Form 4 Biology Notes on Cell Structure & Function PDF

2020 SPM EXAM TIPS NOTES – BIOLOGY FORM 4 CHAPTER 2 CELL STRUCTURE AND CELL ORGANISATION Cellular components Structure Function a. Plasma membrane  two lipid layer  controls the movement of Animal &...

2020 SPM EXAM TIPS NOTES – BIOLOGY FORM 4 CHAPTER 2 CELL STRUCTURE AND CELL ORGANISATION Cellular components Structure Function a. Plasma membrane  two lipid layer  controls the movement of Animal & plant cell  Thin membrane around the substances in & out of the cell (contain protein & lipid) cytoplasm, elastic and semi-  tough & rigid – to maintain permeable the shape of cell b. Cell wall  formed from cellular  give support & protection for *(plant cell only) secretion, mainly cellulose the cell  Rigid & not elastic  maintain fixed shape  Fully permeable  If a cell is put in the solution containing cellulase, cellulase will hydrolyse cellulose in the cell wall to produce glucose. c. Cytoplasm  jelly-like medium  give shape (animal & plant cell)  provide medium for metabolic reaction d. Nucleus  containing a nucleolus,  control activities of cell (animal & plant cell) chromosomes, and  contain genetic material / nucleoplasm information in the form of chromosomes (determine : - type of protein synthesis by ribosome/characteristics inherited by offspring) e. Chromosomes  contain genetic material  Carry genetic information of (DNA) the cell in its DNA  determine the characteristics 1 WWW.ANDREWCHOO.EDU.MY  inherited by the offspring f. Rough endoplasmic  flat sealed sac  synthesis & transport protein reticulum (RER)  ribosome attached by ribosome to Golgi  High density at stomach apparatus  Large amount in gut cells g. Smooth endoplasmic  tubular rather than flat  site of lipid synthesis reticulum (SER)  does not have ribosome  synthesis & transport lipids  large amount in liver & testes h. Golgi apparatus  Consist of a stack flattened  modify protein that receive membranous sacs with small from ER & package into (animal & plant cell) vesicles budding off the specific secretion such as edge. enzyme, hormone, pigment  Packs modified products into vesicle & transport to plasma membrane to be secreted. i. Mitochondrion  spherical or oblong shape  site to generate energy for the  smooth outer membrane and activities of the cell (site of the ‘power house’ folded inner membrane cellular respiration) (cristae)  Energy in the form of ATP  possess by guard cell, spongy (adenosine triphosphate) mesophyll cell and palisade molecules. mesophyll cell  Cristae – to increase surface  contain respiratory enzyme area for the efficiency of cellular respiration j. Vacuoles  single membrane  Storage of food substances & (plant cell only)  Filled with cell sap which waste products exert pressure outwards on  Turgor pressure in cellulose the cell pushes the cell contents &  contain cellulose plasma membrane against  Contain cell sap (consist cell wall of amino acid, water ,  provided supports to sugar solution & mineral herbaceous plants salt) k. Chloroplast  Contain grana  Contain chlorophyll which (plant cell only)  Fill with stoma (jelly like absorb light energy for matrix) photosynthesis  Biconvex discs  Give green colour for the leaf  double membrane  Contain pigment, chlorophyll and enzymes l. Granum  inside the chloroplast  Contain chlorophyll which carries out light reaction of photosynthesis  photolysis of water occur 2 WWW.ANDREWCHOO.EDU.MY m. Centrioles  Cylinder organelle  cell division (mitosis & (animal cell only) meiosis)  Produce spindle fibres n. Ribosome  freely in the cytoplasm  Site of protein synthesis  Attached to the RER o. Lysosome  Sac-like organelle  Generally break down food & foreign materials  Digest protein, lipids & carbohydrates  Eliminate worn out organelles (mitochondria)  Contain hydrolytic enzyme which will hydrolse damage organelles, absorb useful nutrients, elimate the waste. MOVEMENT OF SUBSTANCES ACROSS THE PLASMA FORM 4 CHAPTER 3 MEMBRANE 1. Passive transport / Faciliated diffusion Active transport SIMILLARITY (i) allow movement of molecules through semi permeable plasma membrane (ii) need specific carrier protein to permit molecules to pass through. DIFFERENCES down concentration gradient Direction of movement against the concentration gradient Semi-permeable membrane Occurs through Membrane of living cells   water Type of molecules glucose, mineral salts, K and Na which moves Absent Energy requirement Present Pore or carrier proteins Carrier molecules Carrier proteins Concentration gradient Driving force Lower to higher concentration Facilitated diffusion Active transport either causes continues until reaches a continue until accumulation or excretion of dynamic equilibrium molecules from cells not necessary Living membrane necessary 3 WWW.ANDREWCHOO.EDU.MY  The Structure of the Plasma Membrane 1. 2. The structure of the cell membrane the phospholipid bilayer with related to its permeability: simple diffusion. (i) Larger molecules such as glucose, (a) Semi-permeable. amino acids and small inorganic (b) Selectively permeable to small   ions such as K , Na and Ca can 2 molecules such as water, oxygen only diffuse through the cell and carbon dioxide membrane by facilitated diffusion with the help of (c) Does not permit the movement of transport proteins such as carrier large molecules across it. proteins and pore protein. (d) Consists of a phospholipid bilayer. 2 (j) Ca transported across the (e) The polar hydrophilic heads of membrane against the phospholipids are facing concentration gradient called outwards and attracted to active transport. Active transport watery surroundings. requires carrier proteins & (f) The non-polar hydrophobic fatty energy from ATP produced in acid tails are facing inwards, 2 mitochondria. The Ca and ATP away from water. bind to the carrier protein and (g) Water molecules attracted ATP splits into ADP and hydrophilic heads and small phosphate ion to produce enough to pass across the energy, the carrier protein phospholipid bilayer from the changes its shape and moves the region of higher water 2 Ca into the cell. concentration to low. they are (k) The phospholipid bilayer is not small enough to pass through rigid; it has flexible and dynamic (h) Lipid-soluble molecules such as structure. fatty acids, glycerol and vitamins A, D, E and K can diffuse through 4 WWW.ANDREWCHOO.EDU.MY  Movement of substances across the plasma membrane in everyday life 1. Hypotonic, hypertonic and isotonic solutions Explanation Animal cell Plant cell Potato strip a Isotonic. solution, total concentration of solutes in the The concave disc No change in length extracellular shapes of the RBC A plant cell has normal and shape. Water fluids is equal to remain unchanged. shape & pressure. the intracellular enters and leaves at (do not experienced *does not lose or gain mass the same rate. fluid. Water any change) molecules move into and out at the same rate. b Hypertonic. solution, the extracellular fluid has more solutes than the The strip decreases intracellular Red blood cells *(draw diagram) in length, becomes fluid so that shrivel up. The A plant cell loses pressure flaccid and curves water molecules volume of cell as the plasma membrane decrease. The RBC inwards with the diffuse out by pulled away from the cell experienced epidermal layer on osmosis , wall and become flaccid. causing the cell crenation. the outside. Water The plant cell experienced to shrink. moves out from the plasmolysis. cells at the cut surface by osmosis. Water cannot leave through the epidermal layer because of the layerof cuticle. 