biology form 4 chapter 7.docx

138

SULIT 4551/2Chapter 7:Respiration 2014

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 respirationS:Aerobic respiration

11 2

(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

1

12

(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 RProcess RGlucose lactic acids + energyProcess SGlucose +Oxygen Carbon dioxide + water +2898 kJReactant- 1mProduct -1m

2

211 4

Module Biology Trial Paper Collection SULIT ©All Right Reserved 4551/2

138


(e) Explain process P and Process Q / Explain the cellular respiration process that occurs in individual P and Q Process PF1 - aerobic respiration.P1 - glucose is completely oxidized/breakdown in the presence ofoxygenP2 - releases more energy/2898 kJ of energy ( per mole of glucose)E3-Produce carbon dioxide and waterProcess QF2 - Anaerobic respirationP3 - 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

1111

11

11 6

(f) Anaerobic respiration in cellsExplain the condition of a person after completing a 100 meter race in 12 seconds 2F-the person is panting /higher breathing rateE1-As he is in oxygen debt//anaerobic respirationE2-Much lactic acids is produced (in his muscle cells)E3-Causes muscle cramp Any 2

1111 4

(g)

Explain this statementF1 - (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 ATPP1-( The increase in heated beat rate ) is to deliver more glucose to muscle cellsP2-To induce extra energy cellular respiration P3-To remove more carbon dioxide from the muscle cells Any 2

1

111 4

(h)

Based on the above statement the condition faced by the athleteOxygen 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 immediatelyE3-To oxidize lactic acids Any 2E

111 2


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

After completing vigorous exercise, an athlete will gasp heavily

138


(a) Explain how the oxygen intake by the athlete returns to the normal level at the 25thminuteP1-Lactic acid has been removed from the muscle P2-The lactic acids has been converted to energy/convert to glucose

11 2

(b) Explain the condition of a person after completing a 100 meter race in 12 seconds F-the person is panting /higher breathing rateE1-As he is in oxygen debt//anaerobic respirationE2-Much lactic axids is produced ( in his muscle cells)E3-Causes muscle cramp Any 2

1111 2

(c) Explain the usage of cell W in bread making industry F1-Carbon dioxide releasedE1-Traps in the doughE2-Causes the dough to rise

111 2

(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 /conditionP2-For yeast cells to reproduced //yeast cell die

1

1 2

(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

1

1

1

11

4


138


(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 vigorousExercise. 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 cellDuring-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 sufficientDuring-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 waterDduring -Incomplete breakdown of glucose produce lactic acid

A+B=1m E=1m (Any 1 E)

3

3

3

3

4


138


(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 dioxideE3-Less ATP/2 ATP is produceF1 and any of E1/E2/E3

1111 2

(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 capillaryP2-Cells P contain a lot of mitochondriaP3-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 mitochondriaP6-1 molecule of glucose will produce 38 molecule ATP/ More ATPP7-water and carbon dioxide are released as waste material in this process

111111111

8

(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 6During 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.

1111

111111 6


138


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

(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

1111

1

1

16

(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 =

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

11

1


Molecule X + 2ATP

138


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

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

11111

11 10

7.2the respiration structure and breathing mechanism in human and animal


(a) Adaptation of the respiratory structuresState two characteristic shown by the respiratory surface of animal(common characteristic)P1-the respiratory surface is moistP2-Cells lining respiratory structure are thin P3-Thr respiratory structure has a large surface area

111 3

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 1Name the respiratory system shown in diagram 2.1Tracheal system 1 1


P:Air sacQ: Muscle

R:TracheoleS: Trachea

T: Spiracle

138


State the function of the following(i) Chitin

support the tracheal/prevent the tracheal form collapsing (ii) Air sacSpeeds up the movement of gases exchange to and form tissue during vigorous body movement

1

1 2


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 gasesP2-Tips of tracheoles have thin permeable wall and contain fluid in which respiratory gases can dissolvedP3-Terminal ends the tracheoles remains moist which allows teh gases to be dissolved

