ii Must Know iii – viii Chapter 1 Organisation of Plant Tissues and Growth NOTES 1 Paper 1 2 Paper 2 6 Reinforcement and Assessment of Science Process Skills for Paper 3 (Practical Test) 10 Chapter 2 Leaf Structure and Function NOTES 11 Paper 1 13 Paper 2 17 Reinforcement and Assessment of Science Process Skills for Paper 3 (Practical Test) 21 Chapter 3 Nutrition in Plants NOTES 23 Paper 1 24 Paper 2 27 Chapter 4 Transport in Plants NOTES 31 Paper 1 32 Paper 2 36 Chapter 5 Response in Plants NOTES 40 Paper 1 41 Paper 2 45 Chapter 6 Sexual Reproduction in Flowering Plants NOTES 50 Paper 1 51 Paper 2 55 Chapter 7 Adaptations of Plants in Different Habitats NOTES 59 Paper 1 60 Paper 2 62 Contents Chapter 8 Biodiversity NOTES 65 Paper 1 66 Paper 2 71 Chapter 9 Ecosystem NOTES 75 Paper 1 78 Paper 2 82 Chapter 10 Environmental Sustainability NOTES 86 Paper 1 89 Paper 2 92 Chapter 11 Inheritance NOTES 97 Paper 1 99 Paper 2 104 Chapter 12 Variation NOTES 109 Paper 1 110 Paper 2 112 Chapter 13 Genetic Technology NOTES 115 Paper 1 116 Paper 2 118 Answers 120
MUST KNOW Important Definition Important Definition (Chapter 8) 8 @ Pan Asia Publications Sdn. Bhd. Important Definition (Chapter 2) 2 @ Pan Asia Publications Sdn. Bhd. Important Definition (Chapter 10) 10 @ Pan Asia Publications Sdn. Bhd. Important Definition (Chapter 4) 4 @ Pan Asia Publications Sdn. Bhd. Important Definition (Chapter 12) 12 @ Pan Asia Publications Sdn. Bhd. Important Definition (Chapter 6) 6 @ Pan Asia Publications Sdn. Bhd. Dichotomous Key Leaf Structure and Function Transport in Plants 1. Stomata are the pores found in the epidermis of leaves, stems or other organs. Stoma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plants. 5. +XPDQSRSXODWLRQJURZWKH[SORVLRQLVWKHLQFUHDVHLQWRWDO QXPEHURIKXPDQVOLYLQJLQDFHUWDLQDUHD 1. 3ROOLQDWLRQLVWKHSURFHVVLQZKLFKSROOHQJUDLQVDUH transferred from the anther to the stigma. 2. 'RXEOHIHUWLOLVDWLRQLVWKHIHUWLOLVDWLRQFKDUDFWHULVWLFRI VHHGSODQWVLQZKLFKRQHVSHUPQXFOHXVIXVHVZLWKWKHHJJ QXFOHXVWRIRUPDQHPEU\RDQGDQRWKHUIXVHVZLWKSRODU QXFOHLWRIRUPHQGRVSHUP 1. *XWWDWLRQLVDVHFUHWLRQRIZDWHUGURSOHWVWKURXJKDVSHFLDO VWUXFWXUHDWWKHHQGRIWKHOHDIYHLQVE\KLJKURRWSUHVVXUH 2. 7UDQVORFDWLRQLVDSURFHVVRIWUDQVSRUWLQJRUJDQLF VXEVWDQFHVVXFKDVVXFURVHDPLQRDFLGVDQGKRUPRQHVLQ WKHSKORHPIURPWKHOHDYHVWRRWKHUSDUWVRIWKHSODQWVXFK as the roots and stem. 3. 3K\WRUHPHGLDWLRQLVRQHRIWKHWUHDWPHQWPHWKRGVZKLFK XVHVSODQWVIRUWKHSXUSRVHRIGHJUDGDWLRQH[WUDFWLRQRU HOLPLQDWLRQRISROOXWDQWVIURPVRLODQGZDWHU 1. 9DULDWLRQUHIHUVWRWKHGLIIHUHQFHVLQFKDUDFWHULVWLFVIRXQG ZLWKLQWKHVDPHSRSXODWLRQRUVSHFLHV 2. 7\SHVRIYDULDWLRQffl &RQWLQXRXV YDULDWLRQ LV WKH YDULDWLRQ LQ ZKLFK WKH GLIIHUHQFHVLQWKHFKDUDFWHULVWLFLVQRWGLVWLQFW 'LVFRQWLQXRXV YDULDWLRQ LV WKH YDULDWLRQ LQ ZKLFK WKH GLIIHUHQFHVLQWKHFKDUDFWHULVWLFLVGLVWLQFW 1. 'LFKRWRPRXVNH\LVDWRROXVHGE\WD[RQRPLVWVWRLGHQWLI\ RUJDQLVPVEDVHGRQWKHLUVLPLODULWLHVDQGGLIIHUHQFHV 2. $VDPSOHRIGLFKRWRPRXVNH\ffl 'LFKRWRPRXVNH\ 1a Animals ...................................................................................... Go to 2 E 3ODQWV .......................................................................................... Go to 6 D +DVOHJV ...................................................................................... Go to 3 E 'RHVQRWKDYHOHJV ..................................................................... Go to 5 3a Three pairs of legs ...................................................................... Go to 4 E 0RUHWKDQWKUHHSDLUVRIOHJV ....................................................... Spider D +DVZLQJV ................................................................................%XWWHUÀ\ E 'RHVQRWKDYHZLQJV ........................................................................Ant D +DVVKHOO .........................................................................................Snail E 'RHVQRWKDYHVKHOO ..............................................................Earthworn D +DVVHHGV .................................................................................... Go to 7 E 'RHVQRWKDYHVHHGV ....................................................................... Fern D )ORZHULQJSODQW ........................................................................Hibiscus E 1RQÀRZHULQJSODQW ................................................................ Pine tree Threats to the Environment Variation Pollination and Fertilisation
MUST KNOW Mnemonics & Important Diagrams 0QHPRQLFV(Chapter 1) 13 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 1) 19 @ Pan Asia Publications Sdn. Bhd. 0QHPRQLFV(Chapter 6) 15 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 2) 21 @ Pan Asia Publications Sdn. Bhd. 0QHPRQLFV(Chapter 9) 17 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 3) 23 @ Pan Asia Publications Sdn. Bhd. Organisation of Plant Tissues Zones of Cell Growth Structure of a Flower The Internal Structure of a Leaf Lamina Ecosystem The Internal Structures of Roots MHUU\puts dHOLFLRXVgrapes in vase 0HULVWHPDWLF 0(5,67(0 Dermal 3HUPDQHQW Ground 9DVFXODU Steve, please cRRN egg. 6SHFLHV &RPPXQLW\ 3RSXODWLRQ (FRV\VWHP Plant Parts Pistil or Stamen? Stamen Remember! men - male part of a flower Remember! l (lady) - female part of a flower The stigma (as in mother) is part of the pistil Pistil Primary xylem Primary phloem Pericycle Vascular cylinder Vascular cambium Endodermis Root hair Epidermis Cortex Cuticle Upper epidermis Palisade mesophyll Xylem Spongy mesophyll Vascular bundle Phloem Stoma Lower epidermis Matured tissue Zone of cell elongation Zone of cell differentiation Zone of cell division Shoot apical meristem Zone at the shoot tip Matured tissue Zone of cell differentiation Zones at the root tip Zone of cell elongation Zone of cell division Root cap
