusion of ATP from the mitochondrion.
D Movement of oxygen and carbon dioxide across the membrane of the
alveolus.

Answer : D
Small molecules such as oxygen and carbon dioxide diffuse through the
phospholipid bilayer.





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Biology STPM Chapter 3 Membrane Structure and Transport

Question 11
The diagram below shows a plant cell placed in distilled water. A, B, C and D
represent the direction of pressures exerted on and inside the cell.
Which one of the following represents the cell wall pressure?
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A

Water molecule Semester
D Solute molecule 1
C Tonoplast
Cell membrane


B

Answer : D
Water enters the plant cell by osmosis. As a result, pressure on the cell wall
increases inside the cell, making the cell turgid. The cell wall develops an equal
and opposite pressure to stop water movement into the cell.



Section B Structured Questions

Question 12
The diagram below shows the two adjacent cells. Their values ψ and ψ are
S
P
given in kPa.
Cell A Cell B
ψ = -2000 ψ = -1600
S S
ψ = 1000 ψ = 800
P P
(a) Define osmosis in terms of water potential. (2 marks)
(b) State the direction of water flow between the two cells. (2 marks)
(c) Give one reason for your answer above. (1 mark)
(d) Calculate the values of the pressure potential, ψ and the water potential,
P
ψ of cell A and cell B, when equilibrium is reached. Assume the changes
in solute potential are negligible. (3 marks)









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Biology STPM Chapter 3 Membrane Structure and Transport
Answer :
(a) Osmosis is the passive movement of water molecules from a region
of high-water potential to a region of lower water potential through a
partially permeable membrane.
(b) Cell A = Cell B =
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ψ = ψ + ψ ψ = ψ + ψ P
S
P
S
= -2000 + 1000 = -1600 + 800
= -1000 kPa = -800 kPa
Semester
(c) Water flows down a water potential gradient from a region of high-water
1 potential to a region of low-water potential.
(d) At equilibrium,
Cell A =
ψ = -900 kPa
ψ = -900 + 2000
P
= 1100 kPa
Cell B =
ψ = -900 kPa
ψ = -900 + 1600
P = 700 kPa

Question 13
The diagram below shows two adjacent cells and their values for ψ and ψ
s
p
given in kPa.

ψ = -2000 kPa ψ = -1800 kPa
S S
ψ = +800 kPa ψ = +1000 kPa
P P


Cell P Cell Q
(a) (i) Give the formula to determine water potential of a plant cell. (1 mark)
(ii) Define osmosis in terms of water potential. (2 marks)
(b) State the direction of water movement between the two cells. Explain your
answer. (2 marks)
(c) Calculate the pressure potential, (ψ ) and water potential (ψ) of cell P and
p
cell Q when equilibrium is reached. Assume that changes in solute potential
are negligible. (1 mark)
(d) State two differences between osmosis and sodium-potassium pump.
(2 marks)




60




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Biology STPM Chapter 3 Membrane Structure and Transport
Answer :
(a) (i) ψ = ψ + ψ p
s
(ii) Osmosis is the movement of water molecules from a region of
higher water potential to a region of lower water potential through
a partially permeable membrane.
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(b) ψ cell P = –2000 + 800
= –1200 kPa Semester
ψ cell Q = –1800 + 1000
= –800 kPa 1
Since ψ cell Q is higher than ψ cell P , water flows from cell Q to cell P. This
is because water flows down a water potential gradient from a region
of higher water potential to a region of lower water potential.

(c) At equilibrium,
ψ cell P = ψ cell Q

= –1200 + (–800)
2
= –1000 kPa
ψ for cell P = –1000 + 2000
P
= 1000 kPa
ψ for cell Q = –1000 + 1800
P
= 800 kPa
(d) – Osmosis is a passive process while sodium-potassium pump is an
active process.
– Osmosis does not require energy while sodium-potassium pump
requires energy.
– Movement of water in osmosis is down its concentration gradient
while movement of ions in the sodium-potassium pump is against
its concentration gradient.
– Molecules pass through a semi-permeable membrane while ions are
transported through a carrier protein in sodium-potassium pump.
– Only water molecules are involved in osmosis while the sodium-
potassium pump transports ions.










61




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Biology STPM Chapter 3 Membrane Structure and Transport
Question 14
The diagram below shows two adjacent plant cells, P and Q. The values of their
pressure potential (ψ ) and solute potential (ψ ) are given in kPa.
p s
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= –2 000 = –1800
s s
= 1 000 = 400
p p
Semester
1
(a) State the direction in which water would move betwen the two cells.
Explain your answer. (2 marks)
(b) Complete the table below to show the values of the pressure potential, ψ p
and the water potential, ψ of each cell when equilibrum has been reached.
w
Assume that changes in the value ψ are negligible.
s
Potential Cell P Cell Q
Pressure potential, ψ
p
Water potential, ψ
w
(4 marks)
(c) Then, the cells are transferred into distilled water and allowed to reach
equilibrum. Assume the changes in the values below if ψ are negligible.
s
(i) Give the new value of water potential, ψ of the two cells. (1 mark)
(ii) Give the new value of the pressure potential, ψ of cell P. (1 mark)
P
Answer :
(a) Cell P Cell Q
ψ = ψ + ψ P ψ = ψ + ψ P
w
s
s
w
= –2000 kPa + 1000 kPa = –1800 kPa + 400 kPa
= –1000 kPa = –1400 kPa
Water flows from cell P to cell Q. This is because water flows from a
region of higher water potential to a region of lower water potential.
(b)
Potential Cell P Cell Q
Pressure potential, ψ –1200 + 2000 = 800 –1200 + 1800 = 600
P
Water potential, ψ 1200 1200
w
(c) ψ = 0 kPa (d) 0 = ψ + ψ P
S
w
= –2000 + ψ P
ψ = 2000 kPa
P
62




