Important Questions and Answers — Class 11 Biology Chapter 12 Notes

How to Use This Section

This chapter is examined in a very particular way. Almost every question is "name the enzyme, the intermediate or the complex", "give the number", "which compartment of the cell", or a two-column comparison. There is almost nothing to derive from first principles. The marks sit in exact enzyme names, exact numbers and exact locations, and they are lost by a single swapped word.

That makes this one of the safest chapters in the book to score in, and one of the easiest to bleed marks in. Hexokinase phosphorylates glucose; invertase splits sucrose. Pyruvic acid decarboxylase works in a fermenting yeast; pyruvic dehydrogenase works in the mitochondrial matrix. Glycolysis is in the cytoplasm; the link reaction and the TCA cycle are in the matrix; the ETS and ATP synthase are on the inner mitochondrial membrane. Gross 4 ATP but net 2 in glycolysis. Three $\mathrm{NAD^+}$ reduced and one $\mathrm{FAD^+}$ reduced per turn of the cycle, not four and one. One NADH is worth 3 ATP and one $\mathrm{FADH_2}$ is worth 2. Protons gather in the intermembrane space, not the matrix. $\mathrm{F_0}$ is the channel and $\mathrm{F_1}$ is the headpiece. Every one of those pairs has cost more marks than any reasoning question in the chapter.

Work through the three tiers in order.

Tier 1 - Concept Checks. One fact per question, straight from the chapter - a name, a number or a location. If you cannot answer one of these without stopping to think, go back to the section it came from before moving on.
Tier 2 - Pathway Tracing and Arithmetic. This is where the chapter is actually lost. You are asked how many ATP each stage gave, how many $\mathrm{NADH + H^+}$ each stage gave, where each carbon of the glucose molecule ended up, what the total becomes if one assumption is dropped, and what RQ follows from a given equation. Spend the most time in this tier. The arithmetic is easy; the bookkeeping is what trips people.
Tier 3 - Comparisons and Long Answers. The two-column differences and the joined-up descriptions - respiration against combustion, glycolysis against the Krebs cycle, aerobic against anaerobic respiration, glycolysis against fermentation, aerobic respiration against fermentation, the link reaction, the TCA cycle end to end, the ETS, oxidative phosphorylation against photophosphorylation, the 38-ATP derivation, the amphibolic argument, and RQ across the three substrate types.

One working habit pays in almost every question of this chapter: before you answer, ask which compartment you are standing in and what is being handed to what. Cytoplasm or matrix or inner membrane. Carbon in or carbon out as $\mathrm{CO_2}$ . Coenzyme reduced or coenzyme regenerated. Those three questions alone settle most of the paper.

The twelve chapter-end exercises, which come to sixteen questions once the three parts of exercise 1 and the three parts of exercise 7 are counted separately, are almost all already answered inside the ten teaching sections of this chapter. The last block lists exactly where each one is answered, and works out in full the single part that is not answered anywhere else.

The Facts These Questions Draw On

Gas exchange in a plant. Plants require $\mathrm{O_2}$ for respiration to occur and they also give out $\mathrm{CO_2}$ . Plants, unlike animals, have no specialised organs for gaseous exchange; they have stomata and lenticels for this purpose. They manage without respiratory organs for three reasons: each plant part takes care of its own gas-exchange needs, and there is very little transport of gases from one plant part to another; plants do not present great demands for gas exchange, and roots, stems and leaves respire at rates far lower than animals do; and the distance that gases must diffuse even in large, bulky plants is not great, since each living cell is located quite close to the surface of the plant. Only during photosynthesis are large volumes of gases exchanged, and then availability of $\mathrm{O_2}$ is not a problem in those cells, since $\mathrm{O_2}$ is released within the cell. In stems the living cells are organised in thin layers inside and beneath the bark and have openings called lenticels; the cells in the interior are dead and provide only mechanical support. The loose packing of parenchyma cells in leaves, stems and roots provides an interconnected network of air spaces.

Why respiration is not combustion. The complete combustion of glucose, which produces $\mathrm{CO_2}$ and $\mathrm{H_2O}$ as end products, yields energy most of which is given out as heat:

$\mathrm{C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O} + \text{Energy}$

Heat is useless to a cell. If this energy is to be useful, the cell should be able to utilise it to synthesise other molecules that it requires. So the cell catabolises the glucose molecule in such a way that not all the liberated energy goes out as heat, and the key is to oxidise glucose not in one step but in several small steps, enabling some steps to be just large enough that the energy released can be coupled to ATP synthesis. That is the significance of the step-wise release of energy, and it is the whole story of respiration.

Life without oxygen. There are sufficient reasons to believe that the first cells on this planet lived in an atmosphere that lacked oxygen. Some present-day organisms are facultative anaerobes and in others the requirement for anaerobic condition is obligate. All living organisms retain the enzymatic machinery to partially oxidise glucose without the help of oxygen, and this breakdown of glucose to pyruvic acid is called glycolysis.

Glycolysis - what and where. The term glycolysis has originated from the Greek words glycos for sugar and lysis for splitting. The scheme of glycolysis was given by Gustav Embden, Otto Meyerhof and J. Parnas, and is often referred to as the EMP pathway. In anaerobic organisms it is the only process in respiration. Glycolysis occurs in the cytoplasm of the cell and is present in all living organisms. In this process glucose undergoes partial oxidation to form two molecules of pyruvic acid. In plants this glucose is derived from sucrose, which is the end product of photosynthesis, or from storage carbohydrates. Sucrose is converted into glucose and fructose by the enzyme invertase, and these two monosaccharides readily enter the glycolytic pathway.

Glycolysis - the ten reactions. In glycolysis, a chain of ten reactions, under the control of different enzymes, takes place to produce pyruvate from glucose. Glucose and fructose are phosphorylated to give rise to glucose-6-phosphate by the activity of the enzyme hexokinase. This phosphorylated form of glucose then isomerises to produce fructose-6-phosphate. Subsequent steps of metabolism of glucose and fructose are the same. The fructose 1,6-bisphosphate is split into dihydroxyacetone phosphate and 3-phosphoglyceraldehyde (PGAL) - and from that split onwards every remaining event happens twice, because one six-carbon molecule has become two three-carbon molecules.

Glycolysis - the marked events. There is one step where $\mathrm{NADH + H^+}$ is formed from $\mathrm{NAD^+}$ : when 3-phosphoglyceraldehyde (PGAL) is converted to 1,3-bisphosphoglycerate (BPGA). Two redox-equivalents are removed, in the form of two hydrogen atoms, from PGAL and transferred to a molecule of $\mathrm{NAD^+}$ ; PGAL is oxidised and, with inorganic phosphate, gets converted into BPGA. ATP is utilised at two steps - first, in the conversion of glucose into glucose 6-phosphate, and second, in the conversion of fructose 6-phosphate to fructose 1,6-bisphosphate. ATP is synthesised at two steps - the conversion of BPGA to 3-phosphoglyceric acid (PGA) is an energy yielding process and this energy is trapped by the formation of ATP, and another ATP is synthesised during the conversion of PEP to pyruvic acid.

The glycolytic tally, per one glucose. 2 ATP used, 4 ATP made - that is 2 steps happening twice - so a net gain of 2 ATP; 2 $\mathrm{NADH + H^+}$ made, that is 1 step happening twice; and 2 pyruvic acid. The gross figure 4 and the net figure 2 are different questions, and the difference is the 2 ATP invested at the start.

The three fates of pyruvic acid. What happens to pyruvic acid depends on the cellular need. There are three major ways in which different cells handle pyruvic acid produced by glycolysis: lactic acid fermentation, alcoholic fermentation and aerobic respiration. Fermentation takes place under anaerobic conditions in many prokaryotes and unicellular eukaryotes. For the complete oxidation of glucose to $\mathrm{CO_2}$ and $\mathrm{H_2O}$ , organisms adopt Krebs' cycle, which is also called aerobic respiration, and this requires $\mathrm{O_2}$ supply.

Alcoholic fermentation. In fermentation, say by yeast, the incomplete oxidation of glucose is achieved under anaerobic conditions by sets of reactions where pyruvic acid is converted to $\mathrm{CO_2}$ and ethanol. The enzymes pyruvic acid decarboxylase and alcohol dehydrogenase catalyse these reactions.

$\mathrm{Pyruvic\ acid} \xrightarrow{\text{pyruvic acid decarboxylase}} \mathrm{Acetaldehyde + CO_2}$

$\mathrm{Acetaldehyde + NADH + H^+} \xrightarrow{\text{alcohol dehydrogenase}} \mathrm{Ethanol + NAD^+}$

Lactic acid fermentation. Other organisms like some bacteria produce lactic acid from pyruvic acid. In animal cells also, like muscles during exercise, when oxygen is inadequate for cellular respiration pyruvic acid is reduced to lactic acid by lactate dehydrogenase.

$\mathrm{Pyruvic\ acid + NADH + H^+} \xrightarrow{\text{lactate dehydrogenase}} \mathrm{Lactic\ acid + NAD^+}$

What both fermentations share, and what they cost. The reducing agent is $\mathrm{NADH + H^+}$ , which is reoxidised to $\mathrm{NAD^+}$ in both the processes. Only the alcoholic route releases $\mathrm{CO_2}$ ; lactic acid fermentation releases none. In both lactic acid and alcohol fermentation not much energy is released - less than seven per cent of the energy in glucose is released, and not all of it is trapped as high energy bonds of ATP. Also, the processes are hazardous - either acid or alcohol is produced. The net ATP synthesised when one molecule of glucose is fermented to alcohol or lactic acid is two, because four ATP are synthesised during glycolysis and two are utilised, and the fermentation steps themselves add nothing. Yeasts poison themselves to death when the concentration of alcohol reaches about 13 per cent, so the maximum concentration of alcohol in naturally fermented beverages is about 13 per cent, and beverages of greater alcohol content are obtained by distillation.

Aerobic respiration - the definition and the two locations. Aerobic respiration is the process that leads to a complete oxidation of organic substances in the presence of oxygen, and releases $\mathrm{CO_2}$ , water and a large amount of energy present in the substrate. This type of respiration is most common in higher organisms. In eukaryotes these steps take place within the mitochondria and this requires $\mathrm{O_2}$ . For aerobic respiration to take place within the mitochondria, the final product of glycolysis, pyruvate, is transported from the cytoplasm into the mitochondria. The crucial events are: the complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms, leaving three molecules of $\mathrm{CO_2}$ ; and the passing on of the electrons removed as part of the hydrogen atoms to molecular $\mathrm{O_2}$ with simultaneous synthesis of ATP. What is interesting to note is that the first process takes place in the matrix of the mitochondria while the second process is located on the inner membrane of the mitochondria.

The link reaction. Pyruvate, which is formed by the glycolytic catabolism of carbohydrates in the cytosol, after it enters the mitochondrial matrix undergoes oxidative decarboxylation by a complex set of reactions catalysed by pyruvic dehydrogenase. The reactions catalysed by pyruvic dehydrogenase require the participation of several coenzymes, including $\mathrm{NAD^+}$ and Coenzyme A, with $\mathrm{Mg^{2+}}$ as the cofactor.

$\mathrm{Pyruvic\ acid + CoA + NAD^+} \xrightarrow[\mathrm{Mg^{2+}}]{\text{pyruvate dehydrogenase}} \mathrm{Acetyl\ CoA + CO_2 + NADH + H^+}$

Per pyruvate: 3C in, 1C out as $\mathrm{CO_2}$ , 2C carried on as acetyl CoA, one $\mathrm{NADH + H^+}$ made and no ATP. During this process, two molecules of NADH are produced from the metabolism of two molecules of pyruvic acid, produced from one glucose molecule during glycolysis. The acetyl CoA then enters a cyclic pathway, the tricarboxylic acid cycle, more commonly called Krebs' cycle after the scientist Hans Krebs who first elucidated it.