Important Diagrams (Chapter 8) 32 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 5) 26 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 9) 34 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 6) 28 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 11) 36 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 6) 30 @ Pan Asia Publications Sdn. Bhd. 0RQRK\EULG,QKHULWDQFH MUST KNOW Important Diagrams Nitrogen Cycle Structure of a Flower )RUPDWLRQRI3ROOHQ7XEHDQG0DOH*DPHWHV Energy Flow in the Food Chain Types of Plant Responses 7\SHVRI3ODQW5HVSRQVHV Tropism *URZWKUHVSRQVHLQD SDUWLFXODUGLUHFWLRQV Nastic 5HVSRQVHWRHQYLURQPHQWDO VWLPXOLZLWKRXWGHSHQGLQJRQ WKHGLUHFWLRQRIWKHVWLPXOXV Geotropism *UDYLW\ FDXVHV a response in a SODQW¶VJURZWK +\GURWURSLVP 7KH ZD\ D SODQW JURZV RU EHQGVLQ UHVSRQVHWRZDWHU Thigmotropism 3ODQWVEHQGRUJURZ EHFDXVHRIWRXFK Phototropism 7KH ZD\ D SODQW JURZVRUEHQGVLQ response to light. 3 tall (75%) : 1 drawf (25%) TT (tall) Tt (tall) Tt (tall) tt (dwarf) Gamete Tt (tall) Tt (tall) Gamete TT (tall) tt (dwarf) T T t Tt Tt Tt T T t TT t tt × Fertilisation t T t T t Pollen grain Pollen grain Pollen tube Male gametes (n) Tube nucleus (n) Generative nucleus Mature pollen Ovule Style Male gametes (n) Micropyle Male gametes (n) Tube nucleus Atmospheric nitrogen (N2 ) Lightning (Atmospheric fixation (NO2 – ) Legume plants Nitrogen fixing bacteria (legume root nodules) Rhizobium sp. Ammonia (NH3 ) Azotobacter sp. Nitrosomonas sp. Nitrobacter sp. Ammonium (NH4 +) Decomposers (bacteria/ fungi) Dead organisms, animal waste Nitrite (NO2 – ) Nitrate (NO3 – ) Denitrifying bacteria Fertiliser (NH4 +, NO3 – ) Heat loss Heat loss Heat loss Primary consumer 1 000 kcal Producer 10 000 kcal Secondary consumer 100 kcal Tertiary consumer 10 kcal 90% in respiration, excretion and defaecation 90% in respiration, excretion and defaecation 90% in respiration, excretion and defaecation Energy from Sun Petal Stigma Style Ovary Ovule Peduncle Stamen Anther (male reproductive organ) Carpel (female reproductive organ) Filament Sepal
11 Chapter 2 Leaf Structure and Function NOTES 2.1 Structure of a Leaf 1. The structure of leaf can be divided into two parts, which are the external structure and internal structure. Tip Lamina Midrib Margin Vein Petiole 2. There are two main parts for external structure, there are petiole and lamina. 3. Petiole is the leaf stalk that connects the lamina to the stem of plant. 4. The petiole stretches out into lamina producing a network of middle veins to support the lamina. 5. /DPLQDLVWKHÁDWWKLQVPRRWKJUHHQSDUWRIWKHOHDI 6. /DPLQDLV ÁDWVKDSHGWR SURYLGH D ZLGH VXUIDFHLQ order to expose the cells containing chloroplasts for maximum amount of sunlight. 7. Lamina is thin to allow gases involved in SKRWRV\QWKHVLVWR GLIIXVHHIÀFLHQWO\LQWRDQG RXW RI the leaf. 8. The internal structures of the leaf contain upper epidermis, palisade mesophyll, spongy mesophyll, vascular bundle (xylem and phloem) and lower epidermis. Cuticle Upper epidermis Lower epidermis Palisade mesophyll Spongy mesophyll Xylem Phloem Stoma 9. Upper epidermis is located on the upper surface of the leaf, which is under the cuticle layer. 10. This layer does not contain any chloroplasts and it is transparent so that light can penetrate it. 11. Palisade mesophyll cells are arranged vertically and FORVHO\SDFNHGWRUHFHLYHPD[LPXPOLJKWH[SRVXUH 12. Palisade mesophyll cells are the sites for photosynthesis. 13. Therefore, they contain many chloroplasts. 14. Spongy mesophyll cells DUH LUUHJXODUVKDSHG WKDW increases the internal surface area for gaseous exchange. 15. These cells are loosely arranged and they have many intercellular air spaces. 16. This can ease carbon dioxide and water absorption through the leaves to palisade mesophyll cells during photosynthesis. 17. Spongy mesophyll has less number of chloroplasts than palisade mesophyll. 18. Xylem transports water and mineral salts absorbed from the roots to the leaf. 19. 7KH ZDOOV RI [\OHP DUHOLJQLÀHG DQGWKLFNHQHGWR provide mechanical support and strength to the plants. 20. Phloem transports organic substances which is produced during photosynthesis from the leaves to other parts of the plants. 21. Lower epidermis is located at the lower surface of the leaf. 22. This layer consists of stomata. Each stoma is guarded by a pair of guard cells. 2.2 Main Organ for Gaseous Exchange 1. The mechanism of stomatal opening and closing depends on the conditions of the guard cells whether WXUJLGRUÁDFFLG 2. The condition of the guard cells depends on the potassium ion (K+ ) uptake by the cells or the sucrose concentration in the sap of the guard cells. 3. The accumulation or elimination of potassium ion (K+ ) in the guard cells changes the solute potential. 4. This increases or decreases the water potential in the guard cells. 5. Water is diffused out or into the guard cells through osmosis. 6. This condition determines whether the guard cells are WXUJLGRUÁDFFLG