Chapter 3.indd 62 11/2/21 9:21 AM

Biology STPM Chapter 3 Membrane Structure and Transport

Section C Essay Questions

Question 15
(a) Explain the roles of the cell membrane structures in the transportation of
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substances into the cell.
(b) Explain the role of carbohydrate in the membrane. Semester



Answer : 1
(a) – The cell membrane is semi-permeable.
– The membrane consists of phospholipids and are embedded with
protein molecules.
– The phospholipid bilayer is permeable to steroids, fatty acids and
lipid-soluble molecules as well as very small uncharged molecules
like oxygen, carbon dioxide and water.
– Osmosis involves the diffusion of water molecules through the
semi-permeable membrane from a region of high water potential
to a region of low water potential. Some protein channels called
aquaporins in the membrane increase the rate of diffusion.
– Some proteins form ion channels that follow diffusion of ions such
as K , Ca , Na and Cl down their concentration gradients.
+
–
+
2+
– Some large molecules such as glucose and amino acids cannot move
across the phospholipid bilayer. These molecules move across the
membrane by facilitated diffusion using a protein carrier molecule.
– Carrier proteins / protein pumps are involved in active transport,
whereby energy is required to transport ions or molecules across
the cell membrane against a concentration gradient.
– For example, the sodium-potassium protein pump in the membrane
of a neuron uses energy to pump sodium ions out of the cell and
potassium ions into the cell.
– Endocytosis and exocytosis are active transport processes that
transport material in bulk across the membranes.
– Endocytosis involves the bulk transport of substances into a cell.
– In exocytosis, the vesicle fuses with the cell membrane and the
contents such as enzymes, hormones and mucus in the vesicle are
released out of the cell.







63




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Biology STPM Chapter 3 Membrane Structure and Transport
– In phagocytosis, the cell membrane encloses a particle such as a
bacterium to engulf it into the cell.
– In pinocytosis, the cell surface membrane invaginates inwards
in order for the droplets containing materials to flow inside. The
membrane then pinches off to form pinocytic vesicles.
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– In receptor-mediated endocytosis, ligands such as cholesterol
molecules bind to receptors in coated pits on the cell membrane
which then form coated vesicles by endocytosis.
Semester
1 (b) – Carbohydrates in the membrane exist as short sugar chains
(oligosaccharides)
– These carbohytrates are attached to the proteins and phospholipids
to form glycoproteins and glycolipids respectively.
– Glycoproteins and glycolipids act as antigens, that are identify
markers.
– These molecules enable cells to identify and differentiate themselves
from others.
– This is known as cell to cell recognition.
– Glycoproteins and glycolipids can act as receptors for chemical
signals such as hormones.
– In addition, these molecules are involved in holding the cells together
to form tissues.































64




Chapter 3.indd 64 11/2/21 9:21 AM

Chapter

Biology STPM Chapter 7 Gaseous Exchange
7 Gaseous Exchange






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7.1 Gaseous Exchange in Humans

Section A Multiple-choice Questions


Question 1
The reversible dissociation reaction of oxyhaemoglobin during cellular
respiration is represented by the following equation:
Hb (O )  Hb + 4O 2
2 4
Oxyhaemoglobin Haemoglobin Oxygen
What will happen to the above reaction if the carbon dioxide concentration
Semester
2 in blood increases?
A The reaction shifts to the right
B The reaction shifts to the left
C Carboxyhaemoglobin complex is formed
D Oxyhaemoglobin – CO2 complex is formed

Answer: A
The main function of haemoglobin is to carry oxygen. An increase in partial
pressure of carbon dioxide or a decrease will lower the affinity of haemoglobin
for oxygen. This helps in the unloading of oxygen from oxyhaemoglobin in the
cappilaries of respiring tissues.


Question 2

Which of these statements is about Bohr effect?
A The regulating of constant pH is blood
B The haemoglobin affinity towards carbon dioxide
C The influence of carbon dioxide concentration on the dissociation of
oxyhaemoglobin
D The increase in dissociation rate of carbon dioxide due to the high partial
pressure of oxygen

Answer: C
Bohr effect is the decrease in the oxygen affinity of haemoglobin in response
to decreased blood pH due to increased carbon dioxide concentration in the
blood.



