PENERBIT ILMU BAKTI SDN. BHD.
ii Contents Theme 1: Fundamentals of Biology Introduction to Biology and Laboratory Rules 1 Chapter 1 Paper 1 Objective Questions 1 Cell Biology and Organisation 3 Chapter 2 Paper 1 Objective Questions 3 Paper 2 Section A – Structured Questions 7 Section B & C – Essay Questions 8 Movement of Substances across a Plasma Membrane 11 Chapter 3 Paper 1 Objective Questions 11 Paper 2 Section A – Structured Questions 14 Section B & C – Essay Questions 16 Chemical Compositions in a Cell 18 Chapter 4 Paper 1 Objective Questions 18 Paper 2 Section A – Structured Questions 21 Section B & C – Essay Questions 23 Metabolism and Enzymes 25 Chapter 5 Paper 1 Objective Questions 25 Paper 2 Section A – Structured Questions 28 Section B & C – Essay Questions 30 Chapter 6 Cell Division 34 Paper 1 Objective Questions 34 Paper 2 Section A – Structured Questions 37 Section B & C – Essay Questions 40 Cellular Respiration 42 Chapter 7 Paper 1 Objective Questions 42 Paper 2 Section A – Structured Questions 45 Section B & C – Essay Questions 47 Theme 2: Physiology of Humans dan Animals Respiratory Systems in Humans and Animals 49 Chapter 8 Paper 1 Objective Questions 49 Paper 2 Section A – Structured Questions 52 Section B & C – Essay Questions 53 Nutrition and the Human Digestive System 55 Chapter 9 Paper 1 Objective Questions 55 Paper 2 Section A – Structured Questions 59 Section B & C – Essay Questions 61 Transport in Humans and Animals 63 Chapter 10 Paper 1 Objective Questions 63 Paper 2 Section A – Structured Questions 68 Section B & C – Essay Questions 69 00_Q&A Biology SPM.indd 2 11/01/2023 1:40 PM PENERBIT ILMU BAKTI SDN. BHD.
iii Immunity in Humans 73 Chapter 11 Paper 1 Objective Questions 73 Paper 2 Section A – Structured Questions 75 Section B & C – Essay Questions 76 Coordination and Response in Humans 80 Chapter 12 Paper 1 Objective Questions 80 Paper 2 Section A – Structured Questions 83 Section B & C – Essay Questions 84 Homeostasis and the Human Urinary System 88 Chapter 13 Paper 1 Objective Questions 88 Paper 2 Section A – Structured Questions 91 Section B & C – Essay Questions 92 Support and Movement in Humans and Animals 95 Chapter 14 Paper 1 Objective Questions 95 Paper 2 Section A – Structured Questions 98 Section B & C – Essay Questions 99 Sexual Reproduction, Development and Growth in Humans and Animals 103 Chapter 15 Paper 1 Objective Questions 103 Paper 2 Section A – Structured Questions 106 Section B & C – Essay Questions 107 Theme 1: Physiology of Flowering Plants Organisation of Plant Tissues and Growth 110 Chapter 1 Paper 1 Objective Questions 110 Paper 2 Section A – Structured Questions 114 Section B & C – Essay Questions 116 Leaf Structure and Function 120 Chapter 2 Paper 1 Objective Questions 120 Paper 2 Section A – Structured Questions 125 Section B & C – Essay Questions 128 Nutrition in Plants 131 Chapter 3 Paper 1 Objective Questions 131 Paper 2 Section A – Structured Questions 134 Section B & C – Essay Questions 136 Transport in Plants 138 Chapter 4 Paper 1 Objective Questions 138 Paper 2 Section A – Structured Questions 140 Section B & C – Essay Questions 142 Response in Plants 145 Chapter 5 Paper 1 Objective Questions 145 Paper 2 Section A – Structured Questions 147 Section B & C – Essay Questions 148 00_Q&A Biology SPM.indd 3 11/01/2023 1:40 PM PENERBIT ILMU BAKTI SDN. BHD.
iv Sexual Reproduction in Flowering Plants 152 Chapter 6 Paper 1 Objective Questions 152 Paper 2 Section A – Structured Questions 154 Section B & C – Essay Questions 156 Adaptations of Plants in Different Habitats 158 Chapter 7 Paper 1 Objective Questions 158 Paper 2 Section A – Structured Questions 159 Section B & C – Essay Questions 160 Theme 2: Ecosystem and Environmental Sustainability Chapter 8 Biodiversity 164 Paper 1 Objective Questions 164 Paper 2 Section A – Structured Questions 168 Section B & C – Essay Questions 169 Chapter 9 Ecosystem 173 Paper 1 Objective Questions 173 Paper 2 Section A – Structured Questions 177 Section B & C – Essay Questions 178 Environmental Sustainability 180 Chapter 10 Paper 1 Objective Questions 180 Paper 2 Section A – Structured Questions 183 Section B & C – Essay Questions 184 Theme 3: Inheritance and Genetic Technology Chapter 11 Inheritance 188 Paper 1 Objective Questions 188 Paper 2 Section A – Structured Questions 191 Section B & C – Essay Questions 193 Chapter 12 Variation 195 Paper 1 Objective Questions 195 Paper 2 Section A – Structured Questions 198 Section B & C – Essay Questions 200 Genetic Technology 202 Chapter 13 Paper 1 Objective Questions 200 Paper 2 Section A – Structured Questions 204 Section B & C – Essay Questions 206 SPM Model Test 208 Answers 232 00_Q&A Biology SPM.indd 4 11/01/2023 1:40 PM PENERBIT ILMU BAKTI SDN. BHD.
1 Chapter Form 4 13 Chapter Form 4 1 1 Paper 1 Objective Questions 1.1 Fields and Careers in Biology 1 Which of the following is the study of microorganisms? A Botany B Ecology C Genetics D Microbiology Answer: D Microbiology is the study of bacteria, viruses, fungi, protozoa and algae, which are collectively known as ‘microbes’. 2 Which of the following are the contributions of biology in everyday life? I Architecture II Astronomy III Pharmaceutical IV Food production A I and II B I and III C II and IV D III and IV Answer: D The pharmaceutical industry produces substances used for the treatment and prevention of diseases. Food production produces processed food using microorganisms. 1.2 Safety and Rules in a Biology Laboratory 3 The biology laboratory is a place for learning and research. Which of the following refers to the general laboratory safety rules that must always be practised by students? A Do not work alone in the laboratory without supervision B Do not wash your hands after conducting an experiment C Bring irrelevant items into the laboratory D Bring food and drinks into the laboratory Answer: A Options B, C and D are prohibited in the laboratory because these actions can cause unwanted incidence. 4 Which of the following is the correct sequence of steps in handling chemical spills in the laboratory? P Scoop up the chemical spill using appropriate equipment and dispose it safely Q Inform the teacher R Prevent the chemical spill from spreading using sand S Declare the spill area as a restricted zone A P, Q, R, S C Q, P, R, S B P, R, Q, S D Q, S, R, P Theme 1: Fundamentals of Biology Textbook Pages: 1 – 19 Introduction to Biology and 1 Laboratory Rules CHAPTER 01_Q&A Biology SPM F4.indd 1 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
2 Chapter Form 4 13 Chapter Form 4 1 1 Answer: D When there is a chemical or mercury spill, the first step is to inform the teacher, and follow the instructions given by the teacher until the spill is cleared. 1.3 Communicating in Biology 5 Which of the following terms refer to the plane used in research and observation of an organism’s structure? I Sagittal III Posterior II Frontal IV Longitudinal A I and II C II and IV B I and III D III and IV Answer: A The plane refers to a flat surface of shadow passing through the body. The three planes are the sagittal plane, frontal plane and horizontal plane. 6 A graph shows the relationship between manipulated and responding variables and can be illustrated in the form of line graph, bar chart and histogram. Which of the following correctly describes the bar chart? A The width of the bars must be equal without spaces between the bars. B The width of the bars must be equal with equal spaces between the bars. C The width of the bars must be different without spaces between the bars. D The width of the bars must be different with equal spaces between the bars. Answer: B A bar chart presents data using vertical bars that are not attached to each other. 1.4 Scientific Investigation in Biology 7 Which of the following terms is used by scientists to indicate a condition that is allowed to change during an experiment? A Variable B Conclusion C Hypothesis D Problem statement Answer: A Factors or conditions which influence the outcome of the experiment are called variables and these can be divided into constant (controlled), manipulated (changed) and responding (measured) variables. 8 Which of the following refers to a general statement that can be tested to determine its validity? A Theory B Variable C Hypothesis D Problem statement Answer: C A hypothesis is a general statement about an observed event or a preliminary explanation for it, the validity of which has not been proven. The validity of a hypothesis must be tested by conducting an experiment. 01_Q&A Biology SPM F4.indd 2 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
3 Chapter Form 4 13 Chapter Form 4 2 2 Paper 1 Objective Questions 2.1 Cell Structure and Function 1 Which of the following solutions is used in preparing slides of animal cells? A Iodine solution B DCPIP solution C Benedict’s solution D Methylene blue solution Answer: D Methylene blue solution is used to stain the nucleus and cytoplasm in animal cells, in order to observe these structures under the light microscope. 2 Diagram 1 shows a component found in an animal cell. Diagram 1 What is the function of this component? A Controls cell activity B Generates energy in the form of ATP C Synthesises proteins D Regulates the amount of water in the cell Answer: B The mitochondrion is sphere in shape and consists of two layers of membrane. It is the site that generates energy through glucose oxidation. 3 Which cell component is correctly matched with its feature? Cell component Features A Nucleus Consists of a stack of parallel flattened sacs B Golgi apparatus Membranebound structure with many pores C Ribosome Spherical granule that is small and compact D Centriole Consists of interconnected folded flattened sacs Answer: C Features in options A, B and D refer to Golgi apparatus, nucleus and endoplasmic reticulum respectively. Theme 1: Fundamentals of Biology Textbook Pages: 20 – 43 2 Cell Biology and Organisation CHAPTER 02_Q&A Biology SPM F4.indd 3 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