5 WWW.ANDREWCHOO.EDU.MY  c Hypotonic. solution, the extracellular fluid has less solute than the The strip increases intracellular RBC continue to in length, becomes fluid so that swell until burst. *(based on diagram) turgid and curves water enters a Contents of cells Cell becomes turgid. Water outwards with the cell by osmosis , released into enter into vacuole & size epidermal layer on causing to swell. surroundings. The increase, pushing the outside. Water mes New Roman volume of cell cytoplasm towards cell enters the cells at increase & wall. Cell experienced the cut surface by stretches the cell deplasmolysis. osmosis. The membrane. The epidermal layer has RBC experiences a layer of cuticle haemolysis. which prevents the entrance of water. 2. prevents bacterial growth in the tissues. RBC is mixed with detergent. The This prevent decay of mangoes. detergent dissolves lipids. After 10 minutes, the mixture is examined 5. Too much of fertilizer, potassium under a microscope, no red blood cells nitrate dissolve in soil water. The soil were seen but the mixture turn red water becomes hypertonic to cell sap of and cloudy. Explain why. root hair. Thus, water from cell sap of Detergent dissolve the lipid in the plant cell move out by osmosis into soil. plasma membrane. Plasma membrane The plant plasmolysis & wilts if not disintegrate. Cytoplasm of red blood cell watered with enough water. mix into the solution. Red blood cell is 6. The solution outside the foodstuffs haemolysed. becomes hypertonic to the cell sap of 3. A player is advised to drink an isotonic foodstuffs which created hypertonic drink environment. This causes water move - to balance body fluid out by osmosis & sugar or salt enter the - to replenished the fluid lost cells by diffusion. Cell become through sweating plasmolysed. The dry state of the cells as well as preservatives hinders the 4. Vinegar effective in preservation of action of enzymes, bacteria & fungi of mangoes? decomposition. The foodstuffs last Vinegar which has a low pH. Vinegar longer. diffuses into the tissues of the mangoes. The tissue becomes acidic. Low pH 6 WWW.ANDREWCHOO.EDU.MY  FORM 4 CHAPTER 4 CHEMICAL COMPOSITION OF THE CELL Enzymes 1. General Characteristics of Enzymes ***(answer must be deduced from the diagram) a. Speed up the rate of lower the need of high activation energy & help speed up rate biochemical reactions reactions. b. Not changed or destroyed altered by reactions that catalysed and can be used repeatedly. c. Effective in small quantities small quantity enzyme bring large amount biochemical reactions. d. Reversible reactions Catalyses reactions in two direction e. Denatured by high Above 400C the rate reactions falls rapidly ceasing altogether at temperature about 600C. f. Sensitive to pH change Alterations in ionic changes of acidic & basic groups of enzyme change shape of enzyme g. Enzymes are specific Only molecule substances which matches with molecule of enzyme can bind together the reaction to occur. h. Affected by inhibitors change the active sites i. Required confactors to act Certain type of metallic ions and vitamins 2. The ‘lock and key’ Hypothesis (draw structure) (i) At very low temperature, enzymes are less active. (ii) At optimum temperature 37 0 C, the rate of reaction of the enzyme (a) Enzyme is specific is the highest (activity enzyme (b)The substrate molecule fits to a fixed most active). structure called active site of the (iii) Increasing temperature, increasing enzyme molecule the rate of reaction (kinetic energy (c) The substrate is the ‘key’ that fits of enzyme & substrate molecules the enzyme ‘lock’ to form an produces more collision). enzyme-substrate complex (iv) At very high temperature, (d) The enzyme changes the substrate to enzymes are denatured. products (v) Explain: (e) The product leaves the active site & - Enzymes are proteins. the enzyme is free to catalyse - Higher temperature changes the another reaction shape of the active site of the 3. Factors affects the enzyme activities: enzyme molecules (a) Temperature - Subrates can no longer fit into active site of the enzyme molecules - Subrates cannot be hydrolysed. 7 WWW.ANDREWCHOO.EDU.MY  - The rate of enzyme reaction (i) Increase substrate concentration; decreases. increase the chance of enzyme- (vi) Treatment to avoid apple turn substrate collisions and the rate of brown: reaction increase. (a) soaked in hot/warm water (ii) More chances of collision (b) enzyme denatured by heat between substrate molecules & (c) no chemical reaction /no enzymes for catalytic reaction to oxidation take place (b) pH *[trypsin] (iii) Rate of reaction becomes Constant after maximum point (because all site enzyme attachment are full & becomes limiting factors). (d) Enzyme concentration (i) The optimum pH fastest rate of reaction (ii) Optimum temperature decrease the rate of reaction because bonds maintaining the tertiary shape of (i) No other factors are limiting the the enzyme are broken. rate of reaction (iii) The active site loses shape & the (ii) The substances molecules enzyme is denatured. available is more  More active (iv) Enzyme is denatures by alkali. site is available for enzymes (c) Substrate concentration action Experiment: Aim: Variables: To study the effect of pH on Manipulated: pH of the medium (Use albumen the activity of pepsin suspensions of different concentrations) Problem statement: Responding: Rate of enzyme activity / Time taken What is the effect of pH on the (Measure & record time taken for the cloudiness of the activity of pepsin? albumen suspension to clear by using stopwatch) Hypothesis: Fixed: Temperature of medium, volume and concentration The pepsin enzyme works best of the albumen suspension, volume and concentration of in an acidic medium of pH 3. the pepsin solution. (Using the same enzyme Operational definition for the concentration) hydrolysis of starch Apparatus: Test tubes, test tube rack, stopwatch, thermometer, pH paper, beaker, wire, Bunsen burner, dropper, and 5 ml syringes 8 WWW.ANDREWCHOO.EDU.MY  Materials: 1% pepsin solution, egg albumen suspension, 0.1 M hydrochloric acid distilled water, and 0.1 M sodium hydroxide solution. Technique: Observe the condition of the mixtures before and after 15 minutes. Procedure 1. An albumen suspension is made by mixing the egg white from a egg with 500 ml of distilled water. The mixture is boiled and any bl particles are removed. 2. With the help of a syringe, 5 ml of albumen suspension is placed in each of the test tubes labelled K, L and M. 3. l ml of 0.l M hydrochloric acid and 1 ml of pepsin solution are added 0 test tube K. 4. 1 ml of distilled water and 1 ml of pepsin solution are added to test tube 5. 1 ml of 0.1 M sodium hydroxide solution and 1 ml of pepsin solution are added to test tube M. 6. All the three test tubes are placed in a water bath at 37 °C for 15 minute. 7. The appearance of each mixture is recorded at the beginning and at the end of the experiment. 8. A strip of pH paper is dipped into each test tube and the pH of each mixture is recorded. Conclusion Test tube pH Appearance of the mixture The enzyme pepsin works K 3 cloudy clear best in an acidic medium L 7 Cloudy cloudy of pH 3. The Hypothesis is accepted. M 8 cloudy cloudy Experiment: Aim Variables: To investigate the effect of Manipulated: pH of buffer solution (Use different pH of pH values on the hydrolysis buffer solution) of starch by amylase Responding: Time taken (Measure & record the time taken enzyme. for iodine solution to remain yellow by stopwatch) Constant: Concentration of amylase solution (Fixed at 1%) Hypothesis: Observations: Relationship between the At pH of buffer solution 7 i. At pH of buffer solution 5, rate of amylase activity on (optimum), time taken for time taken for iodine solution the starch & the pH values: iodine solution to remain to remain yellow is 28 The rate of amylase activity yellow is the least. minutes. on starch increases when Operational definition: ii. At pH of buffer solution 7, the pH value of the mixture The hydrolysis of starch by time taken for iodine solution solution increases until its to remain yellow is 2 minutes. optimum rate. A change in 9 WWW.ANDREWCHOO.EDU.MY amylase is breaking down Inferences: pH value will alter charges of starch into simple i. At low pH, rate of amylase on the active site of substances when the time is low. amylase & the substrate. for the iodine solution to ii. At optimum pH, rate of remain yellow is taken & is amylase reaction is at its affected by pH value. maximum. Apparatus: Predict the outcome when same experiment repeated at pH 7 Thermometer, Water bath at a lower temperature of 20%: Materials: The time taken will be more than 2 minutes because 2ml of 1% amylase enzymes are not active at low temperatures when the rates of solution, 3ml of 1% starch enzyme reaction are slow. solution, 5ml buffer solution Procedure: 1. 