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Unit 2 Review, pages 256–263 Knowledge 1. b 2. d 3. a 4. c 5. c 6. b 7. c 8. a 9. d 10. d 11. d 12. d 13. False. The farther away an electron is from the nucleus of an atom, the more potential energy it has. 14. False. Reactions with a negative ΔG are considered spontaneous. 15. False. The more electronegative an atom is, the greater the tendency for it to undergo reduction. 16. False. Cellular respiration can occur in some prokaryotes that do not have mitochondria. 17. False. As reactants progress through the electron transport chain, they lose free energy. 18. False. Fatty acids must be metabolized to acetyl-groups before they can enter citric acid cycle. 19. False. Aerobic cellular respiration provides cells with more energy per gram of glucose than anaerobic cellular respiration. 20. False. The energy that is released when an electron returns to the ground state is less than the energy it absorbed when it was promoted. 21. False. Cyclic photophosphorylation has a positive impact on the ability of chloroplasts to undergo the Calvin cycle because it is a source of additional ATP. 22. True 23. False. The light-dependent reactions of photosynthesis take place within the thylakoid membrane of chloroplasts. 24. False. Photorespiration decreases the efficiency of the Calvin cycle by expending ATP and catabolizing hydrocarbons. 25. (a) v (b) iii (c) i (d) ii (e) iv 26. The equation for the hydrolysis of ATP is: ATP + H2O ADP + Pi Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-3 27. ATP is high in free energy because of its three negatively charged phosphate groups. The phosphate groups crowd together and their close proximity creates a mutual repulsion of their electrons, which makes the bonds unstable and easy to break. During hydrolysis, the bond joining the terminal phosphate group is broken, resulting in the formation of ADP and Pi and the release of free energy. 28. The four stages of cellular respiration are: • Glycolysis: The extraction of some energy from glucose during the formation of pyruvate, leading to a net production of 2 ATP and 2 NADH. • Pyruvate oxidation: The two pyruvates are transported inside the mitochondrion where they are decarboxylated and oxidized to form 2 acetyl groups (which are attached to CoAs), along with 2 CO2 and 2 NADH. • Citric acid cycle: For each acetyl-CoA that enters the cycle, the end result is 2 CO2 as waste, 3 NADH, 3 H+, 1 FADH2, 1 ATP, and 1 CoA that is recycled. • Electron transfer system and oxidative phosphorylation: Oxidation of 10 NADH and 2 FADH2 that are produced from 1 glucose in the above reactions to provide the potential energy for the formation of ATP using electrochemical gradients and ATP synthase. 29. Key steps occurring between the end of glycolysis and the beginning of the citric acid cycle: • Pyruvate in the cytosol is transported into the mitochondrion by a transport protein. • Pyruvate is decarboxylated and oxidized, forming an acetyl group and CO2 as a waste product. • Acetyl group reacts with sulfur atom of coenzyme A, forming high-energy intermediate acetylCoA. • Acetyl-CoA enters the citric acid cycle. 30. The fermentation pathway is used to regenerate NAD+ by oxidizing NADH. This regeneration supplies the NAD+ needed as a reactant in glycolysis. Without a source of NAD+, glycolysis cannot occur. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-4 31. The three possible fates of an “excited” electron in a pigment molecule are: 1) The electron can return to its ground state by emitting a less energetic photon (fluorescence) or by releasing energy as heat. 2) The electron can return to its ground state and it transfers its energy to an electron in a neighbouring pigment molecule. 3) The electron is transferred to an electron-accepting molecule. 32. Three accessory pigments are chlorophyll b, carotene, and xanthophyll. Carotene and xanthophyll are carotenoids. Each accessory pigment absorbs different wavelengths of light, but all transfer their excitation energy to molecules of chlorophyll a. Therefore, they expand the effective range of absorbance of plants by capturing energy from light outside of chlorophyll a’s absorption spectrum. 33. In spring and summer, chlorophyll masks the other pigments present in leaves. In the fall, the dominant green chlorophyll degrades, revealing the other pigments. 34. The function of a photosystem is to convert light energy into chemical energy. This is accomplished by exciting electrons in a chlorophyll a molecule. These excited electrons are then “captured” (and passed on) when the chlorophyll is oxidized. 35. We would not expect plants to grow well in green light. Chlorophyll reflects back green light, so its photons would not be available for the plants to use. 36. The three processes that create a proton gradient across the thylakoid membrane are 1) The protons are taken into the lumen by the reduction and oxidation of plastoquinone, 2) the concentration of protons inside the lumen is enhanced by the addition of two protons for each water molecule that is split, and 3) the removal of one proton from the stroma for each NADPH molecule formed lowers the concentration of protons outside the thylakoid membrane. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-5 37. The three forms, or states, of the chlorophyll molecule P680 are P680, P680*, and P680+. P680* has an excited electron and P680+ has lost an electron and is, therefore, positive and an extremely strong oxidizer. P680 is the neutral form. It occurs in this form after P680+ has oxidized water, but before being excited by a photon. 38. At night, CAM plants open their stomata, absorbing CO2 and releasing O2. The CO2 is fixed into oxaloacetate and converted to malate via the C4 pathway. During the day, the malate that has built up is now oxidized to pyruvate, releasing CO2 that is used to drive the Calvin cycle. Understanding 39. (a) This is an endothermic reaction. (b) As the reaction progresses, the reactants increase in potential energy (absorb energy) and once the high activation energy is achieved, only some of that energy is released as the reaction occurs. The result is a net gain in potential energy in the products compared to the reactants. 40. Sweating is an example of an increase in entropy. As the body sweats, the liquid is spread out over a large area. Sweating cools humans because it involves evaporation of this liquid into the air. Evaporation is an endothermic reaction, in which thermal energy is transferred from skin to sweat and causes sweat to evaporate. 41. (a) Pathway (b) is likely to be catabolic. Pathway (a) is likely to be anabolic. In catabolic pathways, complex molecules are broken down into simpler compounds, thereby releasing free energy (overall ΔG is negative). In anabolic pathways, free energy is consumed to build complicated molecules from simpler molecules (overall ΔG is positive). (b) Pathway (a) is represented in the free energy diagram in Figure 3. 42. (a) This is an exergonic reaction. (b) As the exergonic reaction progresses, free energy is released. The products have less free energy than was present in the reactants. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-6 43. Three major mechanisms that enzymes use to reduce the activation energy of a reaction are: 1) They bring both substrates into contact by binding them to specific sites, which allows them to collide more easily and break bonds more quickly. 2) They can bring a substrate to a charged environment and alter the substrate. 3) They can bend or distort the shape of a substrate, which weakens the chemical bonds. 44. Lipids are a better energy source than proteins or carbohydrates because they contain, almost exclusively, C–H bonds. Molecules with a large number of C–H bonds are high-energy molecules because these bonds are relatively easy to break. The electrons associated with a C–H bond are approximately equidistant from two relatively small nuclei. As a result, the electrons in C–H bonds contain high energy because they can be readily pulled closer to larger and more attractive nuclei—a process that releases energy. 45. Both processes generate ATP, but oxidative phosphorylation is more prominent. Substratelevel phosphorylation is the direct transfer of a phosphate group to an ADP, resulting in the synthesis of ATP. Oxidative phosphorylation involves a sequence of oxidative steps that generate a proton gradient. The free energy of this gradient is subsequently used to synthesize ATP. 46. 47. The proton motive force is established as protons are pumped into the intermembrane space along the electron transport chain. This establishes a chemical and electrical gradient as these protons are positively charged. The proton motive force is instrumental in providing the energy for the work of chemiosmosis and thus, ATP synthesis. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-7 48. The energy released during electron transport causes a difference in H+ concentration between the intermembrane space and the matrix. (The concentration in the intermembrane space is higher than that in the matrix.) This proton gradient is a form of potential energy. 49. The rate of respiration is controlled by key metabolic intermediates (for example, elevated ATP concentrations) through feedback inhibition. When the levels of those intermediates get too high, they inhibit certain enzymes, slowing down cellular respiration. For example, excess ATP negatively regulates phosphofructokinase, an enzyme involved in glycolysis. Stopping glycolysis and the other steps of cellular respiration prevents the cell from producing more ATP than it needs. 50. If your cells were not able to undergo lactate fermentation, you would not be able to exercise as strenuously or sprint as fast. You would need to stop more frequently to rest when the O2 levels in your cells dropped too low. 51. The fermentation process to produce buttermilk breaks down lactose into lactic acid. The acid creates a sour flavour. 52. Fermentation produces only 2 ATP, compared with up to 38 ATP produced in aerobic respiration. Aerobic respiration completely oxidizes the carbon in glucose. In contrast, fermentation produces wastes (for example, ethanol) that still contain significant amounts of energy. 53. Uncoupling occurs when protons leak across the membrane. An uncoupling protein in the inner mitochondrial membrane gives protons an alternative pathway to re-enter the matrix versus oxidative phosphorylation via ATP synthase. The result is that ATP is not produced, but thermal energy is produced. This decreases the number of ATP produced per glucose and, therefore, overall aerobic cellular respiration efficiency. 