Respiration in Plants Class 11 Biology NCERT Solutions

Step-by-step solution

1. 1Take each pair in turn, and for each one choose the points of contrast that are true and few, since a long list of marginal differences is not what earns marks. Three or four contrasts per part is right. The first thing to establish, since all three parts depend on it, is that respiration is a biochemical process inside the cell whereas combustion is a chemical process outside it.
2. 2(a) Respiration and combustion. The differences to give are these. Respiration is a biochemical process that occurs inside the living cell, and it is a series of enzyme-controlled steps catalysed by the enzymes of the cell, whereas combustion is a chemical process that is not a function of the living cell at all, being only a physical reaction of a substance with oxygen that may be brought about in dead matter, and it is not catalysed by the enzymes of the cell in the sense in which the respiratory steps are. Respiration is a slow, controlled and stepwise process, and the energy is released in small instalments so that it can be trapped in the form of ATP, whereas combustion is a rapid, violent and exothermic process and all the energy is released at once as heat. Consequently respiration is energy-efficient, with a part of the energy conserved in ATP whereas combustion converts the energy entirely into heat and none is conserved, which is why a sugar that is oxidised in the cell yields usable energy and the same sugar burnt in air yields only heat. Respiration occurs at a comparatively low temperature and is regulated by the cell, since the rates are controlled by the availability of the substrates, of the ADP and of the oxygen, whereas combustion requires a high temperature to start and once started proceeds to completion. And respiration occurs in living cells only, since it is an enzymatic process, whereas combustion may be brought about in dead matter and needs no living system. It should be added that the ultimate chemical equation for the oxidation of the glucose is broadly the same in both, glucose plus six oxygen giving six carbon dioxide and six water with energy, so the difference lies not in the overall chemistry but in the route, the control and the fate of the energy.
3. 3(b) Glycolysis and the Krebs cycle. The differences to give are these. The site is cytoplasm for the glycolysis and the mitochondrial matrix for the Krebs cycle, since the glycolysis occurs in the cytoplasm and the whole of the rest of the aerobic respiration, that is the pyruvate oxidation and the Krebs cycle, occurs in the mitochondrion. The nature of the pathway is a biochemical one, so it is anaerobic and does not need oxygen, for the glycolysis takes place whether or not the oxygen is present, whereas the Krebs cycle is aerobic and cannot proceed without the oxygen, since the NADH and the FADH2 it forms must be reoxidised by the electron transport system. The substrate is the glucose or the starch broken to glucose for the glycolysis and the acetyl CoA, that is the two-carbon compound formed from the pyruvate by the pyruvate oxidation, for the Krebs cycle. The carbon accounting is that one molecule of the six-carbon glucose gives two molecules of the three-carbon pyruvate in the glycolysis, whereas one turn of the Krebs cycle, which is repeated twice for one glucose, accepts one molecule of the two-carbon acetyl CoA and releases two molecules of the one-carbon carbon dioxide. The place in the sequence is first and last but one, so the glycolysis is the preparatory step and the Krebs cycle is the final common oxidative pathway, and the Krebs cycle is the link between the metabolism of the carbohydrate, the fat and the protein, since the acetyl CoA of the fat and of the protein also enters it. And the yield differs, since the glycolysis gives a net 2 ATP and 2 NADH whereas one turn of the Krebs cycle gives 3 NADH, 1 FADH2 and 1 ATP.
4. 4(c) Aerobic respiration and fermentation. The differences to give are these. The oxygen is required in the aerobic respiration and not required in the fermentation. The completeness of the oxidation is complete in the aerobic respiration, the substrate being fully oxidised to carbon dioxide and water, and partial in the fermentation, the substrate being only partly broken down and the product being an organic compound such as the lactic acid or the ethanol that still contains unreleased energy. The site is the mitochondrion, that is the matrix and the inner membrane, for the aerobic respiration, and the cytoplasm for the fermentation, since the whole of the fermentation, and the glycolysis that precedes it, occurs in the cytoplasm, and the enzymes concerned are soluble. The yield of energy is large in the aerobic respiration, about 38 ATP per glucose in a eukaryotic cell, and very small in the fermentation, a net 2 ATP, so that most of the energy of the substrate is not released but remains locked in the products. The end products are the carbon dioxide and the water in the aerobic respiration, and the lactic acid in the lactic acid fermentation, chiefly in the animal muscle and in the bacterium, or the ethanol and the carbon dioxide in the alcoholic fermentation, in the yeast and in the plant. The relation between the two is that the anaerobic respiration uses the same glycolysis as the aerobic one, since the glycolysis is common to both, and only the fate of the pyruvate differs, in the aerobic respiration it is oxidised completely and in the fermentation it is reduced to the organic end product.