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Biology STPM Chapter 7 Gaseous Exchange

Question 3
Which of these are roles of the carbonic anhydrase enzyme in the process of
gaseous exchange in human beings?
I Increases the release of CO gas into the lung
2
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II Increases the capillary wall’s ability to absorb oxygen
III Increases the concentration of H ions in the red blood cell
+
-
IV Increases the concentration of HCO ions in the red blood cell
3
A I and II C II, III and IV
B I and IV D I, II, III and IV
Answer: D
Carbonic anhydrase is a very efficient enzyme. It is found in the red blood cells
and catalyses the combination of CO and water to form carbonic acid. The
2
carbonic acid then dissociates into H ions and HCO ions. Semester
+
–
3
Question 4 2
The graph below shows three oxygen dissociation curves for three different
animals.


Percentage of saturation of haemoglobin with oxygen P Q R











Partial pressure of oxygen/ pO 2 (kPa)
Which of the following is correct about curves P, Q and R?
A worm in Pigeon Human
waterlogged soil
A P Q R
B P R Q
C Q R P
D R P Q






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Biology STPM Chapter 7 Gaseous Exchange
Answer: B
The haemoglobin in a worm has the highest affinity for oxygen and does not
unload the oxygen until it reaches a very low partial pressure of oxygen. The
pigeon is an active animal, hence the rate of respiration is higher. This means
more oxygen is needed. There is a high level of CO , hence the affinity of
2
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haemoglobin for oxygen decreases. So, oxygen is released into the tissue readily.

Question 5
The graph below represents the oxygen dissociation curves of a mammal at
two different partial pressure of carbon dioxide. What is the advantage of the
effect shown by the two curves to the organism?


100
pCO 2 = 2 kPa
Semester
2 80
Percentage of saturation of haemoglobin with oxygen 60 pO 2 = 10 kPa





40


20


0
4 8 12 16 20
pO 2 /kPa
A Haemoglobin removes more carbon dioxide from active tissues.
B Oxyhaemoglobin is not denatured at low pH values.
C Oxygen is more easily released to the active tissues.
D Haemoglobin is saturated with oxygen at the lungs.



Answer: C
Active tissues require more oxygen and have a high partial pressure of CO since
2
it is a product of respiration. At higher pCO , the oxygen dissociation curve shifts
2
to the right. This shows carbon dioxide reduces the affinity of haemoglobin for
oxygen. This causes oxygen to be released into the active tissues. This is called
the Bohr effect.





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Biology STPM Chapter 7 Gaseous Exchange

Question 6
The graph below shows the oxygen dissociation curve for haemoglobin and
myoglobin. 100

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Percentage of saturation with oxygen (%) 50 x y












0
Partial pressure
5
10
15
20
of oxygen/ kPa
Semester
Which of the following statements concerning the graph above is correct? 2
A Curve x is for haemoglobin and curve y is for myoglobin.
B Curve x is for myoglobin and curve y is for haemoglobin.
C Curve y shows a higher pigment affinity for oxygen.
D Curve x shows a lower pigment affinity for oxygen.
Answer: B
Myoglobin has a higher affinity for oxygen compared to haemoglobin. This
would displace the curve to the left.



Question 7
The following reaction shows the affinity of haemoglobin towards oxygen in
an animal tissue in the conditions of P and Q.
P
Hb + O HbO
2 Q 2
Which of the following pairs of conditions P and Q are correct?
P Q
A High partial pressure of oxygen Low partial pressure of carbon
dioxide
B High partial pressure of oxygen High partial pressure of carbon
dioxide
C High temperature Low hydrogen ion concentration
D Low temperature Low hydrogen ion concentration


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Biology STPM Chapter 7 Gaseous Exchange
Answer: B
In the lungs, when the partial pressure of oxygen is high, haemoglobin readily
combines with oxygen. In active tissues, when the partial pressure of CO
2
increases, the affinity oh haemoglobin for oxygen is decreased and more oxygen
is readily released. This is the Bohr effect.
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Question 8
The graph below shows the oxygen dissociation curves for fish and human.







Percentage of saturation of haemoglobin with oxygen (%)

Semester
2 Fish Human










0 Partial pressure of oxygen/ kPa
Which of the following is the reason the oxygen dissociation curve for fish is
more to the left compared to the curve for human?
A The concentration of oxygen in water is lower than in air.
B At a low partial pressure of oxygen, human haemoglobin is more
saturated compared to fish haemoglobin.
C At a low partial pressure of oxygen, oxygen uptake of fish haemoglobin is
less compared to human haemoglobin.
D Fish haemoglobin has a higher affinity for oxygen.



Answer: D
The fish haemoglobin has an increased affinity for oxygen than human
haemoglobin. This would displace the curve to the left. The initial uptake of
oxygen for fish haemoglobin is higher, till saturation.