54. You would expect the plant stored in the room with a lot of light to contain a higher level of starch than the plant stored in the room with very little light. Starch is produced from sugars generated through photosynthesis, which requires sunlight. When placed in the dark, plants would start consuming more of their starch as their source of food energy. 55. Deciduous trees that have lost their leaves in winter have to rely on the sugars they accumulated while they were actively photosynthesizing during the other seasons. In addition, the low temperature will significantly reduce their level of metabolic activity. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-8 56. Chemiosmosis in both cellular respiration and photosynthesis is made possible by a flow of electrons along an electron transport chain that leads to ATP synthesis by building up a proton gradient across a membrane. In photosynthesis, the source of energy that drives electron transport is light. In respiration, the source of energy is stored chemical energy from energy-rich NADH and FADH2. In both cellular respiration and photosynthesis, chemiosmosis occurs via ATP synthase. 57. (a) The optimal conditions under which photorespiration is likely to occur in C3 plants is in a hot dry environment. (b) The efficiency of photosynthesis decreases under these conditions because rubisco will bind with oxygen more often. The CO2 available is reduced because the C3 plants close, or partially close, their stomata to conserve water; therefore, the carbon dioxide levels within the leaves drop as carbon dioxide is used up during photosynthesis. In addition, at higher temperatures the solubility of CO2 decreases more rapidly than that of O2, further decreasing the CO2 concentration. When the ratio of oxygen to carbon dioxide increases, rubisco is more likely to bind oxygen, resulting in photorespiration, and cell resources are drained. 58. Plants will have many stomata to maximize gas exchange and photosynthesis. Plants with numerous stomata are adapted to, and living in, moist warm environments. However, in environments that are dry and/or very hot, these plants would lose too much water through transpiration if they had many open stomata. Therefore, plant species with fewer stomata are better able to survive in these environments. Analysis and Application 59. (a) You would expect the activity rates to be much higher with the enzyme and to be the highest with the enzyme at the higher temperature. Without an enzyme, hydrolysis is unlikely to occur. (b) Under the most active conditions, you might expect to produce small carbohydrates such as sugars, amino acids, and glycerol and fatty acids if the tissues contained polysaccharides, proteins, and lipids, respectively. 60. (a) The effective efficiency of aerobic cellular respiration is 32.4 %. (b) The rest of the energy is released as thermal energy. 61. (a) The free energy diagram in Figure 5 does not represent the process of cellular respiration because cellular respiration does not require a net input of energy. In cellular respiration, energy stored in glucose is released. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-9 (b) 62. (a) Graph Y is a graph of the free energy in the aerobic cellular respiration electron transport chain. Graph Z is a graph of the free energy in the light-dependent reactions of both photosystems II and I. (b) Graph Y is an exothermic reaction and Graph Z is an endothermic reaction. 63. (a) (b) Yes, this reaction is spontaneous because there is an overall loss in free energy (negative ΔG). (c) The enzyme curve has a lower peak than the curve without the enzyme. The enzymes lowers the activation energy, so the reactant molecule reaches the transition state at a faster rate. 64. Answers may vary. Answers may include: In general, enzymes are beneficial to many industries as catalysts to make different products. They are very efficient (can catalyze many— even millions—of reactions in a very short period of time) and can work at relatively low temperatures. This saves time, energy, and money. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-10 65. The breakdown of sucrose is a spontaneous reaction, meaning that it will continue on its own once it is underway. But, it will not start until enough activation energy is added to the system. At room temperature, this amount of energy cannot be obtained from the surroundings. 66. Both lipids are structurally similar except for the number of C–H bonds. The electrons in C–H bonds contain high energy. Lipid (a) has fewer C–H bonds so it has less available energy. 67. Answers will vary. Glucose and oxygen are the reactants; carbon dioxide and water are the products. The carbon in the glucose molecule is oxidized (oxygen is added), and the oxygen is reduced (hydrogen is added). 68. Controlled oxidation benefits the cell as the energy can be harnessed to drive metabolic reactions as opposed to rapid combustion, in which the energy is given off all at once. 69. (a) One mole of glucose provides 2870 x 0.4 kJ of ATP energy = 1148 kJ. Therefore, the athlete needs 28 000 kJ/1148 kJ/mol of glucose = 24.4 mol of glucose. This is equivalent to 24.4 mol x 180 g/mol = 4.39 kg of glucose. (b) It is unreasonable to think that the athlete would consume over 4 kg of pure glucose before the race. Most athletes replenish their sugars on the course by eating or drinking. They can also use stored body fat as a source of energy. Fat contains more than twice as much energy per gram as glucose does. 70. The different ratios of oxygen to carbon dioxide at inhalation and exhalation are due to respiration. CO2 is a waste product of cellular respiration. The increase in carbon dioxide we exhale comes from the decarboxylation reactions that occur during pyruvate oxidation and the citric acid cycles of cellular respiration. Oxygen levels decrease because an equivalent amount of oxygen is used at the end of the electron transport chain to produce water. 