11

1 2

Structural Adaptation

Explain how structure Q and S increase the efficiency of gaseous exchange in each organism 2F-Consists of million alveoli in lungs and many tracheal Tubes/Tracheole/thin layer/1 cell thickP1-To increase total surface area per volume rate for gaseous exchangeF2-The inner surface of alveolus and tracheoles end consists of tissue fluid moistureP2-To provide moist surface for gas diffusion /to dissolve oxygen /gases for diffusion Any F +P

11

11 2

Breathing mechanism

State how air is drawn from T to S 2P1-By(rhythmic) movements, of the abdominal musclesP2-Decreasing of air pressure inside trachea, ( so the air is drawn in)P3-Gases diffuses into the cells(s)

111

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


138


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 outR4-The trachea branch into fine tubes celled tracheoleR5-The tracheole branch throughout the body and temperature and penetrate into body tissues / muscleBreathing mechanism 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 contractB4-So increase air pressure in side trachea and forces air out through spiraclesB5-Inesct inhale and exhale through rhythmic contraction and expansion of their abdominal musclesB6-the body movement and contraction of abdominal muscle speed up the rate of diffusion of gases from trachea into body cells

1

111

11111

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 cellP3- 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

1111

14

Chitin is a polysaccharide on the outer surface of structure P. Due to the change inthe environment, the insect is unable to form the polysaccharide.Explain how the absence of chitin affects inhalation and the energy production. 6P1- The function of chitin is to prevent trachea from collapsing/sustainthe air pressureP2- During inhalation high pressure air moves into the trachea.P3 -The absent of chitin will cause the trachea / P to collapse / burst /rupture.

1111


138


P4 -Air with oxygen cannot reach tracheal.P5-Body cell cannot get enough oxygen for cellular respirationP6-The insect does not produce enough energy and respire anaerobically. P7-Less energy produced. (Any 6)

111 6


Breathing mechanism

Diagram show a trachea system of and insect Based on the diagram explain the gases exchange between the tracheoles and muscle cellsF-there are concentration gradient of oxygen and carbon dioxide between tracheoles & body cellsE1-(simple) diffusion can take place E2-Oxygen concentration /partial pressure is higher in the tracheoles while the concentration of oxygen is lower in the cellsE3-Oxygen diffuses directly form the tracheoles onto the cells E4-Carbon dioxide concentration is higher in the cells while lower in the tracheolesE5-Carbon dioxide diffuses directly form the cells into the trachoeles

1

11

111 4

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 PP-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.


138


Name the mechanismCountercurrent exchange mechanism 1 1

State two characteristic of X, which makes it a good respiratory structure for fish 2P1-Have lamella and filament to increase total surface areaP2-Numberous blood capillaries for efficient transport of respiratory gases

11 2


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 exchangeP2-The gills filament have numerous thin membrane and covered by net work of capillaries to transport respiratory gasesP3-the surface of gills Is moist which allows the gases to be dissolved

1

1

1 2

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)

11 2

Breathing mechanism

Describe the inhalation in fish E1-th floor of cavity lowersE2-At the same time, the opercular cavity enlarges and operculum closes E3-This lowers the pressure in buccal cavityE4-Water with dissolved oxygen is drawn into the mouth

1111 4

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 cavityP2-opercular cavity enlarges and operculum closesP3 - This lowers the pressure in buccal cavity .P4 - Water with dissolved oxygen is drawn into the mouth.

