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Chapter 7:Respiration 2014
7.1 The respiration process in energy production
No
(a)
Marking scheme
Marks
Aerobic respiration Anaerobic respiration
OR
Process
Respiration equation
S
Glucose +oxygen
R
Glucose
Carbon dioxide +Water +2898 kJ energy
Carbon dioxide +ethanol+210 energy
Name the process R and S
(b)
R:Anerobic respiration
1
S:Aerobic respiration
1
Table 1 shows the respiration equation shown by muscle cells and yeast cells during cellular
respiration
Cell type
Respiration equation
(Smooth) Muscle
cells
Glucose +oxygen
Yeast cells
Glucose
Carbon dioxide +Water +2898 kJ energy
Carbon dioxide +ethanol+210 energy
1
1
(a) Fill in the table by writing in muscle cells or yeast cells that matches with its respiration
equation
(c)
2
State where tissue V(smooth muscle cell) can be found in the body
Blood vessel/alimentary canal/oeosophagus/stomach/uterus/urinary bladder/etc
(d)
2
1
1
Write the equation of process S and R
Process R
Glucose
lactic acids + energy
2
Process S
Glucose +Oxygen
Carbon dioxide + water +2898 kJ
138
2
Reactant- 1m
1
Product -1m
1
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Chapter 7:Respiration 2014
(e)
Explain process P and Process Q / Explain the cellular respiration process that occurs in
individual P and Q
Process P
F1 - aerobic respiration.
P1 - glucose is completely oxidized/breakdown in the presence ofoxygen
P2 - releases more energy/2898 kJ of energy ( per mole of glucose)
E3-Produce carbon dioxide and water
Process Q
F2 - Anaerobic respiration
P3 - glucose is not completely oxidized// the glucose molecules breakdown partially (into lactic
acid)
P4 - releases less energy/150 kJ of energy 9 per mole glucose)
E6-Produce lactic acid
(f)
1
1
1
1
1
1
1
1
6
Anaerobic respiration in cells
Explain the condition of a person after completing a 100 meter race in 12 seconds 2
(g)
F-the person is panting /higher breathing rate
1
E1-As he is in oxygen debt//anaerobic respiration
1
E2-Much lactic acids is produced (in his muscle cells)
1
E3-Causes muscle cramp Any 2
1
4
When a person is resting, the heartbeat rate is 61 to 71 beats per minutes .During vigorous
activity, the heartbeat rate increase to 120 beats per minute
Explain this statement
(h)
F1 - (During the vigorous activity) the muscle cells are in state of oxygen deficiency/oxygen
debt //the blood cannot supply oxygen fast enough to meet the demand for oxygen ATP
1
P1-( The increase in heated beat rate ) is to deliver more glucose to muscle cells
1
P2-To induce extra energy cellular respiration
1
P3-To remove more carbon dioxide from the muscle cells Any 2
1
4
After completing vigorous exercise, an athlete will gasp heavily
Based on the above statement the condition faced by the athlete
Oxygen debt (reject: anaerobic respiration is a process, not a condition)
Explain why
E1-Because of oxygen deficiency//lack of oxygen
1
E2-To get more oxygen immediately
1
E3-To oxidize lactic acids Any 2E
1
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(a)
(b)
(c)
(d)
(e)
Explain how the oxygen intake by the athlete returns to the normal level at the 25th
minute
P1-Lactic acid has been removed from the muscle
1
P2-The lactic acids has been converted to energy/convert to glucose
1
2
Explain the condition of a person after completing a 100 meter race in 12 seconds
F-the person is panting /higher breathing rate
1
E1-As he is in oxygen debt//anaerobic respiration
1
E2-Much lactic axids is produced ( in his muscle cells)
1
E3-Causes muscle cramp Any 2
1
2
Explain the usage of cell W in bread making industry
F1-Carbon dioxide released
1
E1-Traps in the dough
1
E2-Causes the dough to rise
1
2
Explain what happen to the yeast cells if there is too much ethanol produced
P1-( too much ethanol0 causes unsuitable medium /condition //toxic/poisonous medium
/condition
1
P2-For yeast cells to reproduced //yeast cell die
1
2
State the differences between the process that mention I 6(a) (i)
Diagram shows respiratory organs in an insect and human (Prefer)
Aerobic respiration
Anaerobic respiration
D1-Oxidation of glucose in present of
oxygen/ Oxygen is required
D1-Oxidation of glucose in absent of
oxygen / Oxygen is not required
1
D2-Oxidation of glucose is complete/
Complete breakdown of glucose
D2-Oxidation of glucose is not complete/
Incomplete breakdown of glucose
1
D3-Produced higher/large energy/38
ATP/2898 kJ of energy 9 per mole of
glucose)
D4-Produced lower energy /2 ATP/150 kJ
of energy ( per mole of glucose)
1
D4-Produced carbon dioxide and water
D4-Produced lactic acid
1
D5-Occurs in mitochondria
D5-Occur in cytoplasm
1
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Chapter 7:Respiration 2014
(f)
Diagram shows the rate of oxygen intake before, during and after a vigorous Exercise of an
athlete.
(i) Based on the graph, compare the respiration before and during the vigorous
Exercise.