Step-by-step solution

1. 1Define the term, then name the classes of substrate, and then state the commonest one with the reason, since the second half of the question is worth as much as the first.
2. 2Definition: the respiratory substrates are the organic substances that are oxidised in the cells to release the energy required for the cellular activities, so they are the fuels of the cell. They are organic, since they are the carbohydrates, the fats and the proteins, and they are oxidised, being broken down to the simpler substances, ultimately the carbon dioxide and the water, in the process from which the energy is captured as ATP. So a respiratory substrate is the substance that is consumed as the fuel in respiration.
3. 3The classes to name are the three, and the order is the order of preference of the cell. Carbohydrates are the most preferred, the most commonly used respiratory substrate, and the chief carbohydrate is the glucose, which is produced by the hydrolysis of the starch and of the glycogen and of the other polysaccharides, and glucose is also the sugar of the blood, so that the cells have a constant supply of it. Fats are the second choice and are used when the carbohydrate is not available, being stored in the adipose tissue as the triglycerides, and they give more energy per unit mass, about twice as much as the carbohydrate, so they serve as a reserve fuel. Proteins are the last choice and are used only when both the carbohydrate and the fat are exhausted, being the structural proteins of the cell and the enzymes, so that their use for energy is necessarily a last resort.
4. 4The most common respiratory substrate is glucose, and the reasons to give are three. It is the most readily available in the cell, since it is transported in the blood as the glucose of the blood and is constantly supplied to the cells from the digestion of the dietary carbohydrate. It is the substrate for which the whole enzyme machinery of glycolysis is present and has evolved, so it is the direct fuel of the glycolysis. And it is the substrate whose oxidation gives the classical respiratory quotient of 1, since equal volumes of the carbon dioxide and the oxygen are involved, which is the value that the question on the RQ then turns on. The comparison of the energy yields, which closes the answer, is that a gram of carbohydrate yields about 4 kcal, a gram of fat about 9 kcal and a gram of protein about 5.6 kcal, so the fat is the most economical store of energy although the glucose is the most convenient fuel.

Step-by-step solution

1. 1This is a diagram question, so the answer must set out the whole pathway as a scheme, with the energy investment and the energy return shown, and the net yield boxed. Draw it as two phases, since that is how it is examined, and mark the number of ATP at each arrow.
2. 2Begin the diagram with the substrate, one molecule of the six-carbon glucose, entering the pathway, and note that it is first phosphorylated at the sixth carbon by the hexokinase, using one ATP, to the glucose 6-phosphate, and then isomerised to the fructose 6-phosphate, and then phosphorylated again at the first carbon by the phosphofructokinase, using a second ATP, to the fructose 1,6-bisphosphate. This first half is the preparatory phase, or the energy investment phase, sometimes called the pay-up phase, and it costs 2 ATP per glucose, and its purpose is to trap the glucose in the cell, since the phosphorylation makes the sugar charged and unable to leave, and to prime the molecule so that it can be cleaved, and to provide the activation energy for the cleavage.
3. 3Then the cleavage and the energy return. The fructose 1,6-bisphosphate is split by the aldolase into two molecules of the three-carbon triose phosphate, the dihydroxyacetone phosphate and the glyceraldehyde 3-phosphate, and the dihydroxyacetone phosphate is isomerised to a second glyceraldehyde 3-phosphate, so that one glucose gives two molecules of the three-carbon glyceraldehyde 3-phosphate. Each of these two is then oxidised and phosphorylated by the glyceraldehyde 3-phosphate dehydrogenase, using inorganic phosphate and the NAD plus, giving 1,3-bisphosphoglycerate and reducing one NAD plus to NADH plus H plus, so that two NADH are made per glucose, and each 1,3-bisphosphoglycerate is then substrate-level phosphorylated by the phosphoglycerate kinase to 3-phosphoglycerate, giving 2 ATP per glucose, so 4 in all for the two trioses.
4. 4Then the last three steps, which are common to both the aerobic and the anaerobic respiration and are the only stage at which ATP is made in fermentation. The 3-phosphoglycerate is moved to the 2-phosphoglycerate by the phosphoglycerate mutase, and a molecule of water is removed by the enolase to form the phosphoenolpyruvate, the highest-energy compound of the glycolysis, and the phosphoenolpyruvate is then phosphorylated by the pyruvate kinase to pyruvate, giving a second molecule of ATP per triose, so 4 more for the two trioses. The diagram should therefore be drawn as: glucose, with 2 ATP used to reach the fructose 1,6-bisphosphate, then the two triose phosphates, then with 2 NADH formed and 4 ATP made to reach the 2 molecules of pyruvate, then a further 2 ATP, so 2 minus 2 minus 2 plus 2 plus 2 plus 2 equals 2, with the net gain of 2 ATP per glucose and the 2 NADH marked for the aerobic case.
5. 5Box the net yield and state the sites, since these are the marks. The net gain is 2 ATP of direct energy and 2 NADH per molecule of glucose, and the total ATP from the 2 NADH, if they are oxidised by the aerobic route with the NADH dehydrogenase of the matrix, is 2 times 3, that is 6 ATP, so the aerobic yield of the glycolysis is 8 ATP and the anaerobic yield is the 2 ATP alone. The site is the cytoplasm, in the cytosol, and the whole glycolysis is anaerobic, needing no oxygen. And two further points to add, because they are examinable: the glycolysis is the same in the aerobic and the anaerobic respiration, since it is common to both and it is only the fate of the pyruvate that differs, and the glycolysis is a prime example of substrate-level phosphorylation, since the ATP is made by the direct transfer of the phosphate from the substrate, here the 1,3-bisphosphoglycerate and the phosphoenolpyruvate, to the ADP, and not by the electron transport chain.

