Chapter 13: Respiration and Energy Transfer

For most of the nineteenth century, scientists believed that fermentation — the bubbling change that turns sugary juice into wine and dough into risen bread — was something only a living cell could do. It was thought to be tied to the “life force” of yeast itself. Break the cells, and the magic would stop.

In the 1890s, a German chemist named Eduard Buchner set out to do something quite ordinary. He ground up yeast cells with sand, pressed out the juice, and wanted to store this cell-free liquid for later use. Since he could not heat it without spoiling it, he did what any cook would do: he added a thick sugar solution, because sugar is a good preservative.

The juice did not stay preserved. It began to fizz. Bubbles of carbon dioxide rose through a liquid that contained no living yeast cells at all — the sugar was being fermented by something in the juice itself.

That accident showed that fermentation is chemistry, not magic. It is carried out by molecules — enzymes — working in an ordered sequence of steps. Buchner received the Nobel Prize in Chemistry in 1907, and biology gained a new question: exactly what is that sequence of steps, and how does it release the energy that keeps a cell alive?

Diagram

Fig. 13.0 – Opening illustration: a flask of cell-free yeast juice with sugar solution releasing gas bubbles, beside a labelled cell showing cytoplasm and a mitochondrion, with a glucose molecule and an ATP molecule linking the two.

Curiosity Corner

Q1. Why do a sprinter’s legs burn and tire within seconds of a 100 m race, while a marathon runner can keep going for hours without that sudden pain? Q2. Yeast can survive on glucose without any oxygen — so why does the same yeast cell get far more energy out of the same glucose molecule when oxygen is available? Q3. If oxygen is so essential to us, why do biologists say that the primary event in respiration is the removal of hydrogen rather than the addition of oxygen? Q4. Why does a cell break glucose down through roughly thirty small enzyme-controlled steps instead of oxidising it in one quick burst, the way a flame does? Q5. How can Krebs cycle — a pathway meant to break molecules down — also supply the raw material for building fatty acids and amino acids?

By the end of this chapter, you will be able to answer each of these questions using the pathways, energy calculations and control mechanisms you are about to study.

13.1 Formation of ATP

Try Yourself

Which nutrients are used for energy production?
Among carbohydrate, protein and fat, which nutrient is preferred for energy production, and why?
Why do organisms take in oxygen and release carbon dioxide?
What is meant by aerobic and anaerobic respiration?
Which steps are involved in aerobic respiration?

Respiration is a catabolic process. A catabolic process is one in which large, complex organic molecules are broken down into simpler ones, and energy stored in their chemical bonds is released. In respiration, a complex organic substrate such as glucose is oxidised to simple components, and the energy set free is trapped in a usable biological form.

Definition

Respiration is the catabolic process in which a complex organic substrate is oxidised to simple components in order to generate biological energy.

Cellular respiration is carried out in two different ways — anaerobic respiration and aerobic respiration. Both are studied in detail in the sections that follow, but neither makes sense until we understand the molecule in which the released energy is actually stored.

The Energy Currency of the Cell

Energy released during oxidation cannot be handed straight to a muscle fibre or to a growing cell. It must first be packed into a chemical form that any part of the cell can spend. That form is ATP — adenosine triphosphate — which you met while studying biomolecules.

An ATP molecule is built from the nitrogen base adenine, the five-carbon sugar ribose, and a chain of three phosphate groups, as shown in Fig. 13.5. The last two phosphate bonds are energy-rich, which means a large amount of usable energy is released when either of them is broken.

Diagram

Fig. 13.5 – Structure of ATP: adenine base joined to ribose sugar, followed by three phosphate groups in a chain, with the two terminal phosphoanhydride bonds marked as energy-rich bonds.

Structure and Function

ATP molecule (three phosphate groups carrying like negative charges are forced close together → they repel one another, making the terminal bonds unstable and easy to break → energy is released quickly and in a small, convenient packet whenever the cell needs it)

Zoom In / Out

ATP as the cell’s energy carrier (Zoom in: ATP is a single molecule made of a base, a sugar and three phosphates. Zoom out: the same molecule powers every cell, so it ultimately supports contraction of a whole muscle, transport across every membrane and growth of the entire organism.)

Three Ways of Making ATP

The formation of ATP is called phosphorylation, because a phosphate group is added to ADP (adenosine diphosphate). In nature, phosphorylation occurs in three different ways — photophosphorylation, substrate-level phosphorylation and oxidative phosphorylation.

Definition

Phosphorylation is the formation of ATP by the addition of a phosphate group to ADP.

You have already learnt photophosphorylation in photosynthesis, where light energy drives the synthesis of ATP in the chloroplast. The other two types are the ones that operate during respiration, and they differ in where the phosphate comes from.

Substrate-level phosphorylation is a direct phosphorylation of ADP by transfer of a phosphate group from any suitable substrate. It occurs in the cytoplasm of the cell and in the matrix of mitochondria.

Definition

Substrate-level phosphorylation is the direct phosphorylation of ADP by the transfer of a phosphate group from a suitable substrate molecule; it occurs in the cytoplasm and in the mitochondrial matrix.

Oxidative phosphorylation is the phosphorylation of ADP at the cost of energy released during the oxidation of substrates such as \(NADH+H^+\) and \(FADH_2\). Here no phosphate is handed over from a substrate; instead, the energy of oxidation is used to attach inorganic phosphate to ADP. This occurs on the inner mitochondrial membrane only.