The TCA cycle, intermediate by intermediate. The TCA cycle starts with the condensation of the acetyl group with oxaloacetic acid (OAA) and water to yield citric acid. The reaction is catalysed by the enzyme citrate synthase and a molecule of CoA is released. Citrate is then isomerised to isocitrate. It is followed by two successive steps of decarboxylation, leading to the formation of $\alpha$ -ketoglutaric acid and then succinyl-CoA. In the remaining steps of the citric acid cycle, succinyl-CoA is oxidised to OAA, allowing the cycle to continue. The carbon counts are OAA 4C, citric acid 6C, $\alpha$ -ketoglutaric acid 5C, succinyl-CoA 4C, succinic acid 4C and malic acid 4C, with acetyl CoA entering as 2C. A 2C acetyl group joins a 4C acceptor to make a 6C acid, two carbons leave as $\mathrm{CO_2}$ taking it back to 4C, and those four carbons are rebuilt into the acceptor so the cycle can start again.

What one turn of the cycle yields. During the conversion of succinyl-CoA to succinic acid a molecule of GTP is synthesised. This is a substrate level phosphorylation. In a coupled reaction GTP is converted to GDP with the simultaneous synthesis of ATP from ADP. Also there are three points in the cycle where $\mathrm{NAD^+}$ is reduced to $\mathrm{NADH + H^+}$ and one point where $\mathrm{FAD^+}$ is reduced to $\mathrm{FADH_2}$ . So one turn gives 3 $\mathrm{NADH + H^+}$ , 1 $\mathrm{FADH_2}$ , 1 GTP and hence 1 ATP, and 2 $\mathrm{CO_2}$ . The continued oxidation of acetyl CoA via the TCA cycle requires the continued replenishment of oxaloacetic acid, the first member of the cycle. In addition it also requires regeneration of $\mathrm{NAD^+}$ and $\mathrm{FAD^+}$ from NADH and $\mathrm{FADH_2}$ respectively. The summary equation for this phase, which takes place in the mitochondrial matrix, is written per pyruvic acid and bundles the link reaction in with the cycle, which is why it carries 4 $\mathrm{NAD^+}$ and 3 $\mathrm{CO_2}$ rather than the 3 and 2 of the cycle alone:

$\mathrm{Pyruvic\ acid + 4NAD^+ + FAD^+ + 2H_2O + ADP + P_i} \xrightarrow{\text{mitochondrial matrix}} \mathrm{3CO_2 + 4NADH + 4H^+ + FADH_2 + ATP}$

The running tally at the end of the cycle. We have till now seen that glucose has been broken down to release $\mathrm{CO_2}$ and eight molecules of $\mathrm{NADH + H^+}$ and two of $\mathrm{FADH_2}$ have been synthesised besides just two molecules of ATP in the TCA cycle. The eight is made of 2 from the two link reactions and 6 from the two turns of the cycle. Neither $\mathrm{O_2}$ has come into the picture nor the promised large number of ATP has yet been synthesised.

The electron transport system. The following steps in the respiratory process are to release and utilise the energy stored in $\mathrm{NADH + H^+}$ and $\mathrm{FADH_2}$ . This is accomplished when they are oxidised through the electron transport system and the electrons are passed on to $\mathrm{O_2}$ , resulting in the formation of $\mathrm{H_2O}$ . The metabolic pathway through which the electron passes from one carrier to another is called the electron transport system (ETS), and it is present in the inner mitochondrial membrane.

The five complexes, in order. Complex I is NADH dehydrogenase: electrons from NADH produced in the mitochondrial matrix during the citric acid cycle are oxidised by an NADH dehydrogenase (complex I), and the electrons are then transferred to ubiquinone located within the inner membrane. Complex II is the $\mathrm{FADH_2}$ entry point: ubiquinone also receives reducing equivalents via $\mathrm{FADH_2}$ (complex II), which is generated during the oxidation of succinate in the citric acid cycle. Complex III is the cytochrome $\mathrm{bc_1}$ complex: the reduced ubiquinone (ubiquinol) is then oxidised with the transfer of electrons to cytochrome c via the cytochrome $\mathrm{bc_1}$ complex (complex III). Cytochrome c is a small protein attached to the outer surface of the inner membrane and acts as a mobile carrier for transfer of electrons between complex III and complex IV - it is not a complex itself. Complex IV refers to the cytochrome c oxidase complex, containing cytochromes a and $\mathrm{a_3}$ , and two copper centres. Complex V is ATP synthase: when the electrons pass from one carrier to another via complex I to complex IV in the electron transport chain, they are coupled to ATP synthase (complex V) for the production of ATP from ADP and inorganic phosphate. Ubiquinone is the meeting point of complexes I and II.

What each reduced coenzyme is worth. The number of ATP molecules synthesised depends on the nature of the electron donor. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP. Oxidation of one molecule of $\mathrm{FADH_2}$ produces 2 molecules of ATP. $\mathrm{FADH_2}$ is worth less because it enters at complex II and so bypasses complex I, travelling a shorter stretch of the chain.

The role of oxygen, and the name of the process. Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. The presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system. Oxygen acts as the final hydrogen acceptor. Unlike photophosphorylation, where it is the light energy that is utilised for the production of the proton gradient required for phosphorylation, in respiration it is the energy of oxidation-reduction that is utilised for the same process. It is for this reason that the process is called oxidative phosphorylation.

ATP synthase and chemiosmosis in the mitochondrion. The energy released during the electron transport system is utilised in synthesising ATP with the help of ATP synthase (complex V). This complex consists of two major components, $\mathrm{F_1}$ and $\mathrm{F_0}$ . The $\mathrm{F_1}$ headpiece is a peripheral membrane protein complex and contains the site for synthesis of ATP from ADP and inorganic phosphate. $\mathrm{F_0}$ is an integral membrane protein complex that forms the channel through which protons cross the inner membrane. The passage of protons through the channel is coupled to the catalytic site of the $\mathrm{F_1}$ component for the production of ATP. For each ATP produced, $\mathrm{4H^+}$ pass through $\mathrm{F_0}$ from the intermembrane space to the matrix down the electrochemical proton gradient.

The chapter 11 contrast, which is examined constantly. In photosynthesis protons accumulate in the thylakoid lumen; in respiration they accumulate in the intermembrane space of the mitochondrion. Photosynthesis has $\mathrm{CF_0}$ and $\mathrm{CF_1}$ ; respiration has $\mathrm{F_0}$ and $\mathrm{F_1}$ - the same architecture under different names. In photophosphorylation it is light energy that creates the proton gradient; in oxidative phosphorylation it is the energy of oxidation-reduction. What is identical in both is the mechanism: a membrane, a proton gradient across it, a channel component through which the protons return, and a catalytic headpiece on the far side that synthesises the ATP.

The four assumptions behind the balance sheet. It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised; but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions. They are: that there is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another; that the NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation; that none of the intermediates in the pathway are utilised to synthesise any other compound; and that only glucose is being respired, with no other alternative substrates entering the pathway at any of the intermediary stages.

Why the assumptions do not hold. This kind of assumption is not really valid in a living system. All pathways work simultaneously and do not take place one after another. Substrates enter the pathways and are withdrawn from them as and when necessary. ATP is utilised as and when needed. Enzymatic rates are controlled by multiple means. Yet, it is useful to do this exercise, to appreciate the beauty and efficiency of the living system in extraction and storing energy. Hence, there can be a net gain of 38 ATP molecules during aerobic respiration of one molecule of glucose.

The balance sheet itself, per one glucose. Glycolysis, in the cytoplasm: net 2 ATP and 2 $\mathrm{NADH + H^+}$ . The link reaction, in the matrix, happening twice: 0 ATP and 2 $\mathrm{NADH + H^+}$ . The TCA cycle, in the matrix, turning twice: 2 ATP, 6 $\mathrm{NADH + H^+}$ and 2 $\mathrm{FADH_2}$ . Totals: 4 ATP at the substrate level, 10 $\mathrm{NADH + H^+}$ and 2 $\mathrm{FADH_2}$ . Converting: 10 times 3 is 30, and 2 times 2 is 4, so 34 from oxidative phosphorylation, plus the 4 at the substrate level, giving 38. Only 4 of the 38 come from substrate-level phosphorylation - 2 in glycolysis and 2 in the TCA cycle - and the other 34 come from oxidative phosphorylation.

Fermentation compared with aerobic respiration. Fermentation accounts for only a partial breakdown of glucose, whereas in aerobic respiration it is completely degraded to $\mathrm{CO_2}$ and $\mathrm{H_2O}$ . In fermentation there is a net gain of only two molecules of ATP for each molecule of glucose degraded to pyruvic acid, whereas many more molecules of ATP are generated under aerobic conditions. NADH is oxidised to $\mathrm{NAD^+}$ rather slowly in fermentation, however the reaction is very vigorous in the case of aerobic respiration.

The amphibolic argument. Glucose is the favoured substrate for respiration. All carbohydrates are usually first converted into glucose before they are used for respiration. Fats and proteins are broken down to yield energy too, but they do not enter the respiratory pathway at the first step. The four entry points are: fats would need to be broken down into glycerol and fatty acids first; if fatty acids were to be respired they would first be degraded to acetyl CoA and enter the pathway there; glycerol would enter the pathway after being converted to PGAL; and the proteins would be degraded by proteases and the individual amino acids, after deamination, depending on their structure, would enter the pathway at some stage within the Krebs' cycle, or even as pyruvate or acetyl CoA. Since respiration involves breakdown of substrates, the respiratory process has traditionally been considered a catabolic process, and the respiratory pathway a catabolic pathway. But is this understanding correct? What is important to recognise is that it is these very compounds that would be withdrawn from the respiratory pathway for the synthesis of the said substrates. Fatty acids would be broken down to acetyl CoA before entering the respiratory pathway when it is used as a substrate; but when the organism needs to synthesise fatty acids, acetyl CoA would be withdrawn from the respiratory pathway for it. Hence the respiratory pathway comes into the picture both during breakdown and synthesis of fatty acids. Similarly, during breakdown and synthesis of protein too, respiratory intermediates form the link. Breaking down processes within the living organism is catabolism, and synthesis is anabolism. Because the respiratory pathway is involved in both anabolism and catabolism, it would hence be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one.

The respiratory quotient. The ratio of the volume of $\mathrm{CO_2}$ evolved to the volume of $\mathrm{O_2}$ consumed in respiration is called the respiratory quotient (RQ) or respiratory ratio.

$\mathrm{RQ} = \frac{\text{volume of }\mathrm{CO_2}\text{ evolved}}{\text{volume of }\mathrm{O_2}\text{ consumed}}$

RQ has no unit, and the respiratory quotient depends upon the type of respiratory substrate used during respiration. When carbohydrates are used as substrate and are completely oxidised, the RQ will be 1, because equal amounts of $\mathrm{CO_2}$ and $\mathrm{O_2}$ are evolved and consumed respectively:

$\mathrm{RQ} = \frac{6\mathrm{CO_2}}{6\mathrm{O_2}} = 1.0$

When fats are used in respiration, the RQ is less than 1. For tripalmitin:

$\mathrm{2(C_{51}H_{98}O_6) + 145O_2 \rightarrow 102CO_2 + 98H_2O} + \text{energy}$

$\mathrm{RQ} = \frac{102\mathrm{CO_2}}{145\mathrm{O_2}} = 0.7$

When proteins are respiratory substrates the ratio would be about 0.9. What is important to recognise is that in living organisms, respiratory substrates are often more than one; pure proteins or fats are never used as respiratory substrates.