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Biology STPM Chapter 7 Gaseous Exchange

Question 9
Which of the following processes in the red blood cells is catalysed by an
enzyme?
A The reaction between water and carbon dioxide, forming carbonic acid.
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B The reaction between carbon dioxide and haemoglobin, forming
carbaminohaemoglobin.
C The reaction between hydrogen ions and haemoglobin, forming
haemoglobinic acid.
D The dissociation of oxyhaemoglobin, releasing oxygen.


Answer: A
Carbonic anhydrase is a very efficient enzyme found in the red blood cells. It
catalyses the combination of carbon dioxide and water to form carbonic acid,
as well as, the dissociation of carbonic acid, releasing hydrogen carbonate ions Semester
and hydrogen ions.
2
Question 10
Which of the following forms does carbon dioxide mainly transported in
blood?
A As hydrogen carbonate ions
B As carbonic acid
C In solution as dissolved carbon dioxide
D As carbaminohaemoglobin



Answer: A
About 85% of CO produced by tissues combine with water to form carbonic
2
acid. Carbonic acid dissociates to hydrogen and hydrogen carbonate ions. These
ions move out of the red blood cell into the plasma where they combine with
sodium to form sodium hydrogen carbonate.



















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Biology STPM Chapter 7 Gaseous Exchange

Section B Structured Questions

Question 1
The diagram below shows the oxygen dissociation curve for the adult
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haemoglobin at different partial pressures of carbon dioxide (PCO / kPa).
2
Saturation of
haemoglobin with
oxygen / %
100

80

60 pCO = 2 kPa
2
pCO = 5 kPa
2
40 pCO = 8 kPa
2
20
Semester
2
0
0 5 10 15 20
Partial pressure of oxygen / kPa
(a) With reference to the diagram above, explain how the properties of
the haemoglobin molecule cause the oxygen dissociation curve to be
sigmoid. (3 marks)
(b) With reference to the diagram above, describe the effect of the increase
in the partial pressure of carbon dioxide on the percentage saturation of
haemoglobin. (2 marks)
(c) Explain how the increase in the partial pressure of carbon dioxide will
cause the effect shown in the diagram.


Answer:
(a) – oxygen binds to the haem group
– first, the oxygen molecule binds slowly
– this distorts the shape of molecule
– then, it is easier for the next oxygen molecules to bind
– the curve is sleep in the middle
(b) – as the partial pressure of CO increases, the saturation of haemoglobin
2
with O decrease
2
– this shifts the curve to the right
(c) – the increase in CO level releases hydrogen/ hydrogen carbonate form
2
– the decrease pH in the plasma
– this causes the release of oxygen by the haemoglobin/ lowers the affinity
of haemoglobin for oxygen


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Biology STPM Chapter 7 Gaseous Exchange

Question 2
(a) The diagram below shows the lungs in their respective pleural cavities.
Airway
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Lung Lung
P
Pleural
cavity
(i) Sketch in the figure below the appearance of the lungs if the wall of
the thorax were punctured at P. (1 mark)
Airway Semester




P 2



(ii) Give the reason for the appearance of the lungs you have sketched in
the figure above. (1 mark)
(b) The diagram below shows the oxygen dissociation curve for haemoglobin
under different conditions.
Curve A is the normal curve, curve B and C represent the dissociation
curves in conditions of a high level of carbon dioxide or a high level of
toxic carbon monoxide.
100 B
90
Haemoglobin saturation / % 70 A C
80
60
50
40
30
20
10
0
6 12
Partial pressure of oxygen / kpa





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Biology STPM Chapter 7 Gaseous Exchange

(i) Explain why the dissociation curves have a sigmoid or ‘S’ shape.
(1 mark)
(ii) Which of the following curves is due to the presence of a high level of
carbon dioxide. (2 marks)
(iii) Describe the physiological advantage resulting from the altered
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dissociation curve in the presence of a high level of carbon dioxide.
(2 marks)
(iv) Give suggestions on how the effect of carbon monoxide on the
oxygen dissociation curve shown contributes to the toxic effect of the
gas. (2 marks)


Answer:
(a) (i) Airway



P
Semester
2

(ii) – Lungs are only able to inflate when the air pressure in the lungs is
higher than the atmospheric pressure.
– A puncture at P will cause the air pressure in the right lung to
become equal to the atmospheric pressure (it cannot drop below
atmospheric pressure). Hence the right lung becomes deflated
(collapse).
(b) (i) – Initially, the binding of oxygen to haemoglobin is slow.
– But the binding of the first oxygen molecule changes its shape,
exposing other haem groups to oxygen. Thus, making subsequent
binding of oxygen to haemoglobin easier.
(ii) Curve C
(iii) – When the cells are actively respiring, the level of CO is high
2
whereas the oxygen concentration is low. The curve is shifted to
the right (Bohr effect).
– The affinity of haemoglobin for oxygen decreases, enabling oxygen
to be easily released into the active cells.
– The haemoglobin molecule is less saturated for the same partial
pressure of oxygen.
(iv) – Haemoglobin has a higher affinity for carbon monoxide compared
to oxygen.
– Hence, in the presence of carbon monoxide, haemoglobin will be
prevented from combining with oxygen. This will deprive oxygen
from tissues and can be fatal.