Definition

Oxidative phosphorylation is the phosphorylation of ADP using the energy released during oxidation of reduced coenzymes such as \(NADH+H^+\) and \(FADH_2\); it takes place only on the inner mitochondrial membrane.

Compare Table

Spending the ATP

Making ATP is only half of the story. Whenever energy is required for a metabolic process, ATP is hydrolysed. ATP hydrolysis breaks the terminal phosphate bond and releases the energy that is then used for metabolic activities such as muscle contraction, active transport and the synthesis of new molecules.

\[ATP + H_2O \rightarrow ADP + Pi + \text{energy}\]

In this equation \(Pi\) stands for inorganic phosphate. The ADP and phosphate released are not wasted; respiration phosphorylates them back into ATP, so the same molecules are used again and again. This continuous cycle is shown in FC 13.1.

Flowchart

FC 13.1 – The ATP–ADP cycle: oxidation of respiratory substrate → energy released → ADP + Pi phosphorylated → ATP → ATP hydrolysed for metabolic work → ADP + Pi → back to phosphorylation

Real-World Application

After death, a body stops respiring and its ATP supply is not replaced. Muscle fibres need ATP not only to contract but also to release from the contracted state, so once ATP runs out the muscles lock and the body stiffens. This stiffening, called rigor mortis, gradually passes off as muscle proteins themselves begin to break down. It is a direct, visible reminder that ATP must be made continuously — a cell cannot stockpile it.

Side Note

Two points worth remembering as you begin this chapter: maintenance of life requires a continuous supply of energy, and respiration is the process that fulfils this continuous need.

In NEET and board papers, this section is usually tested indirectly — a question will name a particular step of respiration and ask which type of phosphorylation operates there, or will ask which type occurs only on the inner mitochondrial membrane. Fixing the three sites in mind now will save time later.

13.2 Anaerobic Respiration

Anaerobic respiration is the cellular respiration that does not involve atmospheric oxygen. It is also called fermentation. It is completed through two stages — glycolysis, and the incomplete conversion of the products of glycolysis into lactic acid or ethanol.

Definition

Anaerobic respiration (fermentation) is cellular respiration carried out without atmospheric oxygen, in which glucose is broken down by glycolysis and the glycolytic product is then incompletely converted into lactic acid or ethanol.

Notice the word “incomplete”. Because oxygen is absent, the substrate is never oxidised all the way to carbon dioxide and water. The end product still holds a large amount of unreleased chemical energy, which is why anaerobic respiration yields far less energy than aerobic respiration.

Real Incident / Discovery

In the 1850s and 1860s, Louis Pasteur was asked by French wine and beer makers to find out why some of their vats turned sour instead of producing alcohol. Studying the vats under the microscope, he showed that fermentation was the work of living yeast and that souring was caused by other micro-organisms that had got in. He also observed that yeast consumed sugar much faster when air was kept away from it than when air was freely available — an early clue that fermentation is a wasteful way of extracting energy from sugar.

Glycolysis

Glycolysis involves the breakdown of a glucose molecule into two pyruvic acid molecules — hence the name (glykys, sweet; lysis, splitting). This is a common step in anaerobic as well as aerobic respiration, so every pathway in this chapter begins here. It occurs in the cytoplasm of the cell, not inside any organelle. It is completed in two phases — the preparatory phase and the pay-off phase.

Zoom In / Out

where glycolysis happens (Zoom in: glycolysis is a chain of ten reactions among small dissolved molecules. Zoom out: because it needs no membrane and no organelle, it can run in the cytoplasm of every living cell — bacterial, plant or animal.)

The overall process of glycolysis is completed through ten steps, each catalysed by its own enzyme. The full sequence, with the enzymes and the co-factors involved, is summarised in FC 13.2.

Flowchart

FC 13.2 – The ten steps of glycolysis: Glucose (6C, 1M) →(hexokinase, Mg++, ATP→ADP) Glucose-6-phosphate (6C, 1M) →(phosphohexose isomerase) Fructose-6-phosphate (6C, 1M) →(phosphofructokinase, Mg++, ATP→ADP) Fructose-1, 6-diphosphate (6C, 1M) →(aldolase) Glyceraldehyde-3-phosphate (3C, 1M) ⇌ Dihydroxy acetone phosphate (3C, 1M) →(triose phosphate isomerase) Glyceraldehyde-3-phosphate (3C, 2M) →(triose phosphate dehydrogenase, Zn++, 2 H3PO4 added non-enzymatically, 2NAD+ → 2NADH+H+) 1,3-Bisphosphoglycerate (3C, 2M) →(phosphoglycerate kinase, Mg++, 2ADP→2ATP) 3-Phosphoglycerate (3C, 2M) →(phosphoglycerate mutase) 2-Phosphoglycerate (3C, 2M) →(enolase, Mg++, 2 H2O released) 2-Phosphoenolpyruvate (3C, 2M) →(pyruvate kinase, Mg++, 2ADP→2ATP) Pyruvic acid (3C, 2M)

The preparatory phase

The first five steps constitute the preparatory phase. Through these steps, glucose is phosphorylated twice at the cost of two ATP molecules, and a molecule of fructose 1, 6-bisphosphate is formed.