12 7. During day time or in the presence of light, photosynthesis takes place and produces dissolved sugar (sucrose). 8. During night or in the absence of light, sugar in the guard cells converts into starch. 9. Water from plants is lost in the form of water vapour to the surroundings through stomata. 10. When stoma opens widely, the rate of water loss from the plants is high. 11. The opening and closing of the stoma is depend on the turgor pressure of the guard cells. 2.3 Main Organ for Transpiration 1. Transpiration is a process of water loss in the form of water vapour through evaporation from the plants to the atmosphere. 2. The rate of transpiration is affected by light intensity, relative air humidity, temperature and air movement. 2.4 Main Organ for Photosynthesis 1. There are two main stagesLQSKRWRV\QWKHVLVZKLFKDUHOLJKWGHSHQGHQWUHDFWLRQDQGOLJKWLQGHSHQGHQWUHDFWLRQ Light-dependent reactions (occurs in the thylakoids) Light-independent reactions (occurs in the stroma) 1. Photosynthetic pigments on the surface of the thylakoids absorb light energy. 2. Light energy excites electrons in chlorophyll pigments to a higher level. 3. The excited electrons from the chlorophyll go through a series of electron carriers. The energy from the electrons is used to generate energy in the form of ATP. 4. Eventually, these electrons are accepted by the last electron acceptor, which is the NADP+. The NADP+ then combines with H+ from photolysis and forms NADPH which is a reducing agent. 5. Chlorophyll pigment attracts electrons from water via photolysis to become stable. 6. Photolysis is a process whereby water molecules are broken down to form hydrogen ions (H+ ) and hydroxide ions (OH ) in the presence of light energy and chlorophyll. 7. Hydroxide ions lose electrons to form oxygen and water. 1. FDUERQ RUJDQLF FRPSRXQGV À[FDUERQGLR[LGHJDVHVWRIRUP FDUERQRUJDQLFFRPSRXQGV 2. NADPH and ATP from the OLJKWGHSHQGHQWUHDFWLRQUHGXFHV the organic compounds to glucose monomers. 3. Glucose monomers condensate to form starch molecules. Starch granules are stored in the stroma of chloroplasts. Overall reaction for photosynthesis: Light energy 12H2 O + 6CO2 C6 H12O6 + 6O2 + 6H2 O Water Carbon dioxide Chlorophyll Glucose Oxygen Water 2. Environmental factors that affect the rate of photosynthesis are (a) Carbon dioxide concentration (b) Light intensity (c) Temperature 2.5 Compensation Point 1. At the compensation point, the rate of photosynthesis is the same as the rate of respiration. 2. Glucose produced in photosynthesis is used in the respiration of plants. Uptake of CO2 decreases Uptake of CO2 increases Low High Light intensity increases Compensation point
13SOS TIP 2.1 Structure of a Leaf 1. Which of the following are the internal structures of a leaf? I Epidermis III Petiole II Spongy mesophyll IV Lamina A I and II C II and IV B I and III D III and IV 2. Which of the following has no chloroplast? A Guard cell C Palisade mesophyll B Spongy mesophyll D Cuticle 3. Which cell in the plant contains high density of chloroplast? HOTS Applying A Sclerenchyma C Palisade mesophyll B Spongy mesophyll D Epidermis 4. Which of the following properties help the lamina to FDUU\RXWSKRWRV\QWKHVLVLQDPRUHHIÀFLHQWZD\" HOTS Analysing I /DPLQDLVÁDW II Lamina has a layer of cuticle. III Lamina is thin. IV Lamina is shiny. A I and II C II and IV B I and III D III and IV 5. Why are the laminas of the leaf arranged in leaf mosaic structure? HOTS Applying A This allows the leaves to receive sunlight optimally for photosynthesis. B This allows the leaves to have more beautiful arrangement. C This allows the leaves to carry out gaseous H[FKDQJHPRUHHIÀFLHQWO\ D This allows the leaves to reduce water loss. 6. Why there is a layer of cuticle on the upper surface of a leaf? A The cuticle allows gaseous exchange to occur between the leaf and environment. B The cuticle helps the leaf to absorb more sunlight for photosynthesis. C The cuticle prevents the excess loss of water through transpiration. D The cuticle strengthens the upper epidermal cells at the leaf. PAPER 1 Each question has four different answers A, B, C and D. For each question, choose one answer only. 7. Why does the spongy mesophyll cell have an irregular shape? HOTS Analysing A It helps in the absorption of sunlight. B It helps in the absorption of water. C It helps in gaseous exchange. D It helps in the transportation of food. 8. Diagram 1 shows the structure of leaf. A B C D Diagram 1 Which of the parts labelled A, B, C or D is involved actively in photosynthesis? 2.2 Main Organ for Gaseous Exchange 9. Where does gaseous exchange occurs in plants? A Cuticle B Upper epidermis C Stomata D Xylem 10. What cells are responsible to control the stoma? A Guard cells B Upper epidermal cells C Mesophyll cells D Lower epidermal cells 11. In which combination of environmental conditions are the stomata of a plant likely to open? Atmospheric humidity Speed of air Availability of soil water A High Low High B High Low Low C Low High High D Low High Low Question 3: This cell is nearly directly exposed to sunlight and can carry out photosynthesis at a high rate. Question 4: Think of the ways to increase the absorption of sunlight and carry out gaseous exchange. Question 5: Leaf mosaic structure helps the leaves not to overlap to each other. Question 7: Think of the function of the spongy mesophyll cell.