4
1
Before (A)
During (B)
Explanation (E)
Aerobic respiration
Anaerobic respiration
Before-Oxygen Intake is
low/the same as oxygen
required /enough oxygen is
supplied to the cell
3
During-Oxygen required is
more than oxygen intake
2
The muscle are in
normal condition
The muscle are in the
atate of oxygen debt
Before-Oxygen is
sufficient
3
During-Oxygen is
insufficient/oxygen
supplied is less than
oxygen supplied
3
Energy produced is more
/38 ATP
Energy produced is less /2
ATP
Before-complete
breakdown of glucose
(produce more energy )
During-incomplete
breakdown of glucose
(produce less energy)
4
No/less accumulation of
lactic acid in the muscle
High accumulation of
lactic acids in the muscles
Before-complete
incomplete break down of
glucose produce carbon
dioxide and water
Dduring -Incomplete
breakdown of glucose
produce lactic acid
A+B=1m
3
3
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4
E=1m (Any 1 E)
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(g)
Explain what happens to cell w when there is no oxygen
F1-Cell W undergoes anaerobic respiration
1
E1-Glucose break down (partially/incompletely)
1
E2-To produce ethanol, carbon dioxide
1
E3-Less ATP/2 ATP is produce
1
2
F1 and any of E1/E2/E3
(h)
the above process takes place in tissue P in the presence of oxygen .Name and explain the
process
F-Process is called aerobic respiration
1
1
1
P1-Glucose diffuses into cells P from the blood capillary
P2-Cells P contain a lot of mitochondria
P3-Mitochondria ( contain enzymes) for cell respiration //mitochondria carry out cell
respiration
P4-Oxidation of glucose (take placed in mitochondria)
P5-In a series of reaction catalyzed by respiratory enzymes in mitochondria
P6-1 molecule of glucose will produce 38 molecule ATP/ More ATP
P7-water and carbon dioxide are released as waste material in this process
(i)
1
1
1
1
1
1
8
Explain the importance of increased pulse rate during vigorous activity and why it takes several
minutes for the pulse rate to return to normal after activity 6
During vigorous activity,
P1 more blood is sent to the muscles
P2-so that oxygen supply to the muscles is increased
P3-The heart beats faster
P4-to deliver more blood, hence the pulse rate increases
After some time during the activity,
P5-respiration takes place anaerobically
P6-because the maximum rate of oxygen uptake is less than oxygen demand.
P7-there is build up of lactic acid
P8-After activity, a period of recovery is needed to provide the oxygen
P9-so that the lactic acid can be oxidized
and to provide the energy for the recovery of the muscles
1
1
1
1
1
1
1
1
1
1
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Chapter 7:Respiration 2014
(a)
Molecule X + 2ATP
Process Q
- Anaerobic respiration
Molecule X - Lactic acid
P1- Inhale more oxygen by doing fast and deep breathing.
P2-Excess oxygen taken in during inhalation is used to oxidize lactic acid to carbon dioxide
1
1
1
1
and water.
1
P3-This oxidation process takes place in the liver.
P4-Thus the oxygen debt is the amount of oxygen needed to remove the lactic acid from the
muscle cells.
Lactic acid + oxygen
carbon dioxide + water + energy
1
6
1
(b)
P1-The muscle cells of the athlete undergoes anaerobic respiration to produce energy
P2-During intensive physical activity / running / sprinting// when the athlete start running (t =
1
1
0), oxygen requirement increase immediately to produce large amount of energy
P3-The athlete holds his breath for a short period of time // the athlete breath is shallow during
running
P4-The oxygen supplied by breathing between t = 0 minute to 6 minute is insufficient for
cellular respiration
P5-The muscle cells are now in the state of oxygen debt // undergo oxygen deficit
P6-Glucose is broken down incompletely without the presence of oxygen
1
1
1
1
1
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P7- Small amount of energy is released to continue the activity
P8-Lactic acids produced accumulate in the muscle causing the muscular pain and fatigue
P9-The anaerobic respiration occurs in the cytoplasm
P10- (after the activity is over), the athlete breathes faster and deeper to supply more oxygen
P11-Oxygen is used to oxidize the lactic acid into carbon dioxide, water and energy //
converted into glucose and stored as glycogen
1
1
1
1
1
10
7.2the respiration structure and breathing mechanism in human and animal
No
(a)
Marking scheme
Marks
Adaptation of the respiratory structures
State two characteristic shown by the respiratory surface of animal(common characteristic)
P1-the respiratory surface is moist
P2-Cells lining respiratory structure are thin
P3-Thr respiratory structure has a large surface area
1
1
1
3
The respiration structure and breathing mechanism insects
Aspect
Respiratory
structure
Question & Marking Scheme
Marks
The respiration structure and breathing mechanism insects
P:Air sac
Q: Muscle
T: Spiracle
S: Trachea
R:Tracheole
Name the part labeled P ,Q ,Rand S
Which organism has the respiratory structure?
Insect
Name the respiratory system shown in diagram 2.1
Tracheal system
State the function of the following
(i)
Chitin
support the tracheal/prevent the tracheal form collapsing
(ii) Air sac
Speeds up the movement of gases exchange to and form tissue during
vigorous body movement
5
5
1
1
1
1
1
1
2
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Aspect
Question & Marking Scheme
Marks
Explain one adaptation of the respiratory structure in diagram for efficient gaseous
exchange
Structural
Adaptation
P1-The large number of tracheoles provides a large surface for the diffusion of gases
1
P2-Tips of tracheoles have thin permeable wall and contain fluid in which
respiratory gases can dissolved
1
P3-Terminal ends the tracheoles remains moist which allows teh gases to be
dissolved
1
2
Explain how structure Q and S increase the efficiency of gaseous exchange in each
organism 2
F-Consists of million alveoli in lungs and many tracheal Tubes/Tracheole/thin
layer/1 cell thick
1
1
P1-To increase total surface area per volume rate for gaseous exchange
F2-The inner surface of alveolus and tracheoles end consists of tissue fluid moisture
P2-To provide moist surface for gas diffusion /to dissolve oxygen /gases for
diffusion Any F +P
Breathing
mechanism
1
1
2
State how air is drawn from T to S 2
P1-By(rhythmic) movements, of the abdominal muscles
P2-Decreasing of air pressure inside trachea, ( so the air is drawn in)
P3-Gases diffuses into the cells(s)
1
1
1
2
Diagram 7.1 (i), (ii) and (iii) show the respiratory structure of an insect. Describe the
respiratory structure and breathing mechanism of and insect
R-respiratory structure
R1-The tracheal system consists of network of trachea
1
R2-The trachea is lined with chitin to prevent dorm collapsing R3-Spiracles is tiny
opening thet allow air to go in and out
1
R3-spiracles is tiny opening that allow air to go in and out
1
R4-The trachea branch into fine tubes celled tracheole
1
R5-The tracheole branch throughout the body and temperature and penetrate into
body tissues / muscle
1
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Breathing mechanism
B1-When inside inhales, the abdominal muscles relax and spiracles open
1
B2-air pressure inside the trachea decrease and air is drawn in
1
B3-When the insect exhale, the abdominal muscle contract
1
B4-So increase air pressure in side trachea and forces air out through spiracles
1
B5-Inesct inhale and exhale through rhythmic contraction and expansion of their
abdominal muscles
1
B6-the body movement and contraction of abdominal muscle speed up the rate of
diffusion of gases from trachea into body cells
1
8
Breathing
mechanism
Explain the gases exchange between tracheol and body cell.