Step-by-step solution

1. 1List the steps in order with the site of each, since the question asks for both, and then draw the whole thing together as one line with the two sites marked. The sites are the marks, so give them precisely.
2. 2The main steps of the aerobic respiration are four, and the site of each is different, which is precisely why the question asks for both together. Step one is the glycolysis, the Embden-Meyerhof-Parnas pathway, the anaerobic phase common to both respirations, and it occurs in the cytoplasm. Step two is the oxidation of the pyruvate, the link reaction or the transition reaction, in which the two molecules of pyruvate are oxidatively decarboxylated to form two molecules of acetyl CoA, and it occurs in the mitochondrial matrix. Step three is the Krebs cycle, or the citric acid cycle, or the tricarboxylic acid cycle, in which the acetyl CoA is completely oxidised to carbon dioxide, and it occurs in the mitochondrial matrix. Step four is the oxidative phosphorylation, the electron transport system, in which the reduced coenzymes are reoxidised and the energy of the electrons is used to synthesise ATP, and it occurs on the inner membrane of the mitochondrion, at the cristae.
3. 3The whole pathway then condenses into one line, and this is the schematic to write. Glucose, in the cytoplasm, gives 2 pyruvate with 2 ATP and 2 NADH. The 2 pyruvate, in the matrix, give 2 acetyl CoA with 2 NADH and no ATP of their own. The 2 acetyl CoA, in the matrix, give 4 CO2 with 6 NADH, 2 FADH2 and 2 ATP. And the 10 NADH, two of them of glycolytic origin, together with the 2 FADH2, on the inner membrane, give 6 H2O and 34 ATP, that is 38 ATP in all for the eukaryotic cell.
4. 4The two points to add, since they complete the question. The first is that the glycolysis alone occurs in the cytoplasm and needs no oxygen, and it is the only stage that does so, since the link reaction and the Krebs cycle are in the matrix and the oxidative phosphorylation is on the inner membrane, and all three of these require the oxygen. The second is that the compartmentation is functionally significant and not merely structural, because the Krebs cycle and the electron transport system are held apart in the matrix and the membrane respectively, and it is the impermeability of the inner membrane to the NADH that forces the respiratory hydrogen of the cytosol to be transferred in by the shuttle systems, and it is the two hydrogen carriers having their specific entry points, the NADH entering at complex I and the FADH2 at complex II, that is what accounts for the 3 ATP per NADH against the 2 ATP per FADH2, since the FADH2 gives up its electrons at a lower point on the chain and so forges fewer ATP. And a last sentence on the site in general: the reason the whole of the aerobic respiration is in the mitochondrion in a eukaryote is that the respiratory pathway generates the NADH and the FADH2 in the matrix and the protons that the ATP synthase uses are pumped across the inner membrane, and the bacterial cell has no such organelle, so all its aerobic respiration occurs in the plasma membrane and the cytoplasm.

Step-by-step solution

1. 1Draw the cycle as a ring with the entry of the acetyl CoA at one point, the carbon dioxide leaving at two points, the oxaloacetic acid regenerated at the end, and the three reduced coenzymes leaving at three points, since that is the shape of the cycle and every mark depends on the drawing being right. Then account for the turns and the yields.
2. 2The entry. The cycle begins when the two-carbon acetyl CoA, derived from the pyruvate, is received by the oxaloacetic acid, the four-carbon acceptor, in a condensation reaction catalysed by the citrate synthase, to form the six-carbon citric acid, the citrate. So the entry is the condensation of the four-carbon oxaloacetic acid with the two-carbon acetyl CoA to give the six-carbon citrate, and this is the only step in which the carbon of the cycle is increased, and the only step in which the carbon of the acetyl group is introduced.
3. 3The two decarboxylations. The citric acid is converted in a series of steps to the six-carbon isocitric acid, and the isocitric acid is then oxidatively decarboxylated by the isocitrate dehydrogenase, releasing the first carbon dioxide and reducing one NAD plus to NADH plus H plus. The resulting five-carbon alpha-ketoglutaric acid is then oxidatively decarboxylated by the alpha-ketoglutarate dehydrogenase complex, releasing the second carbon dioxide and reducing a second NAD plus to NADH plus H plus, and giving the four-carbon succinyl CoA. So two carbon dioxide are released in the cycle, and both are released as decarboxylations, that is oxidatively. And the point that is often got wrong, so state it carefully, is that these two carbon dioxide do not come from the two carbon atoms of the acetyl CoA that has just entered. The acetyl CoA condenses with the oxaloacetic acid and its two carbons are carried round the cycle in the citrate; in the first turn the two carbon dioxide that leave are the carbons of the four-carbon oxaloacetic acid that accepted the acetyl group, which is why the carbon dioxide and the acetyl group cannot balance in a single turn. It is the continued replenishment of the oxaloacetic acid, with the carbons drawn from the incoming acetyl CoA, that allows the oxidation of the acetyl group to go on, so the two carbon of the acetyl CoA are fully oxidised only over the course of several turns, and that is the whole point of the cycle. It should be noted that the succinyl CoA to succinate step is the one substrate-level phosphorylation of the cycle, the succinyl CoA synthetase transferring the phosphate of the CoA to the GDP, making a GTP or an ATP.
4. 4The regeneration. The succinate is then oxidised to the fumarate by the succinate dehydrogenase, which is a flavoprotein and so passes the hydrogen to the FAD, making the FADH2, and the fumarate is then hydrated to the malate, and the malate is then oxidised, and not decarboxylated, by the malate dehydrogenase, to regenerate the oxaloacetic acid, at the same time reducing a third NAD plus to NADH plus H plus. So the cycle closes, and the oxaloacetic acid is regenerated ready to accept another acetyl CoA, and this is the fourth reduced coenzyme, the FADH2, and the third NADH, of the turn.
5. 5Then the accounting, which is the part the question expects with the scheme. In one turn of the cycle, one molecule of the two-carbon acetyl CoA enters, two molecules of the carbon dioxide leave, and there is formed three molecules of NADH, one molecule of FADH2 and one molecule of ATP or GTP. Since one molecule of glucose gives two molecules of the acetyl CoA, the cycle turns twice for each glucose, and the total for one glucose is therefore 4 CO2, 6 NADH, 2 FADH2 and 2 ATP, in addition to the 2 NADH and the 2 ATP of the glycolysis and the 2 NADH of the link reaction, which makes no ATP of its own. And the whole of it can be written on the diagram as: the oxaloacetic acid, 4C, plus the acetyl CoA, 2C, giving the citric acid, 6C, giving the isocitric acid, 6C, losing a CO2 to the alpha-ketoglutaric acid, 5C, losing a CO2 to the succinyl CoA, 4C, giving a GTP, then the succinate, 4C, the fumarate, 4C, the malate, 4C, and back to the oxaloacetic acid, 4C, so that the carbon count returns to four, the two carbons of the acetyl CoA being still carried in the citrate at the end of the turn.