14SOS TIP Question 12: Consider the processes that help the opening and closing of stomata. Question 14: Think about the concentration of cell sap that affects the guard cell. Question 16: Think about the elasticity of outer cell wall and inner cell wall of guard cell. Question 18: Think about the changing of water content in guard cells. Question 21: Think about the factors affecting the rate of water evaporation. 12. Diagram 2 shows a guard cell and stoma of a terrestrial plant. Q Diagram 2 Which of the following causes the closing of Q? HOTS Applying I Potassium ions move out of the guard cells. II Glucose is produced in the guard cells. III The guard cells become hypotonic to adjacent cells. IV The guard cells swell up and become turgid. A I and II C II and IV B I and III D III and IV 13. Which of the following plants has no stomata on its leaves? A C B D 14. How does the concentration of sucrose in the cell sap of guard cell helps the opening of stoma? HOTS Analysing A The low concentration of sucrose stimulates the water molecules to diffuse out of the guard cells to make the stoma open. B The high concentration of sucrose stimulates the water molecules to diffuse out of the guard cells to make the stoma open. C The low concentration of sucrose stimulates the water molecules to diffuse into the guard cells to make the stoma open. D The high concentration of sucrose stimulates the water molecules to diffuse into the guard cells to make the stoma open. 15. What type of pressure in the guard cells affects the opening and closing of stoma? A Air pressure C Cell pressure B Turgor pressure D Atmospheric pressure 16. Why does the guard cell bend when water molecules diffuse into it? HOTS Analysing A Thick outer cell wall and thin inner cell wall of guard cell B Thin outer cell wall and thick inner cell wall of guard cell C Equal thickness of outer cell wall and inner cell wall of guard cell D No cell wall at guard cell 2.3 Main Organ for Transpiration 17. Which of the following organs are involved in the transpiration process? I Root III Fruit II Stem IV Flower A I and II C II and IV B I and III D III and IV 18. Which of these factor does not contribute to the wilting of plants? HOTS Analysing A The increase of the sucrose concentration in guard cells B The guard cells become turgid C The opening of stomata D The closing of stomata 19. What device can be used to measure the rate of transpiration? A Respirometer C Photometer B Calorimeter D Potometer 20. Which of the following factor affects the rate of transpiration? I Light intensity II Temperature III pH IV Carbon dioxide concentration A I and II C II and IV B I and III D III and IV 21. Which combination of environmental conditions cause the rate of transpiration to increase? HOTS Analysing Atmospheric humidity Speed of air Temperature A Low High Low B Low High High C High Low Low D High Low High
15SOS TIP 22. Diagram 3 shows graph of the effects of light intensity on the rate of transpiration in a plant. Rate of transpiration D C B A Light intensity Diagram 3 Among A, B, C or D, which shows the trend of the graph if the light intensity increases continuously? 23. An experiment is carried out by Fatimah to investigate the rate of transpiration in a plant using a potometer. The diagram shows the apparatus set up and the distance covered by the water column in centimeters. The experiment left for 10 minutes. 0 cm 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Cut shoot Reservoir Ruler Capillary tube Air bubble Diagram 4 Table below shows the position of meniscus at the beginning and the end of the experiment. Position of meniscus at the beginning of experiment (cm) 3.8 Position of meniscus at the end of experiment (cm) 6.5 What is the rate of transpiration of the plant? A 2.7 cm/min C 27 cm/min B 0.027 cm/min D 0.27 cm/min 24. Where should a potometer to be placed to reduce the rate of transpiration of a leafy plant? HOTS Analysing A A bathroom C An open space B A greenhouse D A shady area 2.4 Main Organ for Photosynthesis 25. The following equation shows process P which occurs in chloroplasts. light 24H2 O 24H+ + 24OH– chlorophyll Diagram 5 shows the structure of a chloroplast. C D B A Diagram 5 Which part labelled A, B, C or D is involved in process P? 26. Which statement is correct about photosynthesis? A Photosynthesis only occurs with the presence of light. B Photosynthesis produces starch. C Photosynthesis produces glucose as one of the products. D Photosynthesis releases energy. 27. The following chemical equation shows a type of reaction that occurs in plants. Light energy W + 12H2 O X + Y + 6H2 O Chlorophyll What are W, X and Y? W X Y A 6CO2 C6 H12O6 6H2 O B 6CO2 C6 H12O6 6O2 C 6O2 C6 H12O6 6H2 O D 6O2 C6 H12O6 6CO2 28. What is the conclusion made by Jan-Baptista van Helmont during the contribution of discovery of photosynthesis? A Plant grows using the water in the soil. B Plant grows using the water that is poured into the soil and not the water in the soil itself. C Plant grows by absorbing the nutrients from the soil. D Plant grows by absorbing the nutrients and water from the soil. 29. Which scientist proved that the plants release oxygen using an experiment with a rat and a plant in a bell jar? A Joseph Priestly C Jean Senebier B F. F. Blackman D Robert Mayer Question 23: Calculate using the formula, Rate of transpiration = Distance of air bubble moved Time taken Question 24: Think of the life of this tree. Question 28: Think of the process undergone by the plant.