P1-Partial pressure/concentration of oxygen in the tracheole is higher /than partial
pressure/concentration of oxygen in body cell
1
P2- Oxygen diffuse from tracheole to body cell
1
P3- Partial pressure/concentration of carbon dioxide in the body cell is higher than
partial pressure/concentration of carbon dioxide in tracheole .
P4- Carbon dioxide diffuse from tracheole to body cell
1
1
4
1
Chitin is a polysaccharide on the outer surface of structure P. Due to the change in
the environment, the insect is unable to form the polysaccharide.
Explain how the absence of chitin affects inhalation and the energy production. 6
P1- The function of chitin is to prevent trachea from collapsing/sustain
the air pressure
P2- During inhalation high pressure air moves into the trachea.
P3 -The absent of chitin will cause the trachea / P to collapse / burst /rupture.
P4 -Air with oxygen cannot reach tracheal.
P5-Body cell cannot get enough oxygen for cellular respiration
P6-The insect does not produce enough energy and respire anaerobically.
P7-Less energy produced. (Any 6)
1
1
1
1
1
1
1
6
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Aspect
Question & Marking Scheme
Marks
Breathing
mechanism
Diagram show a trachea system of and insect Based on the diagram explain the gases
exchange between the tracheoles and muscle cells
F-there are concentration gradient of oxygen and carbon dioxide between tracheoles
& body cells
1
E1-(simple) diffusion can take place
1
E2-Oxygen concentration /partial pressure is higher in the tracheoles while the
concentration of oxygen is lower in the cells
1
E3-Oxygen diffuses directly form the tracheoles onto the cells
1
E4-Carbon dioxide concentration is higher in the cells while lower in the tracheoles
1
E5-Carbon dioxide diffuses directly form the cells into the trachoeles
1
4
The respiratory structure and breathing mechanism of fish
Aspect
Respiratory
structural
Marking scheme
Marks
The respiratory structure and breathing mechanism of fish
What is X ?/ Name the respiratory structure of the organism in diagram
Gills/ gill filament
1
1
1
1
1
1
State the function of structure P
P-Speed up the movement of gases to and from the insect’s tissue
The efficiency of gaseous in organism Y is further enhanced by a mechanism.
Name the mechanism
Countercurrent exchange mechanism
State two characteristic of X, which makes it a good respiratory structure for fish 2
147
P1-Have lamella and filament to increase total surface area
1
P2-Numberous blood capillaries for efficient transport of respiratory gases
1
2
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Chapter 7:Respiration 2014
Aspect
Structural
adaptation
Breathing
Mechanism
Breathing
mechanism
Inhalation
Question & Marking Scheme
Marks
Explain one adaptation of the respiratory structure in diagram 1.1 (b) and diagram
1.2 (b) for efficient gaseous exchange
P1-Th e have numerous thin walled lamellae to maximize the surface area for
gaseous exchange
1
P2-The gills filament have numerous thin membrane and covered by net work of
capillaries to transport respiratory gases
1
P3-the surface of gills Is moist which allows the gases to be dissolved
1
Based on the diagram explain how the oxygen is drawn from mouth to X(gill)
P1-Mouth closes
1
P2-The floor of buccal cavity raised (water contain air flow to X)
1
2
Describe the inhalation in fish
E1-th floor of cavity lowers
1
E2-At the same time, the opercular cavity enlarges and operculum closes
1
E3-This lowers the pressure in buccal cavity
1
E4-Water with dissolved oxygen is drawn into the mouth
1
4
Describe the breathing mechanisms in fish.
P1 - When the mouth opens, the floor of the buccal cavity is lowered./Increase the
volume/ space of the buccal cavity
P2-opercular cavity enlarges and operculum closes
P3 - This lowers the pressure in buccal cavity .
P4 - Water with dissolved oxygen is drawn into the mouth.
Exhalation
2
P5 - When the mouth closes, the floor of buccal cavity is raised.
P6 - Water flow through the lamellae and gaseous exchange between
the blood capillaries and water takes place.
P7 - Oxygen diffuses from the flowing water through the gill lamellae into the
blood capillaries.
P8 - Carbon dioxide diffuses from the blood capillaries via the gill lamellae into the
flowing water. Any 4
1
1
1
1
4
1
1
1
1
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Chapter 7:Respiration 2014
The respiratory structure and breathing mechanism of amphibians
Aspect
Question & Marking Scheme
Marks
Respiratory
structural
Name structure X and Y in diagram 3.1
2
1
X: Bucco-pharyngeal
1
Y: Glottis
Structural
adaptation
Breathing
Mechanism
2
Respiratory gases flow in and out through the lungs .Describe the characteristic of
the frog’s lungs
E1-Numerous inner partition to increase the surface area
1
E2-Membrane of lungs are thin and moist to facilitate the efficient diffusion of
respiratory gases
1
E3-Supplied with a rich network of blood capillaries to transport respiratory gases
to the body cells
1
3
Structure Y in diagram 3.1 had been injured .Describe how this condition affect the
respiration of the frog
E1- Glottis unable to open and close
1
E2-Air pressure is not increased /decrease in the bucco-pharyngeal cavity
1
E3-Air cannot be forced into /out the lungs
1
E4-Lung ventilation is not efficient
1
4
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The respiratory structure and breathing mechanism of humans
Aspect
Marking scheme
Marks
Respiratory
structure
Name the parts labeled Y
Y-Alveolus
1
1
1
1
1
1
1
1
What is the function of alveoli?