Step-by-step solution

1. 1Give the abbreviation first, then the location, then the mechanism step by step, then the yields, and then the terminal acceptor. The whole of the question is the mechanism, so the four complexes and the pumping of the protons must be described.
2. 2The ETS is the electron transport system, and it is also called the electron transport chain. It is the fourth and the last step of the aerobic respiration, and it is the machinery by which the energy of the electrons in the reduced coenzymes, the NADH and the FADH2, is converted into ATP, so it is the site of the oxidative phosphorylation. Its location is the inner membrane of the mitochondrion, at the cristae, and in a bacterial cell the plasma membrane, and the reason the membrane is the site is the ATP synthase, which is embedded in it, and the fact that the energy of the electrons is used to pump the protons from the matrix into the intermembrane space, so that an electrochemical gradient is created across the membrane which is then used to drive the synthesis of the ATP.
3. 3The mechanism. The electron of the NADH is handed to the first complex, the NADH dehydrogenase, the complex I or the respiratory complex I, and the FADH2 passes its electron to the FAD of the second complex, the succinate dehydrogenase, the complex II, which is the only component of the chain that does not pump protons, since it is the same enzyme as that of the Krebs cycle and receives the electrons from the FAD of the FADH2 that is formed within the cycle itself, and the two therefore differ in the number of ATP they finally yield. From the first and the second complexes the electrons pass to the ubiquinone or coenzyme Q, which is a mobile and lipid-soluble carrier. From the coenzyme Q they pass to the third complex, the cytochrome bc1 complex, and then to the cytochrome c, another mobile carrier, and then to the fourth complex, the cytochrome c oxidase or the complex IV, and the electron is finally passed to the terminal acceptor, the molecular oxygen, which combines with the protons of the matrix to form water. So the last acceptor in the chain is the oxygen, and this is why the whole of the process is aerobic and why the absence of the oxygen stops the chain.
4. 4Then the energy capture, and this is the crux. The transfer of the electrons from one complex to the next releases energy at each step, and this energy is used to pump the protons from the matrix into the intermembrane space, and the complexes that pump are the first, the third and the fourth, the first pumping four protons per pair of electrons, the third four and the fourth two, so that in the NADH case about ten protons are pumped per pair of electrons, whereas in the FADH2 case the electrons enter at the second complex and bypass the first, so only about six are pumped. The membrane is otherwise impermeable to the protons, so the pumped protons accumulate in the intermembrane space and create an electrochemical or proton motive gradient across the inner membrane, the matrix becoming negative and alkaline relative to the intermembrane space. The protons then flow back into the matrix through the ATP synthase, the F0F1 particle, which spans the membrane, and this flow is the mechanical basis of the oxidative phosphorylation, since the returning protons turn the rotor of the ATP synthase and so drive the catalytic formation of ATP from the ADP and the inorganic phosphate on the F1 head projecting into the matrix. The ATP so made leaves the mitochondrion in exchange for the ADP and the phosphate.
5. 5Then the yields and the closing points, which are examinable. The oxidative phosphorylation yields about 34 ATP per glucose in a eukaryotic cell, since the 10 NADH, 2 of glycolytic and 8 of matrix, give 30 ATP at 3 each and the 2 FADH2 give 4 at 2 each. The whole ETS is a complex of the coenzyme Q, the cytochrome bc1 complex, the cytochrome c and the cytochrome c oxidase, arranged in a definite sequence within the membrane, and the cytochrome c is the one component of the chain that is on the outer surface, the cytochrome c oxidase being on the inner, so the electrons must move across the membrane from one to the other, and the cytochrome c acts as the mobile shuttle between them. And the 3 ATP per NADH against the 2 per FADH2 is not a matter of the number of the electrons but of the point at which they enter the chain, since the FADH2 enters at the second complex and so forgoes the pumping at the first.

