BIOLOGY Form 4 Chapter 7

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Module Biology Trial Paper Collection

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7.1 The respiration process in energy production

No Marking scheme Marks

(a) Aerobic respiration Anaerobic respiration

OR

Process Respiration equation

S Glucose +oxygen Carbon dioxide +Water +2898 kJ energy

R Glucose Carbon dioxide +ethanol+210 energy

Name the process R and S R:Anerobic respiration S:Aerobic respiration

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(b) 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 Carbon dioxide +Water +2898 kJ energy

Yeast cells Glucose Carbon dioxide +ethanol+210 energy

(a) Fill in the table by writing in muscle cells or yeast cells that matches with its respiration equation

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(c) State where tissue V(smooth muscle cell) can be found in the body

Blood vessel/alimentary canal/oeosophagus/stomach/uterus/urinary bladder/etc 1 1 (d) Write the equation of process S and R

Process R

Glucose lactic acids + energy Process S

Glucose +Oxygen Carbon dioxide + water +2898 kJ Reactant- 1m Product -1m 2 2 1 1 4

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(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

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(f) Anaerobic respiration in cells

Explain the condition of a person after completing a 100 meter race in 12 seconds 2 F-the person is panting /higher breathing rate

E1-As he is in oxygen debt//anaerobic respiration E2-Much lactic acids is produced (in his muscle cells) E3-Causes muscle cramp Any 2

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(g)

Explain this statement

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 P1-( The increase in heated beat rate ) is to deliver more glucose to muscle cells

P2-To induce extra energy cellular respiration

P3-To remove more carbon dioxide from the muscle cells Any 2

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(h)

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 E2-To get more oxygen immediately

E3-To oxidize lactic acids Any 2E

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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

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(a) 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

P2-The lactic acids has been converted to energy/convert to glucose

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(b) Explain the condition of a person after completing a 100 meter race in 12 seconds F-the person is panting /higher breathing rate

E1-As he is in oxygen debt//anaerobic respiration E2-Much lactic axids is produced ( in his muscle cells) E3-Causes muscle cramp Any 2

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(c) Explain the usage of cell W in bread making industry F1-Carbon dioxide released

E1-Traps in the dough E2-Causes the dough to rise

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(d) 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

P2-For yeast cells to reproduced //yeast cell die

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(e) 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 D2-Oxidation of glucose is complete/

Complete breakdown of glucose

D2-Oxidation of glucose is not complete/ Incomplete breakdown of glucose

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)

D4-Produced carbon dioxide and water D4-Produced lactic acid D5-Occurs in mitochondria D5-Occur in cytoplasm

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(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

Before (A) During (B) Explanation (E)

1 Aerobic respiration Anaerobic respiration Before-Oxygen Intake is low/the same as oxygen required /enough oxygen is supplied to the cell

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

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 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

E1-Glucose break down (partially/incompletely) E2-To produce ethanol, carbon dioxide

E3-Less ATP/2 ATP is produce F1 and any of E1/E2/E3

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(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

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

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(i) 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

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(a)

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 and water.

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

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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 = 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

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Molecule X + 2ATP

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144 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

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7.2the respiration structure and breathing mechanism in human and animal

(a) 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

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The respiration structure and breathing mechanism insects

Aspect Question & Marking Scheme Marks

Respiratory structure

The respiration structure and breathing mechanism insects

Name the part labeled P ,Q ,Rand S 5 5

Which organism has the respiratory structure?

Insect 1 1

Name the respiratory system shown in diagram 2.1

Tracheal system 1 1

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

1 1 ₂ P:Air sac Q: Muscle R:Tracheole S: Trachea T: Spiracle

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Explain one adaptation of the respiratory structure in diagram for efficient gaseous exchange

P1-The large number of tracheoles provides a large surface for the diffusion of gases P2-Tips of tracheoles have thin permeable wall and contain fluid in which

respiratory gases can dissolved

P3-Terminal ends the tracheoles remains moist which allows teh gases to be dissolved 1 1 1 2 Structural Adaptation

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

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 1 1 1 1 2 Breathing mechanism

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)

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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