This may look wasteful — the cell spends ATP in a pathway meant to earn ATP. Adding phosphate groups, however, does two useful things: it traps the sugar inside the cell, since a phosphorylated sugar cannot slip back out through the membrane, and it makes the molecule unstable enough to be split.

Fructose 1, 6-bisphosphate is then split to form a molecule of glyceraldehyde 3-phosphate and a molecule of dihydroxyacetone phosphate. Both of these molecules are 3-carbon carbohydrates (trioses) and are isomers of each other, which means they contain the same atoms arranged differently.

Only one of the two, glyceraldehyde-3-phosphate, can be carried forward into the next reaction. So dihydroxyacetone phosphate is isomerised to a second molecule of glyceraldehyde-3-phosphate. Thus, two molecules of glyceraldehyde-3-phosphate are formed, and here the first phase, that is the preparatory phase of glycolysis, ends.

From this point onwards every remaining reaction happens twice for each glucose molecule — once for each glyceraldehyde-3-phosphate. This is why the products of the later steps are counted as 2M in FC 13.2.

The pay-off phase

In the pay-off phase, both the molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1, 3-bis phosphoglycerate by oxidation and phosphorylation. Here, phosphorylation is brought about with the help of inorganic phosphate and not ATP.

That detail matters. Because the phosphate is picked up free of cost from the cell’s own inorganic phosphate pool, no further ATP is spent, and the oxidation step also hands two electrons and a proton to \(NAD^+\), reducing it to \(NADH+H^+\).

Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through a series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.

Memory Aid

which steps spend ATP and which steps make it (Two enzymes ending in -kinase spend ATP early — hexokinase and phosphofructokinase. Two enzymes ending in -kinase make ATP later — phosphoglycerate kinase and pyruvate kinase. Spend 2, earn 4.)

Trace It

the ATP account of one glucose molecule in glycolysis 1. Hexokinase phosphorylates glucose — 1 ATP spent (running total: −1). 2. Phosphofructokinase phosphorylates fructose-6-phosphate — 1 ATP spent (running total: −2). 3. The six-carbon sugar splits into two three-carbon molecules, so every step from here counts twice. 4. Phosphoglycerate kinase makes 1 ATP per triose — 2 ATP earned (running total: +2). 5. Pyruvate kinase makes 1 ATP per triose — 2 ATP earned (running total: +4). 6. Gross gain 4 ATP, cost 2 ATP, so the net gain of glycolysis is 2 ATP, plus 2 \(NADH+H^+\) still holding energy.

The overall reaction of glycolysis

\[\text{Glucose} + 2\,ATP + 2\,iP + 4\,ADP + 2\,NAD^+ \rightarrow 2\,\text{Pyruvate} + 2\,ADP + 4\,ATP + 2\,NADH+H^+ + 2\,H_2O\]

In this equation \(iP\) is inorganic phosphate, \(ATP\) and \(ADP\) are adenosine triphosphate and adenosine diphosphate, and \(NAD^+\) is the oxidised coenzyme nicotinamide adenine dinucleotide, which leaves the pathway in its reduced form \(NADH+H^+\).

Numerical 13.1

A yeast culture in an airtight vessel completely converts 30 molecules of glucose through glycolysis alone. Using the overall reaction above, work out how many ATP molecules are produced gross, how many are consumed, what the net ATP gain is, and how many molecules of \(NADH+H^+\) and pyruvate are formed.

Control of glycolysis

Glycolysis is under tight control. Its rate depends upon the requirement of ATP and many other factors. Glycolytic rate control is achieved by complex interplay between ATP consumption, \(NADH_2\) regeneration and regulation of various glycolytic enzymes like hexokinase, PFK-1, pyruvate kinase, etc. Besides, it is also controlled by hormones like glucagon, epinephrine and insulin.

A cell that is resting and already rich in ATP slows the pathway down; a cell that is working hard and running short of ATP speeds it up. This is why control is placed at the enzymes named above — they sit at the irreversible steps of the pathway, and an irreversible step is the safest place to install a switch.

Try Yourself

What is the role of \(Mg^{++}\) and \(Zn^{++}\) in the various steps of glycolysis?
Why are some reactions of glycolysis reversible and some irreversible?
Why is glycolysis considered a biochemical proof of evolution?
Why do athletes such as sprinters have a higher proportion of white muscle fibres?

Side Note

Glycolysis is the only source of energy production in erythrocytes, renal medulla, brain and sperm. Some plant tissues that are modified to store starch, such as potato, also depend mainly on glycolysis for energy production.

What happens to pyruvic acid without oxygen

Glycolysis leaves the cell with a problem. It has used up \(NAD^+\) and produced \(NADH+H^+\), and the cell carries only a small stock of this coenzyme. Unless \(NAD^+\) is regenerated, glycolysis itself must stop. In the absence of oxygen, the cell solves this by dumping the hydrogen onto pyruvic acid or its derivative — and that is exactly what fermentation is. The two possible routes are shown in FC 13.3.

