Pra U STPM Biology Term1 CC039142a

142 5. It is easier for breaking and forming bonds at high temperatures. 6. Too high a temperature will break bonds within the enzyme molecule and changes the shape of the active site. Weak bonds such as hydrogen bonds within the tertiary structure can be easily broken. 7. Enzyme becomes denatured at high temperature with an exception of hot spring bacteria which has a much higher optimum temperature. These enzymes have stronger disulfide bonds to maintain their tertiary structure. Effect of pH The effect of pH on enzyme activities is always a bell shape curve as shown in Figure 4.7. Basic Arginase Pepsin Salivary amylase pH Enzyme activity 1 2 3 4 5 6 7 8 9 10 11 12 Acidic Figure 4.7 The effect of pH on the enzyme activity 1. This indicates that the enzyme is active over a certain range (e.g. 4 – 10 for salivary amylase). 2. There is an increasing activity until optimum pH is reached (e.g. pH 7). 3. This follows by a decreasing activity as pH goes above optimum. 4. Optimum pH is within a narrow range (6.5 – 7.5). The enzyme at this pH is working at maximum rate. 5. This shows that at certain pH range, some enzymes are active. Lower or above the range, the enzymes are not active. 6. Low pH is equivalent to high H+ concentration. As a result, some enzymes are partly denatured. At high pH, equivalent to very low H+, some enzymes are partly denatured too. 7. Hydrogen bonds and ionic bonds within the molecules are broken and their tertiary structures are affected. 8. The active sites are changed and no longer complementary to their substrates. So, the substrate molecules cannot bind to the active sites anymore. 9. The side chain R groups of amino acids at active sites are added with H+ ions or OH– ions as well. This also results in the decrease in reaction rate. Biology Term 1 STPM Chapter 4 Enzymes 4

143 10. Finally, few enzyme-substrate complexes are formed resulting in lower reaction rate. If no enzyme–substrate complex is formed, there is no reaction. Effect of substrate concentration 1. The effect of substrate concentration on the reaction rate is as shown in Figure 4.8. Km [Substrate] V1/2 VM Velocity (v) a c b Figure 4.8 The effect of substrate concentration 2. From the graph, at low substrate concentration, the reaction velocity is proportional to the substrate concentration. This is because the substrate concentration is a limiting factor as the active sites of the enzyme are available. 3. However, when further substrate is added to the reaction mixture, the velocity becomes maximum. This is because the active sites of the enzyme molecules are saturated with the substrate molecules. 4. The substrate molecules are in excess, enzyme concentration is a limiting factor. Effect of enzyme concentration 1. The effect of enzyme concentration on the reaction rate is as shown in Figure 4.9. [Enzyme] Velocity Figure 4.9 2. It is a straight line graph assuming substrate is in large amount and not a limiting factor. Biology Term 1 STPM Chapter 4 Enzymes 4

144 Exam Tips Remember the definition, example and action for each type. 3. This shows a proportional increase in rate as enzyme concentration increases due to proportional increases in active sites. 4. If the substrate concentration is limiting, the rate can be so fast that it is beyond our ability to measure it. Quick Check 1 1. State six ways how enzymes can bring about reactions. 2. Using the different side chains of amino acids as reference, how would you explain the different reactions that can be catalysed by enzymes? 3. Explain two ways how Km of an enzyme can be determined experimentally. 4.3 Cofactors Role of Cofactor 1. Cofactor is an inorganic or organic substance but non-protein, which is required by certain enzymes so as to function efficiently. Without the cofactor, the enzyme functions slowly or it cannot work. 2. Properties of cofactor are as follows: (a) It is not denatured by high temperature, i.e. by boiling. (b) It is an inorganic ion or complex organic compound of medium size. (c) It is not destroyed at the end of the reaction. (d) It is regenerated at the end of the reaction or by another reaction. (e) It may combine with an enzyme forming holoenzyme whereas without one, it is called apoenzyme. (f) It may bind to the allosteric site activating the enzyme. 3. There are three types of cofactors namely activator, coenzyme and prosthetic group. Activator 1. Activator is an inorganic ion that can speed up enzyme reaction. 2. Its roles include the followings: (a) It attaches to an enzyme so that the shape of the enzyme is more suited to attract a substrate. (b) It attaches to a substrate molecule so that the substrate react easier with the enzyme. (c) It helps in the formation of the enzyme-substrate complex so that the reaction is hastened. (d) It can bind to the allosteric site activating the enzyme. 3. Examples of activators include the following: VIDEO Enzyme Cofactors and Coenzymes Students should be able to: (a) explain the roles of cofactors (ion activators, coenzymes and prosthetic groups) in an enzymatic reaction; (b) explain the importance of vitamins and minerals as precursors of coenzymes/cofactors. Learning Outcomes Biology Term 1 STPM Chapter 4 Enzymes 4

145 (a) Chloride ions speed up the action of amylase on the digestion of starch to form maltose. (b) Magnesium or manganese ions speed up hexokinase to react with glucose and ATP to produce glucose phosphate and ADP. Coenzyme 1. It is a non-protein organic compound that helps a certain enzyme, without which, the enzyme may not function. 2. An example is NAD (nicotinamide adenine dinucleotide), which is derived from vitamin niacin. 3. NAD can exist in the oxidised form as NAD or reduced form as NADH. In the oxidised form, it can act as a receptor of H atom whereas in the reduced form it can donate H atom as follows: XH X (H donor) E1 E2 (Reduced product) (Oxidised product) (H Receiver) NAD+ NADH YH2 Y Two enzymes are involved i.e. E1 and E2 as shown above. (a) In example E1 , phosphoglyceraldehyde dehydrogenase converts phosphoglyceraldehyde (PGAL) to diphosphoglyceric acid (DGA). NAD+ acts as a receptor for H atom and is reduced to NADH. (b) In example E2 , lactate dehydrogenase converts pyruvate to lactate. NADH acts as a donor of H atom and is oxidised back to NAD+. 4. The role of NAD+ is to accept oxidised hydrogen atom and reduces itself. The reduced NAD (NADH) can then donate the hydrogen atom away. 5. AMP can be considered as coenzyme binding to the allosteric site of phosphorylase. The enzyme is activated to break down glycogen in the liver cell. Prosthetic group 1. It is a non-protein organic compound bonded to an enzyme by covalent bond forming a conjugated protein. It is considered as a part of the enzyme, without which the enzyme is not functional. 2. Examples include FAD and haem. 3. FAD is derived from vitamin riboflavin and bound to a dehydrogenase such as succinic acid dehydrogenase that changes succinic acid to fumaric acid. The FAD accepts two hydrogen atoms from succinic acid and is converted into FADH2 , forming fumaric acid in the Krebs cycle as follows. Exam Tips Remember how enzymes speed up reactions, their properties and how activator and pH influence them. Role of cofactor (a) Ion activator (eg. Cl– ) – Binds to enzyme so enzyme is in better shape – Binds to substrate so enzyme can react on it faster – Creates better condition to form complex (b) Coenzyme (eg. NAD) – NAD binds to enzyme, enzyme can function – NAD receives H atom from substrate and forms NADH – NADH then passes H atom to electron carrier. (c) Prothetic group (eg. FAD) – FAD is part of succinate dehydrogenase – FAD receives H atoms from succinate to form FADH2 – FADH2 then passes H atoms to electron carrier Summary 2013 Biology Term 1 STPM Chapter 4 Enzymes 4

146 Succinate Fumarate Dehydrogenase-FAD Dehydrogenase-FADH2 Coenzyme QH2 Coenzyme QH2 Then, the FADH2 donates the H atoms to coenzyme Q, one of the carrier in the electron transport chain forming back the FAD that can be reused. This happens at the cristae of mitochondria during the production of ATP. 4. Haem is an organic compound derived from porphyrin that contains iron ion. It acts as a prosthetic group not only in haemoglobin but also in many oxidoreductase including cytochrome oxidase, catalase and other peroxidase. Cytochrome oxidase is the last carrier of the electron transport chain in the inner membrane of mitochondria. It plays an important role in the production of ATP. Importance of Minerals and Vitamin as Precursors of Cofactors 1. Many mineral ions including Cl– for salivary amylase and Mg2+ or Mn2+ for kinase are enzyme activators. 2. Many mineral ions act as precursors of prosthetic groups. 3. An example is iron(II) ion that acts as precursor of haem group found in cytochrome oxidase. 4. Niacin, vitamin B3 , acts as precursor of NAD+ that is a coenzyme. 5. Riboflavin, vitamin B2 , acts as precursor of FAD that is a prosthetic group of succinate dehydrogenase. 4.4 Inhibitors 1. It is an inorganic or organic compound that slows down or stops enzyme reactions. It includes chemicals that denature protein and other chemicals that can bind to a certain enzyme specifically. 2. Generally, inhibitors can be divided into two types i.e. competitive and non-competitive inhibitors. Competitive inhibitor 1. It is a chemical that is similar in structure to the substrate molecule and binds to the active site of the enzyme. It blocks the active site, preventing the binding of the substrate to the active site. Importance of minerals and vitamins as precursors of cofactors 1. Iron ions form haem for prosthetic group of cytochrome oxidase 2. Niacin (B3 ) forms NAD 3. Riboflavin (B2 ) forms FAD Summary Students should be able to: (a) explain the effects of competitive and noncompetitive inhibitions on the rate of enzyme activity of reversible inhibition; (b) relate the LineweaverBurk plot to the effect of inhibition on Km and Vmax values. Learning Outcomes Biology Term 1 STPM Chapter 4 Enzymes 4

147 2. It has the following properties: (a) The inhibitor molecule is similar in structure to the substrate molecule. A classic example is malonic acid that is similar in structure to succinic acid. Malonic acid inhibits succinic acid dehydrogenase. COO– | H–C–H | COO– COO– | H–C–H | H–C–H | COO– Malonic acid Succinic acid (b) It competes with the substrate to bind to the active site of the enzyme. Thus, the active site is blocked by the inhibitor molecule. This is due to the structural similarity with the substrate that enables the inhibitor molecule to bind to the active site. (c) The bond between the inhibitor and the enzyme is reversible. This means that the bond is temporary and the inhibitor can detach itself easily. The enzyme is not destroyed. (d) The inhibition can be reduced by adding more substrate to the reacting mixture. This is because more substrate molecules have a higher chance of binding to the active site compared with that of the inhibitor. (e) The inhibitor usually does not change the conformation of the enzyme. If it does change, it is only temporary. The inhibitor merely prevents the substrate from binding to the active site. (f) The maximum velocity of the enzyme reaction is not changed but the Michaelis-Menten constant becomes bigger. The bigger the KM, the lower the affinity between enzyme and substrate thus, the slower the reaction as more substrate molecules are needed to saturate the enzyme. The Vmax of the reaction remain the same. This can be seen in the Michaelis-Menten plot and the Lineweaver-Burk plot as shown below. KM increases Velocity [Substrate] 1 – 2 Vmaximum Vmax Without inhibitor With inhibitor With more inhibitor –1/ KM increases 1 – v 1 – [S] 1 – Vmax Without inhibitor With inhibitor With more inhibitor Figure 4.10 Effects of inhibitor concentration on KM Exam Tips Remember the definition, examples and action for each type of inhibitors. 2015 Biology Term 1 STPM Chapter 4 Enzymes 4