Tier 1 - Concept Checks

Question 1

Q. Do plants breathe? Name the structures a plant uses for gaseous exchange.

Answer. The answer is not quite so direct. Yes, plants require $\mathrm{O_2}$ for respiration to occur and they also give out $\mathrm{CO_2}$ , so plants have systems in place that ensure the availability of $\mathrm{O_2}$ . But plants, unlike animals, have no specialised organs for gaseous exchange. They have stomata and lenticels for this purpose.

Be careful with the wording. The claim is no specialised organs, not no gas exchange. Gas exchange certainly happens; it simply happens without an organ built for it.

Question 2

Q. Give the three reasons why plants can get along without respiratory organs.

Answer. The chapter gives them in order, and they are asked as a set.

Each plant part takes care of its own gas-exchange needs. There is very little transport of gases from one plant part to another.
Plants do not present great demands for gas exchange. Roots, stems and leaves respire at rates far lower than animals do. Only during photosynthesis are large volumes of gases exchanged, and each leaf is well adapted to take care of its own needs during these periods - when cells photosynthesise, availability of $\mathrm{O_2}$ is not a problem in these cells, since $\mathrm{O_2}$ is released within the cell.
The distance that gases must diffuse even in large, bulky plants is not great. Each living cell in a plant is located quite close to the surface of the plant.

Learn them as own needs, low demand, short distance.

Question 3

Q. In a woody stem, which cells are alive, where do they sit, and how does air reach them?

Answer. In stems the living cells are organised in thin layers inside and beneath the bark, and they also have openings called lenticels. The cells in the interior are dead and provide only mechanical support. Thus most cells of a plant have at least a part of their surface in contact with air. This is also facilitated by the loose packing of parenchyma cells in leaves, stems and roots, which provide an interconnected network of air spaces.

Question 4

Q. What is a facultative anaerobe, and what is an obligate anaerobe?

Answer. There are sufficient reasons to believe that the first cells on this planet lived in an atmosphere that lacked oxygen, and even among present-day living organisms we know of several that are adapted to anaerobic conditions.

Some of these organisms are facultative anaerobes - they can manage either way, with oxygen or without it.
In others the requirement for anaerobic condition is obligate - they must have no oxygen.

The line that follows from both is the important one: all living organisms retain the enzymatic machinery to partially oxidise glucose without the help of oxygen, and this breakdown of glucose to pyruvic acid is called glycolysis.

Question 5

Q. Where does the word glycolysis come from, who gave the scheme, and what is the pathway also called?

Answer. The term glycolysis has originated from the Greek words glycos for sugar and lysis for splitting - literally sugar splitting. The scheme of glycolysis was given by Gustav Embden, Otto Meyerhof and J. Parnas, and is often referred to as the EMP pathway.

The three names are asked directly, and so is what the letters of EMP stand for - Embden, Meyerhof and Parnas.

Question 6

Q. State the four attributes of glycolysis that get asked as a single line.

Answer.

Glycolysis occurs in the cytoplasm of the cell.
It is present in all living organisms.
In this process, glucose undergoes partial oxidation - the word partial is doing real work here.
It forms two molecules of pyruvic acid.

Add one more that goes with them: in anaerobic organisms, it is the only process in respiration.

Question 7

Q. In a plant, where does the glucose for glycolysis come from, and which enzyme prepares it?

Answer. In plants this glucose is derived from sucrose, which is the end product of photosynthesis, or from storage carbohydrates. Sucrose is converted into glucose and fructose by the enzyme invertase, and these two monosaccharides readily enter the glycolytic pathway.

The enzyme name invertase is the answer being looked for. Do not confuse it with hexokinase, which acts one step later.

Question 8

Q. Name the enzyme that phosphorylates glucose in glycolysis, and say what happens to the product.

Answer. Glucose and fructose are phosphorylated to give rise to glucose-6-phosphate by the activity of the enzyme hexokinase. This phosphorylated form of glucose then isomerises to produce fructose-6-phosphate. Subsequent steps of metabolism of glucose and fructose are the same.

Note that the whole of glycolysis is a chain of ten reactions, under the control of different enzymes, and hexokinase catalyses the first of them.

Question 9

Q. At which single step of glycolysis is $\mathrm{NADH + H^+}$ formed, and what exactly is transferred?

Answer. There is one step where $\mathrm{NADH + H^+}$ is formed from $\mathrm{NAD^+}$ : when 3-phosphoglyceraldehyde (PGAL) is converted to 1,3-bisphosphoglycerate (BPGA).

What is transferred is stated precisely in the chapter: two redox-equivalents are removed, in the form of two hydrogen atoms, from PGAL and transferred to a molecule of $\mathrm{NAD^+}$ . PGAL is oxidised and, with inorganic phosphate, gets converted into BPGA.

One step, but it happens twice per glucose, because the six-carbon molecule has already been split into two three-carbon molecules - fructose 1,6-bisphosphate is split into dihydroxyacetone phosphate and 3-phosphoglyceraldehyde (PGAL).

Question 10

Q. Name the two steps of glycolysis at which ATP is spent and the two at which ATP is made, and give the net.

Answer. ATP is utilised at two steps:

In the conversion of glucose into glucose 6-phosphate.
In the conversion of fructose 6-phosphate to fructose 1,6-bisphosphate.

ATP is synthesised at two steps:

In the conversion of BPGA to 3-phosphoglyceric acid - this conversion is an energy yielding process and the energy is trapped by the formation of ATP.
In the conversion of PEP to pyruvic acid.

Each of the two synthesising steps happens twice, because the molecule has already split in half. So 4 ATP are made, 2 are spent, and the net gain is 2 ATP per glucose. The distractor is always 4 - that is the gross figure.

Question 11

Q. What are the three major fates of the pyruvic acid produced by glycolysis?

Answer. Pyruvic acid is the key product of glycolysis, and what happens to it depends on the cellular need. There are three major ways in which different cells handle pyruvic acid produced by glycolysis:

Lactic acid fermentation
Alcoholic fermentation
Aerobic respiration

Fermentation takes place under anaerobic conditions in many prokaryotes and unicellular eukaryotes. For the complete oxidation of glucose to $\mathrm{CO_2}$ and $\mathrm{H_2O}$ , however, organisms adopt Krebs' cycle, which is also called aerobic respiration. This requires $\mathrm{O_2}$ supply.

Question 12

Q. Name the two enzymes of alcoholic fermentation and write the two reactions they catalyse.

Answer. The enzymes pyruvic acid decarboxylase and alcohol dehydrogenase catalyse these reactions. In fermentation, say by yeast, the incomplete oxidation of glucose is achieved under anaerobic conditions by sets of reactions where pyruvic acid is converted to $\mathrm{CO_2}$ and ethanol.

$\mathrm{Pyruvic\ acid} \xrightarrow{\text{pyruvic acid decarboxylase}} \mathrm{Acetaldehyde + CO_2}$

$\mathrm{Acetaldehyde + NADH + H^+} \xrightarrow{\text{alcohol dehydrogenase}} \mathrm{Ethanol + NAD^+}$

Pyruvic acid decarboxylase removes the carbon as $\mathrm{CO_2}$ ; alcohol dehydrogenase does the reducing.

Question 13

Q. Which enzyme makes lactic acid from pyruvic acid, and in which cells does this happen?

Answer. Other organisms like some bacteria produce lactic acid from pyruvic acid. In animal cells also, like muscles during exercise, when oxygen is inadequate for cellular respiration pyruvic acid is reduced to lactic acid by lactate dehydrogenase.

$\mathrm{Pyruvic\ acid + NADH + H^+} \xrightarrow{\text{lactate dehydrogenase}} \mathrm{Lactic\ acid + NAD^+}$

No $\mathrm{CO_2}$ is released on this route - the three-carbon pyruvate simply becomes three-carbon lactate.

Question 14

Q. In both fermentations, what is the reducing agent and what is recovered? Why does the cell bother?

Answer. The reducing agent is $\mathrm{NADH + H^+}$ , which is reoxidised to $\mathrm{NAD^+}$ in both the processes.

That recovery is the entire purpose. Glycolysis needs a supply of $\mathrm{NAD^+}$ at the PGAL step, and without oxygen there is no other way to get it back, so the cell dumps the hydrogen onto pyruvic acid - as acetaldehyde in yeast, as pyruvate itself in muscle - and carries on splitting sugar. The ethanol and the lactic acid are waste; the regenerated $\mathrm{NAD^+}$ is the point.

Question 15

Q. Give the three numbers that summarise fermentation.

Answer.

Less than seven per cent of the energy in glucose is released, and not all of it is trapped as high energy bonds of ATP.
The net ATP synthesised when one molecule of glucose is fermented to alcohol or lactic acid is two - four ATP are synthesised during glycolysis and two are utilised, and the fermentation steps themselves make none.
Yeasts poison themselves to death when the concentration of alcohol reaches about 13 per cent. So the maximum concentration of alcohol in naturally fermented beverages is about 13 per cent, and alcoholic beverages of alcohol content greater than this concentration are obtained by distillation.

Add the qualitative point that goes with them: the processes are hazardous - either acid or alcohol is produced.

Question 16

Q. Name the enzyme, the cofactor and the coenzymes of the link reaction, and say where it happens.

Answer. Pyruvate, which is formed by the glycolytic catabolism of carbohydrates in the cytosol, after it enters the mitochondrial matrix undergoes oxidative decarboxylation by a complex set of reactions catalysed by pyruvic dehydrogenase. The reactions catalysed by pyruvic dehydrogenase require the participation of several coenzymes, including $\mathrm{NAD^+}$ and Coenzyme A, and $\mathrm{Mg^{2+}}$ is the cofactor.

$\mathrm{Pyruvic\ acid + CoA + NAD^+} \xrightarrow[\mathrm{Mg^{2+}}]{\text{pyruvate dehydrogenase}} \mathrm{Acetyl\ CoA + CO_2 + NADH + H^+}$

Location: the matrix of the mitochondria. Do not confuse pyruvic dehydrogenase with pyruvic acid decarboxylase, which works in the cytoplasm of a fermenting yeast and makes acetaldehyde.

Question 17

Q. What is the first reaction of the TCA cycle? Name the enzyme, the reactants and the product.

Answer. The TCA cycle starts with the condensation of the acetyl group with oxaloacetic acid (OAA) and water to yield citric acid. The reaction is catalysed by the enzyme citrate synthase and a molecule of CoA is released. Citrate is then isomerised to isocitrate.

Three details are easy to lose and all three are asked: water is a reactant in this very first step, the enzyme is citrate synthase, and CoA is released here rather than consumed - it hands over its acetyl group and goes back for another pyruvate. The standard distractor for the first product is isocitrate; citric acid comes first, and is then isomerised.

Question 18

Q. List the intermediates of the TCA cycle named in the chapter with their carbon counts.

Answer.

Intermediate	Carbons
Acetyl CoA, entering the cycle	2C
Oxaloacetic acid (OAA), the acceptor	4C
Citric acid, the first product	6C
$\alpha$ -ketoglutaric acid	5C
Succinyl-CoA	4C
Succinic acid	4C
Malic acid	4C

Citrate is isomerised to isocitrate, then it is followed by two successive steps of decarboxylation, leading to the formation of $\alpha$ -ketoglutaric acid and then succinyl-CoA, and in the remaining steps of the citric acid cycle, succinyl-CoA is oxidised to OAA, allowing the cycle to continue.

Question 19

Q. What does one turn of the TCA cycle yield, and which step is a substrate level phosphorylation?

Answer. There are three points in the cycle where $\mathrm{NAD^+}$ is reduced to $\mathrm{NADH + H^+}$ and one point where $\mathrm{FAD^+}$ is reduced to $\mathrm{FADH_2}$ . During the conversion of succinyl-CoA to succinic acid a molecule of GTP is synthesised. This is a substrate level phosphorylation. In a coupled reaction GTP is converted to GDP with the simultaneous synthesis of ATP from ADP.