Place for gaseous exchange //store the oxygen gas before gaseous exchange occur
State the organ in which the tissue in Diagram 4.1(alveolus) can be found
Lung
State the function of organ stated in
Gaseous exchange//respiration
Respiratory gases flow in and out through the trachea .Describe the characteristic of
trachea
F-Have C-shaped cartilage rings //cartilage rings
1
P1-keep the trachea open permanently
1
P2-Avoid the trachea form collapse when the out side pressure is higher than inside
pressure
1
P3-oxygen can continuously flow through trachea to the alveoli/lung F-1m P-1m
Explain the effects of the breathing mechanism if structure R is unable to function
P1-Structure R is diaphragm.
1
P2-Less/no change in volume in the thoracic cavity/ lung
1
P3-Less/ no change in air pressure in the thoracic cavity/ lung
1
P4-Less/ no air exchange/ less/no intake of O2/ less/no CO2 expelled
1
3
1
4
Resulting difficulty in breathing in and out
Structural
adaptations
State the important characteristic of alveoli to ensure the function in (a) is efficient 1
P1-Have very large total surface area//
1
P2-Have moist surface all the time//
1
P3-have very thin wall/one cell thick Note ( any 1P)
1
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Describe the characteristic of the respiratory structure of human that enable gaseous
exchange to be carried out efficiently
Breathing
mechanism
P1-the ratio total surface area per volume (TSA/V) is high for the exchange of gases
1
P2-the cells lining the respiratory surface is a single layer of cell which is very thin to
allow gases to diffuses easily
1
P3-the respiratory surface is constantly moist to allow gases to dissolved in water
before diffusing in and out of the respiratory surface
1
P4-the respiratory surface is covered with a dense network capillaries to allow rapid
diffusion and transport of gases
1
3
Describe how intercostals muscle and diaphragm can change the volume and pressure
in the thoracic cavity during inhalation
P1-External intercostals muscle contract/internal intercostals muscle relax caused the
ribs cage moves out wards and upwards
1
P2-Diaphgram muscle contract , the diaphragm lower and flattenP3-The volume of
thoracic Cavity increase but the pressure decrease (lower the atmospheric pressure)
1
P3-The volume of thoracic cavity increase but the pressure decrease ( lower the
atmospheric pressure)
1
1
P4-Air forced into the lung//alveolus
3
Describe the breathing mechanism of human
Inhalation:
P1-External intercostals muscle contract//internal costal muscle relax
1
P2-ribcage move upwards and out wards
1
P3-diaphragm contracts/flattens
1
P4-Volume of thoracic cavity increase // pressure of thoracic cavity decrease
1
P5-So air ( form outside) is forced into the lungs
1
Exhalation :
P1-External intercostals muscle relax//internal costal muscle contract
1
P2- ribcage move downwards and inwards
1
P3-diaphragm relax/curved upward
1
P4-Volume of thoracic cavity decrease // pressure of thoracic cavity increase
1
P5-So air ( form inside) is forced out of lungs
1
6
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Constructing a model of human lung study the breathing mechanism in humans
No
Marking scheme
(a)
Marks
Rubber cork
Glass tube
Bell Jar
Balloon
Thin rubber sheet
String
Based on the model of the lungs in Figure 3.1, what are the equivalent structures to the glass
tube and the bell jar in the human respiratory system?
Glass tube:
Trachea /
Bell jar :
Rib cage / ribs
Balloon :
lung
Rubber sheet: diaphragm
(b)
1
1
1
1
4
1
1
The thin rubber sheet represents the diaphragm in the human respiratory system.
What is the function of the thin rubber sheet in the model of the lungs?
To increase / decrease the pressure / volume in the bell jar
(c)
The balloons represent the human lungs.
Explain one characteristic of the balloons which is similar to the human lungs[2 marks]
F- elastic
E- can expand (inhalation) and contract/ decrease in size (exhalation )
(d)
1
1
2
1
1
(c) (i) The string in the model of the lungs is released..
Draw the changes to the balloons in Diagram 3.2 below.
-both balloons decrease in size
(e)
(ii) Observe your drawing in (c)(i).
Explain the relationship between the changes in the model of the lungs you have drawn and the
real human respiratory system.
P1- the string represent the diaphragm
P2- when the diaphragm muscles contract,
P3- the volume of the thorax increase
1
1
1
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P4- this will decrease the thorax pressure
P5- air will be inhale
(f)
1
3
1
1
The percentage of oxygen and carbon dioxide gases in inspired and expired air is
determined by using the J-tube.
Why is the end of the J-tube dipped in potassium hydroxide solution and then followed by
potassium pyrogallol solution? 1
To prevent oxygen gas being absorbed by the potassium pyrogallol solution as it can absorb
both carbon dioxide and oxygen
(g)
(ii) Table 3.3 shows the result of a study on the content of inspired and expired air.
Type of gas
Inspired air / %
Expired air / %
Oxygen
21.0
16.0
Carbon dioxide
0..04
4.0
Nitrogen gas
78.0
78.0
Water vapour
Vary
Saturated
Explain why there is an increase in percentage of carbon dioxide in the expired air.
P1-The concentration of carbon dioxide is higher in the cell body; released from the cellular
respiration
P2-Carbon dioxide diffuses into the blood to be transport to the lungs.