71. (a) Animals with a higher abundance of brown adipose tissue are those that need to maintain body temperature during hibernation (for example, bears) or need to increase thermal energy production (for example, birds in very cold environments or newborn animals). (b) You would expect to find brown adipose tissue around vital organs in animals in order to keep them warm and functioning properly. (c) The electron transport chain in the mitochondria in brown adipose tissue is uncoupled from ATP synthesis and used for thermogenesis in these animals. The energy that is released during electron transport is not converted to ATP energy, but released as thermal energy when the protons re-enter the matrix without passing through ATP synthase. 72. Table 1 Pathway ATP produced Associated ATP Total ATP (per glucose through substrateproduced later, Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-11 molecule) level phosphorylation through oxidative phosphorylation 6 glycolysis 2 8 pyruvate 0 6 6 oxidation citric acid 2 22 24 cycle Total 4 34 38 73. (a) Cyanide is toxic to humans because it inhibits cytochrome oxidase C and, therefore, blocks the process of cellular respiration. Cytochrome oxidase C is complex IV in the electron transport chain, so if it is inhibited ATP synthesis is blocked. Humans cannot survive without the energy generated by cellular respiration. (b) A sub-lethal dose would affect those organs with high-energy demands, such as the brain and heart. 74. Table 2 Aerobic respiration Anaerobic respiration yes no Requires oxygen? Location of chemical cytosol and mitochondria cytosol reactions Amount of ATP 38 2 produced per glucose molecule 75. A human can only survive for a few minutes without oxygen because oxygen is the primary electron acceptor at the end of the electron transport chain and, without it, cells would run out of ATP very quickly and metabolism would shut down. Because oxygen is a non-polar gas, it does not dissolve readily in water and cannot be easily stored; therefore, it needs to be obtained continuously from the environment. A human can survive for a few days without water because water is a by-product of cellular respiration and the body can scavenge this water to aid in other metabolic processes. There is a great deal of water produced this way, and a great deal of water already present within the body, so it is not lost quickly. Our bodies produce more water than they use chemically, so they just have to replace water lost by evaporation and excretion. A human can survive for several weeks without food because the body can use stored food molecules and energy-rich compounds (fats, carbohydrates, and proteins). Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-12 76. Answers may vary. Sample answer: Migratory birds probably store energy as fat because, gram for gram, fats contain much more energy than carbohydrates making it a better source of energy. In addition, carbohydrates bind a lot of water, adding to their weight. Migratory birds need to minimize their weight while maximizing their energy stores, making fat a better choice. 77. Answers may vary. Sample answer: An example of feedback inhibition is the thermostat that controls your furnace. When the ambient temperature gets too high, the thermostat switches off the furnace. When the ambient temperature drops too low, the thermostat turns the furnace on. 78. (a) If there was a net production of oxygen in an ecosystem, you would also expect glucose to accumulate. Oxygen is a product of photosynthesis, and glucose is a product of the same reaction, so if one product accumulates, the other must accumulate as well. (b) A net production of oxygen gas would ultimately increase the total biomass of the ecosystem over time. (c) If the rate of photosynthesis were reduced on a global scale, this would cause an increase in atmospheric CO2 and a decrease in atmospheric O2. Carbon dioxide is a greenhouse gas and oxygen is vital for all animals. 79. Answers may vary. Sample answer: Photovoltaic cells are black in colour in order to absorb as much light energy as possible. Plants do not have black leaves suggesting that they do not need to absorb light across the entire spectrum. That is, light may not be the limiting factor in photosynthesis for plants; therefore, an increase in light absorption would provide no benefit to the plant. In fact, it might be detrimental because it would cause an increase in thermal energy absorbed, which could damage the cells and lower the solubility of CO2. Cell damage may also result from the excess energy absorbed being diverted into reactions that produce toxic compounds. 