Step-by-step solution

1. 1This is the companion question to Q1 and the three pairs overlap with it, so the same discipline applies, but two of the pairs are different from those of Q1 and the differences must be noticed. The aerobic and the anaerobic respiration here is a comparison of whole pathways, the glycolysis and the fermentation is a comparison of the two steps of the anaerobic respiration, and the glycolysis and the citric acid cycle is the same comparison as in Q1(b) but must be written in its own right. Take the contrasts that are true and few.
2. 2(a) Aerobic respiration and the anaerobic respiration. The oxygen is the first point, for the aerobic respiration requires the molecular oxygen as the terminal acceptor of the electron transport chain, whereas the anaerobic respiration does not, and it is the absence of the oxygen that makes the fermentation the only route. The completeness of the oxidation is the second, for the substrate is completely oxidised to the carbon dioxide and the water in the aerobic respiration, whereas in the anaerobic respiration it is only partly broken down and the carbon dioxide is produced without the oxygen being used, so the products are the organic compounds such as the lactic acid or the ethanol that still hold most of the energy. The site is the third, since the aerobic respiration involves the mitochondrion, the matrix and the inner membrane, whereas the anaerobic respiration is confined to the cytoplasm. The yield is the fourth and the most striking, for the aerobic respiration gives about 38 ATP per glucose in a eukaryotic cell and the anaerobic gives a net 2, so the aerobic releases some nineteen times as much energy. The enzymes are the fifth, the oxidoreductase enzymes of the membrane being used in the aerobic respiration whereas in the anaerobic respiration the enzymes are the soluble ones of the cytoplasm. And the sixth point is that the two are related and not alternative, since the glycolysis is the common first phase of both, so the anaerobic respiration is not a different route but the same route with the last two steps replaced by the fermentation, which is the same fact that Q1(c) turns on and which is worth stating again because it is the point of the whole comparison.
3. 3(b) Glycolysis and fermentation. The two are the two parts of the anaerobic respiration and the contrasts between them are what the question wants. The first is the nature: the glycolysis is the normal, universal and obligate pathway of the glucose breakdown, occurring in every living cell of every organism and being an aerobic-independent step, whereas the fermentation is not universal, being confined to particular organisms, the yeasts, the bacteria and the animal muscle, and being an adaptation to the absence of the oxygen. The second is the nature of the products: the product of the glycolysis is the pyruvate, which is a three-carbon acid that still holds much unreleased energy, and is not the end of the matter, whereas the product of the fermentation is the lactic acid or the ethanol with the carbon dioxide, and this is the end of the matter, so the fermentation is the final step of the anaerobic respiration and the glycolysis is the first. The third is the site and the enzymes: the glycolysis occurs in the cytoplasm, its enzymes being the kinases and the dehydrogenases of the pathway, and the fermentation also occurs in the cytoplasm, but with the different and specific enzymes, the lactic acid dehydrogenase and the pyruvate decarboxylase and the alcohol dehydrogenase. The fourth is the energy: the glycolysis yields a net 2 ATP, and the fermentation yields none, the whole ATP of the anaerobic respiration being that of the glycolysis, so the fermentation step itself yields no ATP at all and is therefore not an energy-conserving step in the sense of making any; what it does is reoxidise the NADH formed in the glycolysis back to NAD plus, so that the glycolysis itself can continue. The fifth is the fate of the NADH: in the glycolysis the NADH is made, whereas in the fermentation it is consumed and reoxidised, so the fermentation is the sink for the reduced hydrogen of the glycolysis. And the sixth is the yield of the whole: since the fermentation adds nothing, the 2 ATP of the glycolysis is the whole yield of the anaerobic respiration, and this is the reason the anaerobic respiration is some nineteen times less efficient than the aerobic.
4. 4(c) Glycolysis and the citric acid cycle. The contrasts are those already given in Q1(b), and the important ones are the site, that the glycolysis is in the cytoplasm and the citric acid cycle in the mitochondrial matrix; the nature, that the glycolysis is anaerobic and needs no oxygen and that the cycle is aerobic and cannot proceed without it; the substrate, the glucose for the glycolysis and the acetyl CoA for the cycle; the carbon accounting, two molecules of the three-carbon pyruvate from one six-carbon glucose against one molecule of the two-carbon acetyl CoA and two molecules of the carbon dioxide per turn; the position in the sequence, the glycolysis first and the cycle the final common oxidative pathway and the meeting point of the carbohydrate, the fat and the protein metabolisms; and the yield, a net 2 ATP and 2 NADH against 3 NADH, 1 FADH2 and 1 ATP per turn. And it should be added that the glycolysis occurs in the cytoplasm of every living cell, prokaryote and eukaryote alike, whereas the citric acid cycle is a feature of the aerobic eukaryote, since the bacteria perform the analogous reactions in the plasma membrane and the cytoplasm, so the two are not equally universal.