4 Chapter Form 4 13 Chapter Form 4 2 2 4 Diagram 2 shows a component that surrounds the animal cell. Phospholipid bilayer Diagram 2 What is the importance of this component to an animal cell? A Gives shape to the cell B Site for protein synthesis C Supports and protects the cell D Regulates the movement of substances entering and leaving the cell Answer: D The component shown is the plasma membrane. It separates the intracellular and extracellular environments. Exam Tip The plasma membrane is made of proteins and phospholipids, allowing only certain substances to pass through it. Therefore, the plasma membrane is said to be partially permeable. 5 Diagram 3 shows two important components present in both animal and plant cells. P Q Diagram 3 Which of the following are the correct functions of P and Q? P Q A Carries genetic materials Synthesises proteins B Synthesises proteins Involves in osmoregulation C Synthesises proteins Controls the activities of the cell D Breaks down drugs and toxic substances Controls the activities of the cell Answer: C P points to a small circle outside the interconnected folded sacs (known as rough endoplasmic reticulum). This organelle is called the ribosome and it is the site for protein synthesis. Q is spherical and has small pores on the outer surface. It is a nucleus and it controls all cell activities such as cell division. 6 What is cell wall of plant cells made of? A Phospholipid B Glycoprotein C Cellulose D Pectin Answer: C Cellulose is a type of carbohydrate that is rigid and tough. It provides mechanical support to plant cells. 02_Q&A Biology SPM F4.indd 4 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
5 Chapter Form 4 13 Chapter Form 4 2 2 7 What is the function of the rough endoplasmic reticulum? A To generate energy B To break down food C To synthesise protein and lipid D To transport synthesised protein to Golgi apparatus Answer: D The word ‘rough’ in endoplasmic reticulum (ER) refers to the ribosomes attached to it. Proteins synthesised by ribosomes are transported by the rough ER to the Golgi apparatus in vesicles. Common Error Candidates often think that the rough endoplasmic reticulum synthesises proteins. 8 Which of these components are involved in protein synthesis? I Nucleus II Ribosome III Lysosome IV Smooth endoplasmic reticulum A I and II B I and III C II and IV D III and IV Answer: A The flow of protein synthesis starts from nucleus, followed by ribosome, rough endoplasmic reticulum and Golgi apparatus. 2.2 Living Processes in Unicellular Organisms 9 Diagram 4 shows a type of unicellular organism seen under a light microscope. Diagram 4 What is the name of the unicellular organism shown in Diagram 4? A Paramecium sp. B Azotobacter sp. C Euglena sp. D Amoeba sp. Answer: A Paramecium sp. is long, spherical and has cilia on the outer membrane. 10 Which of the following correctly shows the differences between Amoeba sp. and Paramecium sp.? Amoeba sp. Paramecium sp. A Has one contractile vacuole Has two contractile vacuoles B Does not perform phagocytosis Performs phagocytosis C Lives in salt water Lives in freshwater D Feeds using oral groove Feeds using pseudopodia 02_Q&A Biology SPM F4.indd 5 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
6 Chapter Form 4 13 Chapter Form 4 2 2 Answer: A Both Amoeba sp. and Paramecium sp. live in freshwater, perform and undergo phagocytosis. Unlike Amoeba sp., Paramecium sp. has two contractile vacuoles for osmoregulation. Amoeba sp. feeds by moving its pseudopodia while Paramecium sp. uses cilia to move food into the oral groove. 2.3 Living Processes in Multicellular Organisms 11 Which type of tissue is correctly matched with its function? Type of tissue Function A Muscle tissue Connects bones to bones B Nerve tissue Absorbs shock C Blood tissue Protects internal organs D Adipose tissue Stores energy and acts as an insulator Answer: D • Muscle tissue – contracts and relaxes to enable movement • Nerve tissue – detects and carries information in the form of electrical signal • Blood tissue – Involves in regulation, transportation and protection 12 Which type of cell is correctly matched to the cell component found in abundance in the cell? Type of cell Cell component A Sperm cell Mitochondrion B Palisade mesophyll cell Vacuole C Liver cell Lysosome D Plant meristem cell Golgi apparatus Answer: A Sperm cell requires energy to swim through the uterus into the Fallopian tube to fertilise an ovum. The energy is supplied by mitochondria. 2.4 Levels of Organisation in Multicellular Organisms 13 Diagram 5 shows a level of cell organisation. Diagram 5 Which level of cell organisation is this? A Cell B Organ C Tissue D System Answer: B Liver is a type of organ which consists of various tissues working together to perform a specific function. 02_Q&A Biology SPM F4.indd 6 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
7 Chapter Form 4 13 Chapter Form 4 2 2 Paper 2 Section A – Structured Questions 1 Diagram 1 shows components P, Q and R found in a plant cell. Q P R Diagram 1 (a) Name components P and Q. P : Chloroplast Q: Mitochondrion [2 marks] (b) State the function of component P. Site for photosynthesis [1 mark] Chloroplast contains chlorophyll which traps energy from sunlight for photosynthesis. Exam Tip (c) State the function of component Q. Using one example of human cell, explain why component Q is needed in high density. Q is the site of glucose oxidation process which releases energy in the form of ATP (adenosine triphosphate) molecules. An example of a cell is sperm cell. Component Q (mitochondrion) provides a lot of energy to the sperm’s tail which enables it to swim towards the ovum in the Fallopian tube. [3 marks] (d) Justify why component R is important for the survival of a plant cell. HOTS Analysing Component R is a vacuole which contains cell sap. Cell sap contains water which maintains the turgidity of the plant in order to prevent it from wilting. [2 marks] 02_Q&A Biology SPM F4.indd 7 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
8 Chapter Form 4 13 Chapter Form 4 2 2 Paper 2 Section B & C – Essay Questions 2 Diagram 2.1 shows a unicellular organism that lives in a freshwater habitat. (a) Identify the organism and explain how does this organism feeds. [5 marks] (b) Explain how this organism survives in freshwater without having problems in regulating excessive water that enters it. [5 marks] (c) Diagram 2.2 shows a Paramecium sp., another type of unicellular organism that also lives in freshwater habitat. Diagram 2.2 (i) Explain the differences of life processes carried out by two organisms as shown in Diagrams 2.1 and 2.2. [6 marks] (ii) State one cellular component that Paramecium sp. must have in high density and its function. Explain why. [4 marks] Answers 2 (a) – The organism is Amoeba sp. – It feeds via phagocytosis and has pseudopodium (false feet). – Pseudopodium extends out to trap food particles, forming food vacuole. – Food vacuole fuses with lysosome. – The enzyme lysozyme in the lysosome breaks down food particles into smaller pieces/nutrients. – Nutrients will then be absorbed into the cytoplasm. (b) – Amoeba sp. has a contractile vacuole. – Water enters/diffuses into the cell through the plasma membrane via osmosis. – Water then enters/fills the contractile vacuole. – Contractile vacuole expands until it reaches its maximum size. – Contractile vacuole contracts, expelling the water to the surroundings again. (c) (i) – Movement: Amoeba sp. uses pseudopodium to move forwards. – while Paramecium sp. moves by beating its cilia against water. – Reproduction: Amoeba sp. reproduces by binary fission and spore formation. Diagram 2.1 02_Q&A Biology SPM F4.indd 8 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
9 Chapter Form 4 13 Chapter Form 4 2 2 – while Paramecium sp. reproduces by binary fission and conjugation. – Nutrition: Amoeba sp. feeds by engulfing the food particles using its pseudopodia. – while Paramecium sp. sweeps the food particles using the cilia into the oral groove. (ii) – Mitochondrion – Mitochondrion provides energy. – Paramecium sp. requires high consumption of energy for the contractile vacuole to contract during osmoregulation. – Paramecium sp. also needs energy for the movement of cilia to beat against water. 3 (a) (i) Explain the two main systems found in plants. [5 marks] (ii) Multicellular plants need to increase their height to absorb maximum amount of sunlight for photosynthesis whereas roots need to increase its length for water absorption. At the same time, sufficient cellular requirements such as nutrient, water and minerals need to be transported throughout the plant optimally. Using your knowledge, explain how the density of cellular components in certain plant tissues helps the growth of multicellular plants. [7 marks] (b) Several innovations are widely used in the medical industry to treat and cure medical problems. Diagram 3 shows three degrees of burn on the skin of individuals. First degree burn Second degree burn Third degree burn Diagram 3 (i) Based on Diagram 3, explain how the burned tissues in the third degree burn could cause harm to the patient. [3 marks] (ii) Skin grafting consists of the removal of injured tissue, selection of a donor site, removal of healthy skin, covering of the cleaned burned area, harvesting, removal of graft from the donor site, placing and securing the skin graft over the surgically-cleaned wound so it can heal. Suggest and explain the types of cellular components that are needed in high density during the healing process of skin grafting. HOTS Evaluating [5 marks] 02_Q&A Biology SPM F4.indd 9 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