148 3. Another example is sulphonamide that is similar in structure to paraamino benzoic acid (PABA). PABA is a substrate in the synthesis of folic acid that is required for the survival of certain bacteria. If sulphonamide is taken into our body, it inhibits the production of folic acid in bacteria. The bacteria fails to obtain folic acid and dies. Thus, sulphonamide acts as an antibiotic that kills bacteria without affecting the body enzymes. Non-competitive inhibitor 1. It is a chemical that reduces enzyme reaction by binding to the enzyme and inactivating or destroying the enzyme molecule. 2. It has the following properties: (a) The inhibitor has no structural similarity with the substrate. An example is a heavy metallic ion such as mercury or cyanide ion. (b) It may be bound to the active site or other parts of the enzyme. The other site includes the allosteric site, any part of the polypeptide or even the prosthetic group. (c) The bond between the inhibitor and the enzyme is irreversible, especially if it is attached to the active site. Examples include heavy metallic ions like arsenic (As+), silver (Ag+) and mercury (Hg+) that bind to the sulphydryl (–SH) side chains of the active site. It is reversible but not bound to the active site such as cyanide that binds to the copper ions of cytochrome (oxidase). The reaction slows down as cyanide concentration increases. When saturation is reached, the reaction completely stops. (d) The effect of the inhibitor cannot be reduced by adding more substrate. This is because the inhibitor does not bind reversibly to the active site. It may bind reversibly to sites other than the active site. Substrate may still bind to the active site but the enzyme simply cannot perform its function. (e) The inhibitor changes the conformation of the enzyme. The active site, the tertiary or even the quaternary structure is permanently changed. In allosteric inhibition, the end product binds to the allosteric site when the end product increases in concentration. When the end product is used up, it detaches and the enzyme is active again. (f) The maximum velocity of the reaction is reduced but the KM remains the same. This is because the inhibitor reduces the concentration of the enzyme. The remaining enzyme molecules still have the same affinity for the substrate molecule. Therefore, KM is not changed as shown in the Michaelis-Menten and Lineweaver-Burke plots of Figure 4.11. Exam Tips Remember how to obtain Vmax from a plot of velocity against substrate concentration and derive the enzyme concentration at that point. You should be able to write the MichaelisMenten formula and derive V = Vmax 2 when [S] = Km. Remember also how to get KM, Vmax and the curves from competitive and noncompetitive inhibition from Lineweaver-Burke plot. 2015 Biology Term 1 STPM Chapter 4 Enzymes 4

149 3. Another example of non-competitive inhibition is nerve gas, diisopropyfluoro-phosphate (DFP) used in chemical warfare. It binds to the serine side chain of the enzyme acetylcholinesterase active site irreversibly destroying the enzyme. The enzyme is required for clearing the synapse of acetylcholine so that continual transmission across the synapse can take place. The nerve gas prevents transmission of impulse, thus causing paralysis. This results in death as no impulse can be transmitted to the muscles used for breathing. Effects of Competitive and Non-competitive Inhibitions Table 4.1 Comparison of competitive inhibition and non-competitive inhibition Competitive inhibition Non-competitive inhibition 1. Inhibitor binds to active site Inhibitor binds to other sites 2. Temporary inhibition May be permanent inhibition 3. Can be reversed by adding more substrate Cannot be reversed by adding more substrate 4. Inhibitor blocks the active site Inhibitor changes the shape of active sites 5. Shape of enzyme is not changed Shape of enzyme is usually changed 6. KM becomes bigger and Vmax remains the same KM remains the same and Vmax becomes smaller 7. Lineweaver-Burke plot: X intercept (– 1 KM ) becomes smaller, Y intercept (Vmax) remains the same Lineweaver-Burke plot: X intercept (– 1 KM ) remains the same, Y intercept (Vmax) becomes bigger With inhibitor Velocity Vmax Vmax reduced Without inhibitor With more inhibitor Km [Substrate] 1 [S] – 1 Vmax 1/ v With more inhibitor with inhibitor Without inhibitor increases 1 Km – no change Figure 4.11 Graph of velocity plot against substrate concentration Biology Term 1 STPM Chapter 4 Enzymes 4

150 4.5 Classification of Enzymes 1. According to the International Union of Biochemistry (IUB), enzymes are divided into six groups i.e. oxidoreductase, transferase, hydrolase, lyase, isomerase, ligase/synthetase. Oxidoreductase 1. Oxidoreductase is an enzyme that transfers oxygen, electron or hydrogen ion from a substrate molecule to another molecule. 2. Examples are as follows: (a) Dehydrogenase Dehydrogenase AH2 + B → A + BH2 Dehydrogenase Phosphoglyceraldehyde(–CHO) + NAD+ → Phosphoglyceric acid (–COOH) + NADH (b) Oxidase Oxidase AH2 + O → A + H2 O Cytochrome a (Cu+) + 2H+ + 1 / 2 O2 → cytochrome a (Cu2+) + H2 O Transferase 1. Transferase is an enzyme that transfers special groups such as methyl, acyl, amino or phosphate from one molecule to another. 2. Examples are as follows: AB + C → A + BC (a) Transaminase NH2 CR1 HCOOH + R2 COCOOH → R1 COCOOH + NH2 CR2 HCOOH Amino acid 1 Keto acid 2 Keto acid 1 Amino acid 2 (b) Phosphorylase Glycogen + phosphate → glucose phosphate Hydrolase 1. Hydrolase is an enzyme that catalyses hydrolysis or breaking up of a complex chemical with water. 2. Examples are as follows: AB + H2 O (HOH) → AH + BOH (a) Maltase 2013 2015 Students should be able to: (a) describe enzyme classification according to International Union of Biochemistry (IUB) e.g. oxidoreductase, transferase, hydrolase, lyase, isomerase and ligase. Learning Outcomes Biology Term 1 STPM Chapter 4 Enzymes 4

151 Maltose + H2 O → two glucose (b) Lipase, amylase, peptidase, DNase, RNase and phosphotase Lyase 1. Lyase is an enzyme that adds or removes a group of atoms from a substrate with the breaking of C–C, C–N, C–O or C–S bond. 2. Examples are as follows: AB → A + B (a) Decarboxylase (removal of CO2 ) Pyruvate + coenzyme A + NAD → Acetyl coenzyme A + NADH + H+ + CO2 (b) Carboxylase (fixation of CO2 ) Ribulose biphosphate (RuBP) + H2 O + CO2 → 2 Phosphoglyceric acid (PGA) Isomerase 1. Isomerase is an enzyme that rearranges the internal structure of a certain molecule or changes of one isomer to another. 2. Examples are as follows: AB → BA (a) Phosphoglucomutase Glucose-1-phosphate → Glucose-6-phosphate (b) Phosphohexose isomerase Glucose-6-phosphate → Fructose-6-phosphate Ligase/ synthetase (or synthase – USA) 1. Ligase is an enzyme that catalyses the joining of two molecules with the formation of C–C, C–N, C–O or C–S making use of energy from ATP. 2. Examples are as follows: A + B + ATP → AB + ADP + Pi (a) Aminoacyl tRNA synthetase Glycine + tRNA + ATP → Glycine-tRNA + AMP + pyrophosphate (PPi ) (b) Ligase DNA Short DNA + short DNA + ATP → long DNA + ADP + Pi Classification of enzymes 1. Oxidoreductase (a) AH2 + B → A + BH2 (b) AH2 + O → A + H2 O 2. Transferase AB + C → A + BC 3. Hydrolase AB + H2 O → AH + BOH 4. Lyase AB → A + B 5. Isomerase AB → BA 6. Ligase A + B + ATP → AB + ADP + P Summary Exam Tips Remember the six groups of enzymes and their catalytic actions (STPM 2013). Biology Term 1 STPM Chapter 4 Enzymes 4

152 4.6 Enzyme Technology Enzyme immobilisation 2013 2018 1. It is a technology of binding enzyme to a solid medium so that the enzyme can be used and reused to carry out an industrial process continuously. 2. Examples of enzymes that can be immobilised are as follows: (a) Glucose isomerase is used to convert glucose to fructose that is sweeter in taste. It would be more economical to use fructose for sweetening in industries. (b) Amylase is used to convert starch to maltose. Starch is a more economical source of carbohydrate to produce maltose and then glucose. (c) Proteinase is used to clear liquid of protein so that the liquid looks clearer such as beer. 3. Medium are present in the form of beads or membrane made from polymers. The techniques involved are adsorption, entrapment and covalent coupling as shown below in Figure 4.12. E E Porous polymeric matrix E E E E Semipermeable membrane (a) Adsorption (c) Entrapmant (b) Covalent coupling E Enzyme molecule Solid or porous supports E __ _ _ _ _ _ _ ___ __ _ _ _ _ _ _ _ _ _ __ ++ + _ + + + + + +++ E __ _ _ _ _ _ _ ___ __ _ _ _ _ _ _ _ _ _ ___ E __ _ _ _ _ _ _ ___ __ _ _ _ _ _ _ _ _ _ ___ + + + +++ ++ ++ ++ + + + +++ + + + + + + ++ + + + + + E E E Keys: Figure 4.12 Techniques of enzyme immobilisation Importance and techniques of enzyme immobilisation 1. Importance (a) Enzyme can be reused (b) Enzyme can be controlled easily (c) Enzyme can have longer shelf life (d) Enzyme can work better (e) Products not contaminated by enzyme (f) Operation can be cheaper 2. Techniques (a) Adsorption – bound to surface of membrane by hydrogen bonds (b) Entrapment – bound by porous substance to form beads (c) Covalent coupling – bonded by covalent bond like disulfide bond on solid medium Summary Students should be able to: (a) explain the importance and the main techniques of enzyme immobilisation namely adsorption, entrapment and covalent coupling; (b) explain the application of enzyme immobilisation in the development of biosensors. Learning Outcomes Biology Term 1 STPM Chapter 4 Enzymes 4

153 (a) Adsorption The enzyme molecules are bound to the surface of medium e.g. cellulose membrane by hydrogen bonds. (b) Entrapment The enzyme molecules are trapped or encapsulated in porous solid bead e.g. when mixed with sodium alginate solution and drip into calcium chloride to form solid calcium alginate beads. (c) Covalent coupling The enzyme form covalent bonds (eg. disulfide bonds) with solid medium such as polymer membrane or cross-linking organic molecules like proteins. 4. Procedure for immobilisation is as follows: (a) The enzyme is bound to the medium with the help of certain mineral ions like Ca+. + Ca– Ca+ Enzyme + polymer bead immobilised enzyme (b) The beads are then loaded into a reactor. Immobilised enzyme Substrate Reactor Product Figure 4.13 Continual production of product using immobilised enzymes (c) The substrate in the form of a solution is poured to the top of the reactor or allowed to flow through the enzyme. (d) When the substrate flows through the enzyme, it is converted into product. (e) The product flows out continuously as long as the substrate is allowed to flow through the enzyme. Biology Term 1 STPM Chapter 4 Enzymes 4

154 5. The advantages of enzyme immobilisation in the manufacturing industry are as follows: (a) Immobilised enzyme can be controlled more easily (b) The enzymes have a longer shelf life or more stable due to protection by inert support from proteolysis (c) The enzymes are protected from thermal denaturation as they vibrate less when immobilised (d) The enzymes can be reused as they do not get washed out of the reactor (e) The enzymes are not diluted by the medium like free enzymes (f) Some enzymes work better as they are naturally attached to membrane in cells (g) The products are easily purified as they are free from contamination (h) The system is ideal for continuous process (i) The enzymes are able to operate at wide pH range than in solution (j) The system can be much cheaper to use especially in large amount 6. The disadvantages can be as follows. (a) Immobilisation may alter the shape and active site of enzyme (b) The immobilisation may alter the catalytic ability of the enzymes (c) Enzyme may become detached and so may not function efficiently (d) It might be expensive to immobilise some enzymes so it may not be cost effective Biosensor 1. Biosensor is a device using immobilised enzyme to detect the presence and to estimate the concentration of certain chemicals. Besides enzyme, antibodies, receptor proteins and microorganisms can be used in biosensors. 2. The components of a biosensor are as follows: (a) Immobilised enzyme (or antibody or living cell) is used to react with chemicals such as glucose that is to be detected. glucose C6 H12O6 (glucose) + O2 ⎯⎯→ C6 H10O6 (gluconic acid) + H2 O2 oxidase (b) A transducer converts the product (for example gluconic acid or hydrogen peroxide) of the reaction that is usually reactive to the electrode to become an electric current. (c) An amplifier magnifies this tiny electric current so that it is easier to be processed. (d) Microelectronic processor processes the electric current so that the chemical concentration becomes a numerical reading shown on a screen. Biology Term 1 STPM Chapter 4 Enzymes 4