Flowchart

FC 13.3 – Fate of pyruvic acid: Glucose → glycolysis → Pyruvic acid; if oxygen absent and cell is a muscle cell → lactic acid (NAD+ regenerated); if oxygen absent and cell is yeast → acetaldehyde + CO2 → ethanol (NAD+ regenerated); if oxygen present → pyruvic acid enters the aerobic pathway

Lactic acid fermentation

\[C_6H_{12}O_6 \xrightarrow{\text{Glycolysis}} 2CH_3COCOOH + 2NADH+H^+ \rightarrow 2CH_3CHOHCOOH + 2NAD^+\]

Here \(C_6H_{12}O_6\) is glucose, \(CH_3COCOOH\) is pyruvic acid and \(CH_3CHOHCOOH\) is lactic acid.

In muscles, the \(NADH+H^+\) produced during glycolysis is reoxidised to \(NAD^+\) by donating one proton and two electrons to pyruvic acid, which yields lactic acid. Skeletal muscles usually derive their energy by anaerobic respiration.

After vigorous exercise lactic acid accumulates, leading to muscle fatigue. During rest, however, the lactic acid is reconverted to pyruvic acid and is channeled back into the aerobic respiration pathway. Nothing is thrown away — the energy still locked in lactic acid is recovered later, once oxygen is again available.

Side Note

In Chapter 3, Biomolecules, you read about myoglobin, the oxygen-storing and oxygen-transporting pigment of skeletal muscles. Red (dark) muscles are richer in myoglobin than white (pale) muscles. Red fibres can therefore use the oxygen stored in myoglobin to continue energy production over a prolonged period by aerobic oxidation of glucose, which lets them perform sustained work over a long period. White fibres, on the contrary, produce the energy needed for very fast and severe work by glycolysis, as sufficient oxygen is not immediately available to them for such work. But white muscles accumulate lactic acid and get fatigued in a short time. Thus athletes with a higher proportion of red fibres in their muscles are physiologically better adapted for sustained events such as a marathon or long-distance swimming.

Curiosity Corner

Q1. Why do a sprinter’s legs burn and tire within seconds of a 100 m race, while a marathon runner can keep going for hours without that sudden pain?

A. A sprinter’s effort is powered mainly by white muscle fibres working anaerobically. Their glycolysis runs at a furious rate and pyruvic acid is converted to lactic acid so that \(NAD^+\) can be regenerated. The lactic acid piles up quickly and brings on fatigue. A marathon runner relies on red fibres rich in myoglobin, which supply oxygen steadily so that pyruvate is oxidised aerobically instead of being turned into lactic acid — slower, but sustainable for hours.

Alcoholic fermentation

\[C_6H_{12}O_6 \xrightarrow{\text{Glycolysis}} 2CH_3COCOOH \rightarrow CO_2 \uparrow + 2CH_3CHO + 2NADH+H^+ \rightarrow 2C_2H_5OH\]

Here \(CH_3CHO\) is acetaldehyde and \(C_2H_5OH\) is ethanol.

In yeast, the pyruvate is decarboxylated to acetaldehyde, which means one carbon is removed as carbon dioxide. The acetaldehyde is then reduced by \(NADH+H^+\) to ethanol. Carbon dioxide is also produced in this process. This type of anaerobic respiration is termed alcoholic fermentation.

Accumulation of ethanol by fermentation in a culture of yeast may stop further multiplication and lead to the death of the cells. In effect, the organism is slowly poisoned by its own waste product. In the presence of oxygen, however, yeast can respire aerobically.

Compare Table

Do Not Mix Up

Glycolysis vs Fermentation (Glycolysis is only the first stage — glucose to pyruvic acid — and it happens in anaerobic as well as aerobic respiration. Fermentation is the complete anaerobic process: glycolysis plus the conversion of pyruvic acid into lactic acid or ethanol.)

Real-World Application

When dough is left to rise, the yeast in it ferments the sugars of the flour. The ethanol formed evaporates during baking, but the carbon dioxide is trapped by the elastic gluten network and forms the bubbles that make bread light and porous. The same reaction, run in a closed vessel where the alcohol is kept rather than the gas, is what produces wine and beer.

In examinations, this section is most often tested through sequence-ordering questions on the ten steps, through “which enzyme acts here” matching items, and through assertion–reason questions on why glycolysis needs no oxygen yet still requires \(NAD^+\) to be recycled.

13.3 Aerobic Respiration

Aerobic respiration involves molecular oxygen during the oxidation of glucose. Glucose is completely oxidised in this process, which is operated through the following steps — glycolysis, production of acetyl CoA (connecting link reaction), Krebs cycle, electron transfer chain reaction and terminal oxidation.

Definition

Aerobic respiration is the complete oxidation of a respiratory substrate in the presence of molecular oxygen, carried out through glycolysis, the connecting link reaction, Krebs cycle, the electron transport chain and terminal oxidation.

The word “completely” separates this pathway from fermentation. In anaerobic respiration the carbon of glucose ends up locked in lactic acid or ethanol; here every carbon atom leaves as carbon dioxide and every hydrogen atom finally reaches oxygen to form water.

The first step of aerobic respiration, that is glycolysis, has already been studied in detail. In the case of aerobic respiration, the glycolytic product, pyruvic acid, is converted into acetyl CoA. This process occurs in the cytoplasm in the case of prokaryotes and in the mitochondria in the case of eukaryotes. For the structure of mitochondria, refer to Chapter 5, Cell Structure and Organization.