10 Chapter Form 4 13 Chapter Form 4 2 2 Answers 3 (a) (i) – The plant consists oftwo systems, namely shoot and root systems. – The shoot system consists of stem, leaves, shoots, flowers and fruits. – Stem and twigs support the leaves at a vertical position, allowing maximum absorption of light energy during photosynthesis. – The shoot system transports food and water throughout the plants while the flowers produced help in the pollination process. – The root system consists of all root parts such as roots, tubers and rhizomes. – This system absorbs water and mineral salts, besides providing support to the plant. (Any 5) (ii) – Meristematic tissues consist of actively dividing cells. – One of the meristematic tissues is the apical meristems which are located at the tips of shoots and roots. – Mitochondria are needed to produce a lot of energy in the form of ATP (adenosine triphosphate) molecules. – This energy will be used during cell division/mitosis process to increase the number of shoot/root cells. – Therefore, this will increase the length of shoots for maximum absorption of sunlight/the length of roots for maximum intake of water. – In the leaves, mesophyll palisade and spongy mesophyll cells have high density of chloroplasts. – This helps to increase the absorption of energy from sunlight to perform photosynthesis. – Phloem in the stem and leaves contains high density of mitochondria to supply more energy to the companion cells, which helps the translocation process of organic substances to all parts of plants. (Any 7) (b) (i) – The damaged tissues are epithelial tissue and nerve tissue. – Without the epithelial tissue, pathogens can enter/penetrate into the human skin. – If the nerve tissue is damaged, any changes in the surroundings, such as high temperature cannot be detected, which then might harm the patient. – If the blood tissue is damaged, white blood cells will also be destroyed, causing the level of body defence to decrease. (Any 3) (ii) – A lot of mitochondria. – Mitochondria provide energy to enable mitosis/cell division to occur. – The number of cells such as skin cells will be increased. – Skin grafting also needs more lysosomes. – Lysosomes are involved in phagocytosis to break down damaged and worn out cells, and to kill pathogens. 02_Q&A Biology SPM F4.indd 10 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 3 3 11 Theme 1: Fundamentals of Biology Textbook Pages: 44 – 69 Movement of Substances 3 across a Plasma Membrane CHAPTER Paper 1 Objective Questions 3.1 Structure of Plasma Membrane 1 Diagram 1 shows structure X embedded in the plasma membrane. X Diagram 1 What is X? A Carrier protein B Carbohydrate C Phospholipid D Cholesterol Answer: A Protein molecules embedded inside the plasma membrane, which act as carriers, are called carrier proteins. Protein molecules that form a channel or a pore are called channel proteins. 3.2 Concept of Movement of Substances across a Plasma Membrane 2 Distilled water diffuses into a cucumber slice, causing it to increase in mass. Which process causes this condition? A Osmosis B Active transport C Simple diffusion D Facilitated diffusion Answer: A Osmosis involves passive transports similar to diffusion but it involves only the movement of water molecules. Common Error The movement of water molecules is not a simple diffusion process, but it is an osmosis process. 3 The information below shows the characteristics of process Q. • Requires energy in the form of ATP molecules • Results in the accumulation or elimination of molecules or ions in the cell What is process Q? A Osmosis B Active transport C Simple diffusion D Facilitated diffusion Answer: B Only active transport requires energy to change the shape of the carrier protein in order to move the molecules or ions against the concentration gradient. 03_Q&A Biology SPM F4.indd 11 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 3 3 12 4 Diagram 2 shows the movement of substance X across the plasma membrane. Outside Inside Plasma membrane Substance X Diagram 2 What is substance X? A Glucose B Amino acid C Potassium ion D Carbon dioxide Answer: D Lipid soluble molecules (fatty acid, glycerol, fat soluble vitamins and steroid compounds) and small nonpolar molecules (oxygen and carbon dioxide) can diffuse freely through the plasma membrane by simple diffusion. 5 The information below describes the movement of molecules across the plasma membrane. Molecules move from an area of high concentration to an area of low concentration with the aid of carrier proteins and do not require energy. What is the type of movement of these molecules? A Facilitated diffusion B Simple diffusion C Active transport D Osmosis Answer: A Some large molecules need the help of carrier proteins to move across the plasma membrane. 6 Diagram 3 shows a plant cell undergoing process X after being placed in distilled water. Process X Distilled water Vacuole Diagram 3 Which of the following shows the condition of the cell after 20 minutes? A C B D Answer: B The movement of water molecules involves osmosis. Water molecules move into the vacuole, causing it to expand and exert pressure on the cell wall. 7 Diagram 4 shows mineral ions moving into a root hair cell through process Y. Root hair cell Mineral ions Diagram 4 What is process Y? A Osmosis B Simple diffusion C Active transport D Facilitated diffusion 03_Q&A Biology SPM F4.indd 12 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 3 3 13 Answer: C The mineral ions move against the concentration gradient, which is known as active transport. 3.3 Movement of Substances across a Plasma Membrane in Living Organisms 8 What will happen to the red blood cells after being immersed in distilled water? A Haemolysis C Flaccid B Crenation D Turgid Answer: A When red blood cells (erythrocytes) are placed in a hypotonic solution, water will diffuse into the cells by osmosis, causing the cells to swell and eventually burst. 9 Diagram 5 shows a plant cell undergoing process P after being immersed in solution Q for 20 minutes. Before After Diagram 5 What is process P and the type of solution Q that caused the condition shown above? Process P Solution Q A Plasmolysis Hypotonic B Plasmolysis Hypertonic C Deplasmolysis Hypertonic D Deplasmolysis Hypotonic Answer: B When the plasma membrane pulls away from the cell wall, this shows that water has diffused out by osmosis from the cell and into the external solution. This process is called plasmolysis. Hypertonic solution has high solute concentration and low water potential. Exam Tip Compare the conditions of two different solutions or cells when explaining the movement of substances and then select the correct process. 3.4 Movement of Substances across a Plasma Membrane and Its Application in Daily Life 10 A housewife planted a tomato plant in a pot containing soil which was taken from a mangrove swamp. A few days later, the plant wilted. Which of the following can prevent the plant from wilting? A Add water to the soil B Add fertiliser to the soil C Add more mangrove soil into the pot D Dig the soil to improve aeration Answer: A Watering the plant can reduce the salinity of mangrove soil, making it more hypotonic compared to the cell sap of the root cells. Water diffuses into the cell sap of the root cells by osmosis, causing the vacuole to swell and push towards the cell wall. Cells then become turgid. 