155 Quick Check 2 1. Suggest three ways how actions of cofactors can control the enzymes activities. 2. Suggest whether competitive or non-competitive inhibitor is more suitable for end product inhibition. Explain your answer. 3. In the development of nano-technology in which machine is made into a few nanometers, how would immobilised enzymes be useful to our body? Exam Tips Remember the meaning and examples of uses of enzyme immobilisation and biosensing (e) An example of biosensor is as shown in Figure 4.14. Biosensor 1. Immobilised enzyme acts on substance 2. Product is polar and is changed to current by transducer 3. Amplifier increases current strength 4. Microprocessor converts current to readable value correspondingly 5. So, it can determine the amount of substance Summary Read-out unit BioBeep 12.34 Substrate Recognition layer Transducer Membrane allowing the chemical to go in Immobilised enzyme Product enters the next chamber Transducer Amplifier Microelectronic processor Numerical reading Figure 4.14 Schematic diagram of a biosensor 3. Biosensor can be used to detect the presence and to estimate the concentration of chemical in many fields. (a) In the manufacturing industry, the concentration of raw material and finished product can be monitored by biosensor. For example, during fermentation of beer, the concentration of maltose, sucrose and ethanol produced can be monitored. (b) In the agricultural industry, the concentration of elements or mineral ions in the soil can be monitored. Suitable concentration and quality of fertiliser is added based on this sensing. (c) In the environmental study, the pollution of toxic substances in the soil, river, air or coastal areas can be monitored. This sensing can avoid predictable catastrophes. (d) In the medical field, the monitor of chemical in a patient can be done with biosensor. This can prevent imminent attack of certain diseases or disorders. Moreover, a patient can monitor his own blood sugar with micro-biosensor. If incorporated with automatic insulin injecting device, a diabetis patient can almost live a normal life. 4. Research into the miniaturisation of biosensor to the point that it can be implanted in the vein can lead to the development of ‘biochips’. These multi-purpose ‘biochips’ can detect many chemicals at the same time. If these ‘biochips’ are interfaced with other automatic devices, long distance monitoring and automatic remedies are not impossible in the future. Biology Term 1 STPM Chapter 4 Enzymes 4

156 Objective Questions 1. Which functional groups do not form the active site of an enzyme? I A few amino groups of certain amino acids II The amino group at beginning of the chain III The carboxyl group at the end of the chain IV A few carboxyl groups of certain amino acids V A few free sulphur groups A I and II B I and IV C II and III D III, IV and V 2. Which value is used to calculate MichaelisMenten constant, KM in the graph below that shows Lineweaver-Burke plot for an enzyme-catalysed reaction? II 1 – III I 1 – V IV [S] A IV C II B III D I 3. Which combination is true? Enzyme Coenzyme A Small molecule, a type of protein, specific action and denatured by heat Large molecule, not a type of protein, not specific in action and not denatured by heat B Large molecule, a type of protein, specific action and denatured by heat Small molecule, not a type of protein, not specific in action and not denatured by heat C Small molecule, a type of protein, specific action and denatured by heat Large molecule, a type of protein, not specific in action and not denatured by heat D Large molecule, a type of protein, specific action and denatured by heat Small molecule, a type of protein, not specific in action and not denatured by heat 4. Which pair is the example of a cofactor and an activator? Coenzyme Activator A Fe2+ NAD+ B NAD+ Mn2+ C NAD+ Fe2+ D Fe2+ Mn2+ 5. What is the role of boron that decreases the synthesis of chlorophyll? A Non-competitive inhibitor B Competitive inhibitor C Coenzyme D Cofactor 6. What result will be obtained by a competitive inhibitor on an enzymatic reaction? A A decrease in Vmax B An increase in KM C The denaturation of enzyme D A change in the conformation of the active site STPM PRACTICE 4 Biology Term 1 STPM Chapter 4 Enzymes 4

157 7. Substance Y increases the Michaelis-Menten constant for an enzyme-catalysed reaction. What is substance Y? A An allosteric inhibitor B A competitive inhibitor C An irreversible inhibitor D A non-competitive inhibitor 8. Compared to competitive inhibitor, noncompetitive inhibitor A has constant maximum velocity, Vmax B acts on the allosteric site of the enzyme C has the same shape of substrate molecule D varies its Michaelis-Menten, KM, with the concentration of substrate 9. Which combination is correct? Substance Property A Coenzyme Reduces the rate of enzymatic reaction by changing its structure B Competitive inhibitor Resembles and occupies the same site with substrate C Enzyme activator Binds permanently to the enzyme molecule D Non-competitive inhibitor Does not change the conformation of the enzyme 10. Which is true of penicillin that blocks the formation of bacterial wall by mimicking substrate for the enzyme transpeptidase? A It destroys the three dimensional configuration of the enzyme. B It blocks the substrate from entering the active site. C It binds to the allosteric site of the enzyme D It forms a complex with the substrate 11. Which is true about the initial velocity of enzymatic reactions? A The affinity of the enzyme towards the substrate increases. B The active sites of the enzyme are saturated. C The speed of substrate movement increases. D More substrates bind to the enzyme. 12. The graph below shows the effect of pH on the activities of two enzymes, X and Y. Enzymes X and Y are 2 X Y 4 6 pH 0 8 10 12 Enzyme activity X Y A amylase pepsin B pepsin renin C pepsin amylase D renin pepsin 13. Which is true of non-competitive inhibitor? A It depends on the relative concentration of substrate B It increases the KM value of the reaction C It binds to the active site of the enzyme D It changes the conformation of enzyme. 14. Factors that cause denaturation of enzymes include I competitive inhibitor II extreme pH III very high temperature IV presence of heavy metallic ions A I and III B I and IV C I, III and IV D II, III and IV Biology Term 1 STPM Chapter 4 Enzymes 4

158 15. The table shows the class and action of enzymes. Class of enzyme Actions I Ligase w Redox reactions II Hydrolase x Rearrangements within a molecule III Isomerase y Bond formation between two compounds IV Oxidoreductase z Splitting of a large substrate molecule into smaller products Which combination is correct? I II III IV A y w z x B y z x w C z w y x D z y x w 16. Which class of enzyme catalyses the reaction below? R1 R2 R2 R1 | | | | H–N–C –C–OH + H–C–C–OH → H–N–C–C–OH + H–C–C–OH | | || | || | | || | || H H O H O H H O H O A Isomerase C Transferase B Lyase D Ligase 17. Which class of combination is correct? Class of enzyme Reaction A Lyase Pyruvate → ethanol + CO2 B Isomerase Glucose + ATP → glucose6-phosphate C Transferase Sucrose + water → glucose + fructose D Hydrolase Glucose-1-phosphate → glucose-6-phosphate 18. What is the significance of immobilised enzyme? A To enable enzymes to react at 60 °C B To enable enzymes to be used continuously C To produce biological detergent D To enable enzyme to be purified before use 19. Which is a useful characteristic of enzyme immobilisation? I Enzyme can be easily recovered and reused II Enzyme is more stable III The end-product does not contain enzyme and easily isolated IV The properties of enzyme can be changed so that the surface area for its reaction can be increased A I, II and III B I, II and IV C I, III and IV D II, III and IV 20. Biocemical reactions normally have a temperature coefficient (Q10) of 2 from) 0°C to optimum temperature. Which is true of Q10? A The doubling in rate of reactions for every 10°C rise in temperature B The change in the rate of reaction for every 10°C rise in temperature C The temperature 40°C that promotes maximum activity D At 38°C, the Q10 for biochemical reactions is around 2. Biology Term 1 STPM Chapter 4 Enzymes 4

159 Structured Questions 1. The graph below shows how enzyme is capable of speeding up reaction. Progress of reaction Free energy Overall change in energy (the same in both cases) Catalysed reaction Ea of catalysed reaction (lower) Ea of uncatalysed reaction (higher) Uncatalysed reaction Initial state → intermediate state → final state (substrate) (enzyme-substrate complex) (product) H2 O2 → H2 O + O2 + energy Hydrogen peroxide water oxygen (a) What is activation energy? [1] (b) Why is the activation energy high for an uncatalysed reaction? [2] (c) How is activation energy and Michaelis-Menten related? [2] (d) Explain how enzyme lowers the activation energy of a reaction. [2] (e) Label the enzyme involved in the above reaction and name another enzyme having the same effect. [2] (f) Name an inhibitor that would stop the reaction when added to the above enzyme. [1] Essay Questions 1. (a) How an enzyme accelerate a chemical reaction. Explain. [6] (b) State the four most important properties of enzymes.[4] (c) Based on induced fit hypothesis, describe the mechanism of an enzyme action. [5] 2. Explain the International Union of Biochemistry (IUB) system of enzyme classification with the help of suitable examples. Why is there a need for enzymes to have standard names? [15] 3. (a) Explain what an enzyme inhibitor is. Differentiate between competitive and non-competitive inhibitions using Lineweaver-Burke plot. [5] (b) State five classes of enzyme and state its function. [10] 4. (a) Explain enzyme immobilisation. [2] (b) (i) What is biosensor? [4] (ii) Give two applications of biosensor, and state one example of its usage. [3] (c) Give six advantages of using immobilised enzymes. [6] Biology Term 1 STPM Chapter 4 Enzymes 4

160 1 1. • Oxido-reduction. They can remove or add oxygen, hydrogen or electron to the substrate. • Transferring. They can transfer atom, molecule or group of atoms from one substrate to another. • Hydrolysis. They can use water to split a larger substrate into two smaller products. • Splitting. They can break up a larger substrate into two smaller products. • Isomerisation. They can transfer atom, molecules or group of atoms from one part of a substrate into an other part of the same substrate. • Combining. They can combine two smaller substrates to form a bigger product. 2. • The side chains of amino acids can be divided into polar and non-polar in nature. The polar ones can also be divided into positive charged and negative charged. • The strongly positive ones tend to attract electrons or negative groups away from the substrate molecules. • The strongly negative ones tend to attract hydrogen ions or positive groups away from the substrate molecules. • The non-polar side chains tend to attract and rearrange non-polar groups with substrate molecules. • All these help enzyme to bring about oxidoreduction, transferring, hydrolysis, splitting, isomerisation and combining two substrates. 3. • A graph of velocity of the enzyme reaction is plotted against substrate concentration. From the graph, the maximum velocity is determined, so is its half maximum velocity. Then, the Km i.e. the substrate concentration at half maximum velocity can be read from the graph. • A Lineweaver-Burke plot of 1/v against 1/[s] is drawn. The X intercept is equal to -1/Km. Therefore, Km can then be calculated. 2 1. • Enzyme activities can be controlled by the availability of coenzymes such as NAD. • They can also be controlled by the availability of prosthetic groups such as porphyrin or iron ions. • They can be controlled by the availability of their activators such as magnesium ion. 2. • Non-competitive inhibitor. This is because the end product can bind to the enzyme's allosteric site to inactivate the enzyme. 3. • Through biosensor. The biosensor is so small that it can be implanted into any part of the body for monitoring the presence of chemicals or determining their concentration. These chemicals include glucose, amino acids, fatty acids, cholesterol, any other waste products or exogenous poisons. STPM Practice 4 Objective questions 1. B 2. C 3. B 4. A 5. D 6. B 7. B 8. B 9. B 10. B 11. C 12. C 13. D 14. D 15. B 16. C 17. B 18. B 19. A 20. B Structured Questions 1. (a) Activation energy is the minimum energy required to start a reaction such as a spark is required to start a reaction between oxygen and petrol. (b) Energy in the form of heat is required to raise the energy level of a reactant or reactants so that the bonds within can be broken. Energy is also required to split the reactants or to combine reacting reactants to form a single product. (c) Activation energy is directly proportional to the Km, the substrate concentration to produce half the maximum velocity. Reaction with low activation energy starts easily and less substrate is required to saturate the enzyme active site. This is also true that the enzyme has high affinity for the substrate. (d) Enzyme actively participates in weakening the bonds of the substrate to that it is easily converted into product. Enzyme also provides sites for two reacting substrates to be close to each other so that reaction can occur to join them. (e) The enzyme is catalase. Another enzyme that is exergonic is urease. (f) An inhibitor would be heavy metallic ion such as mercury (Hg+). Essay Questions 1. (a) • Enzyme accelerates a chemical reaction with its active site that binds complementarily with substrate. • Only specific substrate can bind like key into lock. ANSWERS Biology Term 1 STPM Chapter 4 Enzymes 4