80. Answers may vary. Answers should include: • Photon energizes electron in P680 forming P680* • P680* transfers the high-energy electron to acceptor in reaction centre and becomes P680+ • P680+ oxidizes H2O, regenerating P680, and the high-energy electron is transferred from reaction centre to PQ • PQ donates the electron to the cytochrome complex • From cytochrome complex, the electron passes to plastocyanin, which shuttles it to photosystem I • Photon energizes P700, an electron is excited, and P700* forms • P700* transfers its electron to the primary electron acceptor of photosystem I, forming P700+; P700 is regenerated by accepting the electron that came from plastocyanin • The electron that was transferred from P700* is transported by a short sequence of carriers and then transferred to ferredoxin • Ferredoxin is oxidized and the electron passes to NADP+, forming NADP • Similarly, a second electron as well as a proton from the stroma are transferred to NADP by another ferredoxin via NADP+ reductase; NADPH is formed, carrying two high-energy electrons Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-13 81. Both the C4 cycle and the Calvin cycle begin with the reactant, carbon dioxide. But they differ in their first product: 4-carbon oxaloacetate from phosphoenolpyruvate (PEP) in the C4 cycle and two 3-carbon phosphoglycerates from ribulose 1,5-bisphosphate (RuBP) in the Calvin cycle. Another key distinction occurs in the carboxylation reactions where the C4 cycle uses the enzyme PEP carboxylase in lieu of rubisco. The products are very different. The C4 cycle temporarily produces malate (which is then used as a source of CO2 for the Calvin cycle, producing pyruvate which must be regenerated to PEP). The Calvin cycle produces G3P to be used to assemble sugars and other organic compounds. The Calvin cycle hydrolyzes 3 ATP to 3 ADP in one cycle, while the C4 cycle hydrolyzes 1 ATP to 1 AMP (a single ATP molecule is hydrolyzed twice). Since C4 plants undergo both the C4 and Calvin cycle, they utilize more ATP overall (equivalent to 6 additional ATP molecules for each G3P produced). 82. Answers may vary. Sample answer: 83. The second law of thermodynamics states that every time energy is converted to another form, some of it becomes unusable. Therefore, there must be continual input of energy from some source in order to make up for the energy lost every time energy is converted within living organisms (i.e., photosynthesis). For life on Earth, sunlight acts as that energy source. Life on Earth could not survive without the constant input of energy from the sun. If life is to continue on Earth, it requires a supply of outside energy. Life will ultimately cease (through an increase of entropy) unless there is an infusion of new energy; therefore: no sun would mean no life on Earth. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-14 Evaluation 84. Answers may vary. Answers should include: • Both processes use fuel (the car uses gasoline, the body uses food) to generate energy. • Neither process is 100 % efficient and both generate waste thermal energy. • Both involve breaking hydrocarbon bonds and complete oxidation. (In the car this is rapid, in the body this is “controlled” and slower.) • Both are exothermic and exergonic. • Gasoline combustion doesn’t require enzymes. 85. A sperm cell contains up to 100 mitochondria while an egg cell can have in excess of 100 000 mitochondria. Answers to this question may vary. Students’ answers will likely mention the size of the cells (eggs are much larger than sperm), the use of mitochondria to power the flagella in sperm (high energy need), the use of mitochondria to supply the energy used in the cell division of the embryo (high energy need), or the fact that mitochondria are maternally inherited. 86. Answers may vary. Sample answer: A carbohydrate molecule represents a bill of a high denomination that must be exchanged for, or converted into, bills of a smaller denomination (ATP) that are easier to spend on daily transactions. 87. (a) The benefits associated with modifying genes that produce the enzymes involved in lipid metabolism are that less fat would be stored in the body and the body would be able to digest fat instead of storing it. The risks would be the possible digestion of good fats on nerve cells, depletion of energy stores, and an inability to store and process certain vitamins due to inadequate fat levels. (b) Answers will vary. Sample answer: A treatment in which increased enzymatic lipid catabolism in cells was increased would not be recommended because lipids are not just stored in the body as fat. Most of our nerve cells (and brain) are associated with fats (in the myelin sheath) and are essential for proper function. There are many fat-soluble vitamins and essential fatty acids that the body needs and their absence would cause chronic deficiencies. 88. Dr. Luft suspected his patient had a metabolic disorder because she was consuming food that her body was converting to energy, but most of that energy was being dissipated as metabolic thermal energy instead of being converted into ATP. 89. If the ionophore channels across the mitochondrial membrane were to leak, the mitochondria would not be able to generate ATP. Ionophores destroy the proton gradient that drives ATP synthase; therefore, although the cell can continue to generate small amounts of ATP through glycolysis and the citric acid cycle, much of the potential energy is lost in the form of thermal energy. This would cause increases in body temperature and appetite with an inability to gain weight. 