Zoom In / Out

the mitochondrion as the site of aerobic respiration (Zoom out: a mitochondrion is one organelle inside a cell, and cells rich in it — heart muscle, flight muscle, liver — are the ones with the highest energy demand. Zoom in: the organelle has two membranes enclosing an outer chamber and a matrix, and each stage of aerobic respiration is assigned to a particular compartment.)

Conversion of pyruvic acid to acetyl CoA

This is an oxidative decarboxylation reaction. Decarboxylation means removal of a carbon as carbon dioxide, and oxidation here means removal of hydrogen. It is catalysed by a multienzyme complex — the pyruvate dehydrogenase complex (PDH). This enzyme is present in the mitochondria of eukaryotes and in the cytosol of prokaryotes. The reaction is set out in FC 13.4.

Flowchart

FC 13.4 – Connecting link reaction: Pyruvic acid (3c) + CoA-SH + NAD+ →(pyruvate dehydrogenase complex) Acetyl CoA (2c) + CO2 ↑ + NADH+H+

This reaction is called the ‘connecting link’ reaction between glycolysis and Krebs cycle. It is worth pausing on the accounting: the three-carbon pyruvic acid loses one carbon as \(CO_2\) and enters the cycle as a two-carbon acetyl group carried by coenzyme A, while the hydrogen removed reduces \(NAD^+\) to \(NADH+H^+\).

Side Note

The pyruvate dehydrogenase complex needs thiamin (vitamin \(B_1\)) as a co-enzyme. It cannot function in the absence of vitamin \(B_1\). Hence, thiamin deficiency causes pyruvic acidosis and lactic acidosis, which are life-threatening conditions. This is one reason why a balanced diet is very important in the maintenance of health.

Krebs cycle (TCA cycle / citric acid cycle)

Pyruvic acid produced by glycolysis undergoes aerobic oxidation in the mitochondrial matrix through the TCA cycle. This cycle serves as a common oxidative pathway for carbohydrates, fats and proteins. Moreover, some intermediates of the TCA cycle are used in synthesizing important biomolecules such as glutamate and aspartate.

Real Incident / Discovery

In the 1930s, Hans Krebs was studying how minced pigeon breast muscle — a tissue with an enormous appetite for oxygen — used up different organic acids. He noticed that adding a small amount of certain acids caused the tissue to consume far more oxygen than the added acid itself could account for. The only explanation that fitted was that these acids were not being used up but were being regenerated in a closed loop that kept turning. From this reasoning he pieced together the cycle that now carries his name, and he was awarded the Nobel Prize in Physiology or Medicine in 1953.

Before participating in the TCA cycle, pyruvic acid (3c) enters the mitochondrion. Here it is decarboxylated and the remaining 2-carbon fragment is combined with a molecule of coenzyme A to form acetyl-CoA (2c). This reaction is an oxidative decarboxylation process and produces \(H^+\) ions and electrons along with carbon dioxide. During the process \(NAD^+\) is reduced to \(NADH+H^+\).

β-oxidation of fatty acids also produces acetyl-CoA as the end product. Acetyl-CoA from both sources is condensed with oxaloacetic acid (4c) to form citric acid (6c). The complete cycle, with each intermediate and the enzyme that acts on it, is given in Fig. 13.1.

Diagram

Fig. 13.1 – Krebs cycle: Pyruvic acid (3c) + CoA-SH + NAD+ → NADH·H+ + CO2 → Acetyl-CoA (2c); Acetyl-CoA + Oxaloacetate → (citrate synthase, condensation, CoA-SH and H2O released) Citrate (6c) → (aconitase, dehydration, H2O removed) cis-Aconitate (6c) → (aconitase, hydration, H2O added) Isocitrate (6c) → (isocitrate dehydrogenase, dehydrogenation) Oxalosuccinic acid (6c) → (isocitrate dehydrogenase, oxidative decarboxylation, CO2 released, NADH+H+ formed) α-Ketoglutarate (5c) → (α-ketoglutarate dehydrogenase complex, oxidative decarboxylation, CoA-SH taken in, CO2 released, NADH+H+ formed) Succinyl-CoA (4c) → (succinyl-CoA synthetase, substrate-level phosphorylation, GDP(ADP)+Pi → GTP(ATP), CoA-SH released) Succinate (4c) → (succinate dehydrogenase, dehydrogenation/oxidation, FADH2 formed) Fumarate (4c) → (fumarase, hydration, H2O added) Malate (4c) → (malate dehydrogenase, dehydrogenation/oxidation, NADH+H+ formed) Oxaloacetate (4c) → back to condensation with the next acetyl-CoA

Citric acid is oxidised step-wise by mitochondrial enzymes, releasing carbon dioxide. This finally regenerates oxaloacetic acid to complete the cycle. Because oxaloacetate is remade at the end of every turn, a very small quantity of it can keep the cycle running indefinitely — this is exactly the point Krebs realised in his pigeon muscle experiments.

There are four steps of oxidation in this cycle, catalysed by dehydrogenases (oxidoreductases) using \(NAD^+\) or \(FAD^+\) as the coenzyme. The coenzymes are consequently reduced to \(NADH+H^+\) and \(FADH_2\) respectively. These transfer their electrons to the mitochondrial respiratory chain to get reoxidised. One molecule of GTP (ATP) is also generated for every molecule of citric acid oxidised.