Step-by-step solution

1. 1List the assumptions, one by one, and then give the two totals, the ideal 38 and the practical 36, and the reason for the loss, since the question is about the accounting rather than about the pathway. The assumptions are few and each is a point.
2. 2The assumptions, as NCERT states them, are these. First, that the entire NADH produced in the glycolysis is transported into the mitochondrion and is oxidised there, so that it gives three ATP each; but the inner membrane of the mitochondrion is impermeable to the NADH and the reducing equivalents must therefore be carried across by the shuttles, and the malate-aspartate shuttle transfers them in such a way as to yield three ATP, whereas the glycerol 3-phosphate shuttle transfers them to the FAD of the second complex and so yields only two. Second, that the two NADH and the two FADH2 produced in the Krebs cycle give three and two ATP respectively. Third, that the two NADH produced in the oxidation of the pyruvate to the acetyl CoA also give three ATP each, and that the ATP synthesised in the substrate-level phosphorylation, the two in the glycolysis and the two in the Krebs cycle, is taken at its face value. And these are the assumptions, and the point to notice is that the first is a simplification, since the yield from the glycolytic NADH depends upon which shuttle is used and is not always three.
3. 3The two totals follow. Taking the ideal case, in which the malate-aspartate shuttle is assumed, the arithmetic is 2 ATP of substrate-level phosphorylation in the glycolysis, plus 2 times 3 for the 2 NADH of the glycolysis, that is 6, plus 2 ATP and 2 NADH at the link reaction, that is 2 plus 6, plus 6 times 3 for the 6 NADH of the Krebs cycle, that is 18, plus 6 times 2 for the 2 FADH2 of the Krebs cycle, that is 4, giving 2 plus 6 plus 2 plus 6 plus 18 plus 4, that is 38 ATP per glucose. The practical total is less, and NCERT gives 36, because the glycolytic NADH in the muscle of the animal is transferred by the glycerol 3-phosphate shuttle and so yields only 2 ATP each rather than 3, so 4 instead of 6, and the arithmetic is 2 plus 4 plus 2 plus 6 plus 18 plus 4, that is 36. It is worth adding that the prokaryotic total is different again, for the bacterial cell has no mitochondrion and no inner membrane to be impermeable, so the glycolytic NADH is oxidised directly by the plasma membrane at the full value of three, and there is no loss on the transport, and the total is therefore higher.
4. 4Close with the caveat that makes the question a good one. The calculation of 38 and of 36 is an accounting exercise based on the assumption of three ATP per NADH and two per FADH2, and the ratios are in fact not exact, since they are calculated from the assumption of near 4 joules per mole of ATP and of near 2.5 joules per mole of proton, and the stoichiometry of the respiratory chain is now believed to be closer to 2.5 ATP per NADH and 1.5 per FADH2, which would give a figure nearer 30 than 38. So the 38 is the classical textbook figure, obtained on the assumptions just listed, and it should be quoted as such, and the 36 is the figure for the shuttle of the animal muscle, and neither should be treated as a directly measured yield.