R2-The trachea is lined with chitin to prevent dorm collapsing R3-Spiracles is tiny opening thet allow air to go in and out

R3-spiracles is tiny opening that allow air to go in and out R4-The trachea branch into fine tubes celled tracheole

R5-The tracheole branch throughout the body and temperature and penetrate into body tissues / muscle

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B1-When inside inhales, the abdominal muscles relax and spiracles open B2-air pressure inside the trachea decrease and air is drawn in

B3-When the insect exhale, the abdominal muscle contract

B4-So increase air pressure in side trachea and forces air out through spiracles B5-Inesct inhale and exhale through rhythmic contraction and expansion of their abdominal muscles

B6-the body movement and contraction of abdominal muscle speed up the rate of diffusion of gases from trachea into body cells

1 1 1 1 1 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

P2- Oxygen diffuse from tracheole to body cell

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

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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)

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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

E1-(simple) diffusion can take place

E2-Oxygen concentration /partial pressure is higher in the tracheoles while the concentration of oxygen is lower in the cells

E3-Oxygen diffuses directly form the tracheoles onto the cells

E4-Carbon dioxide concentration is higher in the cells while lower in the tracheoles E5-Carbon dioxide diffuses directly form the cells into the trachoeles

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The respiratory structure and breathing mechanism of fish

Aspect Marking scheme Marks

Respiratory structural

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

State the function of structure P

P-Speed up the movement of gases to and from the insect’s tissue 1 1 The efficiency of gaseous in organism Y is further enhanced by a mechanism.

Name the mechanism

Countercurrent exchange mechanism ₁ ₁

State two characteristic of X, which makes it a good respiratory structure for fish 2 P1-Have lamella and filament to increase total surface area

P2-Numberous blood capillaries for efficient transport of respiratory gases

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Structural adaptation

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

P2-The gills filament have numerous thin membrane and covered by net work of capillaries to transport respiratory gases

P3-the surface of gills Is moist which allows the gases to be dissolved

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Breathing Mechanism

Based on the diagram explain how the oxygen is drawn from mouth to X(gill) P1-Mouth closes

P2-The floor of buccal cavity raised (water contain air flow to X)

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Breathing mechanism

Describe the inhalation in fish E1-th floor of cavity lowers

E2-At the same time, the opercular cavity enlarges and operculum closes E3-This lowers the pressure in buccal cavity

E4-Water with dissolved oxygen is drawn into the mouth

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Inhalation 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.

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Exhalation 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

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The respiratory structure and breathing mechanism of amphibians

Respiratory structural

Name structure X and Y in diagram 3.1 2 X: Bucco-pharyngeal Y: Glottis 1 1 2 Structural adaptation

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

E2-Membrane of lungs are thin and moist to facilitate the efficient diffusion of respiratory gases

E3-Supplied with a rich network of blood capillaries to transport respiratory gases to the body cells

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Breathing Mechanism

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

E2-Air pressure is not increased /decrease in the bucco-pharyngeal cavity E3-Air cannot be forced into /out the lungs

E4-Lung ventilation is not efficient

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The respiratory structure and breathing mechanism of humans

Respiratory structure

Name the parts labeled Y

Y-Alveolus 1 1

What is the function of alveoli?

Place for gaseous exchange //store the oxygen gas before gaseous exchange occur 1 1 State the organ in which the tissue in Diagram 4.1(alveolus) can be found

Lung 1 1

State the function of organ stated in

Gaseous exchange//respiration 1 1

Respiratory gases flow in and out through the trachea .Describe the characteristic of trachea

F-Have C-shaped cartilage rings //cartilage rings P1-keep the trachea open permanently

P2-Avoid the trachea form collapse when the out side pressure is higher than inside pressure

P3-oxygen can continuously flow through trachea to the alveoli/lung F-1m P-1m

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3 Explain the effects of the breathing mechanism if structure R is unable to function

P1-Structure R is diaphragm.