1111 4

Exhalation P5 - When the mouth closes, the floor of buccal cavity is raised. 1


138


P6 - Water flow through the lamellae and gaseous exchange betweenthe 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 4

The respiratory structure and breathing mechanism of amphibians


Respiratory structural

Name structure X and Y in diagram 3.1 2X: Bucco-pharyngealY: Glottis

11 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 gasesE3-Supplied with a rich network of blood capillaries to transport respiratory gases to the body cells

11

1 3

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 closeE2-Air pressure is not increased /decrease in the bucco-pharyngeal cavityE3-Air cannot be forced into /out the lungsE4-Lung ventilation is not efficient

1111 4


138


The respiratory structure and breathing mechanism of humans


Respiratory structure

Name the parts labeled YY-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 ringsP1-keep the trachea open permanentlyP2-Avoid the trachea form collapse when the out side pressure is higher than inside pressureP3-oxygen can continuously flow through trachea to the alveoli/lung F-1m P-1m

111

13

Explain the effects of the breathing mechanism if structure R is unable to functionP1-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

1111 4

Structural adaptations

State the important characteristic of alveoli to ensure the function in (a) is efficient 1P1-Have very large total surface area//P2-Have moist surface all the time//P3-have very thin wall/one cell thick Note ( any 1P)

111 2


138


Describe the characteristic of the respiratory structure of human that enable gaseous exchange to be carried out efficientlyP1-the ratio total surface area per volume (TSA/V) is high for the exchange of gasesP2-the cells lining the respiratory surface is a single layer of cell which is very thin to allow gases to diffuses easilyP3-the respiratory surface is constantly moist to allow gases to dissolved in water before diffusing in and out of the respiratory surfaceP4-the respiratory surface is covered with a dense network capillaries to allow rapid diffusion and transport of gases

11

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

1

1

11 3

Describe the breathing mechanism of humanInhalation:P1-External intercostals muscle contract//internal costal muscle relaxP2-ribcage move upwards and out wardsP3-diaphragm contracts/flattensP4-Volume of thoracic cavity increase // pressure of thoracic cavity decreaseP5-So air ( form outside) is forced into the lungs

Exhalation :P1-External intercostals muscle relax//internal costal muscle contractP2- ribcage move downwards and inwardsP3-diaphragm relax/curved upwardP4-Volume of thoracic cavity decrease // pressure of thoracic cavity increaseP5-So air ( form inside) is forced out of lungs

11111

11111 6


138


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 / ribsBalloon : lungRubber sheet: diaphragm

1111 4

(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- elasticE- can expand (inhalation) and contract/ decrease in size (exhalation )

11 2

(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

1 1

(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 diaphragmP2- when the diaphragm muscles contract,P3- the volume of the thorax increase

111


Rubber cork

Glass tube

Balloon

Thin rubber sheet

String

Bell Jar

138


P4- this will decrease the thorax pressureP5- air will be inhale

1 3

(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? 1To 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 16.0Carbon dioxide 0..04 4.0Nitrogen gas 78.0 78.0Water 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 respirationP2-Carbon dioxide diffuses into the blood to be transport to the lungs.

1

1 2

Comparison of respiratory system between human and insect


(a)

State one similarity and one difference of structure P in diagram 2.1 and 2.2Similarity: both wall of P consisting ring to strengthen itDifferences: the wall of P in insect consists of chitin ring while P in human consists of cartilage ring

1

1 2

(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 lungsP1-tracheae of cockroach are branch into 2 bronchi which enter the right and left lungs

1


S

RP

Q

138


P2-Thr trachea of human branched into 2 bronchi which enter the right and left lungsP3-The bronchi of human branched ito smaller tubes called bronchioles which ends in a cluster of sacs called alveoli

11 3

(c)

Explain one similarity and four differences between the respiratory organs of insect and humanSimilaritiesS1-Both of respiratory organs has thin wall/one cell thickE1-Incrase rate of diffusion of respiratory gaseousORS2-Both of respiratory organs has respiratory surface such as alveolus in human and tracheole in an insectE2-Provide a large surface area for the diffusionDifferencesD1-Trachea in human is supported by cartilage and traches in insect is supported by chintin E1-To prevent them form collapsingD2-The wall of alveolus is moist surface but the tracheole has fluid E2-To dissolve the respiratory gasesD3-Alveolus is covered by network of blood capillaries but not for trachoeleE3-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 insectE4-oxygen combine with heamoglobin in (erythrocyte) to form oxyhaemoglobin but not in insectD5-(larger) insect have air sacs but not in humanE5-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