2ml of 1% amylase solution was placed into a boiling tube containing 5ml of buffer solution at pH 6. 2. 2 drops of iodine solution was placed into each groove on a white tile 3. 3ml of 1% of starch solution was added into the boiling tube and the stopwatch is started immediately. 4. Every 2 minutes, a drop of mixture from the boiling tube was dropped into new groove of iodine solution on the white tile by using a clean dropper. 5. The time taken for the iodine solution to remain yellow is recorded. 6. Steps 1-5 are repeated using buffer solution at pH 5, 7, 8 and 9. 7. The observation is recorded in a table. Experiment: Aim: Variables: To study the effect of Manipulated: Temperature of medium temperature on the activity of Responding: Rate of enzyme reaction / Time taken for salivary amylase. hydrolysis of starch Problem Statement: Fixed: Volume of salivary amylase, volume and What is the effect of concentration of the starch suspension temperature on the activity of Apparatus: salivary amylase? tubes, test tube rack, stopwatch, thermometer, spotting Hypothesis: tile, beakers, gauze, tripod stand, Bunsen burner,, dropper, The rate of enzyme reaction and 5 ml syringes increases with a rise in Materials: temperature until it reaches the Saliva, iodine solution, 1% starch suspension, distilled optimum temperature. water and ice cubes Technique: Test the presence of starch using iodine solution and record the time taken for the hydrolysis of starch with a stopwatch. 10 WWW.ANDREWCHOO.EDU.MY  Procedure: 1. The mouth is rinsed and saliva is collected by spitting into a beaker. The saliva is diluted with an equal amount of distilled water. 2. 4 ml of starch suspension is poured into test tube K while 1 ml of saliva solution is poured into test tube L. 3. Both test tubes are placed in a water bath at 0 ° C for 5 minutes. 4. A drop of iodine solution is poured into each groove of a spotting tile. 5. After 5 minutes, as soon as the starch suspension from test tube K i poured into the saliva solution in test tube L, a stopwatch is started. 6. A drop of mixture from test tube L is placed in the first groove of the spotting tile containing iodine solution. 7. The iodine test is repeated every minute for 20 minutes and the time taken when the mixture does not turn blue-black anymore is recorded. 8. Steps 2-7 are repeated for temperatures of 15 °C, 37 °C, 45 °C and 60 °C. 9. The results are recorded and a graph of rate of reaction (1/t) against temperature is plotted. Conclusion: Time taken for the hydrolysis of Temperature Rate of reaction As the temperature starch increases, the rate of 0 Not completed after 20 minutes 0 enzyme reaction increases 15 10 0.10 until JS the highest at 37 3 0.33 37 °C. It then decreases 45 4 0.25 and stops at 60 C. The Hypothesis is accepted. 60 Not completed after 20 minutes 0 FORM 4 CHAPTER 5 CELL DIVISION Mitosis 1. Important/Significant of mitosis: i. forms new cells for growth ii. replace tissues which are damage iii. enables heredity material in parent cell to be passed to daughter cell iv. enables asexual reproduction v. maintain chromosomal number of daughter cell 11 WWW.ANDREWCHOO.EDU.MY 2. Stages in Mitosis: Phase /Description Appearance of cell 1. Prophase (name & explain) - Each chromosome shortens, thickens & becomes visible - Spindle fibres start to form. - Centrioles start to move to opposite pole. - Nuclear membrane & nucleolus broken down 2. Metaphase - Chromosomes arrange themselves at the cell equator / metaphase plate. - Sister chromatids are still attached at the centromeres. - Spindle formation is complete *(shown chromosome arranged in a line) 3. Anaphase - The centromere of each chromosome splits and its chromatids are pulled to opposite poles of the cell by the spindle fibres. (Sister chromatids separate and move to the opposite poles of the cell.) 4. Telophase * cytokinesis - Chromosomes have reached the opposite poles of the cell. - A nuclear membrance is re-formed to surround each set of chromosomes. - Nucleolus is also re-formed. Hence, two daughter nuclei are formed. 4 chromosomes form each daughter cell. *(appear 2 set of identical chromosomes, one in each daughter cell formed) Application of mitosis 2. Cloning of animal : 1. Cloning is the process of artificially creating a new individual that is genetically identical to an existing individual. 12 WWW.ANDREWCHOO.EDU.MY 3. Clonning (an asexual reproductive process/does not involve gamete) Advantages of cloning Disadvantage of cloning (i) Many produced in shorter time. (i) shorter lifespan compared co chose produced sexually. (ii) Short maturation time (ii) Clones are genetically- alike hence there is no generic (iii) Plants with desirable variation. Therefore, when there are any changes in the characteristics similar to the parent environment, they have no adaptations and can be can be produced easily wiped out. (iv) carried out anytime because it (iii) All clones have the same level of resistance towards involves only vegetative parts of the certain diseases hence they can be easily infected with parent plant and requires no the same disease or attacked by the same pest and easily pollinating agents eliminated. 4. Tissue culture widely used in (c) Callus is eventually developed into agriculture to produce new plants of embryos and the later into plantlets. desired qualities. (d) New plantlets formed are then transferred to the soil where they 5. Technique of tissue culture: can grow into adult plants./grow (a) Small piece of tissue are taken out into suitable size from the plant, such as leaves. (b) With suitable sterile culture medium and hormones, the plant cells divide by mitosis to form a callus. Meiosis 1. Meiosis occurs in reproductive organs to produce four daughter cells called gametes, each containing half the chromosome number of the parent cell. 2. Gametes that are produced are haploid (n) whereas the parent cell is diploid (2n). 3. Importance of meiosis: a. Reduce the number of chromosomes of gametes into half haploid to ensure that the number of chromosomes in next generation is diploid same as the parent. b. Causes genetic variation by crossing over or combination of different chromosomes. c. To produce gamete. 13 WWW.ANDREWCHOO.EDU.MY 4. Meiosis I & II Meiosis I (correct sequence) Prophase I (a) Each chromosome condense, thicken and become clear is now seen as a pair of chromatids (two sister chromatids) joined by a centromere. (b) Homologous chromosomes pair up through synapsis. In a bivalent, chromatids exchange genetic material at points called chiasmata The process is known as crossing over (c) Nuclear membrane and nucleolus are broken down. (d) Spindle formation begins. Importance of crossing over: - Exchange of genetic material between homologous chromosome - produce new genetic combination - causes variation - enhance the ability of survival in different environment Metaphase I (a) Homologous chromosomes arrange themselves is pairs at the cell equator/metaphase plate (b) Any chromosome could face any of the two cell poles. This is known as the independent assortment of chromosomes which produce new genetic combination. Anaphase I (a) Homologous chromosomes separate and move to opposite poles of the cell. (number of chromosomes starts to reduce to half) Telophase I (draw) (a) Chromosomes reach the poles. (b) Spindle threads disappear. (c) Nuclear membrane and nucleoli are re-formed. 