17SOS TIP PAPER 2 Section A Answer all questions. 1. Diagram 1 shows a cross section of leaf for a type of aquatic plant. Surface of water Air space Z X Y Diagram 1 (a) Name structures X, Y and Z. [3 marks] (b) (i) Suggest one function of the air space in Diagram 1 which helps in photosynthesis. HOTS Evaluating [1 mark] (ii) State another function of the air space in relation to maintaining the position of the leaf. [1 mark] (c) State two ways in which the structure of this leaf is different from terestrial plant. HOTS Creating [2 marks] 2. Diagram 2 shows an experiment conducted to prove the hypothesis that carbon dioxide is needed in the photosynthesis process. Conical flask B Cotton Leaf A Leaf B Sodium potassium hydroxide Conical flask A Diagram 2 (a) What is the function of solid potassium hydroxide? HOTS Analysing [1 mark] Question 1: (b) (i) Explain the function of air space that is related to photosynthesis. (c) Know and understand the position of the stomata and distribution of chloroplasts in the cells on land plants. Question 2: D ,WLVUHODWHGWRDVSHFL¿FW\SHRIJDV
19SOS TIP Question 4: (a) (ii) Think about the reason for the distribution of the stomata. (b) (iii) Think about the opening and closing of the stomata. 4. Sarah carried out an investigation to determine the number of stomata on the leaves of monocotyledonous plants and eudicotyledonous plants. The following information shows the result of the investigation. The number of stomata on the leaves of monocotyledonous and eudicotyledonous plants are different. (a) (i) State the difference in the number of stomata on the leaves of monocotyledonous plants and eudicotyledonous plants. [1 mark] (ii) Explain your answer stated in (a)(i). HOTS Evaluating [1 mark] (iii) Explain the relationship between the distribution of stomata on monocotyledonous plants leaves and photosynthesis. [1 mark] (b) Diagram 4 shows the image of cell X during Sarah’s investigation. X Diagram 4 (i) What is cell X? [1 mark] (ii) State the function of cell X. [1 mark] (iii) Explain the changes of cell X helps the plant to survive on a hot and warm day. HOTS Creating [2 marks] Section B 5. D %ULHÁ\H[SODLQWKHSDUWVZKHUHWUDQVSLUDWLRQRFFXUVLQSODQWVDQGLWVLPSRUWDQFHWRSODQWV >flPDUNV@ (b) Explain the environmental factors that can affect the rate of transpiration. [12 marks]
20SOS TIP 6. Diagram 5 shows production of carbon dioxide by a green plant. 0 X Volume of carbon dioxide Light intensity Diagram 5 (a) What is point X? Explain the incident that happens at point X. [4 marks] (b) Explain photosynthesis that occurs in the chloroplast of green plants. [14 marks] (c) State the differences between photosynthesis and respiration in plant. [2 marks] Section C 7. (a) The leaf is an important organ for a plant to survive. Explain how the external structure of a leaf helps a green plant during photosynthesis. HOTS Analysing [6 marks] (b) Other than the external structure of the leaf, the chlorophyll in leaf also plays an important role in photosynthesis. Explain how the chlorophyll helps in photosynthesis. [8 marks] F &ODUD QXUWXUHV VRPH ÀVKLQ KHU DTXDULXP 6KHLQVWDOOV DQ HOHFWULF SXPSWR SXPS R[\JHQLQWRWKH ZDWHU RI KHUDTXDULXPWRNHHSKHUÀVKVXUYLYH2QHGD\KHUSURIHVVRU3URI.&YLVLWV&ODUD·VKRXVH+HREVHUYHVWKH FRQGLWLRQRIWKHDTXDULXPDQGKHJLYHVDVXJJHVWLRQWR&ODUDVRWKDWKHUÀVKFDQVWLOOVXUYLYHZLWKRXWWKHKHOSRI electric pump. If you are Prof. KC, what suggestions will you give to Clara? HOTS Creating [6 marks] 8. Greenhouse farming is being used in countries with four seasons. This farming method is used to increase the crops yield. (a) Mr. Mak is a strawberry farmer in Malaysia. He needs your expertise as a designer to design and build an effective greenhouse in the hopes of increasing yield production. Explain your design to Mr. Mak. HOTS Creating [10 marks] (b) Mr. Mak may face some challenges when and after he builds the greenhouse for his crops. State the challenges he may face. HOTS Creating [4 marks] (c) As Mr. Mak’s personal greenhouse designer, suggest ways to help Mr. Mak to solve his challenges. HOTS Creating [6 marks] Question 7: (a) Think about the functions of the petiole, lamina and cuticle. (c) Consider the function of the electric pump in the aquarium. Question 8: (a) Think about the factors affecting the rate of photosynthesis. (b) Think of the materials used to build this effective greenhouse. (c) Consider the challenges mentioned in (b), then think of how to solve it.