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Biology STPM Chapter 7 Gaseous Exchange
Exam Tips
Exam Tips
• The more the dissociation curve of a pigment shifts to the right, the slower the
oxygen binds to the pigment but released easily.
• The more dissociation curve of a pigment shifts to the left, the faster the pigment
binds to the oxygen but released slowly.
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Question 3

P







Semester
2


(a) With reference to the above diagram,
(i) name P and state its function. (2 marks)
(ii) as blood passes through capillaries in the lungs, it becomes
oxygenated.
Explain how the structure of haemoglobin helps in the uptake of in
the lungs. (2 marks)
The graph below shows the oxygen dissociation curves for myoglobin, M and
haemoglobin, H.

Saturation
of pigment 100 M
with oxygen
(%) 80
H
60
40

20
0
2 4 6 8 10 12 14
Partial pressure of oxygen/ kPa








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Biology STPM Chapter 7 Gaseous Exchange

(b) State the tissue where myoglobin is found. (1 mark)
(c) With reference to the graph,
(i) state the percentage saturation of myoglobin and haemoglobin when
partial pressure of oxygen is 2 kPa. (1 mark)
(ii) explain the significance of the difference in percentage of saturation
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that you have shown in (i). (1 mark)
(d) When a person exercises vigorously, the partial pressure of carbon
dioxide in the blood increases. Draw on the graph a dissociation curve for
haemoglobin when the partial pressure of carbon dioxide has increased.
(1 mark)


Answer:
(a) (i) – Haem
– able to combine loosely with a molecule of oxygen
(ii) – When an oxygen molecule combines with the ferrous iron atom of
a single haem unit, it distorts the shape of the haem unit slightly.
Semester
2 – This distortion causes the whole haemoglobin molecule to change
its shape accordingly.
– When oxygen molecule attaches to the second and third haem
groups, more structural changes occur.
– Each change facilitates a much faster uptake of oxygen than the
one earlier.
(b) Skeletal muscle
(c) (i) Myoglobin- 90%
Haemoglobin- 25%
(ii) – Myoglobin has a high affinity for oxygen, hence it retains its oxygen
in the resting cell.
– It only begins to release oxygen when pO is below 20 mmHg,
2
which is during vigorous muscle activity.
– It acts as a store of oxygen in muscles, only releasing the oxygen
when the supplies of oxyhaemoglobin have been exhausted.
(d) The curve is to be drawn to the right of the haemoglobin curve (H).


















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Biology STPM Chapter 7 Gaseous Exchange

Question 4
The diagram below shows the chemical reactions which occur in the blood
during the transportation of carbon dioxide.

Cell/ Tissue
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S
P + O 2 CO 2+ H 2 O
IV
I V R
III
CO 2 + H 2 O HCO 3 – H + Hb O 2

II
Q HbO 2
H 2 CO 3 Semester

(a) State the part of haemoglobin molecule that binds to carbon dioxide.
(1 mark) 2
(b) State the substances labelled P, Q, R and S. (2 marks)
P :
Q :
R :
S :
(c) Explain the events that occur at the stages I, II, III, IV and V. (5 marks)
I:
II:
III:
IV:
V:

Answer:
(a) Amino groups in haemoglobin
(b) P : Glucose
Q : Carbonic anhydrase
R : Haemoglobinic acid
S : Chloride ions
(c) I : Carbon dioxide produced during cellular respiration diffuses out from
the cell into the blood plasma. The carbon dioxide then diffuses into
the red blood cell where it combines with water to form carbonic acid.
This reaction is catalysed by the enzyme carbonic anhydrase.
II : The carbonic acid dissociates into hydrogen ions (H ) and hydrogen
+
carbonate ions (HCO ). This action is catalysed by carbonic anhydrase.
-
3
III : The haemoglobin releases the oxygen. The haemoglobin produced
combine with the hydrogen ions (H ) to form haemoglobinic acid.
+
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Biology STPM Chapter 7 Gaseous Exchange
IV : The hydrogen carbonate ions (HCO ) in the red blood cells diffuse
-
3
out. This causes the red blood cell to become positively charged.
V : Chloride ions diffuse into the red blood cells to neutralise the positive
charge in the cells. This process is called the chloride shift.
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Section C Essay Questions

Question 1

(a) Explain the structure of haemoglobin and how it is adapted to carry
oxygen. (8 marks)
(b) Describe how carbon dioxide is transported by the blood. (7 marks)