14 WWW.ANDREWCHOO.EDU.MY  Meiosis II Prophase II (a) Nuclear membrane is broken down again. (b) Each chromosome is now unpaired. Metaphase II (a) Chromosomes arrange themselves at the metaphase plate. (b) Centromeres spilt. Anaphase II (a) Chromatids (now called chromosomes) move to opposite poles of the cell. (b) Centromeres of chromosomes divie into two and sister chromatids separate to move to the opposite poles. Telophase II (a) Chromosomes reach the poles. (b) Spindle threads disappear. (c) Nuclear membrane and nucleoli are re-formed. Comparison of mitosis and meiosis Similarities : a. Both involve the division of nucleus and cytoplasm. b. Both involve the replication of chromosomes (carry genetic material which is transferred from the parent to offspring). c. Both ensure passing down of generic materials from parent to the off spring Differences between mitosis and meiosis Mitosis Features *Meiosis In somatic cells (animals), at the end In reproductive cells in the ovary, Where it takes place of root and shoot tips (planes) testes and anther (planes) Once times a cell divides Twice daughter cells Two Four produced Same as the parent cell Genetic composition ½ from the parent cell Same number as Parent Half No crossing over Crossing over Takes place at meiosis I No Variation Yes Produces new cells for growth & Produces gametes for sexual replaced old, worn-out ones; reproduction (ovary and testis) ; Importance resulting in increase in size, height: ensures the chromosome number of a and length species is maintained Chromosome Chromosome Homologus Chromosome behavior 15 WWW.ANDREWCHOO.EDU.MY  FORM 4 CHAPTER 6 NUTRITION Balanced diet 1. A balanced diet consists of (all the - stimulates peritalsis and aids nutrient in the correct proportion as per defecation. the Food Guide Pyramid) to meet daily (e) water requiment of the individual - aids in chemical reactions in the (a) Proteins body. - provide all the essential for growth and repair of the damaged 2. Food Pyramid (b) carbohydrates and fats - meet the energy requirement of the body. (c) minerals and vitamis - maintain good health and growth. (d) dietary fibre/roughage Experiment: Aim: Procedure To determining the energy 1. 20 ml of distilled water is put into a boiling tube. value in food samples 2. The test tube is then fitted with cotton wool that has a Materials thermometer in the center. Distilled water, peanuts, 3. The test tube is then clamped to retort stand as shown plasticine in Figure 6.1. Apparatus: 4. The initial temperature of the water is recorded. Retort stand and clamp, 5. A whole peanut is weighed and its weight recorded. thermometer, boiling tube, 6. The peanut is spiked to a pin and mounted on a piece of cotton wool, windshield, plasticine. measuring cylinder, electronic 7. The mounted peanut is then ignited by holding it to the balance, pin and Bunsen flame of a Bunsen burner. Then it is placed under the test burner tube in figure 6.1. 8. The water is stirred gently with thermometer. 9. The final temperature of the water is recorded when the Hut stopped burning. Mass of peanut (g) 0. The energy value of the 5 peanut can b calculated Volume of water 20 using the formula: (ml) Energy value of peanut Mass of water (g) 20 = 4.2 x mass of water x Initial temp 31 increase in temp Final temperature 70 mass of food x 1000 Increase in temp 39 16 WWW.ANDREWCHOO.EDU.MY Conclusion Variables: -1 Manipulated: The energy value / calorific value of the peanut is 6.552 kJ g. Discussion: Food sample 1. All the energy released by the peanut during combustion is (Different type of absorbed by the water. Not all the energy released is used to heat food samples are up the water but some are lost as light and heat to the surroundings. used) Hence the result obtained is less than the actual value. Responding: The 2. To obtain a more accurate result, the following precautions need to change in be taken: temperature (The (a) The water in the boiling tube must be stirred so that heat is increase in evenly distributed to get a better reading of the temperature. temperature is (b)A windshield is used to surround the apparatus to reduce heat recorded by using loss to the surroundings. thermometer) (c) Ensure that the peanut is placed not too far away from the Constant: Mass of bottom of the boiling tube and the flame is not extinguished. food sample (The mass is fixed) Special Diets Requirements Target Diet group Pregnant Energy – rich foods (rice, bread) are needed for the growth of the breasts, uterus, women foetus and placenta Protein – for growth of foetus, formation new tissue Calcium - for bone and teeth formation / prepare the mammary gland to produce milk Phosphorus - formation of strong bones in growing foetus Less fat – less risk of cardiovascular problems minerals - good health fibres – prevent constipation Extra folate obtained from green leafy vegetables are needed for the healthy development of foetus Need vitamin D for absorption of calcium, phosphorus & folid acid to reduce birth deflets Low haemoglobin level in the blood can be increases by an intake of iron (spinach) which helps in forming haemoglobin. Children Children should take plenty of breads, cereal, vegetables and fruits Calcium are highly recommended for healthy bone (calcium and iron) (for example bread and eggs) and teeth development; vitamin C and water are needed for active lifestyle during childhood More active than moderately active adult man Teenagers Boys need more protein and energy than girls Girls need more iron than boys to replace menstrual losses More carbohydrates is needed to provide sufficient energy for their active lifestyle. Vitamin/minerals to maintain good health When insufficient in: Carbohydrate: will become weak & experience frequent fatigue 17 WWW.ANDREWCHOO.EDU.MY  Fibre (vegetables & fruit): have scurvy, rickets, scaly skin and constipation When take large amount of: Fats and cholesterol: capable of causing arteriosclerosis and high blood pressure (heart attack) , can deposit on the inner walls of arteries and narrows the arteries, eventually cause artery blockage Oily food: cause pimples or acne problems Fast food: cause harmful side effects Food Digestion 1. The comparison between the digestion of cellulose in human, ruminant and rodents Human Ruminants Rodents i. chamber i. 4 chamber i. Chamber ii. - ii. Shorter caecum than rodent ii. Long & large caecum (herbivorous have large caecum that contains many bacteria & protozoa which secretes cellulose & hydrolyses to glucose by cellulose. iii. NO enzymes iii. Digest cellulose occurs in iii. Digest cellulose occurs in caecum & produced that can stomach appendix digest cellulose iv. Cellulose aid in iv. Food first goes to rumen, iv. Food goes through the alimentary canal the movement of and then reticulum. Food is twice (the first batch of faeces is eaten food through the then returned to mouth to be and the breakdown products will pass intestinal tract thoroughly chewed and then through the alimentary canal for the swallowed and enter into second time). Re-swallow the digested omasum and finally to cellulose from caecum after left anus. abomasums. v. No nutrient from v. Regurgitate food v. Food cut by teeth cellulose vi. Small caecum / vi. Large caecum / appendix appendix 2. Drawing of villus: *fatty acid dissolve into lacteal 18 WWW.ANDREWCHOO.EDU.MY  Photosynthesis 1. Photosynthesis is process by which ii) This results in the release of energy green plants to synthesize organic from the excited electrons which compound [carbohydrates] from CO2 form ATP (adenosine triphosphate). and H2O in presence of sunlight and iii) Light energy splits water molecules and chlorophyll. into H  and OH  /photolysis of water. iv) H  ions will combine with free 2. The reaction of photosynthesis is as electrons from chlorophyll to form follows: neutral hydrogen atoms. 