161 • Then, an enzyme-substrate complex is formed. • Weak bonds like hydrogen bonds are formed. • The enzyme stresses the substrate molecule weakening bonds of substrate. • The substrate is converted into product that leaves the active site. • Enzyme accelerates reaction by lowering the activation energy. + Enzyme Substrate Enzyme-substrate complex Enzyme Products + Enzyme-products complex • Besides that, the enzyme molecule provides site where two reacting substrate molecules are brought close together and form a product. (b) • All enzymes are globular proteins. • They speed up reactions. • They do not change the nature of the product of reactions that they catalyse. • A small amount of enzyme can bring about a lot of substrate to become products. • All enzyme reactions are reversible. • Enzyme is very specific. • Enzyme activities can be affected by temperature. (c) • The hypothesis states that substrate binds to the enzyme’s active site specifically but loosely. • Only when the substrate binds to the active site, then the enzyme molecule ‘moulds’ around the substrate converting it into a product. • The enzyme molecule forms weak bonds with the substrate. • Bonds within the substrate molecule are broken. • New bonds are formed, converting the substrate into the product. 2. • The IUB system of enzyme classification divides enzymes into six classes, namely oxidoreductase, transferase, hydrolase, lyase, isomerase and ligase. • Oxidoreductase catalyses oxidation and reduction, in which addition or removal of oxygen, hydrogen or electron occurs. An example is succinic dehydrogenase that catalyses the removal of two hydrogen ions from succinic acid to become fumaric acid. • Transferase catalyses the transfer of atom, molecule or group of atoms other than oxygen and hydrogen from one substrate to another. An example is transminase that catalyses the transfer of amino group from an amino acid to a keto acid forming a new amino acid. • Hydrolase is a digestive enzyme that catalyses the hydrolysis of a large molecule to form smaller molecules with the help of water. An example is sucrase that catalyses the digestion of sucrose to form glucose and galactose. • Lyase catalyses the break up of a molecule other than hydrolysis to form smaller molecules. An example of lyase is decarboxylase that catalyses the removal of carbon dioxide from an organic acid like iso-citric acid to form α-keto glutaric acid. • Isomerase catalyses isomerisation that converts one type of isomer to another through internal rearrangement of atoms within a molecule. An example of isomerase is glucose-6-phosphate mutase that converts glucose-6-phosphate to fructose-6-phosphate. • Lastly, ligase catalyses the combination of smaller molecules to form a bigger molecule with the help of ATP. An example is glycogen synthetase that converts glucose with the help of ATP to form glycogen. • There is a need for enzymes to have a standard name because different scientists would call the same enzyme by different names. For example salivary amylase is also called ptyalin. • An international system is also needed to standardise enzyme names so that when a particular enzyme is referred, there is certainty to which it is referred even in any language. 3. (a) • An enzyme inhibitor is any inorganic or organic substance that can decrease the rate or stop an enzyme-catalysed reaction. • Examples are cyanide that stops cytochrome oxidase and malonate that slows down succinate dehydrogenase. • Competitive inhibition produces a straight line graph that passes through the y intercept at a fixed point, similar to that of the enzyme reaction without the inhibitor. The graph is unlike that of non-competitive inhibition that passes through the same x intercept. • The graph is as follow. Non-competitive inhibition __1 v _1 [S] __1 Vmax More inhibitor With inhibitor Without inhibitor 1 KM Competitive inhibition __1 v _1 [S] __1 Vmax More inhibitor With inhibitor Without inhibitor 1 KM Biology Term 1 STPM Chapter 4 Enzymes 4

162 • This is because competitive inhibitor does not change the maximum velocity so the reciprocal of the maximum velocity is constant compared to non-competitive inhibitor which lowers the maximum velocity. • Competitive inhibition increases the Km so the negative of reciprocal of Km will increase in the Lineweaver-Burke plot compared to that of non-competitive inhibition. (b) • First class is oxidoreductase. • Oxidoreductase transfers oxygen atoms, electrons or hydrogen ions from one substrate molecule to another. • Second class is transferase. • Transferase transfers special groups such as methyl, acyl, amino or phosphate from one substrate molecule to another. • Third class is hydrolase. • Hydrolase catalyses the hydrolysis or breaking up of a large molecule using water molecule into smaller molecules. • Fourth class is lyases. • Lyase adds or removes a group of atoms from a substrate molecule with breaking up of C-C, C-N, C-O or C-S bond. • Fifth class is isomerase. • Isomerase catalyses isomerisation i.e. the interconversion of one isomer into another by rearrangement of atoms within a molecule. 4. (a) • It is the attachment of enzyme to solid medium such as polymer. • This is to use the enzyme continuously for industrial process without contaminating the product. (b) (i) • Biosensor is the use of biological material to detect the presence or concentration of chemicals. • The materials used are immobilised enzymes, antibodies or living cells. • The enzyme is allowed to react with the chemical such as glucose that diffuses into a special compartment. • The product of the reaction such as acid can be detected by an electrode. • The electrode acts as transducer with the help of amplifier and processor to give a numerical reading in a screen. (ii) • Biosensor is used in the medical field to detect the concentration of glucose in blood or urine. • It is used in pollution studies to detect toxic substances in water. (c) • Immobilised enzymes have a longer shelf life or more stable due to protection from proteolysis by the inert support. • The enzymes are protected from thermal denaturation as they vibrate less when immobilised. • The enzymes can be released as they do not get washed out from the reactor. • The enzymes are not diluted by the medium like free enzymes. • The products are easily purified as they are free from contamination. • Some enzymes work better as they are naturally attached to a membrane in cell. Biology Term 1 STPM Chapter 4 Enzymes 4

CHAPTER 5 CELLULAR RESPIRATION Concept Map Respiration The need for energy Aerobic respiration Anaerobic respiration in living organisms Importance of energy and respiration The structure of energy carriers: ATP, NADH and FADH2 Stages and location Yeast Muscle cells Glycolysis and net energy produce Link reaction and Krebs Cycle Applications of anaerobic respiration in food industries (bread, Formation of NADH, FADH tapai and yogurt) 2, GTP and ATP in Krebs cycle Oxidative phosphorylation and chemiosmosis in electron transport system Role of NADH, FADH2 and ATP synthase in electron transport system Energy produced in aerobic respiration in liver and muscle cells Effect of cyanide & carbon monoxide on respiration Lipid & protein as alternative energy source Bilingual Keywords Cytochrome – Sitokrom Dehydrogenation – Pendehidrogenan Cyanide – Sianida Decarboxylation – Pendekarboksilan Fermentation – Penapaian

164 The Importance of Respiration and Energy 1. Respiration is important as it is a process of breaking down of food molecules with the release of energy in the form of ATP. These food molecules are carbohydrates, lipids and proteins. 2. Complex food molecules need to be broken down into simple molecules before they are respired. An example is glycogen in the liver and muscles must be broken down into glucose. Glucose is then broken down into a simple organic acid step by step. The simple organic acid is further broken into water and carbon dioxide. 3. Similarly, lipids like triglycerides are first broken into glycerol and fatty acid. Then, they are broken down further into simple organic acids before they are completely broken down. 4. Proteins likewise are first broken down into amino acids. Then, the amno acids are broken down completely into ammonia, carbon dioxide and water. 5. ATP is produced in all cases, step by step in small amounts. Hydrogen atoms are removed at various stages combined with coenzyme such as NAD so that ATP is formed with the help of oxygen to form water. 6. ATP is used by cells as immediate energy donor with inorganic phosphate that is readily released and may bind with other organic molecules. The major part of energy given out is in the form of heat. This heat is used to keep the body temperature high especially for mammals and birds. 7. ATP acts as a link between energy yielding and energy requiring reactions. (a) ATP is used in anabolic reactions i.e. reactions involving building or synthesis of substances in which energy or phosphate group is obtained from ATP. Examples are the synthesis of glycogen (glycogenesis) and Calvin cycle during photosynthesis. (b) It is used in active transport i.e. passage of solutes across membrane against concentration gradient. An example is the absorption of food substances in the small intestine. (c) It is used in contraction of muscles i.e. breaking and formation of bonds between muscle proteins at molecular levels. This enables the muscle fibres to contract when the protein filaments slide over one another. (d) It is used in transmission of impulse i.e. maintenance of sodium pump function after membrane is depolarised, before another impulse can be transmitted in a nerve cell. (e) It is used in bioluminescence i.e. conversion of special molecules that can give out light, as in the tail of fireflies. Remember that respiration of glucose is not a one step reaction and it is not the reverse of photosynthesis. Exam Tips The importance of respiration and energy 1. Respiration produces ATP as a source of energy 2. Energy in the form of ATP is important for (a) anabolic process like synthesis of glycogen (b) active transport to transport food molecules into intestine (c) contraction of muscles for movement (d) transmission of impulse for nervous coordination Summary 2015 5.1 The Need for Energy in Students should be able to: Living Organisms (a) outline the importance of energy and respiration in living organisms; (b) describe the structure of the energy carriers such as ATP, NADH and FADH2 . Learning Outcomes Biology Term 1 STPM Chapter 5 Cellular Respiration 5

165 Exam Tips Remember the meaning of catalysis and how enzyme lowers the activation energy. (STPM 2010 structured question) Structure of Energy Carriers: ATP, NADH and FADH2 ATP 1. The structure of adenosine triphosphate (ATP) is as shown in Figure 5.1. Phosphate groups O– – O O P O O– P O CH2 O O– O O H NH2 P N C C C CH N Adenine N N HC H OH Ribose H OH H O Figure 5.1 Structure of ATP P Adenosine diphosphate (ADP) Inorganic phosphate P + Energy i + P H2O P Adenosine triphosphate (ATP) P P Figure 5.2 Hydrolysis of ATP 2. ATP is a nucleotide consisting of adenine, ribose and three phosphate joined by covalent bonds. This ATP is the same nucleotide used to form RNA. It is constantly produced in the cytoplasm. A larger amount is produced by the mitochondria. Thus, active cells have more mitochondria to release more ATP when it is required. 3. Loss of a phosphate by hydrolysis leads to small packets of energy (30.5 kJ) given out. This reaction is reversible by condensation known as phosphorylation during respiration. The ATP is constantly being used by the cell. Therefore, ATP and inorganic phosphate have to combine to reform ATP and thus, mitohondria have to reabsorb the ADP and inorganic phosphate. 4. ATP is sometimes hydrolysed into AMP and pyrophosphate (PPi). Small packets of energy are also given out for more specific purpose. During other instances, ATP may also be broken into cyclic AMP (cAMP) in the membrane. The cAMP acts as a second messenger rather than to release energy. 5. The ATP molecule is considered small and water soluble, so that it can move around cells easily. ATP needs to be able to move effeciently inside the cell. Biology Term 1 STPM Chapter 5 Cellular Respiration 5

166 NADH 1. The structure of NADH is as shown in Figure 5.3(b). O O O O P O NH2 N+ O H O OH OH H2N + H+ + 2e O N N N N OH Adenosine monophosphate (AMP) Nicotinamide mononucleotide (NMN) OH(P) NAD+ (NADP+) O O P O O O P O NH2 N H H O OH OH H2N O N N N N OH OH(P) NADH (NADPH) O O P – – – – O O C NH2 N Ribose 340 nm O H H (b) Reduced: NADH 2P Adenosine NH2 + H+ + 2 e C – N + Ribose O H (a) Oxidised: NAD+ Nicotinamide 2P Adenosine NAD+ + H+ + 2 e– NADH Figure 5.3 Reduced NAD (NADH) and oxidised NAD+ 2. NADH is a reduced form of NAD+ consisting of two nucleotides i.e. adenine, two ribose, two phosphates, nicotinamide and a H atom. 3. Loss of a hydrogen atom when it comes near to the NADH reductase turns into oxidised NAD+. This occurs in the inner mitochondrial membrane and NADH reductase is the first electron carrier of the electron transport chain. Three ATPs can be formed from one NADH. This occurs during aerobic respiration in which oxygen is required. 4. NADH can pass the hydrogen atom to other compounds like pyruvate in the cytoplasm to form lactate catalysed by lactate dehydrogenase during anaerobic respiration. Similarly, NADH can pass the hydrogen atom to ethanal to form alcohol during alcohol fermentation. Structures of energy carrier 1. ATP: Adenine, ribose and 3 Pi 2. NADH: Nicotinamide adenine, 2 riboses, 2 Pi and 1 H atom 3. FADH2 : Flavin, ribitol, adenine, 2 Pi and 2 H atoms Summary Biology Term 1 STPM Chapter 5 Cellular Respiration 5

167 5. NADH usually binds to several types of dehydrogenases as it is a coenzyme. NADH receives the hydrogen atom from a metabolite during dehydrogenation reaction. FADH2 1. The structure of FADH2 is as shown in Figure 5.4. N N N NH2 N O O OH OH H2CO-PO2-O-P-O– FAD+ H3C H3C N FMN N NH O O CH2 HCOH HCOH HCOH CH2 N N N N NH2 N O O OH OH H2CO-PO2-O-P-O– FADH2 H3C H3C N FMNH2 N NH O O CH2 HCOH HCOH HCOH CH2 N 2e– + 2H+ H H Figure 5.4 Reduced FAD (FADH) and oxidised FAD+ 2. FADH2 is a reduced form of FAD consisting of a flavin, ribitol, adenine, two phosphates and two hydrogen atoms. 3. FADH2 passes the two hydrogen atoms to coenzyme Q becoming oxidised FAD. Coenzyme Q is the second electron carrier in the electron transport chain in the inner mitochondrial membrane. Only two ATPs can be formed from one FADH2 . 4. FADH2 is a prosthetic group bonded to succinate dehydrogenase that is found in the mitochondrial matrix. Biology Term 1 STPM Chapter 5 Cellular Respiration 5