11

111111 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 frogD2-Gill have filament and lamella to increase the surface area, but lung of frog have numerous inner partition to increase the surface areaD3-Gill received oxygen directly form water , but lungs and skin of frog received oxygen form the atmosphere

11

1 2


138


(d) Describe the comparison between the respiratory system in insect and human 8Similarities:F1-The structure of tracheal system and trachea branches into small tubesE1-increase the total surface area of tracheole/alveolus so that increase the efficiency of gases exchangeF2-moist surface on tracheole and alveolusE2-Oxygen and carbon dioxide can be dissolve easily F3-Very thin wall of tracheole and alveolus/one cell thickE3-To ensure the simple diffusion can take place /Increase rate of diffusion of respiratory gaseous

Insects Aspect HumanF4-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

1111

1

1

1

1

1

1

1

1

110


138


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 efficientlyP1-Thin membrane /one cell thick for easily diffusion of respiratory gasesP2-Moist surface for respiratory gases easily dissolve

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

111 3

(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 capillariesP2-Oxygen diffuses form Y into blood capillaries by simple diffusion

11 2

(c) Humans and fish have different respiratory systems, Explain one differences between the respiratory system of human and fish 3F1-the respiratory system of fish of gills while the respiratory system of human consists of a trachea and pair of lungsP1-A fish has four pairs of gills which are covered by operculum//the surface of each gills Filament has many plate –like projections called lamellaP2-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

1

1 3

(d) What are the differences between respiratory system of human and fish?P1-gill is the respiratory organ for fish nut is for humanP2-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

11

11 3


138


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 humanP1-To get oxygen for (cellular) respiration P2-To get rid of/excrete the carbon dioxide

11 2

(b)

Name gas X and Y X : Oxygen

Y : Carbon dioxide

11 2

(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)

1111 2

(d) Name blood vessel P and QP: Pulmonary veinsQ:Pulmonary artery

11 2

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

11 2

(f) Describe the role of blood vessel P in transporting oxygen form alveolus to muscle cellsP1-In the blood, Oxygen form alveolus combine with respiratory pigment/haemoglobin to form oxyhaemoglobin /oxygenated bloodP2-Transport oxygenated blood //oxyhaemoglobin to heart

1

1


138


P3-the heart pump the oxygenated blood to muscle cells via the aorta Any2 1 2

(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 YAlveolus/ Alveoli 1 1

(d) State two ways how the alveolus are adapted for efficient gaseous exchange P1-Thin wallP2-MoistP3-Rich with blood capillary

111 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 exchangeF4-Network of blood capillariesP4-To increase the rate of gases transportation F+P=1mE3 : Increase the surface area for rapid diffusion of gaseous

11

111

1

111

2


138


Notes : F1/2/3 + E 1/2/3 = 2 mark

F1/2/3 = 1 mark

E1/2/3 = O mark

(f) Describe the movement of respiratory gases across structure YP1-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 capillariesP2-Oxygen Diffuses form alveolus into the blood capillariesORP3- 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 capillariesP4- Carbon dioxide diffuses form alveolus into the blood capillaries

1

1

1

1 4

(g) Explain the role of oxygen in the muscle cells F-oxygen oxidized the glucose moleculeE1-Cellular respiration /aerobic respiration takes place in muscle cells E2-ATP/energy releasedE3-Produced carbon dioxide and water as by product/waste productsE4-energy is used for contraction and relaxation of muscle cells/movement of insect

11111 4


(a)

Based on the diagram 3.2 name X and YX: oxygenY: Carbon dioxide

11 2

(b) Name structure P and QX: Red blood cell 1


Q

P

R

138


Y:Alveolus 1 2

(c) Name the complex substances contained in XHaemoglobin 1 1

(d) Explain how the gaseous exchange occur across the alveolus 3P1 : Oxygen diffuse/ moves across /through ( plasma membrane) to blood capillaryP2: 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

11

111 3

(e)