Yield per turn	Number
$\mathrm{NADH + H^+}$	3
$\mathrm{FADH_2}$	1
GTP, and hence ATP	1
$\mathrm{CO_2}$	2

The named step is succinyl-CoA to succinic acid, and the label is substrate level phosphorylation, whose distractor is always oxidative phosphorylation.

Question 20

Q. Where is the ETS, name its five complexes in order, and give what each reduced coenzyme is worth.

Answer. The metabolic pathway through which the electron passes from one carrier to another is called the electron transport system (ETS), and it is present in the inner mitochondrial membrane.

Complex	Name	What it does
I	NADH dehydrogenase	Oxidises NADH produced in the mitochondrial matrix during the citric acid cycle and transfers the electrons to ubiquinone located within the inner membrane
II	The $\mathrm{FADH_2}$ entry point	Ubiquinone also receives reducing equivalents via $\mathrm{FADH_2}$ , which is generated during the oxidation of succinate in the citric acid cycle
III	Cytochrome $\mathrm{bc_1}$ complex	The reduced ubiquinone (ubiquinol) is oxidised with the transfer of electrons to cytochrome c
IV	Cytochrome c oxidase complex	Contains cytochromes a and $\mathrm{a_3}$ , and two copper centres; passes the electrons to oxygen
V	ATP synthase	Coupled to the electron flow for the production of ATP from ADP and inorganic phosphate

Cytochrome c is a small protein attached to the outer surface of the inner membrane and acts as a mobile carrier for transfer of electrons between complex III and complex IV - it is not itself a complex.

The number of ATP molecules synthesised depends on the nature of the electron donor. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP. Oxidation of one molecule of $\mathrm{FADH_2}$ produces 2 molecules of ATP. $\mathrm{FADH_2}$ is worth less because it enters at complex II and bypasses complex I.

Tier 2 - Pathway Tracing and Arithmetic

Question 21

Q. Set out, stage by stage, how many ATP one molecule of glucose yields in aerobic respiration, and say which stage each ATP came from.

Answer. Build the answer in two steps, and never memorise the total on its own.

Step one - the ATP made directly, at the substrate level.

Stage, per one glucose	Where	Substrate-level ATP
Glycolysis	Cytoplasm	2 net - 4 made, 2 spent
Link reaction, running twice	Mitochondrial matrix	0
TCA cycle, turning twice	Mitochondrial matrix	2 - one GTP per turn, at succinyl-CoA to succinic acid
Total at the substrate level		4

Step two - the ATP made by oxidative phosphorylation, using 3 ATP per NADH and 2 ATP per $\mathrm{FADH_2}$ .

Source	Number	ATP each	ATP obtained
$\mathrm{NADH + H^+}$	10	3	30
$\mathrm{FADH_2}$	2	2	4
Subtotal from oxidative phosphorylation			34
Plus substrate-level phosphorylation			4
Net gain per molecule of glucose			38

Hence, there can be a net gain of 38 ATP molecules during aerobic respiration of one molecule of glucose. Read the split before you move on: only 4 of the 38 come from substrate-level phosphorylation - 2 in glycolysis and 2 in the TCA cycle - and the other 34 come from oxidative phosphorylation. That is about 89 per cent of the yield from the electron transport system, which is why the chapter spends so long on it.

Question 22

Q. How many $\mathrm{NADH + H^+}$ does each stage produce per glucose? Explain why the answer to "how many NADH" can be 2, 6, 8 or 10 depending on the stem.

Answer. The counts themselves are short.

Stage, per one glucose	$\mathrm{NADH + H^+}$	$\mathrm{FADH_2}$
Glycolysis	2 - one step, PGAL to BPGA, happening twice	0
Link reaction, twice	2 - one per pyruvate	0
TCA cycle, twice	6 - three points per turn	2 - one point per turn
Total	10	2

Now the trap, which is entirely about the wording of the stem.

"In glycolysis" - the answer is 2.
"In the TCA cycle" - the answer is 6, because the cycle itself has three reduction points and it turns twice.
"By the end of the TCA cycle", or "so far" - the answer is 8, because the two from the link reaction are added in. The chapter uses exactly this figure: eight molecules of $\mathrm{NADH + H^+}$ and two of $\mathrm{FADH_2}$ have been synthesised besides just two molecules of ATP in the TCA cycle.
"Fed into the ETS", or "in the whole of aerobic respiration" - the answer is 10, because the two glycolytic NADH are now included as well.

Six, eight and ten are all correct answers to different questions. Read the stem twice before you count.

Question 23

Q. Follow the six carbon atoms of one glucose molecule. Where does each one leave the cell, and at which stage?

Answer. Every one of the six leaves as $\mathrm{CO_2}$ , but not at the same place, and the examiner asks precisely where.

Stage	Carbons handled	$\mathrm{CO_2}$ released, per glucose
Glycolysis	One 6C glucose becomes two 3C pyruvic acid	0
Link reaction, twice	Each 3C pyruvate loses one carbon and becomes 2C acetyl CoA	2
TCA cycle, twice	Each 2C acetyl group is stripped by two successive decarboxylations	4
Total		6

Read it as a running count. After glycolysis, all six carbons are still in the cell, held as two molecules of pyruvic acid. After the link reaction, two have gone, and four remain as two acetyl groups of 2C each. The tricarboxylic acid cycle exists to strip those four away as well, which it does at two decarboxylation steps per turn - the ones that produce $\alpha$ -ketoglutaric acid and then succinyl-CoA.

Two things follow that are worth stating in an answer. Glycolysis releases no $\mathrm{CO_2}$ at all - that is why an anaerobic yeast releases $\mathrm{CO_2}$ from the fermentation step and not from glycolysis. And the crucial event of aerobic respiration is the complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms, leaving three molecules of $\mathrm{CO_2}$ - three per pyruvate, and therefore six per glucose.

Question 24

Q. Assumption two says the glycolytic NADH is transferred into the mitochondria. Recalculate the net ATP if that assumption is dropped and each glycolytic NADH yields only 2 ATP instead of 3.

Answer. This is the single most useful piece of arithmetic in the chapter, because it shows exactly what each assumption is holding up.

Assumption two states that the NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation. That is what lets us count those two NADH at the full mitochondrial rate of 3 ATP each.

Recount with the assumption dropped.

Source	Number	ATP each	ATP obtained
Substrate-level phosphorylation	4	-	4
$\mathrm{NADH + H^+}$ made inside the mitochondrion	8	3	24
$\mathrm{NADH + H^+}$ made in the cytoplasm by glycolysis	2	2	4
$\mathrm{FADH_2}$	2	2	4
Net gain per molecule of glucose			36

The total falls from 38 to 36, and the whole of the 2 ATP difference comes from the two glycolytic NADH being worth one ATP less each. Notice which figures did not move: the 4 substrate-level ATP are unchanged, and the 8 mitochondrial NADH and 2 $\mathrm{FADH_2}$ are unchanged, because they were made inside the organelle in the first place and never had to be transferred.

The chapter's own figure is 38, and it is 38 precisely because assumption two is being made. If an option offers 36, it is testing whether you know which assumption produces which number.

Question 25

Q. A tissue is respiring one molecule of glucose aerobically. How many molecules of $\mathrm{O_2}$ are consumed, how many $\mathrm{CO_2}$ released, and where in the cell does each gas belong?

Answer. Work from the overall equation:

$\mathrm{C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O} + \text{Energy}$

Six $\mathrm{O_2}$ consumed and six $\mathrm{CO_2}$ released. Now place each of them.

The six $\mathrm{CO_2}$ are released in the matrix of the mitochondria - two in the two link reactions and four in the two turns of the TCA cycle. None in glycolysis, and none in the ETS.
The $\mathrm{O_2}$ is consumed on the inner mitochondrial membrane, at the very end of the electron transport system. The electrons are passed on to $\mathrm{O_2}$ , resulting in the formation of $\mathrm{H_2O}$ . Oxygen acts as the final hydrogen acceptor.

That separation is the answer to a favourite trap. The $\mathrm{CO_2}$ and the $\mathrm{O_2}$ of respiration have nothing to do with each other chemically. $\mathrm{O_2}$ never touches the pyruvate or any carbon atom of the glucose. The carbon leaves as $\mathrm{CO_2}$ during decarboxylations in the matrix; the oxygen collects hydrogen at the end of the chain and becomes water. Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process.

Question 26

Q. A mitochondrial preparation is supplied with 10 $\mathrm{NADH + H^+}$ and 2 $\mathrm{FADH_2}$ and nothing else. How much ATP can it make, and what fraction of the aerobic yield of glucose is that?

Answer. Use the two conversion rates. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP. Oxidation of one molecule of $\mathrm{FADH_2}$ produces 2 molecules of ATP.

From the NADH: 10 times 3, which is 30 ATP.
From the $\mathrm{FADH_2}$ : 2 times 2, which is 4 ATP.
Total: 34 ATP.

That is exactly the oxidative phosphorylation share of the respiration of one glucose molecule, because 10 $\mathrm{NADH + H^+}$ and 2 $\mathrm{FADH_2}$ is precisely what one glucose delivers to the electron transport system. The remaining 4 ATP of the 38 are made at the substrate level and would not appear in this preparation at all, since the preparation was given no glucose, no pyruvate and no acetyl CoA to phosphorylate anything from.

So 34 out of 38, and the split is worth remembering as a sentence: 34 by oxidative phosphorylation, 4 by substrate-level phosphorylation.

Question 27

Q. How many protons must cross $\mathrm{F_0}$ to account for the 34 ATP made by oxidative phosphorylation from one glucose? State the direction of the flow.

Answer. The number to use is for each ATP produced, $\mathrm{4H^+}$ pass through $\mathrm{F_0}$ from the intermembrane space to the matrix down the electrochemical proton gradient.

34 ATP times 4 protons per ATP is 136 protons.

Three parts of that sentence carry marks and all three should appear in your answer.

The count is four protons per ATP.
The direction is from the intermembrane space into the matrix - into the interior of the mitochondrion, not out of it.
The movement is down the electrochemical proton gradient, so the cell spends nothing to move them; it is the gradient, already built by the electron transport system, that is being spent.

And name the parts correctly while you are there. $\mathrm{F_0}$ is an integral membrane protein complex that forms the channel through which protons cross the inner membrane. The $\mathrm{F_1}$ headpiece is a peripheral membrane protein complex and contains the site for synthesis of ATP from ADP and inorganic phosphate. The passage of protons through the channel is coupled to the catalytic site of the $\mathrm{F_1}$ component for the production of ATP.

Question 28

Q. A yeast culture respiring aerobically is sealed off from air and switched to fermentation. Calculate how many times less ATP it now gets per glucose, and say which stages it has lost.

Answer. Aerobically, one glucose gives a net gain of 38 ATP. In fermentation there is a net gain of only two molecules of ATP for each molecule of glucose degraded to pyruvic acid.

38 divided by 2 is 19, so the sealed culture gets nineteen times less ATP from every molecule of glucose it consumes. Put the other way round, it must consume nineteen molecules of glucose to obtain the ATP it previously got from one.

What it has lost, stage by stage:

Stage	Aerobic	After sealing off
Glycolysis, in the cytoplasm	net 2 ATP	net 2 ATP - unchanged
Link reaction	2 $\mathrm{NADH + H^+}$	Does not run
TCA cycle	2 ATP, 6 $\mathrm{NADH + H^+}$ , 2 $\mathrm{FADH_2}$	Does not run
ETS and oxidative phosphorylation	34 ATP	Does not run
Total	38 ATP	2 ATP

Glycolysis is the only stage that survives, which is why the chapter says that in anaerobic organisms it is the only process in respiration. The reason the rest stops is not that those enzymes are missing but that the TCA cycle requires regeneration of $\mathrm{NAD^+}$ and $\mathrm{FAD^+}$ from NADH and $\mathrm{FADH_2}$ , and without oxygen as the final hydrogen acceptor there is nowhere to unload them. Fermentation exists solely to reoxidise $\mathrm{NADH + H^+}$ to $\mathrm{NAD^+}$ so that glycolysis can keep going.