Memory Aid

the order of Krebs cycle intermediates (Citrate → Isocitrate → α-Ketoglutarate → Succinyl-CoA → Succinate → Fumarate → Malate → Oxaloacetate: “Citrate Is Krebs’ Starting Substrate For Making Oxaloacetate.”)

Amphibolic pathway

Though we describe aerobic respiration as a catabolic (oxidative) pathway, this is not entirely correct, especially in the case of Krebs cycle. Various reactions of Krebs cycle are mainly responsible for the step-wise oxidation of the acetyl part of acetyl CoA, leading to the release of energy and \(CO_2\).

However, as per need, acetyl CoA or some other intermediates such as α-ketoglutarate and oxaloacetate are used as precursors for the synthesis of fatty acids, glutamic acid and aspartic acid respectively. Hence, Krebs cycle can correctly be referred to as an ‘amphibolic pathway’, that is, one involving catabolism as well as anabolism.

Definition

An amphibolic pathway is a metabolic pathway that functions both in catabolism (breaking molecules down to release energy) and in anabolism (supplying intermediates for the synthesis of new biomolecules).

Curiosity Corner

Q5. How can Krebs cycle — a pathway meant to break molecules down — also supply the raw material for building fatty acids and amino acids?

A. Because its intermediates are simply pulled out of the loop when the cell needs them. Acetyl CoA is the starting block for fatty acid synthesis, α-ketoglutarate is converted to glutamic acid, and oxaloacetate to aspartic acid. The cycle continues to turn as long as enough oxaloacetate remains to accept the next acetyl group, so the same pathway can break down and build up at the same time — which is what “amphibolic” means.

Side Note

Not only glucose but also amino acids from protein metabolism and fatty acids from lipid metabolism participate in Krebs cycle through acetyl CoA. This is why the body can go on producing ATP even when carbohydrate intake is low.

Electron transport chain (electron transfer system)

Wherever the \(NADH_2\) (\(NADH+H^+\)) and \(FADH_2\) are produced during glycolysis, the connecting link reaction and Krebs cycle, they are oxidised with the help of various electron carriers and enzymes. In other words, the hydrogen collected all through the earlier stages is now cashed in.

These carriers and enzymes are arranged on the inner mitochondrial membrane in the form of various complexes as complex I, II, III, IV and V, as shown in Fig. 13.2.

Diagram

Fig. 13.2 – Electron Transport system (ETS): inner mitochondrial membrane separating inter membrane space from matrix, showing Complex I (NADH dehydrogenase, FMN and Fe-S) receiving NADH+H+, Complex II (succinate dehydrogenase, FAD, succinate → fumarate), mobile CoQ/UQH2 in the membrane, Complex III (cytochrome bc1 with Cyt b, Fe-S, Cyt c1), mobile Cyt c, Complex IV (cytochrome c oxidase with Cyt a → Cyt a3 → Cu) passing electrons to ½O2 + 2H → H2O, and Complex V (F0–F1 ATP synthase) converting ADP + P to ATP; protons pumped out at complexes I, III and IV and returning through F0, with the electrochemical gradient marked

\(NADH+H^+\) is oxidised by NADH dehydrogenase (complex I) and its electrons are transferred to ubiquinone (coenzyme Q, i.e. CoQ) present on the inner membrane of mitochondria. Reduced ubiquinone is called ubiquinol.

\(FADH_2\) is oxidised by complex II (succinate dehydrogenase) and these electrons are also transferred to CoQ. Note that \(FADH_2\) therefore enters the chain one step later than \(NADH+H^+\) — it bypasses complex I altogether, and this single fact explains why it yields less ATP.

During oxidation of \(NADH+H^+\) and \(FADH_2\), electrons and protons are released, but only the electrons are carried forward, whereas the protons are released into the outer chamber of mitochondria. The separation of the two is the whole trick of the system: the electrons travel down the chain while the protons pile up on one side of the membrane, storing energy as a gradient.

Ubiquinol is oxidised by complex III (cytochrome \(bc_1\) complex) and its electrons are transferred to cytochrome C. Cytochrome C is a small, iron-containing protein, loosely associated with the inner membrane. It acts as a mobile electron carrier, transferring the electrons between complex III and IV.

Cytochrome C is oxidised by complex IV, or cytochrome C oxidase, consisting of cytochrome a and \(a_3\). Electrons are transferred by this complex to molecular oxygen. This is terminal oxidation. Reduced molecular oxygen reacts with protons to form a water molecule. The complete route taken by a pair of electrons is summarised in FC 13.5.

Flowchart

FC 13.5 – Electron flow through the ETS: NADH+H+ → Complex I (NADH dehydrogenase) → CoQ (ubiquinone → ubiquinol); FADH2 → Complex II (succinate dehydrogenase) → CoQ; CoQ → Complex III (cytochrome bc1) → Cytochrome C → Complex IV (cytochrome c oxidase, cyt a and a3) → molecular oxygen (terminal oxidation) → O2 combines with protons → H2O

Trace It

a pair of electrons from a molecule of NADH+H+ to water 1. \(NADH+H^+\) hands over its electrons to complex I and is reoxidised to \(NAD^+\), which returns to Krebs cycle to be used again. 2. The electrons pass to ubiquinone in the inner membrane, converting it to ubiquinol; meanwhile protons are pushed out into the outer chamber. 3. Complex III oxidises ubiquinol and passes the electrons to the mobile carrier cytochrome C. 4. Cytochrome C shuttles them along the membrane surface to complex IV. 5. Complex IV hands the electrons to molecular oxygen — this is terminal oxidation. 6. The reduced oxygen picks up protons and becomes a molecule of water, and the electrons’ journey ends.