Step-by-step solution

1. 1Define amphibolic first, because the whole question turns on the word, then give the two directions, and then explain why the same pathway must serve both. The Krebs cycle is the amphibolic pathway, and the argument turns on the fact that the intermediates are used up as biosynthetic precursors as well as being oxidised.
2. 2Definition. An amphibolic pathway is one that serves both the catabolic, or the degradative and energy-releasing, function and the anabolic, or the biosynthetic, function, that is a pathway into which the material enters for the purpose of being broken down and a pathway out of which the intermediates are withdrawn for the building up of larger molecules. The word is used in this question of the respiratory pathway and it is the Krebs cycle that is amphibolic, the glycolysis and the ETS being purely catabolic, and so it is the Krebs cycle that the answer must be about.
3. 3The catabolic role. In the catabolic direction the Krebs cycle is the final common oxidative pathway of the three major classes of the respiratory substrate, the carbohydrate, the fat and the protein, since the acetyl CoA of the fat and of the protein and the pyruvate of the carbohydrate all converge on it, and the cycle completely oxidises the carbon of the acetyl group to the carbon dioxide over the course of several successive turns, so its role in respiration is the release of the energy, the six NADH, the two FADH2 and the two ATP of the two turns, for one glucose, being passed to the ETS to yield the 22 ATP, the six NADH at three each and the two FADH2 at two each, and the whole cycle is thus a catabolic process.
4. 4Then the anabolic role, which is the point of the question. The intermediates of the Krebs cycle are themselves the raw materials of several biosyntheses, and it is their withdrawal for that purpose that makes the cycle amphibolic and that obliges it to be replenished. The two-carbon acetyl CoA, the very entry of the cycle, is the precursor of the fatty acids and of the cholesterol and so of the steroids, and it is worth being precise about where this happens, since the acetyl CoA is made in the matrix of the mitochondrion but the fatty acid synthase and the cholesterol synthesis themselves are cytosolic, so the acetyl CoA has to leave the mitochondrion, chiefly as the citrate, before it can be built up into the fatty acid. The succinyl CoA is the precursor of the porphyrins, so of the haem of the haemoglobin and of the cytochromes and of the chlorophyll, the synthesis of the succinyl CoA being the first committed step of the porphyrin synthesis. The oxaloacetic acid is transaminated to the aspartate, which is one of the twenty amino acids, and so is used in the synthesis of the proteins, and it also gives the oxaloacetate used in the synthesis of the sugars, the pyruvate and the phosphoenolpyruvate coming from it, and so in the gluconeogenesis. And the alpha-ketoglutaric acid is transaminated to the glutamate, another of the twenty amino acids, and so serves in the synthesis of the amino acids and hence of the proteins. So several of the intermediates, directly or after a single transamination, are the starting points of the fatty acids, the cholesterol, the porphyrins and the amino acids, and the withdrawal of these is what makes the pathway serve the biosynthesis.
5. 5Then the consequence, which completes the answer and which is the intellectually satisfying part. If the intermediates are continuously withdrawn for the biosynthesis, the cycle would run out of them and would stop, so the cell must replenish them, and this is done by the anaplerotic or the filling-up reactions, the most important being the carboxylation of the pyruvate to the oxaloacetic acid by the pyruvate carboxylase, with the biotin and the ATP and the carbon dioxide. Note that this is a carboxylation and not a decarboxylation, so it is the opposite of what the link reaction does to the pyruvate, the link reaction removing a carbon as carbon dioxide to give the acetyl CoA while this one adds a carbon as carbon dioxide to give the oxaloacetic acid. And a second group of reactions, the cataplerotic or the emptying-up reactions, removes the intermediates, and the two groups are kept balanced in the cell, so that the cycle runs at a constant rate. So the respiratory pathway is amphibolic because it is not merely a degradative sequence but a metabolic crossroads: it is the point at which the degradation of the food and the synthesis of the new cell material meet, and the balance between the two uses of its intermediates is what keeps the metabolism of the cell in a steady state.

Step-by-step solution

1. 1Give the definition with the formula, then the values for the three substrates so the fats are placed in a table, then the explanation of why the fat has the low value, and then its practical use. The explanation by the oxygen content is the mark, so it must be stated.
2. 2Definition. The RQ, the respiratory quotient, is the ratio of the volume of the carbon dioxide released to the volume of the oxygen consumed during respiration, so RQ equals the volume of the carbon dioxide divided by the volume of the oxygen, the two being measured under the same conditions. It is a dimensionless ratio and it is a measure of the substrate being oxidised, so it is used to identify the respiratory substrate.
3. 3The values. For a carbohydrate the RQ is 1, since glucose plus six oxygen gives six carbon dioxide and six water, so equal volumes of the two gases are exchanged. For a protein the RQ is about 0.9 in the general statement, NCERT giving 0.9, and it is less than 1 because the proteins contain a nitrogen that is excreted rather than respired, and the carbon and the hydrogen being relatively less in proportion than in a carbohydrate, so that less oxygen is required. For a fat the RQ is about 0.7, and it is much less than 1 because the fats are poorer in oxygen than the carbohydrates and much richer in hydrogen, so a great deal more oxygen is required to oxidise them completely than is released as carbon dioxide, the ratio of the two being much below unity. The fats are therefore the least oxygen-demanding substrate per molecule of the carbon dioxide produced, and it should be noted that the value is the theoretical one for the complete oxidation, since an animal respiring a mixed diet gives a value between the extremes.
4. 4The explanation in terms of the composition, which is what the examiner wants. Taking a typical triglyceride such as the tripalmitin, the formula is C51H98O6, and its complete oxidation can be written as 2C51H98O6 plus 145O2, that is 102CO2 plus 98H2O, so the volume of the oxygen consumed is 145 and that of the carbon dioxide released is 102, and the RQ is 102 divided by 145, which is 0.7. This is a worthwhile calculation to include, because it demonstrates the point that the fat requires a very large volume of oxygen, and it explains why the fats are the preferred fuel in the hibernating animal and in the migratory bird, in the deep and the prolonged dives of the whale and of the seal, and in the high-altitude flight, where the oxygen supply is the limiting factor, since the fat gives the most energy for the least oxygen, so that the animals migrate or hibernate on a store of fat. The practical use to close with is the diagnosis, for the RQ of a patient is used to identify the substance being respired in the treatment of diabetes, and the respiratory quotient of about 0.7 in the diabetic coma of the uncontrolled case indicates the oxidation of the fat.