6CO2 + 6H2O  C6H12O6 + 6O2 v) OH  ion combine each other to form H2 O & O2 How a molecule of carbon dioxide in vi)The hydrogen atoms use ATP are the air and water absorbed from the used in dark reaction (occur in roots becomes part of a carbohydrate stroma) to reduce carbon dioxide molecule stored in a leaf of a plant into glucose and water, after going 3. During light reaction (occur in grana), through a series of enzyme catalysed sunlight is absorbed by chloroplast to reactions. Glucose is converted into split water molecules to H  and OH  starch through condensation. ions and to form ATP. During dark reaction, ATP and H  are used to reduce How a carbohydrate molecule stored CO 2 to form starch. in a leaf of a plant can become a 4. Water is needed during the light- starch molecule stored in the root dependent reaction of photosynthesis. 6. Starch stored in a leaf is converted into How green plant produces starch sucrose. Sucrose is then transported by molecules phloem tissue (sieve tube) to the roots. 5. During light reaction, In the roots, sucrose is converted to i) light energy is absorbed by starch molecule to be stored. chlorophyll which excites the 7. Starch is changed to glucose in the electrons of to higher energy levels. hydrolysis process with the presence of The electrons in the excited state can enzyme and water. leave the chlorophyll molecules. 8. Differences LIGHT REACTION DARK REACTION - During the light reaction, chlorophyll absorbs light - No requires light energy - carbon dioxide absorbed - No carbon dioxide absorbed - Gives out oxygen - No gives out oxygen - No form glucose - Form glucose (simple sugars). Similarities - Occur in chloroplast - reactions related to photosynthesis - involve chemical reactions - occur in the presence of light 9. Factor affecting rate of (a) The concentration of CO2 photosynthesis : 19 WWW.ANDREWCHOO.EDU.MY  (i) temperature & light intensity are fixed 11.Function : site of photosynthesis radiant energy (food) (ii) light intensity limiting factor Carbon + Water Glucose + Oxygen Dioxide chlorophyll (b) The intensity of light 12.Chloroplast within the cells of green plants and eukaryotic algae which contains the membranes, photosynthetic pigments (notably chlorophyll), and enzymes necessary (i) temperature & concentration of for photosynthesis. CO2 are fixed. 13.The photosynthesis rate usually peaks at (ii) CO2 concentration limiting noon, unless the weather is too hot that factor the stomata pores start to close to (iii) If light intensity increases, rate prevent water loss. of photosynthesis of water increases. 14.During a hot day, the soil dries out. The (c) Temperature concentration of sugar, salt and amino acids in the soil becomes more concentartion then the cell sap of the root. Therefore, water diffuse out of the root cells by osmosis and plasmolysis occurs. The plant becomes flaccid, 0 0 causing the plant to wilt. (i) Optimum 30 C to 40 C 0 15.On the cool day, there is high amount  < 30 C -> enzyme not active of water in the soil. Water will diffuse 0  > 35 C -> enzyme denatured into the root by osmosis and the plant 10.The higher the number of stomata, the will not wilt. higher the rate of photosynthesis. 20 WWW.ANDREWCHOO.EDU.MY 16. 17. As the light intensity increases, the rate of photosynthesis also increases. X: - all the oxygen produced during photosynthesis is used for respiration. - All the carbon dioxide produced during respiration is used in photosynthesis. - Rate of respiration = rate of photosynthesis Experiment: Aim: Material To investigate the effect of a spring of Hydrilla 0.2% sodium hydrogencarbonate light intensity on the rate solution, distilled water, plasticine of photosynthesis Apparatus Problem statement Filter funnel, test tube, beaker, thermometer, 60 W bulb, How does light intensity stopwatch, ruler affect the rate of Technique photosynthesis? Count the number of gas bubbles released in a fixed time. Variables Procedure Manipulated: Distance of 1. The apparatus is set up as shown in Figure 6.15. light source 2. The water temperature is maintained at 30C. Responding : The number 3. The apparatus is placed at a distance of 50 cm away from of gas bubbles released the light source. Fixed: The type and size 4. After gas bubbles are released at a Constant rate, the 21 WWW.ANDREWCHOO.EDU.MY of plant used,concentration number of bubbles released per minutes counted. of carbon dioxide, time 5. Step 4 is repeated with the apparatus being placed at 40 cm, taken for counting total 30 cm, 20cm to 10 cm from the light source. number bubbles Relationship between total number bubbles & the light Hypothesis intensity: The higher the light When the light intensity increase, total number of bubbles intensity, the higher the increases because the rat eof photosynthesis increases. More rate of photosynthesis. oxygen / gas is produced. Distance from the 10 20 30 40 50 light source (cm) Light intensity 0.1 0.05 0.033 0.25 0.02 Number of gas 21 25 10 6 2 bubbles released Predict total number bubbles released in 5 minutes when two Hydrilla sp. is used: The total number bubbles released increases because two Hydrilla sp. have more leaves & the rate of photosynthesis increases. Conclusion The rate of photosynthesis increases with increase in light intensity. The Hypothesis is accepted. Important:The function of hydrogen carbonate solution is to produce carbon dioxide. Problem related to deficiency and excess of nutrients in the diet 1. Malnutrition means in sufficient of 3. Kwashiorkor nutrient in the diet. (a) do not have proper diet over a period of time. (b) imbalanced diet (c) very thin and health may be affected 2. Marasmus (a) Lack of protein (a) lack of protein & energy-providing (b) Children suffering oedema due to nutrients with wrinkled skin and increased interstitial fluid content marked loss of weight. because there is not enough proteins (b) The child experiences impared in the blood to hold back the water. development physically and mentally. (c) Has weak muscles as well as a bloated stomach (d) The child experiences retarded growth of body and brain. 