Do Not Mix Up

Terminal oxidation vs Oxidative phosphorylation (Terminal oxidation is the last electron transfer — the handing over of electrons to molecular oxygen at complex IV to form water. Oxidative phosphorylation is the synthesis of ATP at complex V using the proton gradient that the electron flow created. One is the end of the electron path, the other is the payment collected on the way.)

Chemiosmosis and ATP synthesis

Protons necessary for this are channeled from the outer chamber of mitochondria into the inner chamber by the \(F_0\) part of the oxysome (complex V) embedded in the inner mitochondrial membrane. This proton channeling by \(F_0\) is coupled to the catalytic site of \(F_1\), which catalyses the synthesis of ATP from ADP and inorganic phosphate. This is oxidative phosphorylation.

As the transfer of protons is accompanied with the synthesis of ATP, this process is named ‘Chemiosmosis’ by Peter Mitchell.

Principle / Law

Chemiosmosis: the energy released during electron transport is used to pump protons across the inner mitochondrial membrane, and the return flow of these protons through the \(F_0\)–\(F_1\) particle drives the synthesis of ATP from ADP and inorganic phosphate.

Real Incident / Discovery

When Peter Mitchell proposed the chemiosmotic hypothesis in the 1960s, most biochemists were searching for a chemical intermediate that would link oxidation to ATP synthesis, and his idea that the link was a proton gradient across a membrane was received with considerable scepticism. Over the following years, experiments showed that an artificially created proton gradient could by itself drive ATP synthesis, and the hypothesis was accepted. Mitchell was awarded the Nobel Prize in Chemistry in 1978.

Structure and Function

inner mitochondrial membrane (it is folded into cristae and is impermeable to protons → a large area is available to hold the respiratory complexes and \(F_0\)–\(F_1\) particles, and protons pumped out cannot leak back except through \(F_0\) → a steep electrochemical gradient is maintained, and its energy is captured as ATP)

Oxidation of one \(NADH+H^+\) leads to the production of 3 ATP molecules, whereas oxidation of \(FADH_2\) leads to the production of 2 ATP molecules. However, the number of ATP produced depends upon the physiological conditions and the source of the respiratory substrate.

Compare Table

Try Yourself

What is the effect of carbon monoxide poisoning on cytochromes?

The ATP balance sheet

Now that every stage is in place, the energy yield of one glucose molecule can be added up, as shown in Table 13.3.

Step of Respiration	Consumption	Substrate level phosphorylation	\(NADH+H^+\)	\(FADH_2\)	Oxidative phosphorylation (total)	Total	Net benefit
Glycolysis	2	4	2 x 3 = 6	—–	6	10	8
Pyruvate → AcetylCoA	—–	—–	2 x 3 = 6	—–	6	6	6
Krebs cycle	—–	1 x 2 = 2	6 x 3 = 18	2 x 2 = 4	22	24	24
Total	2	[6]	30	4	30 + 4 = [34]	[40]	38

Table 13.3 Balance sheet for ATP by aerobic oxidation of 1 glucose molecule

Interpret Data

reading the ATP balance sheet (Read the table row by row, not column by column. Each row is one stage. The “Consumption” column shows only glycolysis spends ATP — 2 molecules. The next three columns show how the stage produced its ATP: directly by substrate-level phosphorylation, or indirectly through the reduced coenzymes it handed to the ETS. Notice the multiplication: every \(NADH+H^+\) is counted as 3 ATP and every \(FADH_2\) as 2 ATP. Krebs cycle turns twice per glucose, which is why 3 \(NADH+H^+\) per turn appear as 6 in the table. The gross total is 40 ATP, and after subtracting the 2 ATP consumed, the net benefit is 38 ATP.)

Curiosity Corner

Q2. Yeast can survive on glucose without any oxygen — so why does the same yeast cell get far more energy out of the same glucose molecule when oxygen is available?

A. Without oxygen the cell can only complete glycolysis, and its net gain is 2 ATP; the reduced coenzyme is spent simply to regenerate \(NAD^+\), and most of the energy of glucose stays locked up in ethanol. With oxygen, the pyruvate is oxidised completely through the connecting link reaction and Krebs cycle, and the reduced coenzymes are cashed in at the electron transport chain — bringing the net yield to 38 ATP per glucose molecule.

Numerical 13.2

Using the values in Table 13.3, calculate the net ATP gain when 4 molecules of glucose are completely oxidised aerobically, and state how many molecules of \(CO_2\) are released in the process.

Summary of aerobic respiration

The three sources of respiratory substrate and the three compartments in which the stages occur are brought together in Fig. 13.4.

Diagram

Fig. 13.4 – Summary of aerobic respiration: cytoplasm showing glucose → glycolysis → pyruvic acid, with amino acids and fatty acids also feeding in; pyruvic acid entering the mitochondrion → acetyl CoA → Krebs cycle in the matrix releasing CO2 and forming NADH+H+ and FADH2; electrons passing to the inner membrane, protons moving from matrix to outer chamber and back, O2 + 2H+ → H2O, and ADP + iP → ATP at the ATP synthase

Significance of ETS

The electron transport system (ETS), or terminal oxidation, generates the major amount of energy in the form of ATP molecules — 34 ATP molecules out of a total of 38 ATP molecules are produced through ETS.
It regenerates oxidised coenzymes such as \(NAD^+\) and \(FAD^+\) from their reduced forms (\(NADH+H^+\) and \(FADH_2\)) for recycling.
It also produces water molecules.
It releases energy in a stepwise manner to prevent damage to cells.