P2-Less/no change in volume in the thoracic cavity/ lung P3-Less/ no change in air pressure in the thoracic cavity/ lung P4-Less/ no air exchange/ less/no intake of O2/ less/no CO2 expelled

Resulting difficulty in breathing in and out

1 1 1 1 4 Structural adaptations

State the important characteristic of alveoli to ensure the function in (a) is efficient 1 P1-Have very large total surface area//

P2-Have moist surface all the time//

P3-have very thin wall/one cell thick Note ( any 1P)

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Describe the characteristic of the respiratory structure of human that enable gaseous exchange to be carried out efficiently

P1-the ratio total surface area per volume (TSA/V) is high for the exchange of gases P2-the cells lining the respiratory surface is a single layer of cell which is very thin to allow gases to diffuses easily

P3-the respiratory surface is constantly moist to allow gases to dissolved in water before diffusing in and out of the respiratory surface

P4-the respiratory surface is covered with a dense network capillaries to allow rapid diffusion and transport of gases

1 1 1 1 3 Breathing mechanism

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

P2-Diaphgram muscle contract , the diaphragm lower and flattenP3-The volume of thoracic Cavity increase but the pressure decrease (lower the atmospheric pressure) P3-The volume of thoracic cavity increase but the pressure decrease ( lower the atmospheric pressure)

P4-Air forced into the lung//alveolus

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Describe the breathing mechanism of human Inhalation:

P1-External intercostals muscle contract//internal costal muscle relax P2-ribcage move upwards and out wards

P3-diaphragm contracts/flattens

P4-Volume of thoracic cavity increase // pressure of thoracic cavity decrease P5-So air ( form outside) is forced into the lungs

Exhalation :

P1-External intercostals muscle relax//internal costal muscle contract P2- ribcage move downwards and inwards

P3-diaphragm relax/curved upward

P4-Volume of thoracic cavity decrease // pressure of thoracic cavity increase P5-So air ( form inside) is forced out of lungs

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Constructing a model of human lung study the breathing mechanism in humans

(a)

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

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(b) 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 1 1 (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 )

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(d) (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

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(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 Rubber cork Glass tube Balloon

Thin rubber sheet String

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153 P4- this will decrease the thorax pressure

P5- air will be inhale

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(f) 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 1 1

(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.

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Comparison of respiratory system between human and insect

(a)

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

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(b) 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

P1-tracheae of cockroach are branch into 2 bronchi which enter the right and left lungs

1 S

R P

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154 P2-Thr trachea of human branched into 2 bronchi which enter the right and left lungs

P3-The bronchi of human branched ito smaller tubes called bronchioles which ends in a cluster of sacs called alveoli

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(c)

Explain one similarity and four differences between the respiratory organs of insect and human Similarities

S1-Both of respiratory organs has thin wall/one cell thick E1-Incrase rate of diffusion of respiratory gaseous OR

S2-Both of respiratory organs has respiratory surface such as alveolus in human and tracheole in an insect

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 E1-To prevent them form collapsing

D2-The wall of alveolus is moist surface but the tracheole has fluid E2-To dissolve the respiratory gases

D3-Alveolus is covered by network of blood capillaries but not for trachoele

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)

D4-Haemoglobin is needed in transport of oxygen nt but in insect

E4-oxygen combine with heamoglobin in (erythrocyte) to form oxyhaemoglobin but not in insect

D5-(larger) insect have air sacs but not in human

E5-to speed up the movement of gases to and form the insect’s tissue D6-in human air enters the lungs through the nostrils but spiracles in insects E6-to allow gases in and out of the body any 4 pairs

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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

D2-Gill have filament and lamella to increase the surface area, but lung of frog have numerous inner partition to increase the surface area

D3-Gill received oxygen directly form water , but lungs and skin of frog received oxygen form the atmosphere

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(d) Describe the comparison between the respiratory system in insect and human 8 Similarities:

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

E2-Oxygen and carbon dioxide can be dissolve easily F3-Very thin wall of tracheole and alveolus/one cell thick

E3-To ensure the simple diffusion can take place /Increase rate of diffusion of respiratory gaseous

Insects Aspect Human

F4-Consists of spiracles, trachea and tracheoles

Respiratory structure Consists of nose trachea, bronchus, bronchioles ad alveolus