168 5.2 Aerobic Respiration Stages of Aerobic Respiration and Their Locations 1. Aerobic respiration consists of glycolysis, link reaction, Krebs cycle and oxidative phosphorylation. 2. Glycolysis occurs in the cytoplasm. Link reaction and Krebs cycle occur in the matrix of mitochondria and oxidative phosphorylation occurs in the inner membrane of mitochondria as shown in Figure 5.5. Glucose Oxygen Glycolysis Krebs (tricarboxylic acid) cycle Pyruvate Hydrogen atoms Hydrogen atoms Electron (hydrogen) transport system ATP Phosphorylation ATP Water Carbon dioxide Substrates Stage of Aerobic Respiration Products In the cytoplasm In the mitochondrion Oxidative phosphorylation (cristae) Figure 5.5 Aerobic respiration Stages of aerobic respiration and locations 1. Glycolysis (glucose → pyruvate) occurs in cytoplasm 2. Link reaction (pyruvate → acetyl CoA) – occurs in matrix of mitochondria 3. Krebs cycle (acetyl coA → CO2 and H2 O) – occurs in matrix of mitochondria 4. Oxidative phosphorylation (NADH and FADH2 → ATP) occurs in cristae Summary 2015 VIDEO Aerobic Respiration Students should be able to: (a) describe the various stages of aerobic respiration and its location in the cells; (b) describe glycolysis, and calculate the net energy produced in glycolysis; (c) describe the various steps involved in the Krebs cycle (including the link reaction); (d) explain the formation of NADH, FADH2 , GTP and ATP during the Krebs cycle; (e) describe oxidative phosphorylation and chemiosmosis in the electron transport system; (f) explain the role of NADH, FADH2 and ATP synthase in the electron transport chain; (g) calculate and explain the net energy produced in aerobic respiration per molecule of glucose in liver and muscle cells; (h) describe the effects of cyanide and carbon monoxide on respiration; (i) explain how lipid and protein act as alternative energy sources. Learning Outcomes Biology Term 1 STPM Chapter 5 Cellular Respiration 5

169 Glycolysis and Net Energy Produced 1. Glycolysis is a process consisting of a series of reactions in which glucose is split and finally forms two molecules of pyruvic acid. Each reaction is controlled by an enzyme. It is a process that takes place at the beginning of aerobic respiration and requires no oxygen. 2. Glycolysis may starts with glucose which comes from starch in plants after hydrolysis or glycogen after glycogenolysis. Glycolysis can be divided into three stages shown in Figure 5.6. Glucose aldolase isomerase glycerate 2-phosphate glycerate 3-phosphate glycerate-1,3-biphosphate phosphoenolpyruvate Enters the link reaction Pyruvate fructose-1,6-biphosphate fructose 6-phosphate glucose 6-phosphate dihydroxyacetone phosphate (3C) glyceraldehyde 3-phosphate (GALP) hydrolysis glycogenolysis ATP hexokinase ADP ATP ADP NAD+ ADP ATP H2O ADP ATP phosphoglyceraldehyde dehydrogenase phosphofructokinase phosphoglucoisomerase phosphoglycerokinase enolase pyruvate kinase NADH + H+ Starch Glycogen Figure 5.6 Summary of glycolysis Glycolysis 1. Phosphorylation of glucose forms fructose bisphosphate 2. 2 ATP are used 3. Fructose bisphosphate is split into 2 triose phosphates 4. Each triose phosphate forms 1 pyruvate. 2 pyruvates are formed per glucose. 5. Each triose phosphate can produce 1 NADH i.e. 3 ATP in liver cells. 6. So 2 NADH per glucose can produce 6 ATP. 7. Each triose produces 2 ATP by substrate level phosphorylation, so 4 ATP are produced 8. Therefore from 1 glucose 2 net ATP + 6 from NADH = 8 ATP are produced Summary 'Glyco' means sugar and 'lysis' means split. Hence, glycolysis means the splitting of sugar. Language Check Language Check 2010 Biology Term 1 STPM Chapter 5 Cellular Respiration 5

170 (a) Phosphorylation of sugar/glucose Initially, phosphorylation of glucose occurs to form fructose bisphosphate. (i) Two molecules of ATP are required to form fructose bisphosphate. (ii) The process is important to raise the energy level of the glucose as it consists of only low energy bonds. (iii) This is to create high-energy carbon-phosphate bond so that more energy can be tapped from it later. (b) Lysis (splitting of fructose bisphosphate) (i) Fructose bisphosphate is split to form glyceraldehyde 3-phosphate and dihydroxy acetone phosphate. They are triose phosphates. (ii) Both molecules are then inter-convertible so one glucose molecule is split into two triose molecules. (c) The two triose phosphates are converted into two molecules of pyruvate through a series of processes. (b) During the formation of two pyruvates, dehydrogenation and phosphorylation occur. (e) The two triose phosphates undergo dehydrogenation in which hydrogen atoms are removed and accepted by coenzyme NAD+ to form two molecules of NADH. The two molecules of NADH are transported into the mitochondria, where two molecules of ATP are required (in most cells except heart, liver and kidney cells). (f) Then, substrate level phosphorylation occurs. (i) Two ATP can be formed through phosphorylation at substrate level from each triose phosphate, in which the phosphate ion is transferred to ADP and catalysed by pyruvate kinase as in Figure 5.6. (ii) This process occurs four times. Four molecules of ATP are produced when two pyruvate molecules are formed. 3. As a result, one molecule of glucose yield 2 ATP, 2 NADH and 2 pyruvate molecules through the glycolysis process. Four ATP produced – two ATP used = two net ATP produced. Krebs cycle 1. Krebs cycle is also known as tricarboxylic acid cycle or citric acid cycle. 2017 2. If oxygen is present, each pyruvate molecule formed from glycolysis enters mitochondria and undergoes oxidative decarboxylation. Pyruvate is broken down to form carbon dioxide, H+ and ATP. The H+ ions are accepted either by NAD+ or FAD and then sent to the electron transport system where they are converted into water with the formation of ATP. The process is summarised in Figure 5.7. Exam Tips Remember that a dehydrogenation reaction occurs each time a pyruvate is formed from triose phosphate. This occurs in the cytoplasm and the NADH formed has to be transported into the mitochondria to produce ATP. This NADH can only produce three ATP in liver cells. Biology Term 1 STPM Chapter 5 Cellular Respiration 5

171 CO2 CO2 CO2 NADH NADH NADH NADH FADH2 GTP Succinate (4C) Pyruvate (3C) Acetyl coenzyme A (2C) -ketoglutarate (5C) Krebs cycle Oxaloactate Citrate (6C) Link reaction Figure 5.7 Summary of Krebs cycle 3. Each step of the process is catalysed by an enzyme and the whole process is as shown in Figure 5.8. (a) First, link reaction occurs when pyruvate is converted into acetyl coenzyme A (acetyl CoA). Pyruvate undergoes oxidative decarboxylation in which carbon dioxide is removed and dehydrogenation takes place. Coenzyme A and NAD+ are required to form acetyl coenyme A and NADH. (b) Then, formation of citrate occurs starting the Krebs cycle. Acetyl coenzyme A that consists of two carbon atoms then combines with a four-carbon oxaloacetate to form a six-carbon citrate. (c) Regeneration of oxaloacetate occurs to complete the cycle. (i) Citrate undergoes a series of reactions through α-ketoglutarate and succinate to regenerate oxaloacetate. (ii) In this cyclic process, decarboxylation takes place twice, dehydrogenation takes place four times and formation of GTP from GDP and phosphate, once. (iii) During the dehydrogenation process, three times NAD+ and one time FAD are used as hydrogen acceptors forming NADH and FADH2 respectively. Link reaction and Kreb cycle 1. Link reaction occurs when pyruvate is oxidatively decarboxylated to acetyle CoA 2. Acetyl coA reacts with oxaloacetate to start Krebs cycle. 3. Citrate is converted back to oxaloacetate through α-ketoglutarate and succinate 4. 4 NADH (3 x 4 ATP) are formed from dehydrogenation from 1 pyruvate 5. 1 FADH2 (1 x 2 ATP) is formed 6. 1 GTP (1 x 1 ATP) is formed 7. Therefore, 1 pyruvate can give 15 ATP 8. 1 glucose can give 30 ATP + 8 from glycolysis = 38 ATP Summary 2013, 2015 2011 Biology Term 1 STPM Chapter 5 Cellular Respiration 5

172 Pyruvate COO Acetyl-CoA _ COO_ COO_ CH2 COO_ H COO_ COO_ COO_ NADH, H+ ,CH2 NADH, H+ ,CO2 NAD+ -ketoglutarate dehydrogenase Isocitrate dehydrogenase Aconitase NAD+ COO_ COO_ CHO H C COO_ H CH2 COO COO _ _ CH2 CHO COO_ CH2 COO_ CH2 CH2 C O COO_ CH2 CH2 C O S-CoA CH2 CH2 COO_ C HO C C C _ OOC H H O CH2 CO2 H2 O H2 O + CoA Coenzyme A Pyruvate dehydrogenase NAD+ NAD+ FAD GTP ADP ATP GDP, Pi NADH NADH, H+ FADH2 Oxaloacetate Malate Fumarate Succinate Isocitrate -ketoglutarate Succinyl-CoA Citrate Citrate synthetase Malate dehydrogenase Krebs Cycle Fumarase Succinate dehydrogenase Succinate thiokinase CoA Figure 5.8 The Krebs cycle Formation of NADH, FADH2 , GTP and ATP 1. The formation of NADH, FADH2 , GTP and ATP in Krebs cycle is as shown in Figures 5.7 and 5.8. 2. NADH is produced in three reactions when hydrogen atom is removed each time from the intermediate during dehydrogenation catalysed by dehydrogenase. NAD+ acts as a coenzyme to accept the hydrogen atom each time. 3. FADH2 is produced when succinate is dehydrogenated to fumarate catalysed by succinate dehydrogenase. FAD is the prosthetic group of succinate dehydrogenase and acts as a cofactor to accept two hydrogen atoms. 4. GTP is formed at substrate level phosphorylation from GDP and inorganic phosphate when an intermediate releases energy catalysed by an enzyme. Formation of NADH, FADH2 , GTP and ATP Stages NADH FADH2 ATP Glycolysis 2 – 2 * Link reaction (2×) 2 – – * Krebs cycle (2×) 6 2 2 Total (per glucose) 10 2 4 (* From two pyruvate molecules) Summary Biology Term 1 STPM Chapter 5 Cellular Respiration 5

173 5. A free ATP thus can be formed from GTP with the help of a transferase. The phosphate group is transferred to ADP to form ATP. Therefore, this ATP is considered as formed by substrate level phosphorylation. Oxidative Phosphorylation and Chemiosmosis 1. Oxidative phosphorylation is the formation of ATP from ADP and phosphate in which oxygen is required. 2. It occurs inside the mitochondria, in the inner membrane or cristae, which involves the electron transport system. 3. The electron transport system is also known as cytochrome or respiratory chain. It is a functional unit consisting of a series of coenzymes and enzymes in which H+ and electrons are transferred from one to another and finally accepted by oxygen as shown in Figure 5.9. Oxidative phosphorylation and chemiosmosis 1. NADH (FADH2 ) passes H atom to electron transport chain 2. H atom splits into H+ and electron 3. H+ is pumped into intermembrane space 4. Electron is finally accepted by O2 and H+ to form water 5. H+ ion gradient is created 6. H+ diffuse through ATP synthase to form ATP from ADP and Pi Summary H+ H+ 2H + — O2 H2O H+ H+ H+ H+ ELECTRON TRANSPORT CHAIN ATP SYNTHASE Cytochrome oxidase Cytochrome NADH reductase reductase NADH + (Carrying electrons from food) Protein complex of electron carriers Inner mitochondrial membrane Intermembrane space Inner mitochondrial membrane Mitochondrial matrix NAD+ ADP + P i ATP 1 2 Cyt c Q Figure 5.9 The electron transport system 2013 Biology Term 1 STPM Chapter 5 Cellular Respiration 5