03_Q&A Biology SPM F4.indd 13 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 3 3 14 Paper 2 Section A – Structured Questions 1 Diagram 1.1 shows a plant cell that has been immersed in a concentrated sucrose solution. P: Cell wall Q: Plasma membrane Diagram 1.1 (a) (i) Label P and Q in Diagram 1.1. [2 marks] (ii) State the differences in characteristics between P and Q. P/Cell wall is fully permeable to any substances while Q/plasma membrane is selectively permeable. [1 mark] (b) Any food can be kept for a long time if water is removed from the food. The absence of water prevents the growth of bacteria. Diagram 1.2 shows a fish soaked in a saturated salt solution. Saturated salt solution Diagram 1.2 Explain how the saturated salt solution can remove water from the fish tissues. The saturated salt solution is hypertonic to the fish tissues. Water molecules from the fish tissues diffuse into the saturated salt solution by osmosis. The fish tissues then lose water and become dehydrated. [3 marks] 03_Q&A Biology SPM F4.indd 14 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 3 3 15 2 Diagram 2.1 shows the movement of sodium ions across the plasma membrane of the root hair cell. Diagram 2.2 shows the movement of oxygen across the plasma membrane of an alveolus. Alveolus Blood capillary Oxygen Sodium ions Root hair cell Diagram 2.1 Diagram 2.2 (a) Name the process of the movement of sodium ions in Diagram 2.1. Active transport [1 mark] (b) (i) Name the process of the movement of oxygen in Diagram 2.2. Simple diffusion/Passive transport [1 mark] (ii) Explain how oxygen moves from the alveolus into the blood capillary. Oxygen diffuses/moves across the plasma membrane from the alveolus into the blood capillary. From higher concentration of oxygen in the alveolus to lower concentration of oxygen in the blood capillary/ down the concentration gradient, partial pressure of oxygen in the alveolus is higher than the partial pressure of oxygen in the blood capillary. [2 marks] (c) Explain two differences of the processes that you have stated in 2(a) and 2(b)(i). HOTS Analysing Active transport Simple diffusion / Passive transport The movement of sodium ions against the concentration gradient The movement of oxygen down the concentration gradient This process requires energy This process does not require energy [2 marks] 03_Q&A Biology SPM F4.indd 15 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 3 3 16 (d) Diagram 2.3 shows the condition of a plasmolysed plant cell after being immersed in a type of solution. Before immersion After immersion Diagram 2.3 (i) What is the solution? Distilled water [1 mark] (ii) Explain the process that has occurred to the plasmolysed plant cell. The solution is hypotonic to the cell sap//Cell sap is hypertonic to the solution. Water molecules from the solution diffuse into the cell sap by osmosis, causing the vacuole to expand and push the cytoplasm towards the cell wall. The cell is said to be deplasmolysed. [2 marks] Paper 2 Section B & C – Essay Questions 3 (a) Diagram 3 shows the condition of two solutions of different concentrations separated by a selectively permeable membrane after 30 minutes. Selectively permeable membrane 5% sucrose solution 15% sucrose solution After 30 minutes Diagram 3 (i) What does selectively permeable membrane mean? [2 marks] (ii) Explain the process that took place in the U-tube after 30 minutes. [6 marks] (b) Substances move across the plasma membrane through passive transport and active transport. (i) One of the types of passive transport is facilitated diffusion. State another two types of passive transport. [2 marks] 03_Q&A Biology SPM F4.indd 16 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 3 3 17 (ii) Explain the similarities and differences between active transport and facilitated diffusion with respect to the movement of molecules across the cell membrane. [10 marks] Answers 3 (a) (i) The membrane that allows free movement of certain molecules to pass through it, but prevent or limit the movement of other molecules. (ii) – 5% sucrose solution is hypotonic to 15% sucrose solution – Selectively permeable membrane only allows water molecules to pass through it. – Water molecules move from 5% sucrose solution to 15% sucrose solution by osmosis. – Water molecules move through the selectively permeable membrane. – After 30 minutes, the level of 15% sucrose solution in the U-tube is higher compared to in the beginning. – Meanwhile, the level of 5% sucrose solution in the U-tube is lower at the end of the experiment compared to in the beginning. (b) (i) Simple diffusion and osmosis (ii) Similarities: – Both need carrier proteins in the cell membrane to bind with molecules/ions/substances. – Both transport certain molecules/ions/substances across the cell membrane. – Both occur in living cells because carrier proteins need/can change shape. – Both occur across the selectively permeable membrane. Differences: Facilitated diffusion Active transport Down the concentration gradient Against the concentration gradient Molecules move from a higher concentration area to a lower concentration area Molecules move from a lower concentration area to a higher concentration area Molecules move in both directions across the cell membrane Molecules move in one direction across the cell membrane Molecules can move through pore proteins and carrier proteins Molecules move through carrier proteins only Does not require energy from the cell Requires energy from the cell Occurs until a dynamic equilibrium is achieved There are accumulation and disposal of molecules or ions 03_Q&A Biology SPM F4.indd 17 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 4 4 18 Paper 1 Objective Questions 4.1 Water 1 Which of the following is an inorganic compound? A Water B Lipid C Protein D Carbohydrate Answer: A Water is an inorganic compound, consisting of hydrogen (H) and oxygen (O) elements. Organic compounds are chemical compounds that contain carbon elements. 2 Water can absorb a lot of heat without increasing its temperature. What will happen if water loses this function? HOTS Analysing A Fluid cannot be transported throughout the body B Proteins in the cytoplasm of cells will be destroyed C Digested food cannot be transported to the cells D Simplest sugar formed unable to dissolve in the blood Answer: B During hot weather, body receives heat from the surroundings, resulting in the rise of the body temperature. The increase in temperature destroys the protein molecules as they are sensitive to heat. 4.2 Carbohydrates 3 Which of the following is a polysaccharide? A Maltose B Fructose C Glycogen D Galactose Answer: C Polysaccharides are sugar polymers such as cellulose, starch and glycogen. Exam Tip Examples of • monosaccharides: glucose, fructose and galactose • disaccharides: maltose, sucrose and lactose • polysaccharides: cellulose, starch and glycogen 4 Which of the following is a nonreducing sugar? A Sucrose B Maltose C Lactose D Glucose Answer: A Sucrose is a non-reducing sugar because it does not reduce the blue copper(II) sulphate solution to brick red precipitate of copper(I) oxide. Theme 1: Fundamentals of Biology Textbook Pages: 70 – 83 4 Chemical Compositions in a Cell CHAPTER 04_Q&A Biology SPM F4.indd 18 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 4 4 19 5 Which of the following are examples of carbohydrates? A Glycogen and wax B Lactose and steroid C Cellulose and glycogen D Dipeptide and cellulose Answer: C Glycogen, lactose and cellulose are categorised as carbohydrates, while wax and steroid are lipids. Dipeptide is a form of protein. 4.3 Proteins 6 What are the products of the hydrolysis of a dipeptide? A Dipeptide + water B Amino acid + water C Polypeptide + water D Amino acid + amino acid Answer: D Hydrolysis of a dipeptide will form two amino acid molecules. 7 Which of the following are the importance of proteins? I Build new cells and repair damaged tissues II Synthesis of ribonucleic acid (RNA) III Form building blocks of bones, muscles, cartilage, skin, hair and nails IV Act as a heat insulator A I and II B I and III C II and IV D III and IV Answer: B Option II refers to nucleic acids while option IV refers to fats. 4.4 Lipids 8 Diagram 1 shows the hydrolysis of a fat molecule that occurs in an animal digestive system. 3H2 O + + P Diagram 1 What is P? A Lipids B Glycerol C Fatty acids D Triglyceride Answer: C Hydrolysis of triglycerides will form one molecule of glycerol and three molecules of fatty acids (P). Exam Tip The reverse process of hydrolysis is known as condensation where a triglyceride molecule is formed. 9 X is a type of lipid that does not contain fatty acids. What is X? A Fat B Wax C Steroid D Phospholipid Answer: C Steroids are the only lipids that do not contain fatty acids. 04_Q&A Biology SPM F4.indd 19 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 4 4 20 4.5 Nucleic Acids 10 Diagram 2 shows the structure of a nucleotide. X Diagram 2 What is X? A Fatty acid B Phosphate group C Nitrogenous base D Deoxyribose sugar Answer: C Each nucleotide consists of a pentose sugar (5-carbon sugar), a nitrogenous base and a phosphate group that are combined together through the condensation process. 11 Diagram 3 shows the molecular structure of a compound. Diagram 3 Which of the following has the above structure? A Deoxyribonucleic acid (DNA) B Chromosome C Nucleotide D Gene Answer: A DNA consists of two polynucleotide chains that are intertwined in opposite directions, forming a double helix structure. 12 Which of the nitrogenous bases are complementary to each other? A Adenine Guanine B Cytosine Guanine C Cytosine Thymine D Adenine Cytosine Answer: B The pairing of the nitrogenous bases is as follows: Adenine (A) pairs with thymine (T) and guanine pairs with cytosine (C). 04_Q&A Biology SPM F4.indd 20 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 4 4 21 Paper 2 Section A – Structured Questions 1 Diagram 1 shows the formation of compound X when two monosaccharide molecules, which are glucose and fructose undergo process P. + Process P Glucose Fructose Compound X + water Diagram 1 (a) (i) Name compound X and process P. Compound X: Sucrose Process P : Condensation [2 marks] (ii) Explain process P. Two monosaccharide molecules combine through condensation to form one unit of disaccharide called sucrose. The condensation process involves the elimination of a water molecule. [2 marks] (b) Carbohydrates are classified into three main types, namely monosaccharides, disaccharides and polysaccharides. Table 1 shows three examples of food which consist of carbohydrates. Food Potatoes Sugar cane HONEY Honey Type of carbohydrate Polysaccharides Disaccharides Monosaccharides Table 1 Write the type of carbohydrate for each food in the spaces provided in Table 1. [2 marks] 04_Q&A Biology SPM F4.indd 21 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 4 4 22 (c) Sucrose is a non-reducing sugar and does not react with Benedict’s solution. Suggest one way to convert sucrose into a reducing sugar. Sucrose needs to be broken down (hydrolysed) into its component monosaccharides, which are glucose and fructose. Both glucose and fructose are reducing sugars. [2 marks] Common Error All disaccharides are reducing sugars, except sucrose. Maltose and lactose are reducing sugars. 