E4-Air enters through spiracles into tracheoles

Air enter through nose into lungs/alveolus

F5-Tracheoles directly contact with the muscle cells

Alveolus is surrounded by a large network of blood capillaries

F6-Trachea is reinforced/ supported with ring of chitin

Trachea is reinforced/ supported with ring of cartilage

E6-Prevent the trachea form collapsing due to different air pressure

P5-Prevent the trachea form collapsing

F7-Does not have red blood cell to transport oxygen

Oxygen transportation Has red blood cells to transport oxygen through blood vessels

E7-Oxygen is not transported in the body

Oxygen is transported by red blood cells around the body F8-Oxygen diffuses directly

form the respiratory tructure into the cells

The diffusion of oxygen into the cells

Oxygen needed to be

transported into the cells and then diffuses into the cells E8-Carbon dioxide is directly

released form the cells into tracheoles

Product of respiration Carbon dioxide produced diffuses into the blood

capillary then transported into the lungs 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 10

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Comparison of respiratory system between human and fish

(a)

Explain three adaptation from structure show in diagram 2 (b)(ii) to carry out its function efficiently

P1-Thin membrane /one cell thick for easily diffusion of respiratory gases P2-Moist surface for respiratory gases easily dissolve

P3-Numerous blood capillaries for efficient transport of respiratory gasesAny 2

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(b) 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 P2-Oxygen diffuses form Y into blood capillaries by simple diffusion

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(c) 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

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

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

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(d) What are the differences between respiratory system of human and fish? P1-gill is the respiratory organ for fish nut is for human

P2-gill have filament and lamella to increase the surface area, but lung have alveoli to increase the surface area

P3-gill touch /surrounded by water

P4-Gill receives oxygen directly from water, but lung received oxygen form atmosphere via trachea , bronchus and bronchioles

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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

(a) State the importance of gaseous exchange in human P1-To get oxygen for (cellular) respiration

P2-To get rid of/excrete the carbon dioxide

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(b)

Name gas X and Y X : Oxygen Y : Carbon dioxide

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(c) 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)

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(d) Name blood vessel P and Q P: Pulmonary veins

Q:Pulmonary artery

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(e) State the function of blood vessel P and Q P: Carries deoxygenated blood to lungs Q: carries oxygenated blood back to heart

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(f) 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

P2-Transport oxygenated blood //oxyhaemoglobin to heart

P3-the heart pump the oxygenated blood to muscle cells via the aorta Any2

1

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(a)

State the process by which gaseous exchange takes place across alveolus1

(Simple) diffusion 1 1

(b) Explain how the process occurs

F-Partial pressure of oxygen /carbon dioxide in the air of the alveolus is higher than in blood capillary

1 1

(c) Gaseous exchange takes place across structure Y Name structure Y

Alveolus/ Alveoli 1 1

(d) State two ways how the alveolus are adapted for efficient gaseous exchange P1-Thin wall

P2-Moist

P3-Rich with blood capillary

1 1

1 2

(e) 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

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

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 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 P2-Oxygen Diffuses form alveolus into the blood capillaries

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

1

1 4

(g) Explain the role of oxygen in the muscle cells F-oxygen oxidized the glucose molecule

E1-Cellular respiration /aerobic respiration takes place in muscle cells E2-ATP/energy released

E3-Produced carbon dioxide and water as by product/waste products

E4-energy is used for contraction and relaxation of muscle cells/movement of insect

1 1 1 1

1 4

(a)

Based on the diagram 3.2 name X and Y X: oxygen

Y: Carbon dioxide

1

1 2

(b) Name structure P and Q X: Red blood cell Y:Alveolus

1

1 2

(c) Name the complex substances contained in X

Haemoglobin 1 1

Q P R

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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 ) F1-Oxygen diffuses from alveolus into blood capillaries

E1-Oxygen concentration /partial pressure in alveolus is higher than in blood capillaries F2-Carbon dioxide diffuses from blood capillaries to the alveolus

E2-Carbon dioxide concentration /partial pressure in the blood capillaries is higher than in alveolus MAX:2

1 1 1

1 2

(f) Explain how the red blood cell accepts oxygen form alveolus and transfer to the cell P1-Oxgen diffuses into the blood plasma

P2-Combine with haemoglobin

1

1 2

(g)

.