1
1
2
Comparison of respiratory system between human and insect
No
Marking scheme
Marks
(a)
S
R
P
Q
State one similarity and one difference of structure P in diagram 2.1 and 2.2
Similarity: both wall of P consisting ring to strengthen it
Differences: the wall of P in insect consists of chitin ring while P in human consists of cartilage
ring
(b)
1
1
2
Humans and cockroach have different respiratory system .Explain one difference between the
respiratory system of human and a cockroach
F1-Respieratory structure of cockroach consists of trachea and spiracles while the respiratory
structure of human consists of a trachea and a pair of lungs
1
P1-tracheae of cockroach are branch into 2 bronchi which enter the right and left lungs
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P2-Thr trachea of human branched into 2 bronchi which enter the right and left lungs
1
P3-The bronchi of human branched ito smaller tubes called bronchioles which ends in a cluster
of sacs called alveoli
1
3
(c)
Explain one similarity and four differences between the respiratory organs of insect and human
Similarities
1
S1-Both of respiratory organs has thin wall/one cell thick
E1-Incrase rate of diffusion of respiratory gaseous
1
OR
1
S2-Both of respiratory organs has respiratory surface such as alveolus in human and tracheole
in an insect
1
E2-Provide a large surface area for the diffusion
Differences
D1-Trachea in human is supported by cartilage and traches in insect is supported by chintin
1
1
E1-To prevent them form collapsing
D2-The wall of alveolus is moist surface but the tracheole has fluid
1
E2-To dissolve the respiratory gases
D3-Alveolus is covered by network of blood capillaries but not for trachoele
1
E3-T provide a large surface area for rapid diffusion of gases 9 to and form the alveoli0 in
human but tracheole direct contact to the tissue ( and organs)
1
D4-Haemoglobin is needed in transport of oxygen nt but in insect
1
E4-oxygen combine with heamoglobin in (erythrocyte) to form oxyhaemoglobin but not in
insect
1
D5-(larger) insect have air sacs but not in human
1
E5-to speed up the movement of gases to and form the insect’s tissue
1
D6-in human air enters the lungs through the nostrils but spiracles in insects
1
1
E6-to allow gases in and out of the body any 4 pairs
10
What differences between the respiratory system of frog and fish
D1-Gills is the respiration organ for fish but lung and skin ids for frog
1
D2-Gill have filament and lamella to increase the surface area, but lung of frog have numerous
inner partition to increase the surface area
1
D3-Gill received oxygen directly form water , but lungs and skin of frog received oxygen form
the atmosphere
1
2
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(d)
Describe the comparison between the respiratory system in insect and human 8
Similarities:
1
F1-The structure of tracheal system and trachea branches into small tubes
E1-increase the total surface area of tracheole/alveolus so that increase the efficiency of gases
exchange
F2-moist surface on tracheole and alveolus
1
E2-Oxygen and carbon dioxide can be dissolve easily
1
F3-Very thin wall of tracheole and alveolus/one cell thick
1
E3-To ensure the simple diffusion can take place /Increase rate of diffusion of respiratory
gaseous
Insects
Aspect
Human
F4-Consists of spiracles,
Respiratory structure
Consists of nose trachea,
trachea and tracheoles
bronchus, bronchioles ad
alveolus
E4-Air enters through
Air enter through nose into
spiracles into tracheoles
lungs/alveolus
F5-Tracheoles directly contact
Alveolus is surrounded by a
with the muscle cells
large network of blood
capillaries
F6-Trachea is reinforced/
supported with ring of chitin
E6-Prevent the trachea form
collapsing due to different air
pressure
F7-Does not have red blood
cell to transport oxygen
Trachea is reinforced/
supported with ring of
cartilage
P5-Prevent the trachea form
collapsing
Oxygen transportation
E7-Oxygen is not transported
in the body
F8-Oxygen diffuses directly
form the respiratory tructure
into the cells
E8-Carbon dioxide is directly
released form the cells into
tracheoles
The diffusion of
oxygen into the cells
Product of respiration
1
Has red blood cells to
transport oxygen through
blood vessels
Oxygen is transported by red
blood cells around the body
Oxygen needed to be
transported into the cells and
then diffuses into the cells
Carbon dioxide produced
diffuses into the blood
capillary then transported into
the lungs
1
1
1
1
1
1
1
1
1
1
10
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Comparison of respiratory system between human and fish
No
Marking scheme
Marks
(a)
Explain three adaptation from structure show in diagram 2 (b)(ii) to carry out its function
efficiently
(b)
(c)
(d)
P1-Thin membrane /one cell thick for easily diffusion of respiratory gases
1
P2-Moist surface for respiratory gases easily dissolve
1
P3-Numerous blood capillaries for efficient transport of respiratory gasesAny 2
1
3
Y is the respiratory surface in human, explain how gaseous exchange occurs between structures
Y and blood capillary
P1-t he partial pressure of oxygen in Y is higher than in blood capillaries
1
P2-Oxygen diffuses form Y into blood capillaries by simple diffusion
1
2
Humans and fish have different respiratory systems, Explain one differences between the
respiratory system of human and fish 3
F1-the respiratory system of fish of gills while the respiratory system of human consists of a
trachea and pair of lungs
1
P1-A fish has four pairs of gills which are covered by operculum//the surface of each gills
Filament has many plate –like projections called lamella
1
P2-the trachea of human branched into 2 bronchi which enter the right and left lungs//The
bronchi of human branched into smaller tubes called bronchioles which ends in a cluster of sac
called alveoli
1
3
What are the differences between respiratory system of human and fish?