21SOS TIP Reinforcement and Assessment of Science Process Skills for Paper 3 (Practical Test) 1. You are required to conduct an experiment to study the effect of carbon dioxide concentration on the rate of SKRWRV\QWKHVLV7KHIROORZLQJÀJXUHVKRZVWKHDUUDQJHPHQWRIWKHDSSDUDWXV Retort clamp Thermometer 0.04 M sodium hydrogen carbonate solution Beaker Boiling tube Bulb Distilled water Paper clip Hydrilla sp. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ruler Diagram 1 Carry out the experiment using the following steps. (i) Attach a paper clip under the stem cutting of Hydrilla sp. and put them into the boiling tube. (ii) Pour 5 ml of 0.04 M sodium hydrogen carbonate solution using a measuring cylinder into the boiling tube. (iii) Clamp the boiling tube vertically to the retort stand. (iv) Light a 60 W bulb at a distance of 10 cm from the Hydrilla sp. plant. (v) Count and record the number of air bubbles released in 5 minutes. Take three readings to obtain an average value. (vi) Repeat steps ii to v using the other concentrations of sodium hydrogen carbonate solution. (vii) Record the results in Table 1. [4 marks] Table 1 Concentration of sodium hydrogen carbonate solution (M) Number of air bubbles released in 5 minutes Rate of photosynthesis (minute–1 1 2 3 Average ) 0.04 0.08 0.12 (a) State the hypothesis for this experiment. [1 mark] (b) State the variables for this experiment. (i) Fixed variable (ii) Manipulated variable (iii) Responding variable [3 marks] (c) State the function of thermometer in this experiment. [1 mark]
22SOS TIP (d) (i) State one observation in this experiment. [1 mark] (ii) Give one inference based on the observation in (d)(i). [1 mark] H %DVHGRQWKHUHFRUGHGREVHUYDWLRQSORWWKHJUDSKVRIWKHUDWHRISKRWRV\QWKHVLVDJDLQVWWKHFRQFHQWUDWLRQ of sodium hydrogen carbonate solution. [3 marks] (f) Why is the sodium hydrogen carbonate solution being used, instead of distilled water in the experiment? [1 mark] If you are unable to carry out this experiment, you can answer this using the observation provided in the sample data shown in Table 2. Table 2 Concentration of sodium hydrogen carbonate solution (M) Number of air bubbles released in 5 minutes Rate of photosynthesis (minute–1 1 2 3 Average ) 0.04 5 6 7 6 1.2 0.08 11 8 11 10 2.0 0.12 12 12 12 12 2.4
120 Answers CHAPTER 1 Paper 1 1. C 2. B Vascular tissues, including xylem and phloem are involved in the transportation process in the plant. 3. C 4. D 5. A ;\OHPYHVVHOVDQG VFOHUHQFK\PDWLVVXHKDYHOLJQL¿HGFHOO walls to support the plants. 6. B Parenchyma cells are cells that are loosely arranged and have thin cell walls with many chloroplasts for photosynthesis. 7. A 8. D Meristematic cells present at the root tip helps the plant to grow and root hair cells at the root tip helps the plant to absorb water and minerals. 9. B Vacuolation occurs to elongate the cells for the growth of plant. 10. C The formation of secondary xylem is one of the processes in secondary growth of plants. 11. A P is phloem tissue that helps to transport the product of photosynthesis in plants. 12. C Cork cambium contains meristematic cells which carry out cell division to produce cork. 13. D Zone I is the zone of cell division which contains meristematic cells to carry out mitosis to produce new cells for growth. 14. B 15. B Lateral meristem is the vascular cambium that helps to increase the girth of the woody plant. 16. D 17. A 18. C When secondary growth starts to happen in the plant, primary growth is not halted but continues to be carried out. 19. A 20. B 3ULPDU\[\OHPKDVOLJQL¿HGFHOOZDOOVWRSURYLGHVXSSRUWWR the plants. 21. C 22. D 23. B A cork is made up of the inner soft bark that is produced by the spongy cambium, which is a type of vascular cambium. 24. D 25. A 26. A 27. B Biennial plants complete the growth season and the reproduction season. Examples are cabbage, red carrot and silver cock’s comb. 28. C 7KHWUHHLVDPDQJRWUHHZKLFKLVFODVVL¿HGDVDSHUHQQLDO plant. 29. C P is the stage of germination and the food stored in the cotyledons is used. Q is the stage of aging and eventually, the plant will die. 30. D Grass is a perennial plant and does not have the same curve as watermelon plant which is an annual plant. Other than that, it has several repeated sigmoid curves in its growth curve. It can SURGXFHÀRZHUDQGIUXLWPDQ\WLPHV Paper 2 Section A 1. (a) Primary growth (b) P – Zone of cell differentiation Q – Zone of cell elongation R – Zone of cell division (c) (i) This process enables the cells at this zone to increase their size through the vacuolation process. (ii) Cortex of parenchyma, endodermis 2. (a) Structure P helps to protect the zone of cell division. (b) (i) Dry mass 1 mark 1 mark 1 mark 0 Time (week) (ii) At the beginning of the graph, the dry mass is decreasing because the stored food in the seed is used for respiration to form leaves and roots during the germination of seed. Once the leaves start to carry out photosynthesis, growth continues in a sigmoid curve. At the end of the graph, the dry mass decreases again when seeds are dispersed before death. 3. (a) Tissue P: Xylem vessel/Xylem Tissue Q: Sieve tube/Phloem (b) Mitosis (c) Tissue P (xylem vessel/xylem) is a continuous hollow WXEHZKLFKDOORZVDFRQWLQXRXVXSZDUGÀRZRIZDWHUDQG minerals. Tissue Q (sieve tube/phloem) is a vertical series of elongated cells with a connection of the sieve plate that enables the transportation of food in all directions in a plant. (d) Tissue P [\OHPYHVVHO[\OHP KDVOLJQL¿HGWKLFNFHOOZDOOV to make it hard enough to provide mechanical support to the plant. 4. (a) Annual plant – W Biennial plant – X Perennial plant – Y (b) (i) Dry mass (g) Time (week) Correct labelling of x and y-axis = 1 mark Correct curve pattern = 1 mark LL 7KHJURZWKFXUYHKDVWZRVLJPRLGFXUYHV7KH¿UVW sigmoid curve shows the growth season of the red carrot plant, and the second sigmoid curve shows the reproductive season of the red carrot plant. Chapter 1