174 Exam Tips Remember that 36 molecules of ATP are produced per glucose in aerobic respiration in muscles, and 38 molecules in liver, kidney and heart cells. Remember that ATP is the energy source in cell and is formed by substrate level and oxidative phosphorylation. 4. ATP is produced by a process called chemiosmosis. (a) NADH or FADH2 from dehydrogenation reactions move to the electron transport chain that are embedded in the inner mitochondrial membrane. (b) NADH reductase accepts H atom from NADH and H atom is split into H+ ion and electron. H+ ion is then transported into the intermembrane space. The electron is passed to ubiquinone, then to a protein-cytochrome complex called bc1 complex, then to another carrier called cytochrome c and finally to a cytochrome oxidase complex. (c) The cytochrome oxidase needs oxygen to function in which it gets the H+ from the inner matrix to form water i.e. O2 + 4H+ + 4 e – → 2H2 O. (d) bc1 complex also functions as a proton pump, pumping H+ from the matrix to the inter-membrane space. Thus, more H+ ions are found in the inter-membrane space than the matrix creating a H+ concentration gradient. (e) FADH2 passes its H atoms to cytochrome reductase of the electron transport chain instead of NADH reductase. (f) Chemiosmosis occurs when the H+ re-enters the matrix down H+ gradient through a channel protein functioning also as an enzyme called ATP synthetase or ATP synthase. ATP is produced from ADP and phosphate through this diffusion process.This formation of ATP is called oxidative phosphorylation as oxygen is required. Role of NADH, FADH2 and ATP Synthase 1. The role of NADH is to pass the hydrogen atom to the first electron carrier (NADH reductase) of the electron transport chain and 3 ATP can be produced during oxidative phosphorylation. NAD+ formed (reoxidised) is transported back into the cytoplasm and can be used to accept hydrogen atom for dehydrogenase enzymes during glycolysis (reused). 2. The role of FADH2 is to pass the hydrogen atom to the second electron carrier (cytochrome reductase) of the electron transport chain forming 2 ATP during oxidation phosphorylation. FAD is formed (reoxidised) and can accept hydrogen atom for succinate dehydrogenase (reused). 3. The role of ATP synthase is acting as enzyme to catalyse the formation of ATP from ADP and inorganic phosphate when hydrogen ions diffuse across it in the cristae. It also acts like a channel protein in the inner mitochondrial membrane for the H+ ions to diffuse back into the matrix from the intermembrane space. So, it carries out chemiosmosis. Biology Term 1 STPM Chapter 5 Cellular Respiration 5

175 Net Energy Produced From one pyruvate, four NADH, one FADH2 and one GTP are produced. Total ATP formed is calculated as follows: 4 NADH can produce 4 × 3 ATP = 12 ATP 1 FADH2 can produce 1 × 2 ATP = 2 ATP 1 GTP can produce 1 × 1 ATP = 1 ATP Therefore, total ATP produced per pyruvate = 15. Total ATP produced per glucose is equal to 30 and together with 8 ATP produced in glycolysis, a total of 38 ATP can be produced in liver cells. In muscles, when NADH produced in glycolysis enters mitochondria, 1 ATP is used so each can produce 2 ATP. Therefore, only 36 ATP are produced per glucose in muscles. 2018 Role of NADH, FADH2 and ATP synthese 1. NADH produces 3 ATP each 2. FADH2 produces 2 ATP each 3. ATP synthase catalyses production of 1 ATP Summary Net energy produced per glucose 1. Liver cells = 38 ATP 2. Muscles = 36 ATP Summary Mitochondrion Krebs Cycle CCC CC Pyruvate Acetyl CoA C Glucose Profits 2 ATP Glycolysis CCC Link reaction 2 NAD ATP 4CO2 *4-6 2 NAD 6 ATP GTP ATP 2 ATP 6 NADH 18 ATP 4 2 Total = 36 – 38 ATP FADH2 CC 2CO2 Figure 5.10 Total ATP yield per glucose in aerobic respiration Biology Term 1 STPM Chapter 5 Cellular Respiration 5

176 The Effects of Cyanide and Carbon Monoxide on Respiration 1. Cyanide (a) Cyanide stops oxidative phosphorylation. Thus, it is a respiratory inhibitor which stops cellular respiration. (b) It inhibits the enzyme cytochrome oxidase of the electron transpot chain by non-competitive inhibition. It binds to the copper ion (Cu2+) that is the prosthetic group of the enzyme. Consequently, Cu2+ cannot accept the electrons from cytochrome c. (c) The whole process of oxidative phosphorylation cannot take place. No ATP will be formed. (d) Cyanide is very poisonous, a few parts per million can kill a person as it prevents the formation of ATP, which is essential for life processes especially muscle contraction for breathing and heartbeat. 2. Carbon monoxide (a) Carbon monoxide (CO) also inhibits oxidative phosphorylation and aerobic respiration in general. (b) It acts as a poison, which has high affinity for iron ion of the haemoglobin to form carboxyhaemoglobin and binds to it irreversibly. Its binding capacity is at least 200 times more than that of oxygen. (c) Once bound, the haemoglobin molecule cannot function as a carrier for oxygen. Even small amounts from cigarette smoke or exhaust fumes can tie up almost 50% of the haemoglobin of the body. (d) This impairs oxygen delivery to the cells and lead to hypoxia, a fall in partial pressure of oxygen in arterial blood. (e) In higher CO concentration, no oxygen is carried to the cells and no oxidative phosphorylation can take place. If no ATP can be formed, the person dies immediately. Lipids and Proteins as Energy Sources Lipid 1. Lipid in the form of triglycerides can be respired when carbohydrates are in short supply. 2. Triglycerides are hydrolysed into glycerol and fatty acids catalysed by lipase. This reaction occurs in the liver, muscles and fatty cells. 3. Glycerol is converted into glyceraldehydes phosphate (triose phosphate) and then is converted into pyruvate using enzymes from part of glycolysis. 4. Fatty acids can undergo beta oxidation and is then converted into acetyl CoA to enter Krebs cycle. Each acetyl CoA can release 12 ATP. 5. Each gram of triglyceride can produce twice as much of ATP as a gram of carbohydrate. Exam Tips Remember how cyanide and carbon monoxide can affect respiration and can kill (STPM 2010 essay question). The effects of cyanide and CO on respiration (a) Cyanide – It is a respiratory inhibitor – It binds irreversibly on last carrier of electron transport – With few ppm, oxidative phosphorylation cannot occur – No ATP is produced (b) CO – It is a poisonous gas – It binds irreversibly on haemoglobin – With high concentration, all haemoglobin become carboxy-haemoglobin – No ATP can be produced as oxidative phosphorylation cannot occur without oxygen carried by haemoglobin Summary 2014 Biology Term 1 STPM Chapter 5 Cellular Respiration 5

177 Protein 1. In extreme starvation, proteins can be respired. Even our muscles proteins can be respired under such condition. 2. Proteins are hydrolysed into amino acids in the liver and many cells catalysed by proteinase. 3. The amino acids are deaminated into keto acids by deaminase. The amino group is removed and used to form urea i.e. a waste product. 4. Keto acids can be converted into pyruvate, acetyl CoA and organic acids in the Krebs cycle. They are then decarboxylated and dehydrogenated through the Krebs cycle like those from glucose. 5. Each gram of protein can also provide more energy than each gram of carbohydrate. Quick Check 1 1. Why is respiration important? 2. In the liver, kidney and heart cells, one mole of glucose can produce 38 moles of ATP compared to the usual 36 moles in other cells. Why? 5.3 Anaerobic Respiration Anaerobic respiration is the breakdown of glucose producing energy in the absence of oxygen. Anaerobic Respiration in Yeasts and Muscles In Yeast 1. Yeast is a facultative anaerobe, it can live in the presence of oxygen. The process of anaerobic respiration is also known as fermentation. 2. Its biochemistry is as shown in Figure 5.11 and as follows: (a) Initially, glycolysis takes place as in aerobic respiration in which pyruvate is produced. Two ATP and 2 NADH are produced per glucose molecule. (b) Krebs cycle does not occur. When there is no oxygen, oxidative phosphorylation cannot occur. There is no NAD+ as it is in the form of NADH and is not oxidised. Pyruvate cannot undergo oxidative decarboxylation to form acetyl coenzyme A to enter the Krebs cycle. (c) Ethanol is produced. The pyruvate accumulates and stimulates the production of enzyme pyruvate decarboxylase that converts it to ethanal and carbon dioxide. The ethanal is then converted into alcohol by another enzyme, ethanol dehydrogenase as follows: Lipid and protein as energy source 1. Triglyceride – Glycerol → triosephosphate – Fatty acid → acetyl CoA 2. Protein → amino acids → organic acids Summary 2015 2017 Students should be able to: (a) explain the anaerobic respiration in yeast and muscle cells; (b) describe the applications of anaerobic respiration in food industries (bread, tapai, and yogurt). Learning Outcomes Biology Term 1 STPM Chapter 5 Cellular Respiration 5

178 2NAD+ Alcohol dehydrogenase 2NADH 2NADH + 2H+ 2NAD+ Glucose (6C) 2x Pyruvate (3C) 2x ethanol (2C) 2x ethanol (2C) 2ADP + 2P2 2ATP CO2 CH3 COCOOH CH3 COCHO + CO2 CH3 CH2 OH NADH NAD+ Pyruvate Ethanal Ethanol Figure 5.11 Fermentation in yeast 4. The ethanol produced is poisonous to cells and thus kills the cells. In fermentation, the yeast can withstand about 10% of alcohol in the medium. 5. Total ATP produced in yeast is two moles of ATP per mole of glucose and the alcohol still contains energy. In Muscles 1. Anaerobic respiration occurs in muscles that actively contract like the leg muscles where the supply of oxygen is not enough. This occurs temporarily for a short period of time when we start to exercise and heart rate is still slow. Prolong anaerobic respiration in muscles result in cramps especially in the leg muscle. 2. Its biochemistry is as follows: (a) Glycolysis occurs like that in the yeasts and Krebs cycle cannot take place too. (b) In animals, the accumulation of pyruvate stimulates the enzyme that changes it straight to lactate. A certain amount of lactate is always produced during exercise. CH3 COCOOH + NADH → CH3 CHOHCOOH + NAD Pyruvate Lactate 2NAD+ Lactate dehydrogenase 2NADH 2NADH 2NAD+ Glucose (6C) 2x Pyruvate (3C) 2x Lactate (3C) 2ADP + 2Pi 2ATP Figure 5.12 Anaerobic respiration in muscle Biology Term 1 STPM Chapter 5 Cellular Respiration 5

179 3. The total amount of ATP produced is two moles per mole of glucose used and is supplied to the muscles for contracting. 4. If anaerobic respiration continues for a longer period, the lactate that accumulates can cause muscular cramps. This happens because the lactate is acidic, denatures the proteins and stops the muscles from contracting. 5. After anaerobic respiration, the lactate is converted in the liver to form glucose through Cori cycle. It requires oxygen. That accounts for the heavy breathing after exercise to repay the oxygen ‘debt’ that is ‘owed’ to the body by forming lactate. Differences between anaerobic respiration of yeasts and muscles The differences are summarised in the table below. In yeast In muscles 1. Where It occurs during fermentation when starch is added with yeast. It occurs in skeletal muscles especially the leg muscles. 2. When It occurs when there is no oxygen. It occurs in short oxygen supply when muscles are starting to contract. 3. Carbon dioxide formed Carbon dioxide is formed from decarboxylation of pyruvate. No carbon dioxide is formed. 4. Number of reaction steps Glycolysis forms pyruvate, then to ethanal and ethanol in 2 steps. Glycolysis forms pyruvate and straight to lactate, in 1 step 5. End products Ethanol is poisonous and kills cells and is of no use to yeasts. Lactate is acidic, will not kill cells and is still useful. 6. Process usefulness It is useful to survive in anaerobic conditions. It is useful to sustain muscle contractions. 7. Recycle of end product Ethanol is not recycled Lactate is recycled to form glucose or glycogen. Table 5.1 The differences between anaerobic respiration in yeasts Application of Anaerobic Respiration in Food Industries 1. It is used in making bread and spongy cookies. The process of fermentation produces carbon dioxide. Starchy dough left to ferment will produce carbon dioxide gas and if heated in the oven, the dough will rise to produce the spongy nature as in breads or cookies. Remember that in anaerobic respiration of both yeasts and animal, there is an abundance of NADH + H+, which are used for reductive reactions for the formation of ethanol and lactic acid. (2009 STPM structured question) Exam Tips Anaerobic respiration in yeast and muscles 1. Yeasts – produce ethanol + CO2 , in 2 steps and ethanol can kill yeast 2. Muscles – produce lactate in 1 step and lactate can cramp muscles Summary Application of anaerobic respiration in industries 1. Bread – CO2 produced expands on heating raising bread 2. “Tapai” – Yeast produces enzymes to digest starch to sugar and ethanol 3. Yogurt – Milk reacts with lactic acid produced by bacteria Summary Biology Term 1 STPM Chapter 5 Cellular Respiration 5