2 Diagram 2.1 shows a type of nucleic acid found in the nucleus of a cell. Y is a part of the two polymer chains comprising of nucleotide monomers. A T C G G C Chromosome Nucleus Cell Gene Y Diagram 2.1 (a) (i) State five elements that form nucleic acid. Carbon, hydrogen, oxygen, nitrogen and phosphorus [2 marks] (ii) Based on Y, state the type of nucleic acid. Deoxyribonucleic acid (DNA) [1 mark] Common Error Candidates often confuse DNA and RNA in terms of their structure. DNA consists of two polynucleotide chains that are intertwined in opposite directions and form a double helix while RNA is a single polynucleotide chain which is shorter than DNA. 04_Q&A Biology SPM F4.indd 22 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 4 4 23 (b) Diagram 2.2 shows a nucleotide monomer which is the building block of the nucleic acid. S T Phosphate group Diagram 2.2 (i) Identify the structures of the nucleotide. Nitrogenous base: T Deoxyribose sugar: S [2 marks] (ii) There are four types of T, namely adenine, cytosine, guanine and thymine. Each T will form a hydrogen bond with its complementary pair. Match the correct pair of T. Adenine (A) Guanine (G) Cytosine (C) Thymine (T) [2 marks] (c) Explain a problem that may arise if a cell does not have nucleic acid. HOTS Evaluating Hereditary material in the form of DNA which determines certain traits cannot be passed down to the next generation/the synthesis of protein will not occur (Any 1) [1 mark] Paper 2 Section B & C – Essay Questions 3 (a) Organic compounds are chemical compounds which contain the element carbon. (i) Give four examples of organic compounds. [4 marks] (ii) Based on your answer in 3(a)(i), explain the importance of each organic compound in cells. [10 marks] (b) Fats can be categorised into saturated fatty acids and unsaturated fatty acids. Saturated fatty acids are also known as saturated fats while unsaturated fatty acids are called unsaturated fats. Compare and contrast saturated and unsaturated fats. HOTS Analysing [6 marks] 04_Q&A Biology SPM F4.indd 23 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 4 4 24 Answers 3 (a) (i) Carbohydrates, proteins, lipids, nucleic acids (ii) Importance of carbohydrates: – Glucose is a source of energy. – Plant cells store energy in the form of starch and animal cells store energy in the form of glycogen. – Cellulose in the plant cell wall provides structural support, protecting and maintaining the shape of plant cells. Importance of proteins: – Proteins are used to build new cells and repair damaged tissues. – Proteins are used to synthesise enzymes, hormones, antibodies and haemoglobin. – Proteins form building blocks such as keratin in the skin, collagen in bones and myosin in muscle tissues. Importance of lipids: – Fats function as reserved energy and heat insulator for animals, and as a liner to protect internal organs. – Wax is an important component in cuticles and sebum secreted by human skin. – Glycolipids ensure the stability of the plasma membrane and help in the cell identification process. – Cholesterol is important in steroid hormone synthesis. Importance of nucleic acids: – Nucleic acids act as a carrier of hereditary material and a determinant of traits in organisms. – DNA contains genetic codes for protein synthesis. (Any 10) (b) Similarities: – Both contain carbon, hydrogen and oxygen elements – Both contain glycerol and fatty acids – Both are nonpolar molecules Differences: Saturated fats Unsaturated fats Exist in solid form at room temperature Exist in liquid form at room temperature Fatty acids only have single bonds between carbon atoms Fatty acids have at least one double bond between carbon atoms Do not form chemical bonds with additional hydrogen atoms because all bonds between carbon atoms are saturated Double bonds can still receive one or more additional hydrogen atoms because carbon atoms are unsaturated. Found mostly in animals Found mostly in vegetables Contain more cholesterol Contain less cholesterol (Any 6) 04_Q&A Biology SPM F4.indd 24 11/01/2023 1:41 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 4 13 Chapter Form 4 5 5 25 Theme 1: Fundamentals of Biology Textbook Pages: 84 – 97 5 Metabolism and Enzymes CHAPTER Paper 1 Objective Questions 5.1 Metabolism 1 Which of the following processes is related to the breakdown of complex food particles into simple substances? A Metabolism B Catabolism C Anabolism D Growth Answer: B Catabolism refers to the process of breaking down complex substances into simpler ones to release energy for organisms to use. Common Error Candidates are always confused about the terms used to explain the process of breaking down and synthesising molecules. 2 Diagram 1 shows a chemical reaction which occurs inside a cell. Hydrolysis X Y Diagram 1 What do X and Y represent? X Y A Product Substrate B Substrate Product C Enzyme Substrate D Anabolism Catabolism Answer: B The substrate, X was broken down into simpler (smaller) substances, which are referred to as the product that is obtained from the completion of the reaction. 5.2 Enzyme 3 Diagram 2 shows the fusion of X with Y during the production of extracellular enzyme. X Y Diagram 2 What will happen if X does not fuse with Y? A Proteins will not be synthesised B Proteins will not be modified C Proteins will be hydrolysed D Proteins will be denatured 05_Q&A Biology SPM F4.indd 25 11/01/2023 1:42 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 1 110 Theme 1: Physiology of Flowering Plants Textbook Pages: 2 – 25 Organisation of Plant Tissues and 1 Growth CHAPTER Paper 1 Objective Questions 1.1 Organisation of Plant Tissues 1 Which of the following form vascular tissues? I Meristematic tissues II Parenchyma tissues III Phloem tissues IV Xylem tissues A I and II C II and IV B I and III D III and IV Answer: D Vascular tissues transport substances to all parts of the plants. Exam Tip Tissue organisation in plants is simplified as below: Apical meristems Lateral meristems Epidermal tissues Ground tissues Vascular tissues Plant tissue Meristematic tissues Permanent tissues Parenchyma tissues Collenchyma tissues Sclerenchyma tissues Xylem tissues Phloem tissues 1.2 Meristematic Tissues and Growth 2 Which of the following is categorised as meristematic tissues? A Phloem B Epidermis of the leaf C Cambium in the stem D Parenchyma tissues Answer: C There are two types of meristematic tissues, namely apical meristem and lateral meristem. Lateral meristem consists of vascular cambium and cork cambium which can be found in the stem of a plant. 3 Which plant tissues undergo cell division actively? A Sclerenchyma tissues B Meristematic tissues C Parenchyma tissues D Xylem tissues Answer: B Meristematic tissues are undifferentiated living tissues which are responsible for plant growth. Hence, cell division occurs actively in this type of tissues. Exam Tip Meristematic tissues undergo cell division and differentiation to form permanent tissues. 01_Q&A Biology SPM F5.indd 110 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 1 111 4 Diagram 1 shows the longitudinal section of an eudicot shoot tip. Zone III Zone II Zone I Diagram 1 Which of the following is correct about Zone II and Zone III? Zone II Zone III A Large nucleus Small nucleus B Large vacuole Small vacuole C Presence of xylem Absence of xylem D Actively dividing cells Cells have reached maximum size Answer: B Zone III is the zone of cell division. There are many small vacuoles present in the cells in this zone. Zone II is the zone of cell elongation. Vacuolation occurs in this zone, whereby the small vacuoles fuse to form large vacuoles. Zone I is the zone of cell differentiation where permanent tissues such as xylem and phloem are formed. 5 Diagram 2 shows the crosssection of an eudicot stem. A B C D Diagram 2 Which labelled part, A, B, C or D, is formed during secondary growth and provides mechanical support to the plant? Answer: C Lignin on the walls of the xylem provides mechanical support to plants. Secondary growth gives rise to secondary xylem which is the part labelled C. 6 Diagram 3 shows the crosssection of a plant stem that has undergone secondary growth. I II Cambium ring Diagram 3 What are represented by parts I and II? I II A Secondary xylem Primary phloem B Secondary xylem Secondary phloem C Secondary phloem Primary xylem D Secondary phloem Secondary xylem Answer: A The cells in the cambium ring divide inwards to form secondary xylem, which is labelled as I. The primary phloem is pushed towards the epidermis, which is labelled as II. 01_Q&A Biology SPM F5.indd 111 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 1 112 Common Error Candidates are often unsure about the parts that undergo primary and secondary growths and therefore identifying the new cells formed at the tips of the vascular bundle wrongly. Exam Tip Vascular cambium divides to form new cells, hence can promote growth. The new cells in the inner side of the cambium ring are modified into secondary xylem and the new cells in the outer side are modified into secondary phloem. 