121 Section B 5. (a) The cell that exists in zone P is the meristematic cell. [1] The size of the cell is small. [1] It has a thin cell wall. [1] It has a large nucleus. [1] The vacuole is absent or there are only small vacuoles present in the cell. [1] There are no intercellular spaces between the meristematic cells. [1] (b) Zone P is the zone of cell division. [1] Cells divide actively by mitosis to produce similar cells to increase the number of new cells. [1] Zone Q is the zone of cell elongation. [1] Cells become enlarged by absorbing more water and minerals to increase the size of vacuole. [1] Zone R is the zone of cell differentiation. [1] Cells differ from each other to form specialized cells such as epidermal cell, mesophyll cell, xylem tissue and phloem tissue. [1] (c) (i) A B Epidermal cells are covering the outer layer of the stem. Bark or cork is covering the outer layer of the stem. [1] No wood in the stem. Wood is found in the stem. [1] No formation of cork cambium beneath the epidermis. Formation of cork cambium beneath the epidermis. [1] No joining of vascular cambium in the vascular bundles. Joining of vascular cambium in the vascular bundles to form a cambium ring. [1] Primary xylem is present in the vascular bundles. Primary xylem and secondary xylem are present in the vascular bundles. [1] Primary phloem is present in the vascular bundles. Primary phloem and secondary phloem are present in the vascular bundles. [1] (ii) The growth of perennial plant is a sigmoid curve. [1] It is because secondary growth of perennial plant is disturbed by the factors such as light, water and surrounding temperature. [1] In spring, high light intensity and surrounding temperature encourage the plant to grow rapidly. [1] In autumn, the perennial plant grows in lower rate due to lower light intensity. [1] In winter, the growth rate of perennial plant is very slow because the vascular cambium and the lateral meristem is almost in dormant condition as a result of low light intensity and very low temperature. [1] 6. (a) The age of woody plants can be determined by calculating the total number of annual growth rings. [1] It can shown in the cross Section of its trunk. [1] The rate of secondary growth vascular cambium is affected by the seasons in temperate countries. [1] The vascular cambium is dormant during the winter season. [1] In spring season, the vascular cambium becomes more active. [1] The more active vascular cambium forms larger and thinner-walled xylem vessels. [1] During the autumn season, the vascular cambium becomes less active. [1] The less active vascular cambium forms narrower but more OLJQL¿HG[\OHPYHVVHOV>@ The wood part formed during spring has a lighter colour while the wood part formed during autumn has a darker colour. [1] The annual growth ring is formed from the alternating pattern of light and dark regions. [1] (b) The continuous growth in plants occur at meristems at the root and shoot apex. [1] The continuous growth in plants can help the plant to undergo primary and secondary growth. [1] Meristematic cells can continuously divide by mitosis to produce new cells. [1] These meristematic cells are located at the zone of cell division. [1] So, new cells keep getting added while old cells lose the capacity to divide and constitutes the plant body. [1] (c) The growth of the apical shoot can be prevented by the continuous growth of the lateral shoot. [1] As Ms. Chang suggested to the farmer to trim the mango tree, the apical shoot will be cut off. [1] This allows the lateral shoot to grow actively and continuously. [1] Hence, the height of the mango tree can be controlled without affecting the growth of the mango tree. [1] Other than that, it will also not affect the production of mango tree yield. [1] Section C 7. (a) Growth is the process of increasing the size and number of cells that take place during the life history of an organism. [2] (b) (i) Grass shows unlimited growth, but grasshoppers show limited growth. [1] Grass grows throughout its lifetime, but grasshoppers have a limited growth period. [1] Grass grows continuously but grasshoppers grow intermittently to maximise its size. [1] Grass grows by cell elongation through vacuolation, but grasshoppers grow by increasing its number of body cells. [1] Grass does not show ecdysis, but grasshoppers show ecdysis growth. [1] (ii) Height (m) 0 Time (week) Correct labels at x and y-axis = 1 mark Correct curve pattern = 1 mark The growth curve of a paddy plant shows a sigmoid curve. [1] It is an annual plant. [1] The growth curve shows a reduction because the stored food in the cotyledon is used for the germination of the seed to produce leaves. [1] After the leaves are developed, the paddy plant can carry out photosynthesis and the parameter of the paddy plant increases. [1] Next, the growth rate becomes constant because the paddy plant is already matured. [1] Finally, the paddy plant undergoes the aging process, and the parameter will reduce. [1] (iii) Dry mass is a more suitable parameter to measure the growth of plants. [1] Chapter 1