90. The form of mitochondria is well-adapted to its function. Its inner membrane is folded to increase the surface area where electron transfer and ATP synthesis can occur. It has an intermembrane space to allow for the formation of a proton gradient, it has its own DNA, and it can divide to increase in numbers when needed. 91. The number of mitochondria in an organ is proportional to the activity level of the organ. In a mitochondrial homunculus, the high-energy organs would be the largest (e.g., the heart, brain, liver, kidneys, major muscles), while the skin and bone cells (osteocytes) would be the smallest. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-15 92. Our cells produce pyruvate through glycolysis. Pyruvate is then oxidized to acetyl-CoA, which feeds the citric acid cycle. The reasoning they might claim this is that it appears that simply adding more pyruvate would provide more fuel for the citric acid cycle, so cells can generate more energy, allowing muscles to do more work. Another reason could be to increase the amount of pyruvate during lactate fermentation, which occurs during strenuous activity. During this process, pyruvate is converted into lactate, which regenerates the NAD+ needed for glycolysis. An increased amount of pyruvate may lead to faster NAD+ regeneration, which in turn would allow glycolysis to proceed at a faster rate and increase ATP production. To evaluate the claims, a controlled study could be conducted that would compare the endurance of two groups: those taking the supplements and those not taking them. 93. If you modified the sequence of protein complexes along the electron transport chain, it would not generate ATP. The electron carriers in the chain are organized according to electronegativity. As a consequence, electrons move along the chain spontaneously from the less-electronegative complexes to the more-electronegative complexes, releasing energy as they go. The electrons would not flow along the chain if a less-electronegative complex followed a more-electronegative complex. Therefore if the protein sequences of the complexes were modified, it may affect their ability to bind the cofactors that are responsible for electron transfer. This would stop the electron transport chain and no ATP would be generated. Alternatively, sequence modification could also disrupt the ability of complexes I and IV to pump protons and the creation of the concentration gradient needed for chemiosmosis. 94. Answers may vary. Sample answer: Taking walk breaks early (before fatigue sets in) and often would be preferable over slow-run breaks. Walking during a long distance run helps muscles conserve resources. When a muscle is used continuously, it fatigues relatively soon, forcing the runner to slow down later in the run or experience pain. Mixing walking and running distributes the workload among a number of muscles, increasing overall performance capacity. Walking would provide greater recovery than slow running, while not significantly affecting overall speed. During strenuous exercise, muscle cells consume ATP faster than oxygen can be supplied to the electron transport chain for ATP production by oxidative phosphorylation. This forces cells to use anaerobic lactate fermentation as a source of ATP, leading to an accumulation of lactate molecules. Walking is not as strenuous as running and allows the body to recover; lactate is transported to the liver and is processed back into pyruvate. Overall performance would improve (faster finish time) and fatigue and muscle damage from overuse would be reduced. 95. Answers may vary. Sample answer: The plants were able to survive in the darkness because they were able to use their glucose stores as a source of energy via cellular respiration. The amount stored was different for each of the plants, explaining why some were able to survive for days and some for weeks. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-16 96. (a) Answers may vary. Sample answer: Question (yes=0/no=1) C3 plant C4 plant CAM plant Requires a lot of moisture? 0 1 1 (drains water supply) Requires a lot of light? 0 0 1 (drains electricity resources) Grows quickly? 0 0 1 (takes up a lot of space) SUM 0 1 3 RANK 3 2 1 (b) Answers may vary. Sample answer: A CAM plant may be the best plant for the space station as it requires the least moisture and is accustomed to living in dry environments (like on the space station), it does not need prolonged light exposure like C3 and C4 plants, and it is slow growing so it would not take up a lot of space. (c) Answers may vary. Sample answer: Although it would take many plants (and take up a lot of space) to generate a significant amount of oxygen, a better choice to generate oxygen on the space station may be the C4 plants. They can handle the drier environment, but they grow more quickly, and are better at fixing CO2. Many are edible. Ideally, there should be a separate environment with high moisture and low oxygen so that the plants could more readily use the CO2 from the astronauts and release O2 to be used by them. 97. The levels of G3P should increase as light intensity increases, to a point. Although G3P is produced by the light-independent reactions of photosynthesis, these reactions require the ATP and NADPH produced by the light-dependent reactions. At the point where further increases in light intensity no longer result in an increase in G3P, it is likely that the production is limited by the availability of CO2 or the maximum speed of the Calvin cycle, which is dependent on enzyme activity/concentrations and temperature. 