7 Diagram 4 shows the crosssection of two stems from plants A and B. A B Diagram 4 What are the differences between the stems of plants A and B? A B A Growth occurs longitudinally Growth occurs radially B Has woody tissues Does not have woody tissues C Presence of cork cambium Absence of cork cambium D Bark is thick Bark is thin Answer: A The stem of plant A shows its vascular bundles are divided into two parts, which shows that it undergoes primary growth. The stem of plant B has cambium ring and its vascular bundles are divided into four parts, which proves that it undergoes secondary growth. 8 Diagram 5 shows the crosssection of an eudicot stem undergoing secondary growth. IV III I II Diagram 5 Which labelled parts, I, II, III or IV, are involved in secondary growth? A I and II B I and III C II and IV D III and IV Answer: C Secondary growth occurs at the lateral meristems, which consist of the vascular cambium (part labelled II) and the cork cambium (part labelled IV). Common Error Candidates often assume that all types of plant undergo secondary growth. 01_Q&A Biology SPM F5.indd 112 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 1 113 9 The age of plants in the temperate climate regions can be determined based on the annual growth rings in the stem as shown in Diagram 6. Lighter ring Darker ring Diagram 6 Why do some rings appear lighter? A Lighter colour of rings indicates the formation of secondary xylem with thicker walls B Lighter colour of rings indicates the formation of vascular cambium C Lighter colour of rings indicates the formation of secondary xylem which is bigger and has thinner walls D Lighter colour of rings indicates the formation of secondary xylem instead of the formation of secondary phloem Answer: C Lighter rings show that the secondary xylem formed is bigger and has thinner walls during the spring season when the water supply and sunlight are adequate. However, in summer, the secondary xylem formed is smaller and has thicker walls. The xylem tissues formed during summer are darker in colour. 10 Which of the following are the economic importance of plants with secondary growth? I To make perfume II To make furniture III As a source of protein IV To make high quality jewellery A I and II B I and III C II and IV D III and IV Answer: A The wood and bark of some plants can produce resin and oil that can be commercialised as perfume. Secondary growth produces strong and hard wood which are suitable to manufacture furniture. 1.3 Growth Curves 11 Why do grasses live longer than paddy plants? A Grasses have flowers but paddy plants do not have flowers B Grasses undergo secondary growth but paddy plants only undergo primary growth C Grasses can live anywhere but paddy plants can only grow in certain areas D Grasses are perennial plants but paddy plants are annual plants Answer: D Paddy plants are annual plants. This explains their short lifespan, usually within a year. Grasses are perennial plants. Therefore, they have long lifespan and can live for more than two years. 01_Q&A Biology SPM F5.indd 113 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 1 114 Paper 2 Section A – Structured Questions 1 Diagram 1 shows two types of permanent tissues, P and Q, found in a plant. Q X Y P Diagram 1 (a) Identify the types of permanent tissues labelled P and Q. P: Epidermal tissues Q: Vascular tissues [2 marks] Guard cells and root hair cells are modified epidermal cells. Exam Tip (b) Tissue Q consists of structures X and Y. (i) What is structure X? Xylem [1 mark] (ii) Explain how structure X is adapted to its function. Structure X (or xylem) is made up of dead cells without cytoplasm and nucleus and forms a long continuous hollow tube. This feature allows structure X (or xylem) to transport water and mineral salts from the roots to all parts of the plant. [3 marks] (c) Plant tissues are divided into permanent and meristematic tissues. State one difference between permanent and meristematic tissues. HOTS Analysing Permanent tissues are mature tissues that have undergone differentiation while meristematic tissues are actively dividing tissues. [1 mark] 01_Q&A Biology SPM F5.indd 114 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 1 115 2 Diagram 2.1 shows the longitudinal section of an eudicot root tip. A, B, C and D are the zones of cell growth. A: Matured tissues B: Zone of cell differentiation C: Zone of cell elongation D Diagram 2.1 (a) Name zones A, B and C in Diagram 2.1. [3 marks] (b) Based on Diagram 2.1, (i) identify which zone comprises actively dividing cells through mitosis. Zone D [1 mark] (ii) explain two differences between zones C and D. HOTS Analysing Zone C is the zone of cell elongation while zone D is the zone of cell division. Cells in zone C increase in size through vacuolation while cells in zone D increase in number through mitotic division. Each cell in zone C has one large vacuole while each cell in zone D has numerous small vacuoles. [2 marks] (c) Diagram 2.2 shows two types of organisms, a bird and a cherry tree. Diagram 2.2 01_Q&A Biology SPM F5.indd 115 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 1 116 What is the difference between the growth of bird and cherry tree? HOTS Analysing The growth in bird occurs throughout its body while the growth in cherry tree occurs in parts that contain meristematic tissues. [1 mark] Common Error It is a common misconception that growth takes place in all parts of plants and animals. Plants show continuous growth throughout their life but the process only occurs at the apical and lateral meristematic tissues. Paper 2 Section B & C – Essay Questions 3 (a) Diagram 3.1 shows the cross-section of an eudicot root. Cortex X Y Z Diagram 3.1 (i) Explain the function of structures X and Y. [2 marks] (ii) Structure Z separates structures X and Y during secondary growth. Explain the process that takes place in structure Z which enables the root to increase in size during secondary growth. [4 marks] (iii) Predict what will happen to the plant if structure Z is not formed. HOTS Analysing [4 marks] (b) Diagram 3.2 shows an eudicot root that has undergone secondary growth. Cortex Secondary phloem Vascular cambium Secondary xylem Primary xylem Diagram 3.2 01_Q&A Biology SPM F5.indd 116 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 1 117 Based on Diagrams 3.1 and 3.2, compare and contrast the primary growth and secondary growth in roots of eudicot plants. HOTS Analysing [10 marks] Answers 3 (a) (i) – X (or phloem) transports organic substances / products of photosynthesis away from the leaves to other parts of the plant. – Y (or xylem) transports water and mineral ions from the roots to the leaves. (ii) – Structure Z is the vascular cambium. – Vascular cambium cells divide actively by mitosis to form a complete ring called cambium ring. – The cells in the cambium ring which divide inwards form the secondary xylem. – The cells in the cambium ring which divide outwards form the secondary phloem. (iii) – Structure Z consists of actively dividing cells. – The plant roots will not be able to carry out secondary growth. – Secondary phloem and secondary xylem cannot be formed. There will not be an increase in the circumference / diameter of the roots. – The plant is unable to sustain stability. (b) Similarities: – Both growths increase the size of roots permanently. – Both growths involve cell division at meristematic tissues. – Both growths involve elongation / growth of cells. – Both growths occur in woody plants. Differences: Primary growth Secondary growth Occurs in the apical meristematic tissues Occurs in the lateral meristematic tissues Occurs in younger roots Occurs at mature roots Results in longitudinal growth Results in radial growth Increase in the length of roots Increase in the thickness or circumference of roots Only has primary xylem Has both primary and secondary xylems Only has primary phloem Has both primary and secondary phloems 01_Q&A Biology SPM F5.indd 117 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 1 118 4 (a) Diagram 4.1 shows the growth curve of onion plants. 