Explain how gaseous exchange occurs during respiration in Diagram 4.1 (in human )F1-Oxygen diffuses from alveolus into blood capillariesE1-Oxygen concentration /partial pressure in alveolus is higher than in blood capillariesF2-Carbon dioxide diffuses from blood capillaries to the alveolusE2-Carbon dioxide concentration /partial pressure in the blood capillaries is higher than in alveolus MAX:2

1111 2

(f) Explain how the red blood cell accepts oxygen form alveolus and transfer to the cellP1-Oxgen diffuses into the blood plasmaP2-Combine with haemoglobin

11 2

(g)

.

CO2

O2

R

S

Based on the diagram , explain the exchange of respiratory gases

1


138


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.

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

1

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 activitiesDiagram 7 (ii) (a)) shows a boy watching television Diagram m 7 (ii(b)) shows a man is chased by a fierce dogDiagram 7 (ii(c)) shows a man climbing a mountain


138


Explain the effect of the 3 different situations towards the physiological process that occur in organ X as shown in diagram 7 (ii)


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

(Relaxing)

F1-At rest, the respiratory rate is normal /12-20 breaths per minuteP1-The partial pressure of O2and CO2 are normal

11 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 increaseP2-It’s needed because the demand of a higher respiration rate in cellsP3-In order to oxidize more glucoseP4-To produce more energyP5-(then), rapid muscles contraction (as a responded to the dangerous situation /running)

11111 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 decreaseP6-Thus, partial pressure of O2becomes lowerP7-Causes a drop in the oxygen level in bloodP8-(the person will face difficulty in breathingP9-So, the person will experience headache/nausea/dizziness

1

1111 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


138


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

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

11

1

1

1

111

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 4E1 : 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

1111111 4


138


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

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 tarE1-Causes lugs cancerF2-cigarette smoke contain acidic gases

111 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

1111 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 monoxideE2-Bind with haemoglobin to form carboxyhaemoglobinE2-Less oxygen combine with haemoglobinE4-Tobacco tar will be deposited/logged /accumulate (inside the lungs)E5Reduce diffusion of oxygenE6-Haet fom the smoke m

111111


138


E7-Dry the surface of the alveoliE8-Oxygen cannot be dissolved Any 4

11 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

11

111

1 4

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 lenticelsS2-Stomata can be found in epidermis of leaves. the stem of herbaceous plantsS3-Lenticels can be found on the stems and root of plantsExplanation P1-When stomata open, they connect the air space (within the leave) to atmosphereP2-Oxygen form the atmosphere diffuses into the air spaces P3-then dissolves in the film of water around the mesophyll cellsP4-So the concentration of oxygen in the cells becomes lower than in the air spacesP5-Thus, oxygen diffuse continuously form air space to the cell P6-During daytime, carbon dioxide that is produced during respiration is used in

photosynthesisP7-The excess carbon dioxide diffuses into the air spaces and then through stomata into

atmosphere

111

11111

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.


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

138


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 of spongy mesophyll (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

11

1

11 4

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 increaseP2-The plant carries out anaerobic respiration

11

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

1

111


Pore M

Epidermal cell

Guard cell

Pore MBroken epidermis

Cork tissue

138


P6-(the air space) allow oxygen to penetrate through to the cells of roots ( growing in the absence of oxygen)

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 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)

111111

111

1

1

11 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.


138


Based on the graph, explain how light intensity and the structure in diagram 6.2 affect the rate of water loss 10F1 : 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.

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

11111

11111111 10


138


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)

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

1111

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 4.2D1-Alveolus LeafD2-Carry out transpiration Carry out photosynthesisD3-Absent of chlorophyll Presence of chlorophyll

1

1

1 2

Extra Question

Diagram 7.1 shows how the respiratory gases are transported in the human body


138


(i) Based on Diagram 7.1, explain how the transport of oxygen and carbon dioxide takes place in the body cells

Aspect Marking schemeTransport 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


biology form 4 chapter 7.docx

Documents