Based on the diagram , explain the exchange of respiratory gases P1-Respiratory surfaces in human are alveoli.

P2-The concentration of oxygen in the alveoli is higher than its concentration in the blood capillaries.

P3-Oxygen in the alveoli diffuses into the blood capillaries.

1 1 1 CO2 O2 R S

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161 P4The concentration of carbon dioxide in the blood capillaries is higher than its concentration

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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162

Diagram 7 (ii) (a)) shows a boy

watching television (Relaxing)

F1-At rest, the respiratory rate is normal /12-20 breaths per minute P1-The partial pressure of O2and CO2 are normal

1 1 2 Diagram m 7 (ii(b)) shows a man is chased by a fierce dog (In fear)

F2-When a person is in fear, breathing rate increase

P2-It’s needed because the demand of a higher respiration rate in cells P3-In order to oxidize more glucose

P4-To produce more energy

P5-(then), rapid muscles contraction (as a responded to the dangerous situation /running) 1 1 1 1 1 5 Diagram 7 (ii(c)) shows a man climbing a mountain

(At high altitude)

F3-( in mountain climbing) as the altitude increase, the atmospheric pressure of decrease

P6-Thus, partial pressure of O2becomes lower P7-Causes a drop in the oxygen level in blood P8-(the person will face difficulty in breathing

P9-So, the person will experience headache/nausea/dizziness

1

1 1 1

1 5

The regulatory mechanism of carbon dioxide content in the body

(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

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163 P2-Thus , carbon dioxide reacts with water to form carbonic acids

P3-(due to high level of co2 in blood ), its results in a drop im the pH value of the blood ( and)/cerebrospinal fluid

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

P6-Pespiratory muscle to contract and relax faster

P7-breathing and ventilation rates faster

P7-Breathing and ventilation rates increase

P8-Excess CO2is eliminated from the body

P9-CO2concentration /pH value so blood return to normal levels Any 7p

1 1 1 1 1 1 1 1 1 7

(b) 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.

Before running Right after he finishes running After 10 minutes running Percentage of carbon dioxide (%) 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

1 1 1 1 1 1 1 ₄

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7.5 the importance of maintaining a healthy respiratory system

(a) Explain how smoking can harm the respiratory system in human F1-Cigarette smoke contain tar

E1-Causes lugs cancer

F2-cigarette smoke contain acidic gases

1 1

1 2

(b) 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

P

Notes : F1 + any two Es

1 1 1

1 2

(c) 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

P2-the body cells lack oxygen //Less oxygen is transported to the body cells

1

1 2

(d) 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

E2-Bind with haemoglobin to form carboxyhaemoglobin E2-Less oxygen combine with haemoglobin

E4-Tobacco tar will be deposited/logged /accumulate (inside the lungs) E5Reduce diffusion of oxygen

E6-Haet fom the smoke m E7-Dry the surface of the alveoli E8-Oxygen cannot be dissolved Any 4

1 1 1 1 1 1 1 1 4

(e) Explain the effects of smoking on the human respiratory system.

P1-Carbon monoxide competes with oxygen to bind with haemoglobin and forms 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. P3- BENZO-(α)-PYRENE is carcinogenic chemical that can cause cancer.

P4-Nicotine can stimulate the production of cancer cell in trachea and lung. P5-Heat and dryness irritation the lungs and can lead to laryngitis

1 1 1 1 1 1 4

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7.6Respiration in plants

The intake of oxygen by plants for respiration

(a)

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

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 1 1 1 1 1 1 1 1 1 7

(b) 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.

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

1 4

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

Pore M Epidermal cell Guard cell Pore M Broken epidermis Cork tissue

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Respiration and photosynthesis in plants

(a) 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

1 1

2

(b) 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) 1 1 1 1 1 6 (c) 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 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) 1 1 1 1 10 (d) 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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168 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

(a) 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

Energy Not required light energy Required in form of light

1 1 1 1

1 4

(b) 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

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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 Marking scheme

Transport of oxygen 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 carbon dioxide

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