Answer:
(a) – Haemoglobin is a conjugated protein
– with a quaternary structure.
Semester
2 – It consists of four polypeptide chains, two alpha and two beta chains
coiled to form a spherical structure.
– Each polypeptide chain contains a prosthetic group, haem.
– Within each haem group there is a ferrous iron atom
– Each haem group is able to combine with an oxygen molecule. Hence,
there are 4 haem groups in a haemoglobin molecule; therefore each
haemoglobin molecule can combine with 4 oxygen molecules.
high partial pressure of oxygen
Hb + 4O HbO 4
2
Haemoglobin Oxygen low partial pressure of oxygen Oxyhaemoglobin
– In the lung alveolar capillary, where the partial pressure of O is high,
2
when the first oxygen molecule binds to a haem group, it causes a change
in shape of the haemoglobin so that the affinity of the other haem groups
for oxygen increases. This allows rapid binding with oxygen in the lungs.
Haemoglobin shows cooperativity.
– In the tissue, when the partial pressure of O is low and partial pressure
2
of CO is high, oxygen dissociates readily from haemoglobin.
2
(b) Carbon dioxide is transported in the blood in three ways:
(i) – about 5% carbon dioxide dissolves in the blood as a physical
solution
(ii) – about 10% carbon dioxide combines with the amino acid group
(NH ) in the protein part of the haemoglobin molecule to form
2
carbaminohaemoglobin





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Biology STPM Chapter 7 Gaseous Exchange
(iii) – about 85% carbon dioxide is transported in the form of hydrogen
carbonate ions (HCO )
-
3
– carbon dioxide from the active tissue cells diffuses from the tissue
fluid through the blood capillaries into the blood plasma and then
into the red blood cells
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– In the red blood cells, carbon dioxide combines with water to form
carbonic acid.
– The reaction is catalysed by carbonic anhydrase.
Carbonic anhydrase
CO + H O H CO
3
2
2
2
Carbon Water Carbonic
dioxide acid
– The carbonic acid dissociates into hydrogen ion (H ) and a
+
hydrogen carbonate ion (HCO ), catalysed by carbonic anhydrase.
-
3
H CO H + HCO 3 – Semester
+
2
3
Carbonic Hydrogen Hydrogen
acid ion carbonate ion
– The hydrogen carbonate ions (HCO ) diffuse out of the red blood 2
–
3
cell into the blood plasma. The red blood cell membrane is more
permeable to anions than to cations such as Na and K . In order to
+
+
maintain electrical neutrality, the chloride ions (Cl ) diffuse from
–
the blood plasma into the red blood cell. This process is called a
chloride shift.
– The increase in concentration of H ions in the red blood cell
+
lowers the pH. This causes the oxyhaemoglobin to dissociate into
oxygen and haemoglobin. Oxygen is released from the red blood
cells into tissues for respiration.
– The hydrogen ions (H ) combine with haemoglobin to form
+
haemoglobinic acid (HHb). Hence, haemoglobin acts as a buffer.
– In lungs, these processes are reversed. Then carbon dioxide diffuses
out into the alveolus from the blood capillaries.













163




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Biology STPM Chapter 7 Gaseous Exchange
7.2 Breathing Cycle

Section A Multiple-choice Questions


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Question 1
During breathing movements of a human, which of the following causes a
change from inspiration to expiration?
A The arrival of deoxygenated blood in the lungs.
B The stimulation of stretch receptors of the inflated alveoli.
C A low partial pressure of CO detected by the chemoreceptors of the
2
carotid bodies and aortic bodies.
D A high pH of the arterial blood.


Answer: B
During inspiration, as the lungs expand, stretch receptors on the wall of the
Semester
2 alveoli and bronchioles are stimulated. Nerve impulses are sent via vagus nerve
to expiratory centre in the medulla. This causes a change from inspiration to
expiration.


Question 2

The increase in concentration of carbon dioxide is detected by the
chemoreceptors located in the
A intercostal muscles
B carotid and aortic bodies
C diaphragm
D bronchiole tubes


Answer: B
The peripheral chemoreceptors are located in the carotid and aortic bodies,
which are sensitive to high CO concentration and low arterial blood pH.
2

Question 3
Which of the following factors is the most effective in increasing the breathing
rate in a mammal?
A A lack of oxygen in the tissues.
B A lack of oxygen in the blood.
C An excess of carbon dioxide in the lungs.
D An excess of carbon dioxide in the blood.



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Biology STPM Chapter 7 Gaseous Exchange
Answer: D
The breathing is accelerated when the partial pressure of carbon dioxide in the
blood increases.


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Section B Structured Questions


Question 1
Figure 7 below shows the structure involved in human breathing control
mechanism.

Cerebral hemisphere
(voluntary control of
breathing) Semester
Pons
J Cerebellum
2
K


L
M






Figure 7
(a) State the structures labelled J, K, L and M. (2 marks)
J :
K :
L :
M :
(b) Explain the role of J and M in the control of breathing. (1 mark)
J :
M :
(c) State the two main stimuli which affect the rate of breathing. (1 mark)
(d) Explain what happens to the breathing rate as we exercise. (2 marks)
(e) Describe briefly how the breathing rate is brought back to normal.
(2 marks)