180 2. It is used to make “tapai”, fermented glutinous rice. The cooked rice is added with yeast and left to ferment in a tight container without air. The fermentation is only allowed for a few days so some rice is hydrolysed into sugar and very little alcohol. 3. It is used in the production of yoghurt. Milk added with Lactobacillus will be converted into yoghurt. The bacteria convert the lactose in the milk to glucose before fermenting it to lactic acid. Exam Tips Remember the examples of the use of fermentation in industries. Quick Check 2 1. Why anaerobic respiration cannot be sustained for a long period of time in both animals? STPM PRACTICE 5 Objective Questions 1. Which of the following processes in a cell does not require ATP molecules? A Movement of chromosomes during cell division B Synthesise of enzymes from amino acids C Intake of steroid hormones into the cell D Flow of cytoplasm within the cell. 2. Which combination is not correct? Enzyme Action A Enolase Conversion of phophoenolpyruvate to 2-phosphoglycerate B Phosphofructokinase Transfer of a phosphate group to fructose-6- phosphate C Phosphoglycerokinase Transfer of a phosphate group from 1,3-bisphosphoglycerate to ADP D Phosphoglucoisomerase Conversion of glucose-6-phosphate to fructose-6-phosphate 3. The number of ATP used from one molecule of glucose during the energy investment phase of glycolysis is A 0 C 4 B 2 D 36 4. Pyruvic acid is the A final product of glycolysis B three-carbon molecule that accumulates when oxygen decreases C three-carbon molecule that loses one carbon when the oxygen supply is adequate D four-carbon molecules that can enter Krebs cycle 5. Which sequence of carboxylic acids is found in Krebs cycle? A Oxaloacetate→citrate→succinate→α– ketoglutarate→malate→fumarate B Oxaloacetate→citrate→fumarate→α– ketoglutarate→succinate→malate C Oxaloacetate→citrate→α–ketoglutarate →succinate→fumarate →malate D Oxaloacetate→citrate→α–ketoglutarate →succinate→malate→fumarate Biology Term 1 STPM Chapter 5 Cellular Respiration 5

181 6. At which stage ATP is produced by substrate level phosphorylation in cellular respiration? I electron transport system II transition reaction III Krebs cycle IV glycolysis A I and II C II and III B I and IV D III and IV 7. From the list of substances below, which combination is true of Krebs cycle? i O2 ii CO2 iii ADP iv ATP v NAD vi NADH Requirements Products A i, iii, vi ii, iv, v B i, iii, v ii, iv, vi C ii, iii, v i, iv, vi D ii, iv, vi i, iii, v 8. What is X in the diagram below that shows a simple scheme in the electron transport system? X H2 X 2H+ X is oxidised 2e– 2H+ 2e Oxidative – reactions Reductive reactions A NADH C NADP+ B FAD D Coenzyme A 9. Which process is true of respiration? I Dehydrogenation occurs in glycolysis II Pyuruvate changes to acetyl coenzyme A by releasing CO2 and formation of NADH III Decarboxylation and dehydrogenation occurs in Krebs cycle IV During oxidative phoshorylation, oxygen is reduced to water A I and II C II, III and IV B I, III and IV D I, II, III and IV 10. Which part does chemiosmosis occur in the mitochondrion? A Matrix C Cristae B Cisterna D Plasmalemma 11. P, Q and R represent the biochemicals during the breakdown of glucose in the cell. Q Glucose → → → P ↓ R Which combination is correct? P Q R A Malate Pyruvate Ethanol B Pyruvate Lactate Acetyl CoA C Acetyl CoA Ethanol Lactate D Fructose bisphosphate Dihydroxyacetone phosphate Glyceraldehyde phosphate 12. Which is true of the number of molecules of ATP, NADH and FADH2 produced during I glycolysis? II oxidative decarboxylation? III Krebs cycle? IV electron transport chain? I II III IV A 2 ATP 1 NADH 1 NADH 1 ATP 3 NADH 1 FADH2 20 ATP B 2 ATP 2 NADH 2 NADH 2 ATP 6 NADH 2 FADH2 34 ATP C 2 ATP 2 NADH 2 NADH 2 ATP 6 NADH 2 FADH2 38 ATP D 2 ATP 1 NADH 1 NADH 1 ATP 3 NADH 1 FADH2 17 ATP Biology Term 1 STPM Chapter 5 Cellular Respiration 5

182 13. During aerobic respiration, which is not produced after pyruvate enter the mitochondria? A CO2 B ATP C NADH D Acetyl coenzyme A 14. Which reaction occurs during the breakdown of pyruvate to produce NADH and carbon dioxide in aerobic respiration? A Citrate α-Ketoglutarate B Malate Oxaloacetate C Oxaloacetate Malate D Succinate Fumarate 15. How many molecule of ATP can be formed from two molecules of α–ketoglutarate through the Krebs cycle when each molecule can produce 2 molecules of NADH, 1 molecule of GTP and 1 molecule of FADH2 ? A 2 C 12 B 9 D 18 16. Which is not the common metabolic intermediate in aerobic breakdown of glucose and fatty acids? A Pyruvate B Acetyl CoA C Glyceraldehyde-3-phosphate D Citrate 17. Which combination is correct about the substances in the catabolism of food reserves is correct? Glycogen I II IV III V Glucose-1-phosphate Glucose-6-phosphate Pyruvic acid Acetyl coenzyme A Krebs cycle Electron transport chain Glyceraldehyde phosphate Hydrolysis I II III IV V A Protein Amino acid Fatty acid Fat Glycerol B Protein Amino acid Fat Fatty acid Glycerol C Fat Glycerol Fatty acid Protein Amino acid D Fat Fatty acid Glycerol Protein Amino acid 18. The reaction for the conversion of pyruvate to acetyl CoA is shown in the diagram below. Y Pyruvate + X Acetyl CoA + CO2 Name X and Y? X Y A ATP Reduced NAD B ATP Oxidised NAD C CoA Reduced NAD D CoA Oxidised NAD 19. What reaction converts pyruvic acid to lactic acid during anaerobic respiration? A Reduction B Oxidation C Dehydrogenation D Decarboxylation 20. Why animal cells are unable to carry out alcohol fermentation? A They are unable to withstand the accumulation of alcohol in the cells B They are active cells thus, the cells require energy from lactate fermentation C Do not have vacuoles to accommodate the acetyldehyde produced D Do not have enzymes to convert pyruvate into acetyldehyde Biology Term 1 STPM Chapter 5 Cellular Respiration 5

183 21. Which is true of the fermentation process? A ATP is produced by substrate level phosphorylation. B The final electron is accepted by oxygen. C NADH + H+ are produced. D FADH2 is oxidised. 22. Beside carbon dioxide, what are also produced from the root of a plant undergoing respiration when it is in a submerged condition? A Ethanol and ATP B Lactic acid and ATP C ATP and reduced NAD D Ethanol and reduced NAD 23. A mole of fat always produces more ATP than that of carbohydrate because A fat is heavier than carbohydrate B fat can be hydrolysed only in aerobic condition C oxidation of fat produces more acetyl coenzyme A D oxidation of carbohydrate produces more ATP only without going through electron transport chain 24. How many ATP are produced from the oxidation of one molecule of NADH in the presence of potassium cyanide? A 0 C 2 B 1 D 3 25. Which is used by dehydrogenase enzymes during a reduction process? A FAD C Cytochrome B Pyruvate D Coenzyme A Structured Questions 1. The diagram shows a cycle biochemical reactions of an aerobic metabolism. Acetyl Co-A Citrate Isocitrate ɑ-ketoglutarate Succinate X(4C) Fumarate Malate Oxaloacetate Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8 Co-Ash (a) Name this biochemical pathway. [1] (b) What is the compound X? (c) (i) Name the process which occurs in Step 4. [2] (c) Describe briefly reaction in Step 4. [2] (d) What is the enzyme that catalyses in Step 6? [2] Biology Term 1 STPM Chapter 5 Cellular Respiration 5

184 2. The process of anaerobic respiration is as shown below. H O 1 C | H–2C–OH | HO– 3C–H | H–4C–OH | H–5C–OH | H–6C–OH | H Glucose H O C | C=O | H–C–H | H Pyruvic acid H | O | H–C–H | H–C–H | H Ethanol 2NAD+ 2NADH Energy X Molecule Y CO2 (a) Name energy X and molecule Y. [2] (b) Name the process that converts glucose into pyruvic acid. [1] (c) Name the overall process shown above. [1] (d) Where does the process occur in the cell involved? [1] (e) State three types of industry that made use of the above process. [3] (f) State the products formed if this process occurs in our muscles. [2] Essay Questions 1. (a) Describe the steps in the Krebs cycle and state the amount of high energy molecules produced per glucose molecule. [9] (b) How does cyanide affect cellular respiration? [6] 2. (a) What are the differences between aerobic respiration and fermentation in cells? [8] (b) Describe how can a person obtain energy during starvation. [7] 3. (a) Outline how energy could be produced from fats and proteins in the human body without going into the details of Krebs cycle and glycolysis. [9] (b) Describe how 6C from fatty acids could produce more energy in the form of ATP as compared to 6C from carbohydrates. [6] Biology Term 1 STPM Chapter 5 Cellular Respiration

185 1 1. It produces energy in the form of ATP. The ATP is used in processes that require energy such as anabolic process, active transport, muscle contraction, nervous transmission and bioinflorescence. Besides that, respiratory process produces many intermediary products e.g. pyruvate and α-ketoglutarate, which can be converted to form many substances in the body. 2. In all cells, the total ATP produced per mole of glucose is 30 whereas that of glycolysis is either 6 or 8 moles depending on how the 2 NADH get to the electron transport chain from cytoplasm into the mitochondria. If the NADH gives up its proton to FAD forming FADH, then only 2 ATP can be formed and glycolysis produces 6 ATP. If the NADH gives up its proton to NAD on the other side of the membrane as in liver, kidney and heart cell, then glycolysis produces 8 ATP and a total of 38 ATP per mole of glucose. 2 1. In animals, the product is lactic acid formed inside the muscle and becomes too acidic, causing cramp after a certain amount is accumulated. STPM Practice 5 Objective Questions 1. C 2. A 3. B 4. A 5. C 6. D 7. B 8. B 9. D 10. C 11. D 12. D 13. B 14. A 15. D 16. A 17. C 18. C 19. C 20. D 21. C 22. A 23. C 24. A 25. C Structured Questions 1. (a) Krebs cycle (b) Succinyl – CoA (c) (i) Dehydrogenation (ii) Decarboxylation occurs with one molecule of carbon dioxide removed. Dehydrogenation occurs with NAD+ becomes reduced into NADH. GTP is formed from GDP + Pi when energy is given out / substrate level phosphorylation with the formation of ATP. (d) Succinate dehydrogenase 2. (a) X is ATP; Y is NAD+ (b) Glycolysis (c) Fermentation (d) Cytoplasm (e) Wine-making industries; baking industries and vinegar-making industries (f) Lactic acid Essay Questions 1. (a) • Krebs cycle starts from acetyl CoA that combines with oxaloacetate and water to form 6C citrate catalysed by synthetase. • Citrate is converted to isocitrate catalysed by aconitase. • Isocitrate undergoes oxidative decarboxylation forming α–ketoglutarate, carbon dioxide and NADH catalysed by dehydrogenase. • α–ketoglutarate reacts with coenzyme A and dehydrogenation forming succinyl CoA and NADH catalysed by dehydrogenase. • Succinyl CoA undergoes substrate level phosphorylation forming succinate and GTP catalysed by thiokinase. • Succinate undergoes dehydrogenation forming fumarate and FADH2 catalysed by dehydrogenase. • Fumarate reacts with water to form malate catalysed by fumarase. • Malate undergoes dehydrogenation to regenerate oxaloacetate and forming NADH catalysed by dehydrogenase. • So, one acetyl Co-A can form 3 NADH (= 9 ATP), one FADH2 (= 2 ATP) and one GTP (=1 ATP) so forming 12 ATP. • Thus, one molecule of glucose can form two acetyl Co-A thus can form 24 ATP molecules through Krebs cycle. 2. (a) • Aerobic respiration needs oxygen, whereas fermentation may not need oxygen. • Aerobic respiration involves oxidative phosphorylation whereas fermentation only involves substrate level phosphorylation. • Aerobic respiration produces higher ATP per unit mass of substrate than fermentation. • Aerobic respiration produces water and carbon dioxide, whereas fermentation may not have the products. • Aerobic respiration does not produce alcohol whereas fermentation may produce alcohol. • Aerobic respiration does not produce lactic acid, whereas fermentation may produce lactic. • Aerobic respiration involves chemiosmosis, whereas fermentation does not involve chemiosmosis. ANSWERS Biology Term 1 STPM Chapter 5 Cellular Respiration 5