98. Warming temperatures and very dry climates might also have led to the evolution of C4 plants. Reflect on Your Learning 99. Answers will vary. Sample answer: I learned that all foods are broken down and their energy extracted by the various reactions of cellular respiration. Whether they are carbohydrates, fats, or proteins, food molecules enter the cellular respiration pathway at various points and lead to the production of ATP. 100. Answers will vary. Answers should include: Photosynthesis and cellular respiration complement each other, fitting together like the pieces of a puzzle. Photosynthesis uses light energy to fix carbon dioxide and store energy in food molecules, creating oxygen in the process. Cellular respiration breaks down these food molecules to extract their energy. It uses up oxygen and releases carbon dioxide, making carbon once again available for fixation during photosynthesis. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-17 101. Answers will vary. Sample answer: 102. Answers will vary. Sample answer: Changes in temperature and the concentration of gases associated with climate change would affect photosynthetic organisms. For example, an increase in CO2 would decrease photorespiration, which would be especially beneficial to C3 plants. On the other hand, high temperatures would cause plants to close their stomata to decrease water loss. This would prevent gas exchange and increase photorespiration. At higher temperatures the solubility of CO2 decreases more rapidly than that of O2 and photorespiration would increase even more. Under these conditions, C4 and CAM plants may survive better. 103. Answers will vary. Sample answer: As a result of studying this unit, I would like to learn more about the careers of baker or chef and food scientist. I am a vegetarian, and I’m interested in how to get energy from plants into people in creative ways. Research 104. (a) Kleptoplasty (“stealing” of chloroplasts) occurs in nature when an alga is eaten and partially digested, but the plastid is left intact and functional (for varying amounts of time) within the host. Sacoglossan sea slugs can capture, and use, intact functional chloroplasts from algae. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-18 (b) The prime advantage of kleptoplasty is that the host can get energy from sunlight (photoautotrophy). In a lab setting, sea slugs can sustain themselves completely by photosynthesis for up to ten months. The chloroplasts may also offer the host camouflage protection. Disadvantages are that it is not permanent and it is not hereditary. Associations must constantly be remade. (c) Answers will vary. Sample answer: Kleptoplasty would be a benefit to humans if we ran out of food (but we would be green). 105. Answers may vary. Articles should include the following information. • The discovery of photosystems was a case of scientists building on one another’s research. In 1772, Joseph Priestly discovered that plants could produce oxygen in the absence of light. Cornelius van Niel elucidated the mechanism of photosynthesis further in 1931. In 1931, Robin Hill showed that the two phases of photosynthesis occurred in two different places in a plant cell. Louis Duysens and Robin Hill (with Fay Bendall) independently proposed 2 photosystems in 1960. • Chemiosmosis was first described by Peter Mitchell in 1961. His idea was not accepted until Andre Jagendorf showed significant evidence that it was true. Mitchell eventually did win the Nobel Prize for his work in 1978. 106. (a) Retinal is a photopigment found in photoreceptor cells of the eye. It is associated with opsin proteins. Retinal changes shape when it absorbs light, allowing the opsin to function as light receptor. (b) Animals obtain retinal from fatty acid esters of retinol (vitamin A) or certain carotenoids (provitamin A). For example in humans, a provitamin A carotenoid, β-carotene, is split into two molecules of retinol (vitamin A), which are oxidized into retinal. (c) Retinal is involved in light absorption in the eye so it is not surprising that is synthesized from a photosensitive pigment. (d) Certain xanthophyll carotenoids (for example, lutein) are found in the retina. These yellow pigments absorb blue light. Thus, they may protect the retina from the ionizing effect of blue light and oxidative stress. (e) Melanopsin is a photopigment found in ganglion cells of the inner retina. Like other photopigments of the retina, it consists of an opsin protein and the pigment retinal. Melanopsin is involved in the regulation of the circadian rhythm (biological clock) and the pupillary reflex (pupil contraction in response to light). 107. (a) Microbes are preferred to chemical treatment because they are more environmentallyfriendly. They use their enzymes to break down or sequester organic matter and nutrients, which would otherwise require harsh chemicals. (b) Both aerobic and anerobic microbes are used. In aerobic treatment, microbes break down organic matter into carbon dioxide. In anaerobic treatment, different microbes participate in the sequential breakdown of organic matter, ultimately producing methane and carbon dioxide. (c) Using microbes to break down organic matter and nutrients such as nitrogen and phosphorus prevents them from being released into the environment, where they may stimulate growth of microorganisms detrimental to plant and animal life (eutrophication). As a benefit, anaerobic digestion produces a biogas (containing methane and carbon dioxide) than can be used in generators to produce electricity or in boilers to produce heat. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-19 (d) Answers may vary. Sample answers: • biomining • bioremediation at nuclear sites • food processing • metal refining 108. (a) Chemoautotrophs are bacteria or archaea that obtain energy from inorganic compounds. They can be found living in hostile environments such as deep-sea vents and hot springs. (b) Photoautotrophs utilize light energy from the sun and chemoautotrophs obtain energy through oxidation of inorganic (non-carbon) compounds from the environment. Like photoautotrophs, chemoautotrophs obtain their carbon from carbon dioxide. (c) Examples of compounds oxidized by chemoautotrophs to supply hydrogen are hydrogen sulfide, H2S, and ammonia, NH3. (d) An example: in the colourless sulfur bacteria, oxygen accepts an electron from hydrogen sulfide, H2S, creating water and sulphur. The energy from this reaction is used to reduce CO2 to create sugars. Examples of other electron acceptors besides oxygen used by chemoautotrophs include sulfate, SO42–, and carbon dioxide. 109. (a) Answers will vary. Answers should be based on: • Coral and zooxanthellae (photosynthetic algae) have a symbiotic relationship and could not survive without the other. • Green sea slugs use chloroplasts from the algae it eats and can produce chlorophyll like a plant. • The Australian orchid Rhizanthella gardneri grows underground, cannot photosynthesize (but still retains chloroplasts), and receives all its nutrients from a mycorrhizal fungus associated with another plant. Mycorrhizal fungi form symbiotic associations with the roots of plants, and the orchid gets its food by parasitizing this relationship. (b) Answers will vary. Sample answer: The line between plants and animals is blurring because of the vast diversity of life. Plants and animals have adapted to live in many situations we would not typically expect. 110. (a) Answers may vary. Reports should follow the guides of the questions given. Some topics that may be included: • To make ethanol from starch, the corn is either dry ground or wet milled then cooked at high temperature with an added enzyme. It is then fermented and, finally, distilled. • Problems with current corn starch ethanol production is that it takes a lot of energy to make (electricity at many distilleries comes from coal plants), it takes a lot of water to grow crops, and it diverts corn and agricultural land from providing food. • Cellulosic ethanol is produced from non-edible, structural plant materials rather than starch. The raw material is abundant and varied (versus corn). Production steps are similar to that of starch ethanol following pretreatment and involve 1) hydrolysis to break the long, complex sugars down into simple sugars, 2) fermentation, and 3) distillation. Additional steps to remove non-fermentable residual materials are also included. • Currently, corn is easier and less expensive to process into ethanol versus cellulosic ethanol (high cost due to enzymes needed to break down cellulose), but it is cheaper to produce cellulose biomass (less energy, less fertilizer, less herbicides, less soil erosion, better soil fertility) than corn. • Studies show ethanol from cellulose reduces greenhouse gas emission by 90 % versus cornbased ethanol, which reduces emissions 10–20 %. • Cellulose is not used for food (corn is a food crop) and the entire plant can be used. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-20 (b) Answers may vary. Choosing cellulosic ethanol may benefit the Ministry of the Environment more because it is a cleaner burning fuel and produces far less greenhouse gas emissions. Choosing ethanol from corn starch may benefit the Ministry of Industry because there are already a number of refineries using this method across Canada. 111. (a) The centre of a tumour may be low in oxygen (called tumour hypoxia) because it has grown so large that the capillaries cannot reach the centre (to deliver oxygen), or the cells of the tumour have proliferated so much that they “overwork” the local oxygen supply. (b) The bacteria activate the drug by expressing an enzyme that converts the inactive form of the anti-cancer drug to its active form. (c) The bacteria are being genetically altered to express more efficient drug-converting enzymes and to produce higher quantities of them. Therefore, the non-toxic inactive form of the anticancer drug can be injected in higher concentrations into the body, but it will only convert to the active, cancer-killing form where the bacteria are located—in the centre of the tumour. (d) Answers may vary. At the time of this publication, this technique was not yet in clinical trials (testing on humans). It is still in its testing phase, but researchers have recently made improvements to the enzyme DNA that makes it much more efficient at turning the inactive anticancer drug into its active form. Clinical testing is planned for 2013. (e) Answers may vary. Some of the first articles were published in the British Journal of Cancer and Cancer Research. Copyright © 2012 Nelson Education Ltd. Unit 2: Metabolic Processes U2-21