Note also that fermentation accounts for only a partial breakdown of glucose, whereas in aerobic respiration it is completely degraded to $\mathrm{CO_2}$ and $\mathrm{H_2O}$ , and that less than seven per cent of the energy in glucose is released in fermentation.

Question 29

Q. One molecule of sucrose arriving in a root cell is completely respired. Work out the ATP yield.

Answer. Do it in two moves, and state the enzyme.

Move one - break the sucrose. Sucrose is converted into glucose and fructose by the enzyme invertase, and these two monosaccharides readily enter the glycolytic pathway. Glucose and fructose are phosphorylated to give rise to glucose-6-phosphate by the activity of the enzyme hexokinase, and after isomerisation to fructose-6-phosphate, subsequent steps of metabolism of glucose and fructose are the same.

Move two - respire two hexoses instead of one. Each hexose gives a net gain of 38 ATP, so:

$38 \times 2 = 76$

One sucrose yields 76 ATP, on the same assumptions that give 38 for glucose. Alongside it, 12 $\mathrm{CO_2}$ are released and 12 $\mathrm{O_2}$ are consumed, and 20 $\mathrm{NADH + H^+}$ and 4 $\mathrm{FADH_2}$ pass through the electron transport system.

The point of the question is the first move. All carbohydrates are usually first converted into glucose before they are used for respiration - a disaccharide is not fed into glycolysis as it stands.

Question 30

Q. Define RQ and work it out for the complete oxidation of glucose.

Answer. The ratio of the volume of $\mathrm{CO_2}$ evolved to the volume of $\mathrm{O_2}$ consumed in respiration is called the respiratory quotient (RQ) or respiratory ratio.

$\mathrm{RQ} = \frac{\text{volume of }\mathrm{CO_2}\text{ evolved}}{\text{volume of }\mathrm{O_2}\text{ consumed}}$

Now take the equation for the complete oxidation of glucose and read the two coefficients off it:

$\mathrm{C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O} + \text{Energy}$

Six molecules of $\mathrm{O_2}$ go in and six molecules of $\mathrm{CO_2}$ come out, so:

$\mathrm{RQ} = \frac{6\mathrm{CO_2}}{6\mathrm{O_2}} = 1.0$

When carbohydrates are used as substrate and are completely oxidised, the RQ will be 1, because equal amounts of $\mathrm{CO_2}$ and $\mathrm{O_2}$ are evolved and consumed respectively. RQ is a pure ratio of two volumes, so it has no unit. Keep both qualifiers - the substrate must be a carbohydrate and it must be completely oxidised.

Question 31

Q. Work the RQ for tripalmitin from its equation, and explain from the two formulas why a fat must give a value below 1.

Answer. The equation is:

$\mathrm{2(C_{51}H_{98}O_6) + 145O_2 \rightarrow 102CO_2 + 98H_2O} + \text{energy}$

Two molecules of tripalmitin need 145 molecules of $\mathrm{O_2}$ and give out only 102 molecules of $\mathrm{CO_2}$ . Put the $\mathrm{CO_2}$ coefficient over the $\mathrm{O_2}$ coefficient:

$\mathrm{RQ} = \frac{102\mathrm{CO_2}}{145\mathrm{O_2}} = 0.7$

More oxygen goes in than carbon dioxide comes out, so the fraction must be less than one.

Now the reason, which is the part worth marks. Glucose is $\mathrm{C_6H_{12}O_6}$ - six carbons carrying six oxygens of their own. Tripalmitin is $\mathrm{C_{51}H_{98}O_6}$ - fifty-one carbons carrying only six oxygens. A fat brings far less oxygen of its own to the reaction, so far more has to be supplied from outside, the denominator of the RQ grows, and the ratio falls below 1. When fats are used in respiration, the RQ is less than 1.

Question 32

Q. Germinating castor seeds and germinating wheat grains are set up in separate respirometers. Predict the RQ of each, and say what the measured value will really look like.

Answer. Identify the stored food first, because the respiratory quotient depends upon the type of respiratory substrate used during respiration.

Seed	Stored food being respired	Expected RQ
Germinating castor	Fat	Less than 1, about 0.7
Germinating wheat	Carbohydrate, as starch converted to glucose	1.0, if completely oxidised

So the castor respirometer will show markedly less $\mathrm{CO_2}$ evolved than $\mathrm{O_2}$ consumed, while the wheat respirometer will show the two volumes nearly equal.

Now the caution that the question is really testing. What is important to recognise is that in living organisms, respiratory substrates are often more than one; pure proteins or fats are never used as respiratory substrates. So the castor reading will not be a clean 0.7 and the wheat reading will not be a clean 1.0 - each will be a mixed value, pulled towards the protein figure of about 0.9 by whatever protein is being respired alongside. Report the values as approximate and say why.

Question 33

Q. A cell is respiring a fat, and another is respiring a protein. State the entry point of each product into the respiratory pathway, and name the preparation each needs first.

Answer. Other substrates can also be respired, but they do not enter the respiratory pathway at the first step.

Substrate	Broken down first into	Enters the respiratory pathway as
Fats	Glycerol and fatty acids	Not as fat - each half enters at its own point
Fatty acids	Acetyl CoA	Acetyl CoA, that is after glycolysis and after the link reaction
Glycerol	PGAL	PGAL, a glycolysis intermediate
Proteins	Amino acids, released by proteases, then deaminated	At some stage within the Krebs' cycle, or even as pyruvate or acetyl CoA

Read the three sentences in the order the chapter gives them. Fats would need to be broken down into glycerol and fatty acids first. If fatty acids were to be respired they would first be degraded to acetyl CoA and enter the pathway there, so a respired fatty acid skips the whole of glycolysis. Glycerol would enter the pathway after being converted to PGAL, which sits in the middle of glycolysis, so glycerol joins the road much earlier than a fatty acid does. The proteins would be degraded by proteases, and the individual amino acids, after deamination, depending on their structure, would enter the pathway at some stage within the Krebs' cycle, or even as pyruvate or acetyl CoA - there is no one entry point for protein, because there is no one amino acid.

Two mistakes cost marks here every year: swapping glycerol and fatty acid, and forgetting deamination. Glycerol goes to PGAL and the fatty acid goes to acetyl CoA, never the other way round, and an amino acid cannot enter the pathway carrying its amino group.

Question 34

Q. Oxygen is withdrawn from a cell that was respiring aerobically. Trace, in order, what stops and why, given that no step of glycolysis or of the TCA cycle uses $\mathrm{O_2}$ directly.

Answer. The paradox in the question is real and the chapter states it plainly: although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. Nothing in glycolysis or in the TCA cycle consumes an oxygen molecule. Yet everything halts. Here is the order.

The electron transport system stops first. Oxygen acts as the final hydrogen acceptor, so with no oxygen there is nothing to pass the electrons to, and the carriers of the inner membrane stay reduced.
Oxidative phosphorylation stops with it. No electron flow means no protons pumped into the intermembrane space, no proton gradient, and no flow back through $\mathrm{F_0}$ , so ATP synthase, complex V, makes nothing. The 34 ATP of the aerobic yield are gone at this point.
$\mathrm{NAD^+}$ and $\mathrm{FAD^+}$ are never regenerated. The presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system. With the chain blocked, the reduced coenzymes cannot be emptied.
The TCA cycle grinds to a halt. The continued oxidation of acetyl CoA via the TCA cycle requires regeneration of $\mathrm{NAD^+}$ and $\mathrm{FAD^+}$ from NADH and $\mathrm{FADH_2}$ respectively, and that regeneration has just been cut off. The link reaction stops for the same reason, since it too needs $\mathrm{NAD^+}$ .
Glycolysis alone can continue - but only if it is given a way to reoxidise its own NADH. Glycolysis needs $\mathrm{NAD^+}$ at the PGAL step. The cell supplies it by fermentation: pyruvic acid is converted to $\mathrm{CO_2}$ and ethanol by pyruvic acid decarboxylase and alcohol dehydrogenase, or pyruvic acid is reduced to lactic acid by lactate dehydrogenase, and in both the reducing agent $\mathrm{NADH + H^+}$ is reoxidised to $\mathrm{NAD^+}$ .

The yield falls from 38 ATP to 2. The one-line version of the whole answer: oxygen is used only at the last step, but every earlier step depends on that last step to hand back its oxidised coenzymes.

Tier 3 - Comparisons and Long Answers

Question 35

Q. Differentiate between respiration and combustion, and explain the significance of the step-wise release of energy in respiration.

Answer. Start with the anchor difference: both oxidise glucose to the same end products, but combustion does it in one uncontrolled step and loses the energy as heat, while respiration does it in many small enzyme-controlled steps and traps some of the energy as ATP.

The complete combustion of glucose, which produces $\mathrm{CO_2}$ and $\mathrm{H_2O}$ as end products, yields energy most of which is given out as heat:

$\mathrm{C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O} + \text{Energy}$

Point of comparison	Combustion	Respiration
Number of steps	One step	Several small steps
Control	Uncontrolled	Enzyme controlled at every step
Form the energy takes	Most given out as heat	Coupled to ATP synthesis at some steps
Usefulness to the cell	Heat is useless to a cell	The ATP can be spent to synthesise other molecules the cell requires
Where it happens	Outside living systems, in a flame	Inside living cells, in the cytoplasm and the mitochondria
Rate	Very fast, and the temperature rises sharply	Slow and controlled, at the temperature of the cell
Intermediates	None that are isolable	A long series - pyruvic acid, acetyl CoA, the acids of the Krebs cycle
End products	$\mathrm{CO_2}$ and $\mathrm{H_2O}$	$\mathrm{CO_2}$ , $\mathrm{H_2O}$ and energy as ATP

Now the significance of the step-wise release, which is the same argument stated positively. Heat is useless to a cell. If this energy is to be useful, the cell should be able to utilise it to synthesise other molecules that it requires. The strategy the plant cell uses is to catabolise the glucose molecule in such a way that not all the liberated energy goes out as heat. The key is to oxidise glucose not in one step but in several small steps, enabling some steps to be just large enough that the energy released can be coupled to ATP synthesis.

Three consequences are worth listing in a long answer.

Energy is released in packets small enough to be trapped as ATP instead of lost as heat, which raises the efficiency of the cell enormously.
The cell is not damaged, because releasing the whole of the energy of a glucose molecule at once would raise the temperature destructively.
The release can be regulated, since each step has its own enzyme and enzymatic rates are controlled by multiple means, so the cell makes ATP as and when it is needed.

How this is done is, essentially, the story of respiration.

Question 36

Q. Distinguish between glycolysis and the Krebs cycle.

Answer. The anchor difference to open with: glycolysis is a linear chain of ten reactions in the cytoplasm that only partially oxidises glucose, while the Krebs cycle is a closed cycle in the mitochondrial matrix that completely oxidises the acetyl group it is handed.