P1-gill is the respiratory organ for fish nut is for human
1
P2-gill have filament and lamella to increase the surface area, but lung have alveoli to increase
the surface area
1
P3-gill touch /surrounded by water
1
P4-Gill receives oxygen directly from water, but lung received oxygen form atmosphere via
trachea , bronchus and bronchioles
1
3
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7.3Gaseous exchange across the respiratory surfaces and transport of gases in humans
The process of gaseous exchange across the surface of the alveolus and blood capillaries and between the
tissue capillaries and the body tissue cells
No
(a)
Marking scheme
Marks
State the importance of gaseous exchange in human
P1-To get oxygen for (cellular) respiration
1
P2-To get rid of/excrete the carbon dioxide
1
2
(b)
Name gas X and Y
1
X : Oxygen
Y : Carbon dioxide
(c)
1
Explain the difference between the concentration of gas x and Y in blood vessel Q
F1 : The concentration of gas X in blood vessel Q is lower than gas Y
E1 : Oxygen has been used by the body cells /cellular respiration
E2 : (Cellular respiration) produces gas Y
E3 : to be sent to the lung (to be excreted)
(d)
(e)
(f)
2
1
1
1
1
2
Name blood vessel P and Q
P: Pulmonary veins
1
Q:Pulmonary artery
1
2
State the function of blood vessel P and Q
P: Carries deoxygenated blood to lungs
1
Q: carries oxygenated blood back to heart
1
2
Describe the role of blood vessel P in transporting oxygen form alveolus to muscle cells
P1-In the blood, Oxygen form alveolus combine with respiratory pigment/haemoglobin to form
oxyhaemoglobin /oxygenated blood
1
P2-Transport oxygenated blood //oxyhaemoglobin to heart
1
P3-the heart pump the oxygenated blood to muscle cells via the aorta Any2
1
157
2
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(a)
State the process by which gaseous exchange takes place across alveolus1
(Simple) diffusion
(b)
1
1
1
1
1
Explain how the process occurs
F-Partial pressure of oxygen /carbon dioxide in the air of the alveolus is higher than in blood
capillary
(c)
1
Gaseous exchange takes place across structure Y
Name structure Y
Alveolus/ Alveoli
(d)
State two ways how the alveolus are adapted for efficient gaseous exchange
P1-Thin wall
P2-Moist
P3-Rich with blood capillary
(e)
1
1
1
2
Explain how the alveolus is structured to increased the efficiency of gaseous exchange
F1 : Alveolus has thin wall ( one cell thick)
E1 : Gaseous can diffuse in and out through the wall more efficiently / Quick /easy gases
diffusion
1
F2 : The (inner) surface of the alveolus is moist
E2: Allowing oxygen to dissolve first before diffusing out
F3 : A large number of alveoli /The (outer surface) of the alveolus is covered by a network of
blood capillaries
1
1
1
P1-Large total surface area per volume for gaseous exchange
F4-Network of blood capillaries
P4-To increase the rate of gases transportation F+P=1m
E3 : Increase the surface area for rapid diffusion of gaseous
Notes : F1/2/3 + E 1/2/3 = 2 mark
F1/2/3 = 1 mark
E1/2/3 = O mark
1
1
1
1
1
2
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(f)
Describe the movement of respiratory gases across structure Y
P1-Partial pressure of oxygen on alveolus is higher than the partial pressure of oxygen in the
blood capillaries//oxygen concentration is higher in alveolus than in the blood capillaries
1
P2-Oxygen Diffuses form alveolus into the blood capillaries
1
OR
P3- Partial pressure of carbon dioxide on alveolus is higher than the partial pressure of oxygen
in the blood capillaries/Carbon dioxide oxygen concentration is higher in alveolus than in the
blood capillaries
P4- Carbon dioxide diffuses form alveolus into the blood capillaries
(g)
1
1
4
Explain the role of oxygen in the muscle cells
F-oxygen oxidized the glucose molecule
1
E1-Cellular respiration /aerobic respiration takes place in muscle cells
1
E2-ATP/energy released
1
E3-Produced carbon dioxide and water as by product/waste products
1
E4-energy is used for contraction and relaxation of muscle cells/movement of insect
1
No
Marking scheme
4
Marks
(a)
R
P
Q
Based on the diagram 3.2 name X and Y
(b)
(c)
X: oxygen
1
Y: Carbon dioxide
1
2
Name structure P and Q
X: Red blood cell
1
Y:Alveolus
1
2
1
1
Name the complex substances contained in X
Haemoglobin
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(d)
Explain how the gaseous exchange occur across the alveolus 3
P1 : Oxygen diffuse/ moves across /through ( plasma membrane) to blood capillary
P2: From higher (oxygen ) concentration ( in alveolus )to lower concentration ( in blood
capillary)
P3: On the other hand the partial pressure of carbon dioxide is lower in the air of the alveoli
compared to the blood capillaries.
P4: Carbon dioxide diffuses out of the blood capillaries into the alveoli.
P5 : expelled through the nose or mouth into the atmosphere
1
1
1
1
1
3
(e)
Explain how gaseous exchange occurs during respiration in Diagram 4.1 (in human )
(f)
F1-Oxygen diffuses from alveolus into blood capillaries
1
E1-Oxygen concentration /partial pressure in alveolus is higher than in blood capillaries
1
F2-Carbon dioxide diffuses from blood capillaries to the alveolus
1
E2-Carbon dioxide concentration /partial pressure in the blood capillaries is higher than in
alveolus MAX:2
1
2
Explain how the red blood cell accepts oxygen form alveolus and transfer to the cell
P1-Oxgen diffuses into the blood plasma
1
P2-Combine with haemoglobin
1
(g)
2
CO2
O2
R
S
.
Based on the diagram , explain the exchange of respiratory gases
P1-Respiratory surfaces in human are alveoli.
1
P2-The concentration of oxygen in the alveoli is higher than its concentration in the blood
1
160
capillaries.
1
P3-Oxygen in the alveoli diffuses into the blood capillaries.
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P4The concentration of carbon dioxide in the blood capillaries is higher than its concentration
1
in the alveoli.
P5-Carbon dioxide diffuses from the blood capillaries of the lungs into the alveoli.