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Biology STPM Chapter 7 Gaseous Exchange
Answer:
(a) J : medulla oblongata
K : vagus nerve
L : carotid body
M : aortic body
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(b) J : Has respiratory centre that contains the central chemoreceptors which
are stimulated by the high concentration of hydrogen ion (H ). It then
+
sends impulses to the diaphragm and intercostal muscles to increase
the rate of breathing.
M : These are specialised cells with chemoreceptors that detect the increase
in carbon dioxide or hydrogen ion (H ) concentration in the blood and
+
sends impulses to the respiratory centre.
(c) – Concentration of carbon dioxide.
– Concentration of hydrogen ions (H ).
+
(d) – As we exercise, the breathing rate increases.
– Exercise increases cellular respiration.
– This increases the concentration of carbon dioxide or hydrogen ion (H )
+
Semester
2 in the blood which is then detected by the chemoreceptors.
– Nerve impulses from the chemoreceptor are then sent to the respiratory
centre that will increase the rate of breathing.
– Excess carbon dioxide is expelled.
(e) – As more carbon dioxide diffuses from the blood into the alveoli, the
concentration of CO in the blood decreases.
2
– When the concentration of CO returns to normal and less H ions
+
2
are produced, the chemoreceptors will not be stimulated, hence the
breathing rate returns to normal.


Section C Essay Questions


Question 1
(a) Describe briefly the adaptations of the alveolus for gaseous exchange in
the lungs. (3 marks)
(b) Explain how the mechanism of breathing is controlled in a human.
(12 marks)












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Biology STPM Chapter 7 Gaseous Exchange
Answer:
(a) – The alveoli provide a large surface area for gaseous exchange as there are
numerous alveoli (about 700 million alveoli).
– The inner surface of the alveolus is always moist, allowing the gases to
dissolve easily before diffusion can occur.
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– The alveolar walls are thin and permeable allowing the gases to diffuse
across it easily.
– The alveoli are surrounded by a network of blood capillaries for gaseous
exchange.
(b) – The involuntary control of breathing is carried out by respiratory centres
in the medulla oblongata.
– The respiratory centres are divided into the inspiratory centre that
controls inspiration and expiratory centre that stimulates expiration.
– The peripheral chemoreceptors, carotid and aortic bodies, in the carotid
arteries and aorta are stimulated by the high partial pressure of carbon Semester
dioxide and the concentration of H .
+
– The central chemoreceptors in the medulla are stimulated by high
concentration of H . As the partial pressure of CO in the artery 2
+
2
increases, some of the molecules diffuse into the cerebrospinal fluid
surrounding the medulla. As a result, the concentration of H increases
+
and pH decreases.
CO + H O ⎯⎯→ H CO ⎯⎯→ H + HCO 3 -
+
2
3
2
2
– Nerve impulses from the peripheral and central chemoreceptors are sent
to the inspiratory centre.
– The inspiratory centre sends out nerve impulses via the phrenic nerve
to the diaphragm and via the thoracic nerve to the external intercostal
muscle causing increase in rate of contraction. This increases rate of
breathing.
– As the lungs expand, stretch receptors in the walls of alveoli and
bronchioles are stimulated.
– Nerve impulses are sent via vagus nerve to the expiratory centre in the
medulla.
– This inhibits inspiration. Expiration occurs.
– The lungs are no longer stretched. Stretch receptors are no longer
stimulated.
– Expiratory centre becomes inactive. Inspiration begins again.












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Biology STPM Chapter 7 Gaseous Exchange
7.3 Gaseous Exchange in Plants

Section A Multiple-choice Questions


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Question 1
Which of the following condition causes the opening of a stoma?
A The influx of potassium ions from the guard cells
B The decrease in the concentration of glucose in the guard cells
C The increase in the concentration of carbon dioxide in the guard cells
D The increase in the concentration of the abscisic acid when plants are
exposed to stress


Answer: A
According to the potassium ion accumulation hypothesis, the opening of the
stomata is associated with the influx of K ions into the guard cells from the
+
Semester
2 epidermal cells


Section B Structured Questions


Question 1
(a) With the aid of a diagram, describe the structure of stomata.
(b) Explain the mechanism of stomatal opening and closing.



Answer:
(a) – The stomata consist of a pair of guard cells.
– that surround a small pore known as the stomatal aperture
– The guard cells are kidney-shaped.
– The guard cell contains chloroplasts, nuclei, dense cytoplasm, vacuole
and mitochondria.
– It has a thicker, less elastic inner wall and a thinner outer wall.
– There are cellulose microfibrils that are radially orientated in the cell
walls of the guard cells.
– The guard cells are joined at the ends.
– The epidermal cells surrounding these cells are called subsidiary cells.