Point of comparison	Glycolysis	Krebs cycle
Other names	The EMP pathway, after Gustav Embden, Otto Meyerhof and J. Parnas	The tricarboxylic acid cycle, or the citric acid cycle, named after Hans Krebs who first elucidated it
Where it occurs	In the cytoplasm of the cell	In the matrix of the mitochondria
Oxygen required	No - it runs in aerobic and anaerobic conditions alike	Yes, indirectly - it needs $\mathrm{NAD^+}$ and $\mathrm{FAD^+}$ handed back by the oxygen-dependent ETS
Shape of the pathway	A linear chain of ten reactions, under the control of different enzymes	A cycle, whose acceptor OAA is regenerated at the end of every turn
Starting material	Glucose, 6C	Acetyl CoA, 2C, condensing with OAA, 4C
End product	Two molecules of pyruvic acid, 3C each	OAA is regenerated; the carbon leaves entirely as $\mathrm{CO_2}$
Oxidation of glucose	Partial	Complete, for the carbon that reaches it
$\mathrm{CO_2}$ released	None	2 per turn, that is 4 per glucose
Occurs in	All living organisms	Only in organisms carrying out aerobic respiration, and in eukaryotes only in the mitochondria
Reduced coenzymes per glucose	2 $\mathrm{NADH + H^+}$	6 $\mathrm{NADH + H^+}$ and 2 $\mathrm{FADH_2}$ , from two turns
ATP per glucose	4 made, 2 spent, net 2	2, one GTP per turn at succinyl-CoA to succinic acid

Two points to add underneath the table, because they are what the examiner is checking.

First, the word partial. In glycolysis glucose undergoes partial oxidation to form two molecules of pyruvic acid - most of the energy of the glucose is still sitting in those two pyruvate molecules, which is exactly why there is a whole chapter after glycolysis.

Second, why a cycle at all. The continued oxidation of acetyl CoA via the TCA cycle requires the continued replenishment of oxaloacetic acid, the first member of the cycle. Because OAA goes in at the top and comes out at the bottom, one molecule of it can process acetyl group after acetyl group - a linear pathway would need a fresh acceptor for every acetyl group.

Note for the exercises: the source asks this same comparison a second time, under the name "Glycolysis and Citric acid Cycle". It is answered again as Question 47 below.

Question 37

Q. Distinguish between aerobic respiration and anaerobic respiration.

Answer. Aerobic respiration is the process that leads to a complete oxidation of organic substances in the presence of oxygen, and releases $\mathrm{CO_2}$ , water and a large amount of energy present in the substrate. Anaerobic respiration is the incomplete oxidation of glucose under anaerobic conditions, of which fermentation is the example this chapter works through.

Point of comparison	Aerobic respiration	Anaerobic respiration
Oxygen	Required	Not required - it proceeds in the absence of $\mathrm{O_2}$
Oxidation of glucose	Complete	Incomplete, only a partial breakdown
Site in a eukaryote	Glycolysis in the cytoplasm, the rest within the mitochondria	Entirely in the cytoplasm
Stages	Glycolysis, the link reaction, the TCA cycle, the ETS and oxidative phosphorylation	Glycolysis, followed by the fermentation steps
End products	$\mathrm{CO_2}$ and $\mathrm{H_2O}$	Ethanol and $\mathrm{CO_2}$ , or lactic acid
Net ATP per glucose	38	2
Energy of glucose released	A large amount	Less than seven per cent
Oxidation of NADH to $\mathrm{NAD^+}$	Very vigorous	Rather slow
Final hydrogen acceptor	$\mathrm{O_2}$ , which is reduced to $\mathrm{H_2O}$	An organic molecule - acetaldehyde or pyruvic acid itself
Hazard	None; the end products are harmless	Hazardous - either acid or alcohol is produced
Found in	Most common in higher organisms	Many prokaryotes and unicellular eukaryotes; yeast; muscle short of oxygen

The sentence to close with is the one about oxygen's real job: the presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system. Oxygen acts as the final hydrogen acceptor, and an anaerobic organism has to use an organic molecule for that job instead, which is why it can never oxidise the substrate completely.

Question 38

Q. Distinguish between glycolysis and fermentation.

Answer. Say the relationship first, because it is what half the marks are for: fermentation does not replace glycolysis, it follows it. Glycolysis takes glucose to pyruvic acid; fermentation is what an anaerobic cell then does with that pyruvic acid.

Point of comparison	Glycolysis	Fermentation
Position in the pathway	The first stage, common to every organism	Follows glycolysis, in the absence of oxygen
Starting material	Glucose	Pyruvic acid, the product of glycolysis
End product	Two molecules of pyruvic acid	Ethanol and $\mathrm{CO_2}$ , or lactic acid
Where it occurs	In the cytoplasm	In the cytoplasm
Oxygen	Not required; it runs whether or not oxygen is present	Takes place under anaerobic conditions
Enzymes named	Hexokinase, and ten reactions under the control of different enzymes	Pyruvic acid decarboxylase and alcohol dehydrogenase, or lactate dehydrogenase
ATP made	4 made, 2 spent, so a net 2	None at all - the fermentation steps themselves make no ATP
$\mathrm{NAD^+}$ and NADH	$\mathrm{NAD^+}$ is reduced to $\mathrm{NADH + H^+}$ at the PGAL step	$\mathrm{NADH + H^+}$ is the reducing agent and is reoxidised to $\mathrm{NAD^+}$
$\mathrm{CO_2}$ released	None	Yes in alcoholic fermentation; none in lactic acid fermentation
Occurs in	All living organisms	Many prokaryotes and unicellular eukaryotes, yeast, and muscle short of oxygen

The link between the two columns is the coenzyme. Glycolysis reduces $\mathrm{NAD^+}$ and cannot continue unless it gets it back; fermentation gives it back. The reducing agent is $\mathrm{NADH + H^+}$ , which is reoxidised to $\mathrm{NAD^+}$ in both the processes. So the net 2 ATP that is often credited to fermentation is really glycolysis's ATP - four ATP are synthesised during glycolysis and two are utilised, and fermentation adds nothing to that tally.

Question 39

Q. Distinguish between aerobic respiration and fermentation, giving the three points the chapter makes.

Answer. The chapter states the comparison in three sentences, and all three are asked - students routinely give the first two and lose the mark on the third.

Fermentation accounts for only a partial breakdown of glucose, whereas in aerobic respiration it is completely degraded to $\mathrm{CO_2}$ and $\mathrm{H_2O}$ .
In fermentation there is a net gain of only two molecules of ATP for each molecule of glucose degraded to pyruvic acid, whereas many more molecules of ATP are generated under aerobic conditions.
NADH is oxidised to $\mathrm{NAD^+}$ rather slowly in fermentation, however the reaction is very vigorous in the case of aerobic respiration.

Point of comparison	Fermentation	Aerobic respiration
Breakdown of glucose	Only partial	Completely degraded to $\mathrm{CO_2}$ and $\mathrm{H_2O}$
Net ATP per glucose	Only two	Many more - a net gain of 38
Oxidation of NADH to $\mathrm{NAD^+}$	Rather slow	Very vigorous
Oxygen	Not required - anaerobic	Required
Site in a eukaryote	Cytoplasm	Within the mitochondria, after glycolysis in the cytoplasm
End products	Ethanol and $\mathrm{CO_2}$ , or lactic acid	$\mathrm{CO_2}$ and water
Energy of glucose released	Less than seven per cent	A large amount
Common in	Yeast, some bacteria, muscle short of oxygen	Higher organisms

Put the two totals next to each other and the reason for the gap is the first point. Fermentation gets 2 ATP; aerobic respiration gets 38 from the same molecule of glucose. Fermentation stops at pyruvic acid or its products and never touches the energy still locked in those carbon skeletons, while aerobic respiration takes the glucose all the way down to $\mathrm{CO_2}$ and $\mathrm{H_2O}$ . The ethanol a yeast throws away is still a fuel; the $\mathrm{CO_2}$ and water a root cell throws away are not.

Question 40

Q. Explain the link reaction in full - what it is called, where it happens, what catalyses it, and what it yields per glucose.

Answer. The link reaction is the bridge between glycolysis in the cytoplasm and the Krebs cycle in the mitochondrion, and it is the reason the two can be treated as one pathway at all.

Getting the pyruvate in. For aerobic respiration to take place within the mitochondria, the final product of glycolysis, pyruvate, is transported from the cytoplasm into the mitochondria.

What happens there. Pyruvate, which is formed by the glycolytic catabolism of carbohydrates in the cytosol, after it enters the mitochondrial matrix undergoes oxidative decarboxylation by a complex set of reactions catalysed by pyruvic dehydrogenase. The reactions catalysed by pyruvic dehydrogenase require the participation of several coenzymes, including $\mathrm{NAD^+}$ and Coenzyme A.

$\mathrm{Pyruvic\ acid + CoA + NAD^+} \xrightarrow[\mathrm{Mg^{2+}}]{\text{pyruvate dehydrogenase}} \mathrm{Acetyl\ CoA + CO_2 + NADH + H^+}$

Take the name apart and the reaction explains itself. Oxidative - hydrogen is removed and handed to $\mathrm{NAD^+}$ , which becomes $\mathrm{NADH + H^+}$ . Decarboxylation - a carboxyl carbon is removed and leaves as $\mathrm{CO_2}$ . $\mathrm{Mg^{2+}}$ is the cofactor, and Coenzyme A picks up the two-carbon fragment that is left.

The bookkeeping, per pyruvate and then per glucose.

Per one glucose	Amount
Pyruvate entering the matrix	2
$\mathrm{CO_2}$ released	2 - one carbon of each 3C pyruvate
Acetyl CoA formed	2, each of 2C
$\mathrm{NADH + H^+}$ formed	2
ATP formed	0

During this process, two molecules of NADH are produced from the metabolism of two molecules of pyruvic acid, produced from one glucose molecule during glycolysis. So the link reaction makes no ATP whatever. Its entire yield is two $\mathrm{NADH + H^+}$ and two molecules of acetyl CoA - and at 3 ATP per NADH those two coenzymes are worth 6 ATP later, which is more than the whole of glycolysis nets.

Where the carbons stand afterwards. Two of the six carbons of glucose have left as $\mathrm{CO_2}$ , and four remain, as two acetyl groups of 2C each. The tricarboxylic acid cycle exists to strip those four away as well.

Then the handover. The acetyl CoA enters a cyclic pathway, the tricarboxylic acid cycle, more commonly called Krebs' cycle after the scientist Hans Krebs who first elucidated it.

Finally, keep the two similar enzyme names apart. Pyruvic dehydrogenase works in the mitochondrial matrix and makes acetyl CoA. Pyruvic acid decarboxylase works in the cytoplasm of a fermenting yeast and makes acetaldehyde.

Question 41

Q. Give a schematic account of the citric acid cycle from start to finish, with the intermediates, the carbon counts and everything one turn yields.

Answer. The cycle is named three ways - the tricarboxylic acid cycle, the citric acid cycle and Krebs' cycle, after the scientist Hans Krebs who first elucidated it - and it runs in the matrix of the mitochondria.

Trace it in order.

The TCA cycle starts with the condensation of the acetyl group with oxaloacetic acid (OAA) and water to yield citric acid. The reaction is catalysed by the enzyme citrate synthase and a molecule of CoA is released.
Citrate is then isomerised to isocitrate.
It is followed by two successive steps of decarboxylation, leading to the formation of $\alpha$ -ketoglutaric acid and then succinyl-CoA.
In the remaining steps of the citric acid cycle, succinyl-CoA is oxidised to OAA, allowing the cycle to continue. During the conversion of succinyl-CoA to succinic acid a molecule of GTP is synthesised.

Now the same thing as a table, which is the form to draw in an answer book.

Intermediate	Carbons	What happens at or after it
Acetyl CoA, entering	2C	Condenses with OAA and water
Oxaloacetic acid (OAA)	4C	The acceptor; citrate synthase catalyses the condensation and CoA is released
Citric acid	6C	The first product; isomerised to isocitrate
$\alpha$ -ketoglutaric acid	5C	Formed by the first decarboxylation; one $\mathrm{CO_2}$ leaves
Succinyl-CoA	4C	Formed by the second decarboxylation; a second $\mathrm{CO_2}$ leaves
Succinic acid	4C	Formed from succinyl-CoA, with a GTP made
Malic acid	4C	Oxidised in the last step to regenerate OAA

Follow the carbons and the cycle stops being a list of names. A 2C acetyl group joins a 4C acceptor to make a 6C acid; two carbons then leave as $\mathrm{CO_2}$ , taking it back down to 4C; and those four carbons are rebuilt into the acceptor so the whole thing can start again.