0 10 3 20 0 40 50 0.4 0.8 1.2 1.6 2.0 2.4 Time (Week) Dry mass (g) Diagram 4.1 (i) Explain the shape of the graph shown in Diagram 4.1. [6 marks] (ii) Do all plants have the same growth curve as shown in Diagram 4.1? Justify your answer. [5 marks] (b) Diagram 4.2 shows the cross-section of an eudicot stem structure which undergoes secondary growth. Cork Cork cambrium Cortex Phloem (dark areas) Summer wood Spring wood Primary xylem Secondary xylem Pith Wood (3 annual rings) Vascular cambrium Bark Diagram 4.2 (i) Discuss the economic importance of plants that undergo secondary growth. [4 marks] (ii) Mr David planted a few pine trees in his backyard. A pine tree can grow more than 3 metres in height. Mr David plans to control the height of the trees so that they are not too tall. Based on your knowledge in biology, suggest a way on how he can do this. Justify your answer. HOTS Analysing [5 marks] 01_Q&A Biology SPM F5.indd 118 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 1 119 Answers 4 (a) (i) – Two combined sigmoid curves shows that the onion plants are biennial plants. – The first sigmoid curve represents the first growth season. – During this season, plants produce leaves so that photosynthesis can take place. – Food produced during photosynthesis is stored in the tubers. – Second sigmoid curve is the second growth season. – The food stored is used to produce reproductive structures such as flowers and seeds (reproduction). (ii) – No, not all plants have the same shape of growth curve as the onion plants. – Besides biennial plants, there are another two types of plants, which are annual and perennial plants. – Graph of annual plants only have one sigmoid curve. – Annual plants have only one life cycle within a year. – Perennial plants have a series of small combined sigmoid curves. – Each curve represents the growth within a year. – Perennial plants have longer lifespan and can live for more than two years. (Any 5) (b) (i) – Secondary growth makes wood strong and hard which are suitable to construct buildings, furniture and supporting structures. – The presence of annual rings in the stems makes the furniture look visually appealing. – Resin and oil from wood or bark can be commercialised as varnish, adhesive substance, perfume and medicine. – Fruits from woody plants can be commercialised to generate income. (ii) – The height of the trees can be controlled by trimming the tips of plant shoots which consist of apical meristematic tissues. Trimming should be carried out periodically. – Apical meristematic tissues allow trees to grow longitudinally to increase their height. – However, when the apical meristematic tissues in the shoots are removed, this allows lateral meristematic tissues in the shoots to grow actively. – Lateral meristematic tissues contributes to the increase in diameter of the trees. – This method keeps the height of the trees under control. 01_Q&A Biology SPM F5.indd 119 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 2 120 Theme 1: Physiology of Flowering Plants Textbook Pages: 26 – 57 2 Leaf Structure and Function CHAPTER Paper 1 Objective Questions 2.1 Structure of a Leaf 1 Diagram 1 shows the external structure of a leaf. P Q Diagram 1 What are structures P and Q? Structure P Structure Q A Lamina Petiole B Lamina Leaf C Petiole Lamina D Cuticle Leaf Answer: C The flat, thin and green surface is the lamina while the stalk that attaches the leaf to the stem is a petiole. 2 Diagram 2 shows the arrangement of leaves of a plant which grow without overlapping each other. Diagram 2 Why are the leaves arranged this way? A To provide space between the leaves B To receive maximum amount of sunlight C To receive maximum amount of carbon dioxide D For the leaves to receive equal amount of nutrients Answer: B This arrangement is called leaf mosaic, which allows the leaves to be spread out in an optimal position for photosynthesis. 3 Diagram 3 shows the crosssection of a leaf lamina. Upper epidermis Cells X Diagram 3 Why are cells X arranged vertically and closely-packed? A To store a lot of water B To release a lot of oxygen C To receive maximum light energy D To absorb a lot of carbon dioxide 02_Q&A Biology SPM F5.indd 120 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 2 121 Answer: C Palisade mesophyll cells (cells X) are located right below the upper epidermis and they contain a lot of chloroplasts. This arrangement enables more cells to fit into a smaller space. 4 What is the importance of cuticle in the lower and upper epidermis of a leaf? A To support the leaf B To transport water easily C To prevent excessive loss of water D To ensure gaseous exchange occurs efficiently Answer: C Cuticle is a waxy and waterproof layer found on the surface of the leaf to reduce excessive loss of water through evaporation. 5 Diagram 4 shows cells P, Q, R and S found in the leaf lamina. Q R S P Diagram 4 Which cells, P, Q, R or S, transport water and organic substances? Transport water Transport organic substances A R S B Q R C S R D S P Answer: A R is xylem, which is shown by the thick, lignified cell wall in the diagram. Xylem transports water and mineral ions from the roots to other parts of the plant. Phloem is located below the xylem and it transports organic substances produced during photosynthesis. Exam Tip Xylem and phloem are found in the vascular bundle. Xylem faces the upper surface while phloem is found closer to the lower surface. 2.2 Main Organ for Gaseous Exchange 6 What causes the opening of the stoma? A Water diffusing out from the guard cells B Potassium ions moving out from the guard cells C High sucrose concentration in the guard cells D No difference in the water potential in the guard cells Answer: C Concentration of sucrose in the guard cells is high, causing water potential to decrease. Water diffuses into the guard cells through osmosis until they become turgid and curve outwards, thus opening up the stoma. 02_Q&A Biology SPM F5.indd 121 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 2 122 7 Diagram 5 shows a model which works similarly as the mechanism of stomatal opening and closing. Y-tube Balloons Thread Space Diagram 5 Which structures in a plant are represented by the balloons and the space between the balloons? Balloons Space between the balloons A Stoma Lower epidermis B Upper epidermis Stoma C Guard cells Stoma D Stoma Mesophyll cells Answer: C The balloons represent the guard cells which control the opening and closing of the stoma. The space between the balloons represents the stoma. 8 A student found out that the distribution of stomata is higher on the lower epidermis than the upper epidermis of a balsam leaf. Why is the stomatal distribution different in the balsam leaf? A More water is transported to the lower epidermis B The lower epidermis is not exposed to direct sunlight C Less gaseous exchange occurs in the lower epidermis D More potassium ions accumulate in the lower epidermis Answer: B More stomata are found in the lower epidermis to prevent excessive loss of water through evaporation during carbon dioxide uptake. 2.3 Main Organ for Transpiration 9 Why is transpiration necessary in plants? I To prevent absorption of mineral salts from soil water II To ensure organic substances are transported throughout the entire plant III To produce a pulling force to move water upwards and continuously in the xylem vessel IV To provide a cooling effect to plants A I and II B I and III C II and IV D III and IV Answer: D Water absorbs heat energy and releases it into the atmosphere during evaporation. This process creates a pulling force which attracts water molecules upwards inside the xylem vessels in order to replace the water molecules that are lost through evaporation. 02_Q&A Biology SPM F5.indd 122 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 2 123 10 The lower the relative air humidity in the surrounding atmosphere, the faster the water vapour escapes from the stomata. Which graph represents this relationship? A Rate of transpiration Relative air humidity Rate of transpiration Relative air humidity Rate of transpiration Relative air humidity Rate of transpiration Relative air humidity B Rate of transpiration Relative air humidity Rate of transpiration Relative air humidity Rate of transpiration Relative air humidity Rate of transpiration Relative air humidity Rate of transpiration C Relative air humidity Rate of transpiration Relative air humidity Rate of transpiration Relative air humidity Rate of transpiration Relative air humidity D Rate of transpiration Relative air humidity Rate of transpiration Relative air humidity Rate of transpiration Relative air humidity Rate of transpiration Relative air humidity Answer: D When the relative air humidity is high, air is saturated with water vapour. This reduces the evaporation of water from leaves through the stomata. Thus, reducing the rate of transpiration. 11 Which environmental factors affect the rate of transpiration in plants? I Light intensity II Temperature III Air density IV Soil pH A I and II C II and IV B I and III D III and IV Answer: A Air movement and relative air humidity are two other factors that affect the rate of transpiration. 12 Diagram 6 shows the set-up of apparatus for a potometer. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Leafy twig Cork Capillary tube Water Screw clip Air bubble as marker Diagram 6 Why must the leafy twig be cut underwater? A To transport water efficiently B To allow water to escape faster to the surroundings C To prevent air bubbles from entering the xylem vessel D To increase the movement of water molecules Answer: C Cutting the twig underwater ensures continuous flow of water in the xylem vessel. Air bubbles that enter the xylem vessel interrupt the movement of water from the tip of the twig to the leaves. 02_Q&A Biology SPM F5.indd 123 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 2 124 2.4 Main Organ for Photosynthesis 13 Which of the following is needed during the light-dependent reaction in photosynthesis? A Carbon dioxide B Glucose C Oxygen D Water Answer: D Water undergoes photolysis, where water molecules are broken down to form hydrogen ions and hydroxide ions in the presence of light energy and chlorophyll. 