P6-Blood leaving the blood capillary of the lungs has higher concentration of oxygen and lower
concentration of carbon dioxide
1
1
6
7.4 The Regulatory mechanism in respiration
The human respiratory response and rate of respiration in different situation
Diagram 7 (ii) shoes 3 different situation of human activities
Diagram 7 (ii) (a)) shows a boy watching television
Diagram m 7 (ii(b)) shows a man is chased by a fierce dog
Diagram 7 (ii(c)) shows a man climbing a mountain
Explain the effect of the 3 different situations towards the physiological process that occur in organ X as shown in
diagram 7 (ii)
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Aspect
Diagram 7 (ii) (a))
shows a boy
watching television
Marking scheme
Marks
F1-At rest, the respiratory rate is normal /12-20 breaths per minute
1
P1-The partial pressure of O2and CO2 are normal
1
F2-When a person is in fear, breathing rate increase
1
P2-It’s needed because the demand of a higher respiration rate in cells
1
P3-In order to oxidize more glucose
1
P4-To produce more energy
1
P5-(then), rapid muscles contraction (as a responded to the dangerous situation
/running)
1
F3-( in mountain climbing) as the altitude increase, the atmospheric pressure of
decrease
1
P6-Thus, partial pressure of O2becomes lower
1
P7-Causes a drop in the oxygen level in blood
1
P8-(the person will face difficulty in breathing
1
P9-So, the person will experience headache/nausea/dizziness
1
2
(Relaxing)
Diagram m 7 (ii(b))
shows a man is
chased by a fierce
dog
(In fear)
Diagram 7 (ii(c))
shows a man
climbing a
mountain
(At high altitude)
5
5
The regulatory mechanism of carbon dioxide content in the body
No
Marking scheme
Marks
(a)
30 breath per minute while the heartbeat rate increase to 120 beats per minute .Explain how the
body During vigorous activities such as swimming running and aerobic the breathing rate
increase to about regulates the carbon dioxide content in human body 7
P1-during vigorous exercise , the partial pressure of carbon dioxide increase //rate of cellular
respiration increase
1
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P2-Thus , carbon dioxide reacts with water to form carbonic acids
1
P3-(due to high level of co2 in blood ), its results in a drop im the pH value of the blood
( and)/cerebrospinal fluid
1
P4-The drop in pH is detected by (central) Chemoreceptor’s (in the medulla oblongata
P5-Send the nerve impulse to the respiratory centre / (which is in turn sends nerve impulse to)
diaphragm and intercostals muscles
1
P6-Pespiratory muscle to contract and relax faster
1
P7-breathing and ventilation rates faster
1
P7-Breathing and ventilation rates increase
1
P8-Excess CO2is eliminated from the body
1
P9-CO2concentration /pH value so blood return to normal levels Any 7p
(b)
1
1
7
1
1
1
1
1
1
1
4
In an experiment, a boy takes part in an 800 meter event track. His exhaled air was obtained
three times which were before running, right after he finished running and 10 minutes after
running to determine the percentage of carbon dioxide. Table 3.1 shows the result of the
experiment.
Percentage of carbon
dioxide (%)
Before running
Right after he finishes
running
After 10 minutes
running
4%
7.5%
4%
Based on the table 3.1, Explain how the percentage of carbon dioxide is returned to normal
after 10 munites running 4
E1 : The high concentration of carbon dioxide
E2 : decreases the blood pH
E3 : Detected by central chemoreceptor and/ peripheral chemoreceptor
E4 : Impulses are sent to the respiratory centre
E5 : (Impulses are sent to) the cardiac and respiratory muscles
E6 : Increase the heart beat and breathing rate
E7 : To remove excess carbon dioxide (so that the of carbon dioxideis returned tonormal)
Notes : Choose any three Es
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7.5 the importance of maintaining a healthy respiratory system
No
(a)
(b)
Marking scheme
Marks
Explain how smoking can harm the respiratory system in human
F1-Cigarette smoke contain tar
1
E1-Causes lugs cancer
1
F2-cigarette smoke contain acidic gases
1
2
Explain why does this occur?
F1 : Cigarette smoke contains carbon monoxide
E1 : (Carbon monoxide) has higher affinity to bind with hemoglobin compared to oxygen
E2 : forms carbaminohaemoglobin
E3 : Therefore, less oxygen will bind with hemoglobin to be transported in blood vessel
1
1
1
1
2
P Notes : F1 + any two Es
(c)
(d)
(e)
Explain why carbon monoxide is poisonous to the body cells
P1-C02 has higher affinity to bind with heamoglobin the with oxygen //CO2 reduce the ability
of haemoglobin to combine with oxygen
1
P2-the body cells lack oxygen //Less oxygen is transported to the body cells
1
2
Smoker do not realize that they destroy their respiratory organ during smoking, Explain how
this habit will affect the intake of oxygen efficiency
E1-Carbon monoxide
1
E2-Bind with haemoglobin to form carboxyhaemoglobin
1
E2-Less oxygen combine with haemoglobin
1
E4-Tobacco tar will be deposited/logged /accumulate (inside the lungs)
1
E5Reduce diffusion of oxygen
1
E6-Haet fom the smoke m
1
E7-Dry the surface of the alveoli
1
E8-Oxygen cannot be dissolved Any 4
1
4
Explain the effects of smoking on the human respiratory system.
P1-Carbon monoxide competes with oxygen to bind with haemoglobin and forms
1
1
carboxyhaemoglobin. It reduces the supply of oxygen to the cells.
P2Nitrogen dioxide can dissolve in mucus to form an acidic medium which erodes lung tissue.
1
P3- BENZO-(α)-PYRENE is carcinogenic chemical that can cause cancer.
1
P4-Nicotine can stimulate the production of cancer cell in trachea and lung.
1
164
P5-Heat and dryness irritation the lungs and can lead to laryngitis
1
4
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Chapter 7:Respiration 2014
7.6Respiration in plants
The intake of oxygen by plants for respiration
No
(a)
Marking scheme
Marks
Like animals, plants also respire aerobically to obtain energy for metabolism . They derive
most their energy from cellular respiration .during cellular respiration, the plants cells take
in oxygen and release carbon dioxide
Based on the above statement, describe the intake of oxygen by the plants for respiration
S1-The intake occurs by diffusion mainly through stomata and lenticels
S2-Stomata can be found in epidermis of leaves. the stem of herbaceous plants
S3-Lenticels can be found on the stems and root of plants
1
1
Explanation
P1-When stomata open, they connect the air space (within the leave) to atmosphere
P2-Oxygen form the atmosphere diffuses into the air spaces
P3-then dissolves in the film of water around the mesophyll cells
P4-So the concentration of oxygen in the cells becomes lower than in the air spaces
P5-Thus, oxygen diffuse continuously form air space to the cell
P6-During daytime, carbon dioxide that is produced during respiration is used in
photosynthesis
P7-The excess carbon dioxide diffuses into the air spaces and then through stomata into
atmosphere
(b)
1
1
1
1
1
1
1
7
Diagram 6.1 shows the surface view of lower epidermis in a leaf of a plant.
Diagram 6.2 shows part of cross section of a woody stem.