What one turn yields.

Yield per turn	Number
$\mathrm{NADH + H^+}$	3
$\mathrm{FADH_2}$	1
GTP, and hence ATP	1
$\mathrm{CO_2}$	2

There are three points in the cycle where $\mathrm{NAD^+}$ is reduced to $\mathrm{NADH + H^+}$ and one point where $\mathrm{FAD^+}$ is reduced to $\mathrm{FADH_2}$ . The GTP is made at the succinyl-CoA to succinic acid step, and this is a substrate level phosphorylation - in a coupled reaction GTP is converted to GDP with the simultaneous synthesis of ATP from ADP.

What keeps it turning. The continued oxidation of acetyl CoA via the TCA cycle requires the continued replenishment of oxaloacetic acid, the first member of the cycle. In addition it also requires regeneration of $\mathrm{NAD^+}$ and $\mathrm{FAD^+}$ from NADH and $\mathrm{FADH_2}$ respectively. Nothing in the cycle itself uses $\mathrm{O_2}$ , but without oxygen there is nowhere to unload the reduced coenzymes, so the cycle grinds to a halt.

The summary equation for this phase, which takes place in the mitochondrial matrix, is written per pyruvic acid and includes the link reaction, which is why it carries 4 $\mathrm{NAD^+}$ and 3 $\mathrm{CO_2}$ rather than the 3 and 2 of the cycle alone:

$\mathrm{Pyruvic\ acid + 4NAD^+ + FAD^+ + 2H_2O + ADP + P_i} \xrightarrow{\text{mitochondrial matrix}} \mathrm{3CO_2 + 4NADH + 4H^+ + FADH_2 + ATP}$

Remember that the cycle turns twice per glucose, so per glucose it gives 6 $\mathrm{NADH + H^+}$ , 2 $\mathrm{FADH_2}$ , 2 ATP and 4 $\mathrm{CO_2}$ .

Question 42

Q. Explain the electron transport system.

Answer. State what it is for first. By the end of the TCA cycle glucose has been broken down to release $\mathrm{CO_2}$ , and eight molecules of $\mathrm{NADH + H^+}$ and two of $\mathrm{FADH_2}$ have been synthesised besides just two molecules of ATP in the TCA cycle - the energy is not lost, it is stored in a different currency. The following steps in the respiratory process are to release and utilise the energy stored in $\mathrm{NADH + H^+}$ and $\mathrm{FADH_2}$ . This is accomplished when they are oxidised through the electron transport system and the electrons are passed on to $\mathrm{O_2}$ , resulting in the formation of $\mathrm{H_2O}$ .

The definition and the address. The metabolic pathway through which the electron passes from one carrier to another is called the electron transport system (ETS), and it is present in the inner mitochondrial membrane.

Follow the electron through the complexes, in order.

Complex I, NADH dehydrogenase. Electrons from NADH produced in the mitochondrial matrix during the citric acid cycle are oxidised by an NADH dehydrogenase (complex I), and the electrons are then transferred to ubiquinone located within the inner membrane.
Complex II. Ubiquinone also receives reducing equivalents via $\mathrm{FADH_2}$ (complex II), which is generated during the oxidation of succinate in the citric acid cycle. This route skips complex I altogether.
Complex III, the cytochrome $\mathrm{bc_1}$ complex. The reduced ubiquinone (ubiquinol) is then oxidised with the transfer of electrons to cytochrome c via the cytochrome $\mathrm{bc_1}$ complex (complex III).
Cytochrome c. Cytochrome c is a small protein attached to the outer surface of the inner membrane and acts as a mobile carrier for transfer of electrons between complex III and complex IV. It is not a complex itself.
Complex IV. Complex IV refers to the cytochrome c oxidase complex, containing cytochromes a and $\mathrm{a_3}$ , and two copper centres. This is where the electrons finally reach oxygen.
Complex V, ATP synthase. When the electrons pass from one carrier to another via complex I to complex IV in the electron transport chain, they are coupled to ATP synthase (complex V) for the production of ATP from ADP and inorganic phosphate.

What the chain yields. The number of ATP molecules synthesised depends on the nature of the electron donor. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP. Oxidation of one molecule of $\mathrm{FADH_2}$ produces 2 molecules of ATP. $\mathrm{FADH_2}$ is worth less because it enters at complex II and so bypasses complex I - it travels a shorter stretch of the chain, so less energy is released along its path.

The role of oxygen. Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. The presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system. Oxygen acts as the final hydrogen acceptor. Take oxygen away and the chain backs up: the carriers stay reduced, $\mathrm{NAD^+}$ and $\mathrm{FAD^+}$ are never regenerated, and glycolysis and the TCA cycle come to a halt.

Close with the point of the whole system. The ETS is where the hydrogen collected by glycolysis and the citric acid cycle is finally handed to oxygen, and the energy of that handover pays for 34 of the 38 ATP of aerobic respiration.

Question 43

Q. What is oxidative phosphorylation? Distinguish it from photophosphorylation.

Answer. Oxidative phosphorylation is the synthesis of ATP from ADP and inorganic phosphate, carried out by ATP synthase (complex V) on the inner mitochondrial membrane, using a proton gradient built by the energy of oxidation-reduction released as electrons pass along the electron transport system.

The name is the definition. Unlike photophosphorylation, where it is the light energy that is utilised for the production of the proton gradient required for phosphorylation, in respiration it is the energy of oxidation-reduction that is utilised for the same process. It is for this reason that the process is called oxidative phosphorylation. The word "oxidative" is doing all the work in the name - it tells you where the energy for the phosphorylation came from.

The mechanism, in order. The energy released during the electron transport system is utilised in synthesising ATP with the help of ATP synthase (complex V). Electrons pass along the carriers of the inner mitochondrial membrane; the energy released as they do so is used to move protons across that membrane, so that they accumulate in the intermembrane space; a proton gradient is built; and the protons then flow back into the matrix through ATP synthase, and that flow is what makes the ATP. This complex consists of two major components, $\mathrm{F_1}$ and $\mathrm{F_0}$ . The $\mathrm{F_1}$ headpiece is a peripheral membrane protein complex and contains the site for synthesis of ATP from ADP and inorganic phosphate. $\mathrm{F_0}$ is an integral membrane protein complex that forms the channel through which protons cross the inner membrane. For each ATP produced, $\mathrm{4H^+}$ pass through $\mathrm{F_0}$ from the intermembrane space to the matrix down the electrochemical proton gradient.

Now the comparison with chapter 11, which is what the marking scheme wants alongside the definition.

Point of comparison	Photophosphorylation	Oxidative phosphorylation
Where it happens	The thylakoid membrane of the chloroplast	The inner mitochondrial membrane
Energy that builds the proton gradient	Light energy	The energy of oxidation-reduction
Where protons accumulate	In the thylakoid lumen	In the intermembrane space of the mitochondrion
Where protons flow back to	The stroma	The matrix
The channel component	$\mathrm{CF_0}$	$\mathrm{F_0}$
The catalytic headpiece	$\mathrm{CF_1}$	$\mathrm{F_1}$
Terminal electron acceptor	$\mathrm{NADP^+}$	$\mathrm{O_2}$ , which becomes $\mathrm{H_2O}$
Occurs during	Photosynthesis, in a green cell in the light	Respiration, in every aerobic cell, light or dark

Three lines to say out loud when you revise. In photosynthesis protons accumulate in the thylakoid lumen; in respiration they accumulate in the intermembrane space of the mitochondrion - the stroma and the matrix are where the protons go back to, not where they gather. Photosynthesis has $\mathrm{CF_0}$ and $\mathrm{CF_1}$ ; respiration has $\mathrm{F_0}$ and $\mathrm{F_1}$ - same architecture, different names. In photophosphorylation it is light energy that creates the proton gradient; in oxidative phosphorylation it is the energy of oxidation-reduction.

And what is identical. A membrane, a proton gradient across it, a channel component through which the protons return, and a catalytic headpiece on the far side that synthesises the ATP. Chemiosmosis is one idea, used twice: build a gradient with whatever energy you have, then let it collapse through ATP synthase.

Question 44

Q. Derive the net gain of 38 ATP for one molecule of glucose, and state the assumptions on which the calculation rests and why they do not really hold.

Answer. It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised; but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions, and there are four of them.

There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
None of the intermediates in the pathway are utilised to synthesise any other compound.
Only glucose is being respired - no other alternative substrates are entering the pathway at any of the intermediary stages.

Notice what each one protects. The first lets us add the stages up as if they were steps in a queue. The second lets us count the two glycolytic NADH at the full mitochondrial rate of 3 ATP each. The third stops anything leaking out of the pathway to be built into something else. The fourth stops anything leaking in.

Now the derivation. Step one - collect what each stage produced, per molecule of glucose.

Stage	Where it happens	Substrate-level ATP	$\mathrm{NADH + H^+}$	$\mathrm{FADH_2}$
Glycolysis	Cytoplasm	2 net; 4 made, 2 spent	2	0
Pyruvic acid to acetyl CoA, happening twice	Mitochondrial matrix	0	2	0
TCA cycle, turning twice	Mitochondrial matrix	2	6	2
Totals		4	10	2

Step two - convert the reduced coenzymes using the rates from the electron transport system.

Source of ATP	Number	ATP each	ATP obtained
Substrate-level phosphorylation	4	-	4
$\mathrm{NADH + H^+}$ oxidised through the ETS	10	3	30
$\mathrm{FADH_2}$ oxidised through the ETS	2	2	4
Net gain per molecule of glucose			38

Where the 4 substrate-level ATP were made, exactly: BPGA to 3-phosphoglyceric acid and PEP to pyruvic acid, each happening twice in glycolysis - that is the gross 4, of which 2 are spent at the start, leaving a net 2; then succinyl-CoA to succinic acid, happening twice in the TCA cycle, for 2 more.

Why the assumptions do not hold. This kind of assumption is not really valid in a living system, and the chapter gives one reason against each.

All pathways work simultaneously and do not take place one after another.
Substrates enter the pathways and are withdrawn from them as and when necessary.
ATP is utilised as and when needed.
Enzymatic rates are controlled by multiple means.

Why do the exercise at all, then? Yet, it is useful to do this exercise, to appreciate the beauty and efficiency of the living system in extraction and storing energy. Hence, there can be a net gain of 38 ATP molecules during aerobic respiration of one molecule of glucose.

Read that conclusion as it is written - there CAN BE a net gain of 38 ATP, not "there is". And note the consequence of dropping assumption two on its own: if the glycolytic NADH is not transferred and yields 2 ATP each instead of 3, the total falls to 36, which is why 36 is the standard distractor.

Question 45

Q. Discuss the statement: the respiratory pathway is an amphibolic pathway.

Answer. Because the respiratory pathway is involved in both anabolism and catabolism, it would hence be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one. That is the conclusion; the marks are in the argument that gets you there, and in the examples.

Start with the received view. Since respiration involves breakdown of substrates, the respiratory process has traditionally been considered a catabolic process, and the respiratory pathway a catabolic pathway. Glucose goes in, $\mathrm{CO_2}$ and $\mathrm{H_2O}$ come out, and big molecules become small ones. But is this understanding correct?

Step one - where substrates enter. Glucose is the favoured substrate for respiration, and all carbohydrates are usually first converted into glucose before they are used for respiration. Fats and proteins are broken down to yield energy too, but they do not enter the respiratory pathway at the first step. The four entry points are the heart of the answer.