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Chapter 7.indd 168 11/2/21 9:27 AM

Biology STPM Chapter 7 Gaseous Exchange




Chloroplast Epidermal cell
(subsidiary cell)
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Outer wall of guard cell
Stoma
Nucleus
Inner wall of guard cell



(b) Stomatal opening
– During the day, chloroplast in the guard cells generate ATP by
photophosphorylation during photosynthesis. Semester
– The ATP generated is hydrolysed to provide energy to drive the proton-
pump in the cell membrane of the guard cells.
– Hydrogen ions (H ) are pumped out of the guard cells. 2
+
– This causes the cells to become negatively charged.
– So, the potassium ions (K ) diffuse from the subsidiary cells through
+
the potassium channel down the electrochemical gradient into the guard
cells.
– Chloride ions (Cl ) enter to balance the charges.
–
– The increase in K concentration causes the water potential of the guard
+
cells to become more negative (lower).
– Then, water from the subsidiary cells enters the guard cells.
– The guard cells become turgid, increasing the turgor pressure.
– The thinner and more elastic outer walls of the guard cells stretch more
compared to the thicker and less elastic inner walls.
– This causes the guard cells to curve outwards. Stoma opens.
Stomatal closure
– During the night, potassium ions (K ) move out of the guard cells into
+
the subsidiary cells.
– The water potential in the guard cells increases.
– Water diffuses out of the subsidiary cells by osmosis.
– Turgor pressure in the guard cells decreases.
– The cells become flaccid. Stoma closes.










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Chapter 7.indd 169 11/2/21 9:27 AM

Answers





STPM Model Paper 1 964/1 Section C
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18. (a)
Section A NH
1. D 2. D 3. C 2
4. B 5. A 6. A N C C N
7. D 8. A 9. D CH
10. A 11. D 12. C O O O HC N C N
13. C 14. A 15. B O—P—O—P—O—P—OCH O
2
O O O H H H H
Section B OH OH
16. (a) Paper chromatography
(b) Spot of concentrated – The structure consists of
chlorophyll extract. a purine base (adenine)
(c) (i) R value of A = 9.0/9.7 = attached to the carbon
f
0.93 atom 1 of a pentose sugar /
R value of C = 6.1/9.7 = ribose.
f
0.63 – Three phosphate groups
(ii) The R value can be are attached at the carbon
f
used as a comparison atom 5 of the pentose
to identify unknown sugar.
compounds in a mixture. – Terminal phosphate group
(d) – Molecular size has high energy level.
– Solubility – The structure can be
– Adsorption broken down/hydrolyse
into ADP +P+energy and
– Adhesion of the is a reversible reaction.
i
macromolecules to the
chromatography paper (b) – Occur in the inner
(choose any two answers) membrane of
mitochondria.
17. (a) Condition : Anaerobic – NADH/FADH is oxidised.
Location : Cytoplasm – Remove H ion/ H split
+
(b) Condition : Aerobic into H ion and electron.
+
Location : Matrix of the – The first electron acceptor
mitochondrion for NADH is NADH
(c) P : Ethanol dehydrogenase and FADH -
Q : Lactic acid succinate dehydrogenase.
(d) The pathways are important – As the electrons flow
because NAD is required for through the carriers, the
+
glycolysis to produce NADH. energy is released.
The NADH is oxidised to
regenerate NAD for glycolysis
+
to continue in living cells.

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Jawapan.indd 534 11/2/21 9:13 AM

Biology STPM Answers
– The oxygen is a final – Glycerate-3-phosphate is
electron acceptor that phosphorylated by ATP
combines with hydrogen to become glycerate-1,
ion/H to form water. 3-diphosphate.
+
– H ions / protons pumped – The glycerate-1,
+
across the membrane into 3-diphosphate is then
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the intermembrane space. reduced by NADPH
– There is a build-up and is converted into
of a proton gradient/ glyceraldehyde-3-phosphate
electrochemical gradient/ (PGAL, triose phosphate),
proton motive force. a 3-carbon sugar.
– The protons are pumped – Some glyceraldehyde-3-
1
back through the ATP phosphate ( PGAL) is
6
synthase. used to assimilate organic
– The energy generated is molecules such as glucose,
used to synthesis ATP from sucrose, starch, proteins
ADP+P. i
– Therefore, for 1 molecule of and lipids.
NADH, 3 ATP while for 1 – The rest of the 5
molecule of FADH, 2 ATP glyceraldehyde ( PGAL)
6
are produced. is converted in a series
19. (a) – Calvin cycle occurs in the of complex reactions
stroma of the chloroplast. to regenerate ribulose
– It involves a series of bisphosphate.
reactions that results in the – The processs requires ATP
reduction of carbon dioxide and a series of accessory
into carbohydrate. enzymes.
– NADPH and ATP, products
of the light-dependent (b) – In C plants, ribulose
3
reaction provides the bisphosphate (RuBP)
reducing power and energy combines with carbon
required in the reactions. dioxide while in C plants
4
– Carbon dioxide from the the carbon dioxide acceptor
atmosphere combines is phosphoenol-pyruvate
with ribulose bisphosphate (PEP).
(RuBP), a 5-carbon – In C plants, this reaction
3
compound to form an is catalysed by ribulose
unstable 6-carbon sugar. bisphosphate carboxylase
– The enzyme involved in (RuBisCo) which in
this reaction is Ribulose C plants, the reaction
4
bisphosphate carboxylase between PEP and carbon
(RuBisCo). dioxide is catalysed by PEP
– The six-carbon compound carboxylase.
immediately splits into two – In C plants, the first
3
molecules of glycerate-3- product of carbon dioxide
phosphate (GP), a 3-carbon fixation is glycerate-3-
compound. phosphate, a 3-carbon


535




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