186 • Aerobic respiration involves mitochondria, whereas fermentation does not involve mitochondria. (b) • From glycogen in the muscles and liver. • When the blood glucose level is low, insulin is stimulated and released into the blood. • Insulin binds to the receptors of liver cells and causes the activation of phosphorylases which will catalyse the formation of glucose from the glycogen. • Further starvation causes the release of adrenaline and corticosteroids that cause the hydrolysis of triglycerides. • The products of hydrolysis of triglycerides are converted into glucose for respiration and maintenance of blood glucose level. • Further starvation may cause the breakdown of muscle proteins which is for the use of respiratory substrates. 3. (a) • Fats in the form of triglycerides can be hydrolysed into glycerol and fatty acids. • Glycerol can be converted into triose phosphate by enzymes in the cytoplasm. • Then, the enzymes of the glycolysis will convert the triose phosphate into pyruvate step by step. • From pyruvate, link reaction occurs in the presence of O2 to form acetyl CoA by oxidative decarboxylation. • β-oxidation breaks down fatty acid into acetyl CoA in the mitochondria. • Proteins are hydrolysed into amino acids and then deaminated in the liver to form keto acids. • Most keto acids are converted into acetyl CoA in the liver cells. • Acetyl CoA undergoes Krebs cycle in the mitochondria by combining with oxaloacetate to form citrate. • Citrate will be regenerated to form oxaloacetate after decarboxylation and dehydrogenation to form CO2 and NADH respectively. • NADH can be used to form ATP by oxidative phosphorylation in the cristae of mitochondria. (b) • 6C fatty acids can produce more energy compared to 6C carbohydrates such as glucose in terms of acetyl CoA production. • 6C fatty acid produces 3 molecules of acetyl CoA (2C) through β-oxidation. • 6C glucose produces only 2 molecules of acetyl CoA (2C) through glycolysis and link reaction. • Each molecule of acetyl CoA is broken down completely in the Krebs cycle. • Each molecule of acetyl CoA produces 3 NADH (×3 ATP), 1 FADH2 (×2 ATP) and 1 GTP (×1 ATP) giving a total of 12 ATP. • Therefore 6C fatty acid produces 3 × 12 = 36 ATP while 6C glucose (carbohydrate) produces only 2 × 12 = 24 ATP. Biology Term 1 STPM Chapter 5 Cellular Respiration 5

6 CHAPTER PHOTOSYNTHESIS 6 Concept Map Photosynthesis Photosynthetic pigments Chemoautotroph Chemosynthesis Light Dependent Reactions Light Independent Reactions Absorption spectrum Autotrophs Photoautotroph Action spectrum Limiting factors of photosynthesis Calvin cycle Photorespiration Photo-activation of chlorophyll a C4 leaf compared to C3 leaf Carbon fixation in C4 plants and CAM plants The metabolism of C3, C4 and CAM plants The cyclic and non-cyclic photophosphorylation Bilingual Keywords Autotroph – autotrof Photoautotroph – Fotoautotrof Chemoautotroph – Kemoautotrof Photorespiration – fotorespirasi Limiting factors – faktor pengehad Light intensity – Keamatan cahaya Primary – Primer Carotenoid – Karotenoid Cyclic – Berkitar Coenyzme – Koenzim Photolysis – Fotolisis Mesophyll – Mesofil Bundle sheath – Salut berkas Pathway – Laluan Fixation – Pengikatan

Biology Term 1 STPM Chapter 6 Photosynthesis 6 188 Autotrophs are organisms which can synthesise their own organic food substances from inorganic chemicals using energy from an external source. Classification of Autotrophs Autotrophs are divided into photoautotrophs and chemoautotrophs. Photoautotrophs 1. Photoautotrophs are organisms that synthesise their organic substances especially carbohydrates from carbon dioxide and water with energy source in the form of light obtained by photosynthetic pigments. The organisms include all plants, algae and bacteria. They carry out photosynthesis as follows. Light 6CO2 + 6H2 O → C6 H12O6 + 6O2 Chlorophyll 2. Examples of photoautotrophic bacteria are cyanobacteria (e.g. Nostoc). (a) These bacteria make use of water and carbon dioxide as raw materials to carry out photosynthesis in the double thylakoid membranes found in concentric rings beneath the surface membrane. (b) They have chlorophyll and phycobillins as photosynthetic pigments to capture light. These pigments result in their bluegreen colour. (c) They have both photosystem I and II that are like those of chloroplasts. (d) The process of light dependent stage and light independent stage are the same as green plants. Light dependent stage: Light ADP + Pi + NADP+ + 2[H] → ATP + NADPH + H+ Photosynthetic pigments Light independent stage: ATP+ NADPH + H+ + 6CO2 → C6 H12O6 + 6H2 O + ADP + Pi + NADP+ 3. Another example is green sulphur bacteria (e.g. Chlorobium). (a) They are green in colour due to the presence of green pigment called bacteriochlorophyll, which is the primary pigment found inside the vesicles or foldings of the plasma membrane. (b) There is no photosystem II, water is not required and no oxygen is produced. 6.1 Autotrophs Students should be able to: (a) classify autotroph into photoautotroph and chemoautotroph; (b) describe photosynthetic pigments; (c) explain the absorption spectrum and action spectrum of photosynthetic pigments. Learning Outcomes VIDEO Photosynthesis

Biology Term 1 STPM Chapter 6 Photosynthesis 6 189 (c) These are anaerobic bacteria making use of hydrogen sulphide or sulphur containing compounds as donors of hydrogen ions or electrons. Carbon dioxide is converted into carbohydrate forming sulphur as by-product. (d) The process is as summarised in the formula as follows: 12H2 S + 6CO2 → C6 H12O6 + 12S + 6H2 O 4. There are also the purple sulphur bacteria (e.g. Chromatium). (a) They are purplish in colour containing red and brown carotenoids mixed with bacteriochlorophyll to trap light. (b) The process of photosynthesis is same as that of green sulphur bacteria. 5. Other examples are purple non-sulphur bacteria (e.g. Rhodospirillum). (a) They are purple in colour and have carotenoids and bacteriochlorophyll to trap light. (b) They make use of hydrogen gas as a donor of hydrogen to reduce carbon dioxide to form carbohydrates. (c) There is a type called photoheterotroph that can make use of carbon containing organic compounds such as alcohols or malate as donors of hydrogen. 24H+ + 6CO2 → C6 H12O6 + 6H2 O Chemoautotrophs 1. Chemoautotrophs synthesise organic substances especially carbohydrates from carbon dioxide and water, making use of energy from chemical reactions. This is the most primitive form of nutrition in ancient bacteria living anaerobically, converting carbon dioxide and hydrogen into methane gas. 2. The usual chemical reactions involved are oxidations of inorganic compounds such as hydrogen sulphide, sulphur, iron(II), ammonia and nitrite. 3. One example is iron bacteria (e.g. Leptothrix). These bacteria make use of energy released from oxidative reactions of iron(II) to iron(III). Fe2+ + O2 → Fe3+ + energy 4FeCO3 + + 6H2 O → 4Fe(OH)3 + 4CO2 + energy 6H2 O + 6CO2 + energy → C6 H12O6 + 6O2 4. Another example is colourless sulphur bacteria (e.g. Thiobacillus). These bacteria make use of energy released from oxidative reactions of sulphur to sulphuric acid. These reactions occur in the presence of oxygen or nitrate. S + O2 (NO3 ) → SO4 2– + energy 2S + 3O2 + H2 O → 2H2 SO4 + energy Exam Tips Remember four types of photosynthetic bacteria (sulphur bacteria in STPM 2010 essay question) and how each one of them differs from the other. Classification of autotrophs Organisms able to synthesise their own food from simple inorganic substances 1. Using light as energy source – photoautotrophs 2. Using chemical reaction as energy source – chemoautotrophs Summary

Biology Term 1 STPM Chapter 6 Photosynthesis 6 190 5. Another common example is nitrifying bacteria. (a) Nitrosomonas converts ammonia or ammonium salts to nitrite as shown in the following equations. NH4 + + 2O2 → NO2 – + 2H2 O + energy 2NH3 + 3O2 → 2HNO2 + 2H2 O + energy It requires oxygen as an electron or hydrogen acceptor and plays an important role in nitrogen cycle to convert organic nitrogen to ammonia and then to nitrite. (b) Nitrobacter converts nitrite to nitrate in the presence of oxygen too. It is the final conversion in nitrifying process so that plants can make use of the nitrate reformed from organic sources. The reaction is shown in the following equations. 2NO2 – + O2 → 2NO3 – + energy 2HNO2 + O2 → 2HNO3 + energy Photosynthetic Pigments 1. Photosynthetic pigments can be divided into primary pigments and accessory pigments. The accessory pigments can transfer their energy to those of primary. 2. The primary pigments consist of P680 and P700 i.e. chlorophyll a that can absorb light of wavelength 680 nm and 700 nm respectively. 3. The accessory pigments consist of chlorophyll a and b other than P680 and P700.They also include carotenoids i.e. carotene and xanthophylls. 4. Besides that, the pigments are organised into photosystems, which have the following characteristics: (a) Photosystem is a cluster of photosynthetic pigments grouped together in the form of particles or lipoprotein globules that span the internal membrane of the thylakoid and integranal lamellae inside the chloroplast. (b) Each photosystem consists of 200-300 molecules of chlorophyll, accessory pigments and proteins held within a protein matrix that act like an antenna for gathering light. (c) Each type of pigment molecules can absorb a photon of light, gets excited and energised, then the energy is passed to another. (d) Finally, as shown in Figure 6.1, the energy is received by the primary pigment that is found in the reaction centre where it can release electron. (e) The electron lost can be restored by a number of ways including from water. (f) There are two photosystems i.e. photosystem I and photosystem II. Photosynthetic pigments Primary pigments (Chlorophyll a) P680 in PS I P700 in PS II Secondary pigments a(670) Chlorophyll b(650) Xanthophyll Carotenoids Carotene Summary

Biology Term 1 STPM Chapter 6 Photosynthesis 6 191 5. Photosystem I has the following characteristics: (a) Photosystem I is a system of pigments in smaller particles that are usually found in larger numbers in the intergranal lamella than the thylakoid membrane. (b) It seems to be the first photosystem evolved and the wavelength of light most strongly absorbed is 700 nm. (c) It has primary chlorophyll a molecule, P700 , in its reaction centre and its components are shown in Figure 6.1. Chlorophyll b 650 Chlorophyll a 670 Chlorophyll a 680 Chlorophyll a 690 P700 Cone representation Globular representation P700 Light Pigment molecule Protein matrix e– e– Light trap I (reaction centre) Figure 6.1 Photosystem I 7. Photosystem II has the following characteristics: (a) Photosystem II is a cluster of pigments in the form of bigger particles, which are found usually in the thylakoid membrane of the grana. (b) It seems to be the later photosystem evolved and the wavelength of light most strongly absorbed is shorter at 680 nm. (c) It has primary chlorophyll a molecule, P680, in its reaction centre and its components are as shown in Figure 6.2. Chlorophyll b 650 Chlorophyll a 670 P680 P680 Light e– Trap II (reaction centre) Cone representation Globular representation e– Figure 6.2 Photosystem II