22 WWW.ANDREWCHOO.EDU.MY 4. Lack of carbohydrates in the diet (d) Cholestrol deposits on the walls of (a) will bring about the onset of diabetes the blood vessels narrowing them mellitus, obesity and dental caries. (e) Contributing to cardiovascular (b) Obese people are rich in saturated disease, suffering from high blood fats ,high level of cholesterol pressure , heart disease and stroke. (c) Health problem like cardiovascular (f) smoking increases rate of disease, high blood pressure, (fat cardiovascular disease. deposited in the lumen of blood (g) Might allso lead to hypertension and vessel) atherosclerosis and high risk diabetes mellitus. in stroke, brain haemorhage, (h) Excess of glucose contain in blood, diabetes mellitus , cancers highly rich in carbohydrate 5. Over – consumption of saturated fats such as animal fats, may: 6. Lack of fibre in diet: (a) Excess storage of energy in the form (a) Constipation can be overcome by of fat/Increases in body weight eating more fruits and vegetables drastically due to energy requirement which is high in fibre. is less than energy intake. (b) Faeces moving to slowly through (b) A health problem will be obesity colon [unbalanced diet ] (c) Colon cancer (c) Increase the blood cholesterol level (d) Haemorrhoids (piles) Eating Habits 2. Good eating habits should be 1. Good eating habits include the accompanied by good choices of food following. particularly processed foods. The a) Eating food at the correct time - factors to consider when choosing food three meals a day are: b) Eating nutritious food / a balanced a) the nutritional values of the food diet b) the freshness of the food c) Eating enough vegetables and fruits c) the type of preservatives and to provide sufficient fibres for colouring used in the food roughage d) Refrain from overeating or eating 3. Unhealthy and bad eating habits leads too little during each meal to health problem like: e) Refrain from taking too much sugar, a) Obesity spicy, oily and salty food f) Drinking enough water (2-3 litres) each day g) Chewing the food before swallowing for easier digestion 23 WWW.ANDREWCHOO.EDU.MY  Obesity due to: - excessive intake of food rich in fats and carbohydrates which results in excess body weight (20% > the actual body weight with respect to height) an obese person can be can easily contract avoided by - Diabetes mellitus - Exercising regularly - Hypertension - Refraining from eating too much food rich in fats - Cardiovascular diseases and carbohydrates (which provides sugar) - Taking only three meals a day b) Gastritis - Lead to malnutrition, lost of Irregular meals gastric juice weight or even death. acting on the gastric wall of an - Treatment: empty stomach mucus covering (i) psychotherapy, of the epithelial lining which forms a (ii) nutritional therapy to restore protective layer is disrupted body mass inflammation of the epithelium of the stomach wall gastritis d) Bulimia is an emotional disorder. stomach ulcers when the breach in - It is characterized by bouts of the stomach develops into a hole. excessive eating followed by - individual will feel pain/discomfort, purging through induced vomiting loss of apetite, nausea, cause loss or misuse of laxatives. weight - This may cause serious injury to - increase risk of stomach cancer the alimentary canal and c) Anorexia nervosa – too thin imbalance of minerals in the body - is a psychological illness of fluid. dislike eating/refuse to eat. - Acid in vomit corrodes teeth & - Due to intense fear of gaining damage oesophagus. weight and starving oneself to - Individuals suffering from bulimia become thin. may experience dehydration, - Leads to excessive loss in amounts hormonal imbalance, irregular of fats and muscles which periods, kidney damage, liver ultimately leads to the disruption diseases and cardiovascular in the functioning of the heart, problems. endocrine and reproductive systems. The Formation of Faeces and Defaecation 1. The colon carries out two functions: (a) reabsorption of 90% water and minerals to form faeces which are a Semi-solidwaste (b) storage and elimination of faeces (defaecation) 24 WWW.ANDREWCHOO.EDU.MY 2. Formation of faeces in a human: 3. A high fibre diet is important because: a) it give bulk to the intestinal contents (a) Faeces are the undigested food residue - increases the volume of the faeces after absorption of nutrient in small - facilitates the grip and the intestine. movement in the large intestine (b) The undigested food contains fibre, - encourages peristalsis and cellulose and bile pigments that enter defaecation. colon from the small intestine after b) help to absorb water passing through the caecum by - absorb water from the undigested peristalsis. food residue (c) Theses materials cause toxicity, - softens the faeces and facilitates therefore need to be removed from the movement of the faeces along our body. the colon and out through the anus. (d) When the undigasted food residue c) prevent constipation reaches the colon, water is absorbed - ease the movement of faeces out of from them thus hardening the residue. the rectum through the anus, thus (e) By the time it arrives at rectum, it preventing constipation. becomes semi-solid and is known as faeces. (f) When it becomes full, defaecation takes place whereby the rectum will contact to puch out the faeces. Technology Used in Food Processing Technology Used in Food Processing Technology Method Food Sample a. Pasteurization Milk is heated to 63C for 30 mins or 70C for 15 seconds. Then cooled immediately to 10°C i. Kill bacterial or 5C & keeps in cool conditions. but not the Advantage: - The flavour & nutrients of - Milk spores the food do not change. - Fruit juices -The bacteria is killed - Wine Disadvantage: - not 100% sterile, can’t keep in a long period because the bacterialspores not killed & can hatch back into bacterial. - unable to kill all types of bacteria b. UHT Milk is heated to 132C for 2 seconds and than i. Ultra heat cools down immediately. treatment. Advantage: UHT kill bacterial including ii. Kill bacterial its spores. Wine including its Disadvantage: The taste of the milk is spores. champed a little and certain vitamins are destroyed. 25 WWW.ANDREWCHOO.EDU.MY c. Dehydration / - Spraying the liquid food into hot air - Coffee powder Drying - Sunning the food - Milk powder i. Passing hot air - Placing food in a vacuum - Shrimps over the food to Advantage: The flavour, vitamins and - fish remove water nutrient of the food is - squids ii. to kill bacteria preserved. anchorvies or dried chillies d. Freezing/ - Keeping food in refrigerators Cooling / Advantage: Flavours and nutrient of the - Vegetables Chilling/ food is not changed - Fruits Refrigeration Disadvantage: Bacterial and spores are not - Cakes killed e. Freezing - Maintain the taste and the nutrients & remove Drying (Deep water from food. - Meat Freezing) - The microorganisms’ growth rate is slowed - Fish (-18C to - down. - Prawn 40C) Disadvantage: Bacteria and spores are not killed f. Irradiation Advantage: i. Kill microorganisms (gamma rays) ii. Prevent shoots from - Onions germinating - Potatoes Disadvantage: Destroy nutrients such as - Rice vitamins & change the taste of the food. g. Canning - First cooked & heated to high temperature to kill the microorganisms then pressure to drive out all the air. Cans containing food are sealed while food is being cooled. - Sardine Advantage: - All bacteria & spores killed. - Jam - kept long period of time - Fruit - keep food sterile Disadvantage: Some of the flavour and nutrient is destroyed (low in quality) f. Vacuum - Keep air out of food to prevent oxidation of - Coffee powder Packaging the food and the growth of bacterial - Groundnuts Advantage: Does not change flavour or - Biscuits destroy the nutrients of the - Fruit food. h. Cooking Cooking at high temperature or boiling for at least 5 minutes Advantages - Can kill microorganisms - Vegetables - Prevent bacterial & fungal - Fresh meat decay Disadvantages - Toxins produced by bacteria are not always destroyed 26 WWW.ANDREWCHOO.EDU.MY  FORM 4 CHAPTER 7 RESPIRATION Aerobic respiration and Anaerobic respiration 1. Aerobic respiration occurs under the 3. Vigorous activity will reduce the pH presence of oxygen gas. Oxygen is of blood due to the increase in carbon carried by the blood to the body cells dioxide. E.g. The oxygen level in the and is used to oxidise glucose in order blood of a mountain climber drops from to produce energy. its normal level. So, the breathing and The equation of aerobic respiration : ventilation rate increases & Glucose + Oxygen Carbon dioxide + respiratory muscles contract and 2898kJ relax faster. 