122 Dry mass refers to the mass after the removal of water in plants by heating the plants in 110°C until constant mass. [1] Dry mass is more suitable than height as a growth parameter because height cannot show the exact quantity of growth in plants as it only considers one dimension of growth in plants. [1] Plants can show their growth at different rates for different dimension, hence dry mass is more suitable as the growth parameter. [1] Dry mass assures that only organic matters in the plants are weighed. [1] Paper 3 1. (a) Procedure: 3ODFHDPXQJ EHDQLQDFRQWDLQHU¿OOHGZLWKZHWFRWWRQ to enable its radicle to grow. 2. Cut the root tip at the length of about 10 mm. 3. Immerse the root into 10 cm3 acetic acid with 30 cm3 ethanol. 4. Slice the root with a blade longitudinally. 5. Rinse the sliced root with distilled water and place it on a glass slide with a drop of distilled water. 6. Drop two drops of aceto-orcein dye to the sliced root. 7. Observe the prepared slide under a light microscope. 8. Draw and label the zone of cell division, zone of cell elongation and the zone of cell differentiation of the root. (b) Zone of cell differentiation Zone of cell elongation Zone of cell divison Root cap (c) At the zone of cell division, there are meristematic cells that are very small in size with large nucleus. At the zone of cell elongation, the cells have large vacuoles. At the zone of cell differentiation, the cells are differentiated into specialized cells. (d) (i) Red (ii) To stain the chromosomes. (e) The aceto-orcein dye has a strong, vinegar-like smell. It is an irritant. The fume chamber is needed to remove the strong smell from the biology laboratory during the experiment. CHAPTER 2 Paper 1 1. A 2. D 3. C Palisade mesophyll cells are the second layer of cells in the leaf which are exposed to sunlight. They need large amounts of chloroplast to carry out photosynthesis. 4. B /DPLQDLV ÀDWWR LQFUHDVHWKH VXUIDFHDUHDWRDEVRUEPRUH sunlight for photosynthesis. Lamina is thin to make gaseous exchange easier and faster. 5. A The leaf mosaic structure helps to reduce the overlapping of leaves and allows the leaves to receive sunlight optimally for photosynthesis. 6. C Cuticle layer is a waterproof layer that helps to prevent water loss from the surface of the leaf. 7. C The irregular shape of spongy mesophyll cell can increase the surface area in gas exchange. 8. B 9. C 10. A 11. A 12. B The movement of potassium ions out of the guard cells causes the guard cells to become hypotonic to adjacent cells that enables the water molecules to move out of the guard cells to make the stoma to close up. 13. C 14. D The high concentration of sucrose causes the cell sap to become hypertonic to stimulate the water molecules to diffuse into the guard cells and make the stoma open. 15. B 16. B Thin outer cell wall is more elastic and thicker inner cell wall is not so elastic, which causes the guard cell to bend. 17. C 18. D The increase of sucrose concentration in guard cells and the guard cells becoming turgid can make the guard cells bent and open the stomata. When the stomata are opened, the rate of transpiration increases, which causes the plant to lose a lot of water. 19. D 20. A 21. B Low humidity of air, high speed of air and high temperature increase the rate of transpiration. Hence, water will be evaporated from the plants at the highest rate. 22. C 23. D Rate of transpiration = 6.5 – 3.8 10 = 0.27 cm/min 24. A A bathroom has high humidity and low speed of air to decrease the rate of transpiration. 25. A 26. C 27. B 28. B 29. A 30. C Light-dependent reaction occurs in thylakoids during the daytime. It releases oxygen and no glucose is formed. Lightindependent reaction occurs in stroma during the day and night-time. It does not release oxygen and glucose is formed. 31. A 40 W lamp gives lower light intensity to the plant. Hence, the rate of photosynthesis will be reduced and less oxygen bubbles will be released. 32. B Light intensity and temperature are the limiting factors in this experiment. They will limit the rate of photosynthesis, even when carbon dioxide concentration increases. 33. D 34. D When light intensity is high, the rate of photosynthesis is high. Hence, the amount of oxygen gas is more than the amount of carbon dioxide gas. 35. C When the plant reaches its compensation point, the rate of photosynthesis and the rate of respiration are the same. Hence, the production of oxygen from photosynthesis will be completely used up by respiration. Paper 2 Section A 1. (a) X: Guard cells Y: Epidermal cell Z: Vascular bundle Chapter 1 – Chapter 2