Broken epidermis
Pore M
Epidermal cell
Cork tissue
Guard cell
Pore M
Explain the gas uptake for respiration through pores M and N in the plant
Through M:
F- (In day time) stoma / M (in the epidermis of the leaf) open
P1-Oxygen from the atmosphere diffuses (through stoma) into intercellular air spaces
ll (and palisade mesophyll)
P2- follow the concentration gradient
Through N:
P3- At the lenticels (N) oxygen from atmosphere diffuses into the air spaces between cork cells
which are loosely arranged
P4- then diffuses into the cells at the stem /and old roots
1
1
1
165
1
1
4
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Respiration and photosynthesis in plants
No
(a)
Marking scheme
Marks
Diagram 6.4 shows the changes in the volume of carbon dioxide absorbed or released by a
plant in different light intensity
State the relationship between light intensity and rate of transpiration
P1-Light increase as the rate of transpiration increase
P2-The plant carries out anaerobic respiration
(b)
(c)
Explain the changes in the volume of carbon dioxide absorbed or released by a plant in
different light intensity
P1-glucose is broken down in the absence of oxygen to release energy produces ethanol, CO2
(and energy)
P2- cells in the roots of rice plants are extremely tolerant of ethanol
P3-Many of the roots are very shallow
P4-the roots use the oxygen which diffuses into the water surface.
P5-Rice stem contain a large number of air spaces
P6-(the air space) allow oxygen to penetrate through to the cells of roots ( growing in the
absence of oxygen)
Explain the relationship between the rate of photosynthesis and the rate of respiration in the
plant at points P, Q, R and S.
At P :
P1-In the dark / low light (intensity), only respiration occurs
P2-hence large quantity of CO2 is produced/released
P3-As light (intensity) increases the quantity of CO2 / produce decreases
P4because part of CO2 produced during respiration is used for photosynthesis
P5-sugar used in respiration more rapidly than it is produced in photosynthesis
At Q:
P6- (At this point of light intensity) all the CO2 release from respiration is reused / equivalent
to CO2 used up during photosynthesis // no net gain or loss in CO2 / sugar produced
P7- rate of photosynthesis is equal to the rate of respiration
P8-this point is called compensation point
P9-net gaseous exchange is zero
1
2
1
1
1
1
1
1
6
1
1
1
1
1
1
1
1
1
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At R:
P10- as light intensity increases, the rate of photosynthesis become faster than / exceed the rate
of respiration
P11-the CO2 needed is obtained from the atmosphere (at the same time) excess O2 is releases
(into the atmosphere)
At S:
P12- is the light saturation point
P13-an increase in light intensity does not increase the rate of photosynthesis // maximum rate
of photosynthesis (Any 8)
(d)
1
1
1
1
10
An experiment on a plant is carried out to study the rate of water loss from 0500 to 0300 the
next day. Graph 6.1 shows the result of the experiment and diagram 6.2 shows the structure of a
stoma and the cells found in the epidermal layer of a leaf.
Based on the graph, explain how light intensity and the structure in diagram 6.2 affect the rate
of water loss 10
F1 : From 0500 to 0170, the rate of water loss increases
E1: Light intensity increases
E2 : stimulates photosynthesis in the guard cells./ (The guard cells) start producing glucose
E3 : This makes energy available for potassium to move into guard cells
E4 by active transport
E5 : (The guard cells) become hypertonic (compared to the cell sap) of the epidermal cells.
1
1
1
1
1
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E6 : Water molecules from the epidermal cells diffuse into the guard cells by osmosis
E7 : Causing the guard cells to bend outwards
E8 : the stoma opens (to allow water to escape to the atmosphere through it)
F2 : From 0170 to 0300, the rate of water loss decreases
E9 : Light intensity decreases / causes the rate of photosynthesis to decrease / soon stop.
E10 : The guard cells become flaccid
E11 : and bend inwards
E12: The stoma closes and this prevent water molecules to escape through it.
Notes : (F1 + any 5 Es) + (F2 + 3 Es)
1
1
1
1
1
1
1
1
10
Comparision between photosynthesis an d respiration
No
(a)
Marking scheme
Marks
Explain the differences between the process in organelle P and Q
Site
Organelle P / mitochondria
Organelle Q/ chloroplast
Process
Respiration
Photosynthesis
Aim /purpose
Released energy
Stores energy
Raw material
Glucose, oxygen
Water, carbon dioxide, light
Products
Energy, water , carbon dioxide
Glucose / starch water and oxygen
1
1
1
1
Energy
Not required light energy
Required in form of light
1
(b)
4
The intake of oxygen by plants for respiration
State two differences between tissues in diagram 4.1 and 4.2
Tissue in diagram 4.1
Tissue in diagram 4.2
D1-Alveolus
Leaf
D2-Carry out transpiration
Carry out photosynthesis
D3-Absent of chlorophyll
Presence of chlorophyll
1
1
1
2
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Extra Question
Diagram 7.1 shows how the respiratory gases are transported in the human body
(i) Based on Diagram 7.1, explain how the transport of oxygen and carbon dioxide takes place in the body cells
Aspect
Transport of oxygen
Marking scheme
P1: The blood circulatory system transport oxygen from the alveoli to the body
cells.
P2: Oxygen combines with the haemoglobin in the red blood cells
P3: to form oxyhaemoglobin (which is unstable.)
P4: Oxygen is carried (in form of oxyhaemoglobin) to the tissues (which have a
low partial pressure of oxygen.)
P5: The (unstable) oxyhaemoglobin breaks down into oxygen and haemoglobin
again.
P6: Oxygen (molecules are) transferred to the body cells
Transport of
P7: Carbon dioxide binds (itself) to the haemoglobin
P8: (and is) transported in the form of carbaminohaemoglobin.
P9: Carbon dioxide is (also) transported as dissolved carbon dioxide (in the blood
plasma.)
P10: Most of carbon dioxide is carried as bicarbonate ions (dissolved in the blood
plasma.)
P11: When the blood carrying carbon dioxide reaches the body cells, the carbon
dioxide diffuses into the blood plasma and combines with the red blood cells.
P12:Carbon dioxide reacts with water to form carbonic acid.
P13:Carbonic anhydrase in the red blood cells catalyse the formation of carbonic
acid.
P14: The carbonic acid then dissociates into a hydrogen ions and bicarbonate ions.
MAXIMUM: 6 marks
carbon dioxide
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