2. During vigorous activities, the heart 4. Anaerobic respiration occurs, when the rate, respiratory rate, rate of gaseous supply of oxygen is not enough. exchange and breathing rate However, lactic acid is produced which increases in order to supply more causes muscles fatigue. In the liver, oxygen rapidly to the muscle cells for lactic acid is broken down into carbon their rapid contraction. When the dioxide and water. This breakdown oxygen supply is insufficient, the process requires extra oxygen to be muscle cells have to carry out anaerobic taken into the body by an increased respiration. breathing rate. 5. Comparisons between aerobic respiration and anaerobic respiration Similarities (c) Energy is stored in ATP molecules./produce (a) Both are cellular respiration. energy (b) Both use glucose as the source of (d) Both occur in plants and animals. energy. (e) Both are catalysed by enzymes. Differences All type of living cells Carried out by Some plant (rice plant) and animal cells, yeast and bacteria Oxygen is used Use of oxygen Oxygen is not used Complete Oxidation of glucose Incomplete (partially oxidised) Carbon dioxide, water Products of Ethanol, carbon dioxide & heat and energy respiration energy (in yeast) or lactic acid and heat energy (in muscle cells) Large amount (2898kJ Amount of energy Small amount (muscle cells – 150kJ, per mole of glucose) released yeast fermentation – 210kJ) 38 molecules of ATP ATP molecules 2 molecules of ATP produced Cytoplasm and Location of reactions Cytoplasm mitochondria No Production of lactic Yes acid 27 WWW.ANDREWCHOO.EDU.MY  The Respiratory Structures and Breathing mechanisms in Organism Protozoa 1. Yeast is a microorganism which carries to ethanol, carbon dioxide and 210 kJ out both aerobic respiration and of energy. anaerobic respiration. 4. Differences between: 2. Anaerobic respiration in yeast is also Muscle cell Yeast called fermentation. Product is lactid Product is acid ethanol Carbon dioxide is Carbon dioxide is 3. Yeast secretes an enzyme, zymase not released released which speeds up the change of glucose Less energy More energy Oxygen debt No oxygen debt Experiment: Aim: Materials: Exhaled air, water, potassium To study the effect of different duration of hydroxide solution activity on the percentage of carbon dioxide Apparatus: beaker, boiling tube, rubber tubes, in exhaled air J-tube Observations: Inferences: i. When resting / At 0 minutes activity, the i. When resting, the rate of respiration is length of air column (after being treated lower, so amount of carbon dioxide with potassium hydroxide solution) is expelled is lower. 9.9cm. ii. After duration of vigorous activity, the ii. After 3 minutes of running on the spot, rate of respiration is faster, so more carbon the length of air column (after being treated dioxide is expelled. with potassium hydroxide solution) is 9.6cm. Variables: Predict the percentage of carbon dioxide released if Manipulated: Activity (Doing experiment collected 10 minutes: different durations of activity) Percentage of CO released will be less because Responding: Length of air column / after 10 minutes the rate of respiration returns to carbon dioxide content (Measure & normal/back to the resting stage. record the length of air column by Operational defination of exhaled air: using a ruler) Exhale air is the change in length of the air column Constant: The initial reading for all due to release of carbon dioxide gas after being the activities (The reading is fixed treated with potassium hydroxide solution and it at 10 cm) is affected by different durations of activity. Hypothesis: Relationship between the time of activity & the As the time duration of activity percentage of carbon dioxide: increases, there is a decreases in the The percentage of carbon dioxide in the air sample length of the air column. increases as the duration of activity increases. The ( Finalv  initialv ) rate of respiration increases and more carboon  100% initialv dioxide is expelled out of the lungs. 28 WWW.ANDREWCHOO.EDU.MY Table: Duration of Initial reading Final reading of The length of Percentage of activity (min) of the length of the length of the air column CO 2 in the air the air column the air column (min) sample (%) (min) (min) 0 10 9.9 0.1 1 1 10 9.8 0.2 2 2 10 9.7 0.3 3 3 10 9.6 0.4 4 Experiment: Aim: Problem Apparatus: To determine the statement: J-tube, ruler, beaker, boiling tube, basin, composition of Does inhaled air rubber tubings oxygen and carbon contain more Materials: dioxide in the oxygen and less Potassium hydroxide solution, alkaline inhaled and exhaled carbon dioxide potassium pyrogallaie solution, water air than exhaled air? Hypothesis: Variables: Inhaled air contains more oxygen and less Manipulated: Type of air sample carbon dioxide than exhaled air. (inhaled and exhaled air) Technique: Responding: Percentage of oxygen and Measure and record the length of air column carbon dioxide in inhaled and exhaled air occupied by oxygen and carbon dioxide in Constant: Initial length of air column, inhaled and exhaled air by using a ruler. temperature Procedure: 1. The screw of the J-tube is turned clockwise until the end. 2. The end of the J-tube is placed in water and the screw is turned anticlockwise to draw a length of 5 cm of water into the J-tube (Figure 7.18).The J-tube is removed from the water. 3. The screw is turned anticlockwise to draw about 10 cm of air column (inhaled air) into the J-tube. 4. The end of the J-tube is placed in the water again and a little more water is drawn into the J-tube to trap the air column in the J-tube. 5. The screw is adjusted so that the air column is in the middle of the J-tube. 6. The J-tube is then immersed in a basin of water for 2 minutes to stabilise the temperature of air column. 7. The length of the air column is measured in the basin of water by ruler. The length of the air column is recorded as a cm. 29 WWW.ANDREWCHOO.EDU.MY 8. The screw is turned clockwise to remove some water in the J-tube so that the air column is about 2-3 mm from the end of the tube labelled A. 9. The end of the J-tube is then immersed in potassium hydroxide solution and the screw is turned anticlockwise to draw about 2-3 cm of the solution into the J-tube. 10. The tube is removed from the solution and the screw is used to move the potassium hydroxide column to and fro several times so that it can absorb the carbon dioxide in the trapped air column. 11. Steps 7 and 8 are repeated. The length of the air column is then measured and recorded as b cm. 12. The screw is turned clockwise to remove the potassium hydroxide solution until about 2-3 mm is left at the end of the tube. 13. Step 10 is repeated by using alkaline potassium pyrogallate solution to absorb oxygen from the air column. 14. The tube is removed from the solution. The alkaline potassium pyrogallate column and air column in the tube are moved to and fro by adjusting the screw. 15. Steps 7 and 8 are repeated. The length of air column is measured and recorded as c cm. 16. Based on the results, the percentage of carbon dioxide and oxygen in the sample of inhaled air column are calculated. 