Step-by-step solution

1. 1Define the term by splitting it, since the word is self-explanatory and the examiner wants it spelled out, then state the site, then the mechanism in short, then the significance, and then the contrast with the substrate-level phosphorylation.
2. 2Definition. Oxidative phosphorylation is the synthesis of the ATP by the phosphorylation of the ADP with the inorganic phosphate, the energy for the phosphorylation being provided by the oxidation of the reduced coenzymes, that is the NADH and the FADH2, through the electron transport system. So the word oxidative refers to the oxidation of the coenzymes and the word phosphorylation to the attachment of the phosphate to the ADP, and the two are coupled, the oxidation of the coenzyme driving the phosphorylation of the ADP. It is the fourth step of the aerobic respiration and it is the major site of the ATP production, yielding about 34 of the 38 ATP of a glucose in a eukaryotic cell.
3. 3The site is the inner mitochondrial membrane, at the cristae, where both the electron transport system and the ATP synthase are located, and in a bacterial cell the plasma membrane, since the bacteria have no mitochondrion. And the site is not incidental, for it is the membrane that makes the process possible, since a membrane is required to hold the proton gradient and the ATP synthase must be embedded in it, which is why an artificial membrane system reconstituting both the respiratory chain and the synthase will carry out the oxidative phosphorylation in the test tube, and this reconstitution is the classic demonstration that no soluble factor is required.
4. 4The mechanism in brief. The electrons of the NADH and the FADH2 are passed down the chain, the complex I, the coenzyme Q, the complex III, the cytochrome c and the complex IV, to the oxygen, which is the terminal acceptor and forms the water, and at three points in the chain the energy released by the transfer is used to pump the protons from the matrix into the intermembrane space, so that a proton motive gradient is set up across the membrane, the matrix becoming negative and alkaline. The protons then return to the matrix through the F0F1 ATP synthase, and the flow turns the rotor and drives the synthesis of the ATP from the ADP and the phosphate. So the sequence is oxidation, the transfer of the electrons; pumping, the creation of the gradient; and then the return of the protons through the synthase, which makes the ATP. It should be noted that the ATP is not made by a direct transfer of the phosphate from the substrate as in the substrate-level phosphorylation of the glycolysis and of the Krebs cycle, but indirectly, through the gradient, and that is the distinction the examiner expects.
5. 5The significance, and the closing contrast. The significance of the oxidative phosphorylation is that it is the mechanism by which the bulk of the energy of the food is converted into the ATP that the cell can use, since the 34 ATP of a glucose are made here against the 4 made by substrate-level phosphorylation, so the oxidative phosphorylation is about eight times more efficient than the substrate-level, and it is the only stage at which the oxygen of the air is consumed, so it is the point at which a cell is sensitive to the lack of the oxygen. And the substrates that support it are the NADH and the FADH2 only, so the oxidative phosphorylation is possible only when there is a supply of them, and this is why the fermentation, which regenerates the NAD plus by the oxidation of the NADH, is what permits the cell to make its 2 ATP in the absence of the oxygen, when the oxidative phosphorylation cannot run at all.

Step-by-step solution

1. 1The answer is that the step-wise release allows the energy to be trapped in ATP rather than lost as heat, and everything else is an elaboration of that. Give the central point first, then the mechanism of the trapping, then the benefits, and then the contrast with combustion.
2. 2The central point. The significance of the step-wise release of the energy in respiration is that it enables the cell to trap the energy of the substrate in the form of the ATP rather than to lose it as heat, and this is the whole difference between respiration and combustion. The energy is released not in one step but in a series of enzyme-controlled steps, each of which releases a small and manageable portion, and at each of a few of those steps the released energy is used to make ATP, so that a very large part of the energy of the substrate is conserved in a form the cell can spend.
3. 3Then the mechanism of the trapping, which is the second point and which the examiner wants. The trapping occurs in two ways. In the substrate-level phosphorylation, which happens twice in the glycolysis and once in the Krebs cycle per turn, the phosphate is transferred directly from a high-energy substrate of the pathway, the 1,3-bisphosphoglycerate and the phosphoenolpyruvate in the glycolysis and the succinyl CoA in the cycle, to the ADP, so the ATP is made without any membrane and without any gradient. And in the oxidative phosphorylation, which happens in the fourth step, the energy is not released at the point of the substrate at all but is released in the transfer of the electrons down the chain, and the complexes of the chain, the complex I to the complex IV of the transport chain together with the complex V, the ATP synthase, which is coupled to it, and with the ubiquinone and the cytochrome c as the mobile carriers, are arranged in such a way that the energy of the transfer at each step can be used to pump the protons, so that a single large release is broken into several smaller releases, and the small releases are the reason the gradient can be built. So the step-wise release is what makes the oxidative phosphorylation possible at all, since a single release could not have been harnessed this way.
4. 4Then the benefits, which is the third part. The cell gains usable energy in the form of the ATP, and 38 molecules of it for each molecule of the glucose, and the whole of the metabolism of the cell is run on the ATP, so without it nothing could proceed. The cell saves the energy that would otherwise be lost as heat, and the step-wise release makes respiration efficient where combustion is not, since combustion of the same glucose liberates the whole energy at once and none of it is conserved. The energy is released in a controlled manner, since each step is catalysed by a specific enzyme, and so the rate of the release can be regulated according to the needs of the cell, by the availability of the ADP, of the substrate and of the oxygen, so that the respiration is switched down in a resting cell and up in an active one. And the intermediates of the pathway, since they are produced in small steps rather than consumed in one, are available as the raw materials of the biosynthesis, which is the amphibolic role, so the step-wise nature of the pathway is what makes it usable for both purposes. And the temperature of the cell is not raised, since the energy is not liberated as a single burst of heat, so the cell can run this chemistry at a constant and survivable temperature, which is the last and the most practical point.