14 Diagram 7 shows the structure of a component found in plant cells. A C B D Diagram 7 Which part, A, B, C or D, involved in the fixation of carbon dioxide gases to form 6-carbon organic compounds? Answer: D The fixation process of carbon dioxide occurs in the stroma which is labelled as D. 2.5 Compensation Point 15 Diagram 8 shows the compensation point of a plant. Compensation point Uptake of CO2 decreases Uptake of CO2 increases Low Light intensity increases High Diagram 8 Which of the following best explains the compensation point in a plant? A The rate of photosynthesis is higher than the rate of respiration in the plant B Carbon dioxide released by the plant is more than the carbon dioxide absorbed C The plant releases excess oxygen to the surroundings D No net gain or net loss of glucose in the plant Answer: D At the compensation point, the rate of photosynthesis is equal to the rate of respiration for a given light intensity. Glucose produced during photosynthesis is used by the plant for respiration. Exam Tip The rate of photosynthesis in plants must exceed the rate of respiration in order for the plant to produce more glucose for its development and growth. 02_Q&A Biology SPM F5.indd 124 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 2 125 Paper 2 Section A – Structured Questions 1 Variegated leaf in Diagram 1.1 was taken from a plant that has been exposed to sunlight. Green White Diagram 1.1 A student tested the presence of starch on the leaf and found that: • starch is present in the green area • starch is absent in the white area (a) Based on the information in Diagram 1.1, (i) state the process that occured in the green area. Photosynthesis [1 mark] (ii) write the word equation for the process stated in 1(a)(i). light energy Water + carbon dioxide ⎯⎯⎯⎯⎯⎯→ glucose + oxygen + water chlorophyll [1 mark] (b) Explain why green and white areas in Diagram 1.1 show different results. The green area contains chlorophylls which absorb sunlight for photosynthesis to produce glucose (starch). The white area does not contain chlorophyll and thus, photosynthesis cannot occur / cannot produce glucose. [2 marks] (c) The graph in Diagram 1.2 shows the relationship between light intensity and the rate of photosynthesis. Light intensity Rate of photosynthesis Diagram 1.2 02_Q&A Biology SPM F5.indd 125 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 2 126 Discuss the effects of different light intensities on the rate of photosynthesis. Rate of photosynthesis increases as the light intensity increases until a point where the rate of photosynthesis becomes constant because carbon dioxide concentration / temperature becomes a limiting factor. [2 marks] 2 Diagram 2 shows the structure of an enlarged stoma (plural: stomata) on the leaf surface. Lower epidermis Stoma Guard cell Diagram 2 (a) Land plants have more stomata in the lower epidermis of their leaves compared to the upper epidermis. Why? The lower epidermis is not directly exposed to sunlight. This prevents excessive loss of water. [2 marks] (b) The opening of stomata is controlled by the water potential in the guard cells. (i) Explain how sucrose concentration in the guard cells affect the water potential. During the day / in the presence of light, photosynthesis takes place in the guard cells. Concentration of sucrose in the guard cells increases, decreasing water potential. [2 marks] Guard cells contain chloroplasts. Therefore, these cells can carry out photosynthesis. Exam Tip (ii) The stomata open according to the changes of water potential in the guard cells. Explain this statement. Water potential decreases when the amount of solutes in the guard cells are high. Water diffuses from epidermal cells into the guard cells via osmosis. Guard cells become turgid and curve outwards to open the stomata. [3 marks] 02_Q&A Biology SPM F5.indd 126 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 2 127 3 (a) Diagram 3.1 shows the distribution of leaves in a plant which prevents them from overlapping with one another. (i) Name the type of leaf arrangement in the plant. Leaf mosaic [1 mark] (ii) Based on your knowledge, identify the type of plant (monocots or eudicots) shown in Diagram 3.1. Justify your answer. Eudicots. Eudicots have leaves with network-like veins. [2 marks] (b) Diagram 3.2 shows the cross-section of a leaf lamina. A Epidermal cell Epidermal cell X Y Stoma Xylem Phloem Vascular tissue Diagram 3.2 (i) State three differences between the cells in structures X and Y. Structure X consists of palisade mesophyll cells while structure Y consists of spongy mesophyll cells. Cells in structure X contain a lot of chloroplasts while cells in structure Y contain less chloroplasts. The cells in structure X are closely arranged while the cells in structure Y are loosely arranged. [3 marks] (ii) Layer A in Diagram 3.2 is waxy, waterproof, thin, transparent and it is located on the leaf surface. Identify layer A and state its functions based on its characteristics. Layer A is cuticle. It is waxy and waterproof to prevent excessive loss of water through evaporation. It is thin and transparent to allow light to pass through the cuticle for photosynthesis. [3 marks] Diagram 3.1 02_Q&A Biology SPM F5.indd 127 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
Chapter Form 5 2 128 Paper 2 Section B & C – Essay Questions 4 (a) Diagram 4 shows how light intensity affects the uptake of carbon dioxide in plants. Uptake of carbon dioxide Dark Bright light Increasing light intensity P Diagram 4 (i) Explain point P, in terms of the release of oxygen into the atmosphere. [3 marks] (ii) In a tropical rainforest, different plants occupy different layers. Ferns are usually found on the rainforest floor, whereas teak trees grow higher than ferns. Compare how ferns and teak trees achieve point P. [3 marks] (b) (i) Plants carry out photosynthesis and respiration. Compare and contrast both processes. HOTS Analysing [8 marks] (ii) Should the rate of photosynthesis be equal to the rate of respiration in plants? Discuss this statement. HOTS Analysing [6 marks] Answers 4 (a) (i) – Point P is the compensation point. – It is the level of light intensity when the rate of photosynthesis is equal to the rate of respiration. – No oxygen is released into the atmosphere during photosynthesis as it has been used for respiration. (ii) – Teak trees achieve point P before ferns. – This is because teak trees are taller than ferns. – Hence, teak trees receive more light than ferns. (b) (i) Similarities: – Both processes take place in living organisms. – Both processes involve the uptake and release of gases. 02_Q&A Biology SPM F5.indd 128 11/01/2023 1:46 PM PENERBIT ILMU BAKTI SDN. BHD.
SPM Model Test 208 SPM Model Test 1 In the field of biology, the study and observation of structures of an organism is done based on planes, sections and directions. What is a frontal plane? A Divides the body into frontal and rear parts B Divides the body into upper and lower parts C Divides the body into left and right parts D Divides the body into symmetrical planes 2 The following statements are about the characteristics of component A in an animal cell. • Small cylindrical structures that exist in pairs • Made up of complex microtubules arrangement What is component A? A Centriole B Ribosome C Lysosome D Mitochondrion 3 Diagram 1 shows a plant cell. DC B A Diagram 1 Which component, A, B, C or D contains green pigments that can absorb sunlight and convert it to chemical energy during photosynthesis? 4 Diagram 2 shows the structure of a plasma membrane. P Diagram 2 What will happen if molecule P is absent? A The phospholipid bilayer will become stronger B No changes to the plasma membrane C There are more channel proteins present within the plasma membrane D The phospholipid bilayer will be more permeable to watersoluble substances Paper 1 Time: 1 hour 15 minutes [40 marks] Instructions: Answer all questions. Each question is followed with four options, A, B, C and D. Choose one answer only. 14_Q&A Biology SPM F5.indd 208 11/01/2023 1:49 PM PENERBIT ILMU BAKTI SDN. BHD.
SPM Model Test 209 5 Diagram 3 shows the absorption of mineral ions by a plant root hair cell. Epidermis of root Soil particle Root hair cell Mineral ions Diagram 3 Which of the following statements is correct about the movement shown in Diagram 3? A The movement requires energy B Mineral ions move with the help of channel protein C The process is known as passive transport D Mineral ions move following the concentration gradient 6 What are the functions of proteins in cells? I Build new cells II Repair damaged tissues III Source of energy in a cell IV Carrier of hereditary information A I and II C II and IV B I and III D III and IV 7 Food X is not good for health because it raises the level of lowdensity lipoprotein (LDL) in the blood. Which of the following is food X? A B C D 8 Which of the following is the correct sequence of organelles in the production of extracellular enzymes? A Ribosomes → Tranpsort vesicles → Golgi apparatus → Secretory vesicles B Golgi apparatus → Tranpsort vesicles → Ribosomes → Secretory vesicles C Ribosomes → Secretory vesicles → Golgi apparatus → Transport vesicles D Transport vesicles → Ribosomes → Golgi apparatus → Secretory vesicles 9 Diagram 4 shows the condition of a shirt before and after washing using a type of washing powder. Oil stain Egg stain Blood stain Vegetable puree Before washing After washing Diagram 4 Which of the following enzymes is most probably used in the washing powder? A Protease C Lipase B Amylase D Cellulase 14_Q&A Biology SPM F5.indd 209 11/01/2023 1:49 PM PENERBIT ILMU BAKTI SDN. BHD.