17. The experiment is repeated using exhaled air. Sample of exhaled air can be prepared in the following ways: (a) Air is blown into the plastic drinking straw to get rid of all the air in the straw. (b) The straw is then placed in a test tube full of water. Air is blown into the straw to collect a sample of exhaled air. (c) The composition of oxygen and carbon dioxide in the exhaled air is determined using the same procedure as in steps 1 to 17. Results: Measurement Inhaled air Exhaled air Length of air column (cm) a d Length of air column after adding potassium b e hydroxide solution (cm) Length of air column after adding alkaline c f potassium pyrogallate solution (cm) Length of carbon dioxide in the air column (cm) (a – b) (d – e) Percentage of carbon dioxide Length of oxygen in the air column (cm) (b – c) (e – f) Percentage of oxygen Conclusion: The hypothesis is accepted. Inhaled air contains more oxygen and less carbon dioxide than exhaled air. 30 WWW.ANDREWCHOO.EDU.MY  Difference of respiration between human and animals 1. Fish Adaptation: (d) Maximise the efficiency of gaseous (a) The process gas exchange of fish is the exchange by the opposite direction of diffusion process. water & blood flow through the gills. (b) During gas exchange, dissolved oxygen (e) The gill filaments have thin will diffuse through the epithelium of gill membrane & covered by net into blood capillaries at filaments of gill cappilaries to transport respiratory whereas carbon dioxide will diffuse from gases. blood capillaries into surrounding water. (f) The surface of gills is moist which (c) Lamella has the highest rate of gaseous allows the gases to be dissolved. exchange. /thin walled / one cell thick (g) Numerous blood capillaries for lamella to maximise the surface area for efficient transport of respiratory gases gaseous exchange Characteristics of gills which make it a good respiratory structure: (a) have lamella and filament to increase total surface area (b) numerous blood capillaries for efficient transport of respirratory gases 2. Insects Adaptation: a. Trachea made up of chitin to prevent from collapsing b. The large number of tracheoles provides a large surface for the diffusion of gases c. Tip of tracheoles have thin permeable walls & contain fluid in which respiratory gases can dissolve d. Terminal ends of the trachea remains moist which allows gases to be dissolved. 31 WWW.ANDREWCHOO.EDU.MY 3. Amphibians *mouth cavity is involved in the breathing of a frog. There are a few adaptations of frogs’ mouth for gaseous exchange: (a) Mouth cavity’s floor always moves up and down to let air enter mouth through nostrils. (b) The lining of mouth cavity is thin and contains many blood capillaries which will increase the diffusion rate of oxygen and carbon dioxide. The structure of the human respiratory system 1. The following activity demonstrate the breathing mechanism in humans INHALATION EXHALATION Respiration structure Inhalation Exhalation Diaphragm Contract & flatten out Relax & return to its original shape (dome shape) Thorax volume Increases Decreases Air Pressure Decrease Increase (cause air flow out) External intercostal muscle Contract Relax Internal intercostal muscle Relax Contract Movement of ribs Upwards & Outwards Downwards & Inwards Air Suck in (High atmosphere Flow out pressure forces air to enter the lungs) 32 WWW.ANDREWCHOO.EDU.MY If the diaphragm is not functioning,. (i) the diaphragm muscles do not contract & will not become flat during inhalation. (ii) No changes in the volume & air pressure of the thoracic cavity. (iii) Air may not be sucked into the lungs effiently. (iv) The person may experience shortness of breath 2. Characteristics of Alveoli for efficient gaseous exchange: a. A large surface The total surface area of the alveoli in the lungs is about 75 m2 area b. Very thin walls This enables rapid diffusion of gases./increases TSA ratio for diffusion occur c. Moist The inner surface each alveolus is lined with a thin layer of fluid. respiratory Gases dissolved in fluid before diffusing across the alveolar wall. surface d. A rich supply of helps to transport gases to and from the alveoli quickly / located blood capillaries close to the blood capillary 3. Important of gas exchanges in a human Gaseous exchange across the surface of the alveolus & blood capillary in the lungs Gaseous exchange occur across the alveoli: - The partial pressure of CO 2 is lower in the air - O 2 diffuse across a plasma membrane to of the alveoli compared to the blood capillaries blood capillary - CO 2 diffuse out of the blood capillaries into - From higher O 2 concentration (in alveolus) the alveoli and expelled through the nose / to lower concentration (in blood capillary) mouth into the atmosphere Comparison between Photosynthesis and Respiration Photosynthesis Respiration C 6 H 12 O 6 + 6O 2 6CO 2 + 6H 2 O + energy Chlorophyll Transport by blood circulatory system to body 6CO 2 + 6H 2 O light C 6 H 12 O 6 + 6O 2 cells Only in green plants All living organisms 33 WWW.ANDREWCHOO.EDU.MY Anabolism Catabolism Presence of light Use at all times Produces glucose and oxygen Produces carbon dioxide and water Needs water and carbon dioxide Needs glucose and oxygen Occurs in chloroplasts Occurs in mitochondria Stores energy Releases energy No ATP production sugar (glucose) molecule are oxidized by oxygen to release energy in the form of ATP FORM 4 CHAPTER 8 DYNAMIC ECOSYSTEM Interaction between Biotic Components in  e.g the Rafflesia sp Relation to Feeding 1. Permanent Interaction (a) Symbiosis -close relationship between two / more different species which live closely together: (ii) Mutualism (i) Commensalism  Relationship between two  One species (the commensal) species in which both benefit. benefits, the other (the host) e.g. bacteria in the digestive neither derives any benefits nor system of human harmed  Nitrogen-fixing bacteria such  e.g. epiphytes grows on as Rhizobium sp. Live in the branches of trees , have thick root nodules of a legume leaves with waxy cuticle which plant. These bacteria fix reducs water loss. It still can nitrogens from the air to produce its own food and may produce nitrate which are still survive when the tree is absorbed by the plant to build dead and tall enough to suport it protein. In return, the in obtaining enough sunlight. leguminous plant provide e.g. the crab and the shelter and food to the barnacles. Barnacles obtain bacteria. protection, transport and leftover food from crab, whereas the crab is neither benefited nor harmed. Barnacles on the shell of crabs (iii) Parasitism are known as epizoites. E.g.  Relationship between two Shark and Remora fish. species in which one Remora fish (commercial) species(the parasites) benefits from the shark (host) benefits and the other(the without bringing any harm to host) is harmed. the shark. 34 WWW.ANDREWCHOO.EDU.MY   Mushroom is a saprophyte which obtains food from dead / decaying organic matter. Mushroom secrete enzymes to digest the complex organic matter into simple sugars which are then absorbed by the mushroom. i. cultured in two separate dishes ii. same amount of food were given to each dish iii. the population initially increased 3. Competition – interaction between and then reached a maximum organisms living together in a habitat and (d) When two species of Paramecium competing for the same resources that are cultured together: limited supply. (a) Intraspecific competition refers to the competition among organisms of the same species. More intense because the members of the same species share the same resources.E.g. plants that grow quickly obtain sufficient light will survive at the expense of others i.cultured in same culture dish (b) Interspecific – competition between ii.fixed amount of food and space different species. Two different iii.P. caudatum lost out to P. Aurelia species compete for limited iv. population of P. caudatum rose resources. The stronger species will initially then decreased awhile survive in the competition. E.g. v. population of P. Aurelia rose and Plants: between maize and paddy finally became dominant species plants. Animals: between two species of Paramecium sp. (c) When two species of Paramecium cultured separately: 35 WWW.ANDREWCHOO.EDU.MY  The Processes of Colonisation and Succession in an Ecosystem 1. Profile of succession in a mangrove swamp: Plants Avicennia sp, Rhizophora sp. Brugulera sp. Pandanus Sonneria sp. sp. Species pioneer species [eg: Successor Successor Terrestial grass] species species species 2. Swamp ecosytem: A : Avicennia S : Sonneratia R : Rhizophora B : Bruguiera