Substrate	Broken down first into	Enters the pathway as
Fats	Glycerol and fatty acids	Not as fat; each half enters at its own point
Fatty acids	Acetyl CoA	Acetyl CoA
Glycerol	PGAL	PGAL, a glycolysis intermediate
Proteins	Amino acids, released by proteases, then deaminated	At some stage within the Krebs' cycle, or even as pyruvate or acetyl CoA

Step two - the reversal. What is important to recognise is that it is these very compounds that would be withdrawn from the respiratory pathway for the synthesis of the said substrates. The same molecule that is the entrance when a substrate is being pulled apart is the exit when that substrate is being put together.

Step three - the fatty acid example, which must be written out in full. Fatty acids would be broken down to acetyl CoA before entering the respiratory pathway when a fatty acid is being used as a substrate. But when the organism needs to synthesise fatty acids, acetyl CoA would be withdrawn from the respiratory pathway for it. Hence the respiratory pathway comes into the picture both during breakdown and synthesis of fatty acids. Acetyl CoA is a two-way door.

Step four - the protein example. Similarly, during breakdown and synthesis of protein too, respiratory intermediates form the link. The Krebs' cycle acids that accept a deaminated amino acid on the way in are the same acids that are withdrawn to build an amino acid on the way out.

Step five - the definitions and the conclusion. Breaking down processes within the living organism is catabolism. Synthesis is anabolism. Because the respiratory pathway is involved in both, it is better considered an amphibolic pathway.

	Catabolic pathway	Anabolic pathway	Amphibolic pathway
What it does	Breaks molecules down	Builds molecules up	Does both
Energy	Releases energy	Consumes energy	Both, depending on the direction
Example from this chapter	Glucose oxidised to $\mathrm{CO_2}$ and $\mathrm{H_2O}$	Acetyl CoA withdrawn to build fatty acids	The respiratory pathway itself

Amphi means both. The intermediates of glycolysis and the Krebs' cycle are not just waste stations on a demolition route; they are the common pool from which the cell draws carbon skeletons when it needs to build. A student who writes only "the pathway does both" without the acetyl CoA example loses most of the marks.

Question 46

Q. Define RQ and give its value for each of the three substrate types, working out the arithmetic in each case.

Answer. The ratio of the volume of $\mathrm{CO_2}$ evolved to the volume of $\mathrm{O_2}$ consumed in respiration is called the respiratory quotient (RQ) or respiratory ratio.

$\mathrm{RQ} = \frac{\text{volume of }\mathrm{CO_2}\text{ evolved}}{\text{volume of }\mathrm{O_2}\text{ consumed}}$

Carbon dioxide evolved is on top; oxygen consumed is at the bottom. RQ is a pure ratio of two volumes, so it has no unit. The respiratory quotient depends upon the type of respiratory substrate used during respiration, which is exactly why the measurement is worth making.

Carbohydrates. When carbohydrates are used as substrate and are completely oxidised, the RQ will be 1, because equal amounts of $\mathrm{CO_2}$ and $\mathrm{O_2}$ are evolved and consumed respectively.

$\mathrm{C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O} + \text{Energy}$

$\mathrm{RQ} = \frac{6\mathrm{CO_2}}{6\mathrm{O_2}} = 1.0$

Fats. When fats are used in respiration, the RQ is less than 1. For tripalmitin:

$\mathrm{2(C_{51}H_{98}O_6) + 145O_2 \rightarrow 102CO_2 + 98H_2O} + \text{energy}$

$\mathrm{RQ} = \frac{102\mathrm{CO_2}}{145\mathrm{O_2}} = 0.7$

Two molecules of tripalmitin need 145 molecules of $\mathrm{O_2}$ and give out only 102 molecules of $\mathrm{CO_2}$ , so the fraction must be less than one. The reason is in the formulas. Glucose is $\mathrm{C_6H_{12}O_6}$ - six carbons carrying six oxygens of their own. Tripalmitin is $\mathrm{C_{51}H_{98}O_6}$ - fifty-one carbons carrying only six oxygens. A fat brings far less oxygen of its own to the reaction, so far more has to be supplied from outside, and the denominator of the RQ grows.

Proteins. When proteins are respiratory substrates the ratio would be about 0.9 - between the two.

Respiratory substrate	RQ value
Carbohydrates, completely oxidised	1.0
Proteins	About 0.9
Fats, for example tripalmitin	0.7, that is less than 1

One caution to end on. What is important to recognise is that in living organisms, respiratory substrates are often more than one; pure proteins or fats are never used as respiratory substrates. A measured RQ from a real tissue is therefore a mixed value, and you should expect something between the clean textbook numbers rather than exactly 1.0 or exactly 0.7.

The commonest error is simply inverting the ratio; if you write oxygen on top you will report 1.43 for a fat instead of 0.7.

The Chapter-End Exercises

There are twelve exercises at the end of this chapter, but exercise 1 asks for three separate comparisons and so does exercise 7, which comes to sixteen questions to write out. Fifteen of the sixteen are already answered in full inside the ten teaching sections of this chapter, and most of them are answered a second time in Tier 3 above. Only one part - exercise 7 (c) - is not answered anywhere else, and it is worked out below as Question 47.

The table tells you exactly where each answer sits, so you can attempt the exercise yourself first and then check your version against a complete one.

Exercise	Answered as
1 (a). Differentiate between respiration and combustion	Question 6 of the Do Plants Breathe? section, and again as Question 35 above
1 (b). Differentiate between glycolysis and Krebs' cycle	Question 14 of the Glycolysis section, and again as Question 36 above
1 (c). Differentiate between aerobic respiration and fermentation	Question 10 of the Fermentation section, and again as Question 39 above
2. What are respiratory substrates? Name the most common respiratory substrate	Question 7 of the The Respiratory Pathway Is Amphibolic section
3. Give the schematic representation of glycolysis	Question 11 of the Glycolysis section
4. What are the main steps in aerobic respiration? Where does it take place?	Question 10 of the Aerobic Respiration and the Link Reaction section
5. Give the schematic representation of an overall view of Krebs' cycle	Question 12 of the The Tricarboxylic Acid Cycle section, and again as Question 41 above
6. Explain ETS	Question 3 of the The Electron Transport System and Oxidative Phosphorylation section, and again as Question 42 above
7 (a). Distinguish between aerobic respiration and anaerobic respiration	Question 12 of the Fermentation section, and again as Question 37 above
7 (b). Distinguish between glycolysis and fermentation	Question 11 of the Fermentation section, and again as Question 38 above
7 (c). Distinguish between glycolysis and citric acid cycle	Question 47 below - the only part not answered anywhere else
8. What are the assumptions made during the calculation of net gain of ATP?	Question 2 of the The Respiratory Balance Sheet section, and again as Question 44 above
9. Discuss: the respiratory pathway is an amphibolic pathway	Question 10 of the The Respiratory Pathway Is Amphibolic section, and again as Question 45 above
10. Define RQ. What is its value for fats?	Question 1 of the The Respiratory Quotient section, and again as Question 46 above
11. What is oxidative phosphorylation?	Question 12 of the The Electron Transport System and Oxidative Phosphorylation section, and again as Question 43 above
12. What is the significance of step-wise release of energy in respiration?	Question 7 of the Do Plants Breathe? section, and again inside Question 35 above

Work all sixteen out on paper before the exam, in your own words and with the tables drawn out, rather than reading the answers and moving on. And notice how the set is built. Six of the sixteen are two-column comparisons, and three of those six are different pairings of the same four ideas - glycolysis, the Krebs cycle, fermentation and aerobic respiration. Learn the four processes properly once, with their location, their oxygen requirement, their products and their ATP yield, and half the exercise set answers itself.

Question 47

Q. Distinguish between glycolysis and the citric acid cycle. This is one of the chapter-end exercises.

Answer. Before the table, clear up two names, because this question looks like a new one and is not.

First, the source asks this same comparison twice, under two different names. Exercise 1 (b) calls it "Glycolysis and Krebs' cycle" and exercise 7 (c) calls it "Glycolysis and Citric acid Cycle". They are the same question, and one answer serves for both.

Second, the citric acid cycle, the Krebs cycle and the tricarboxylic acid (TCA) cycle are three names for the same cycle. The chapter uses all three: the acetyl CoA enters a cyclic pathway, the tricarboxylic acid cycle, more commonly called Krebs' cycle after the scientist Hans Krebs who first elucidated it, and the sections that follow speak of the citric acid cycle throughout. Nothing changes but the label.

So the answer is:

Point of comparison	Glycolysis	Citric acid cycle
Other names	The EMP pathway, after Gustav Embden, Otto Meyerhof and J. Parnas	Krebs' cycle, after Hans Krebs; also the tricarboxylic acid cycle
Where it occurs	In the cytoplasm of the cell	In the matrix of the mitochondria
Oxygen requirement	None directly - it runs in aerobic and anaerobic conditions alike, and in anaerobic organisms it is the only process in respiration	Oxygen is needed indirectly - the cycle requires regeneration of $\mathrm{NAD^+}$ and $\mathrm{FAD^+}$ , which only the oxygen-dependent ETS can supply
Shape of the pathway	A linear chain of ten reactions, under the control of different enzymes	A cycle - the acceptor OAA is regenerated at the end of every turn, so one molecule of it can process acetyl group after acetyl group
Starting material	Glucose, 6C	Acetyl CoA, 2C, which condenses with oxaloacetic acid, 4C, and water
End product	Two molecules of pyruvic acid, 3C each	OAA is regenerated and the carbon leaves entirely as $\mathrm{CO_2}$
$\mathrm{CO_2}$ released	None at all	2 per turn, that is 4 per glucose
Oxidation of glucose	Partial - glucose undergoes partial oxidation to form two molecules of pyruvic acid	Complete - for the carbon that reaches it, every atom leaves as $\mathrm{CO_2}$
Occurs in	All living organisms	Only in organisms carrying out aerobic respiration; in eukaryotes only inside the mitochondria
Reduced coenzymes produced	2 $\mathrm{NADH + H^+}$ per glucose, from one step happening twice	3 $\mathrm{NADH + H^+}$ and 1 $\mathrm{FADH_2}$ per turn, that is 6 and 2 per glucose
ATP yield	4 made and 2 spent, so a net 2 per glucose	1 GTP, and hence 1 ATP, per turn at the succinyl-CoA to succinic acid step, that is 2 per glucose

Two sentences are worth adding under the table, because they are what the comparison is really about.

The word partial is the anchor. Glycolysis only splits the sugar and leaves most of its energy sitting in two molecules of pyruvic acid. The citric acid cycle finishes the job on the carbon it is given, which is why there is a whole chapter after glycolysis and why fermentation, which stops at pyruvate, gets less than seven per cent of the energy of glucose.

And the shape follows from the chemistry. The continued oxidation of acetyl CoA via the TCA cycle requires the continued replenishment of oxaloacetic acid, the first member of the cycle - a cycle regenerates its own acceptor, whereas a linear pathway would need a fresh one for every acetyl group.

Finally, so that you do not think you have missed something: if you have answered exercise 1 (b), you have already answered this one. Exercise 1 (b) is set out as Question 36 above, and the answer here is the same answer with the cycle called by its other name.

How to Use This Section

The Facts These Questions Draw On

Tier 1 - Concept Checks

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Tier 2 - Pathway Tracing and Arithmetic

Question 21

Question 22

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

Question 29

Question 30

Question 31

Question 32

Question 33

Question 34

Tier 3 - Comparisons and Long Answers

Question 35

Question 36

Question 37

Question 38

Question 39

Question 40

Question 41

Question 42

Question 43

Question 44

Question 45

Question 46

The Chapter-End Exercises

Question 47

Great job!