Real-World Application

Newborn babies and hibernating mammals carry deposits of brown adipose tissue, which is packed with mitochondria. In these mitochondria the proton gradient is deliberately allowed to leak back across the inner membrane without passing through the ATP synthase. The energy of the gradient is then released as heat instead of being stored as ATP — which is how a newborn, who cannot shiver effectively, keeps warm. It is a striking demonstration that the proton gradient, and not any chemical intermediate, is what links oxidation to ATP synthesis.

Try Yourself

Aerobic respiration can be demonstrated by two simple experiments.

A. A pinch of dry baker’s yeast suspended in water, or a few ml of the yeast suspension used in a bakery, is added to about 10 ml of 10 percent glucose solution in a test tube (Tube A). The surface of the liquid is carefully covered with oil to prevent contact with air. The test tube is closed tightly with a rubber stopper. One end of a short bent glass tube is inserted through it to reach the air inside the tube. The other end of the glass tube is connected by a polyethylene or rubber tubing to another bent glass tube fitted into a stopper. The open end of this delivery tube is dipped into lime water contained in a test tube (Tube B). Stoppers of both tubes are fitted tightly to prevent leakage of gases. The first test tube is placed in warm water (\(37^0C\)–\(38^0C\)) in a beaker. Lime water gradually turns milky, indicating the evolution of carbon dioxide from the yeast preparation. The level of the lime water in the delivery tube does not rise, showing that there is no decline in the volume of gas in test tube A and consequently no utilisation of oxygen by yeast. The preparation is stored for a day or two. When you open the stopper of tube A, you will notice a smell of alcohol, indicating the formation of ethanol. From this activity it may be inferred that yeast respires anaerobically to ferment glucose to ethanol and carbon dioxide.

B. Seed coats of a few germinating seeds (peas, beans or gram) are removed. The dehulled seeds are then put in a test tube filled with mercury. After closing the test tube with the thumb, it is vertically inverted in a trough of mercury and the thumb is carefully removed. Being lighter than mercury, the seeds rise to the closed upper end of the test tube. No gas is seen at first in the test tube. As germination proceeds, a gas begins to collect at the top of the mercury in the test tube. On introducing a pellet of potassium hydroxide into the tube, it rises to the top and absorbs the gas. The mercury again fills the tube. The potassium hydroxide reacts with carbon dioxide gas to produce potassium carbonate and water. The gas therefore disappears. Evidently, germinating seeds produce carbon dioxide by anaerobic respiration in the absence of oxygen in the mercury column.

Try Yourself

Do plants breathe like animals? If yes, how and why?

For examinations, the parts of this section most often converted into questions are the site of each stage, the number of \(CO_2\) molecules released per glucose, the single substrate-level phosphorylation step of Krebs cycle, the identity of the terminal electron acceptor, and assertion–reason items on why \(FADH_2\) yields less ATP than \(NADH+H^+\).

13.4 Utility of Stepwise Oxidation

You have noted that both anaerobic and aerobic respiration are conducted in many steps. Counting glycolysis, the connecting link reaction, Krebs cycle and the electron transport chain together, a single glucose molecule passes through roughly thirty enzyme-controlled reactions before it is fully oxidised.

You may wonder what could be the utility of a metabolic pathway with so many steps. A chemist could oxidise the same glucose in one stroke by burning it, and the overall equation would be identical. Such stepwise metabolism, however, serves several purposes.

Why the cell prefers many small steps

A stepwise release of the chemical bond energy facilitates the utilisation of a relatively higher proportion of that energy in ATP synthesis. If all the energy of glucose were released at once, most of it would appear as a burst of heat that the cell could neither store nor use — and which would in fact damage its proteins and membranes. Released in small packets, each packet is just about the right size to be trapped in a bond of ATP.
Activities of the enzymes for the different steps may be enhanced or inhibited by specific compounds. This provides a means of controlling the rate of the pathway and the energy output according to the need of the cell. A one-step reaction offers no such handle; a thirty-step pathway offers thirty possible points of control, and you have already seen this in the regulation of hexokinase, PFK-1 and pyruvate kinase.
The same pathway may be utilised for forming intermediates used in the synthesis of other biomolecules like amino acids. Every intermediate that exists even briefly is a molecule the cell can withdraw and build with — which is the very reason Krebs cycle qualifies as an amphibolic pathway.

The contrast between a single-step release of energy and a stepwise release is shown in Fig. 13.6.

Diagram

Fig. 13.6 – Comparison of energy release: on the left, glucose burning in one step with a single steep drop in energy released entirely as heat; on the right, the same total energy drop shown as a staircase of small steps, with several steps labelled as coupled to ATP synthesis.

Do Not Mix Up

Respiration vs Combustion (Both oxidise glucose to \(CO_2\) and \(H_2O\) and both release the same total energy. Combustion is uncontrolled, happens in one step, needs a high temperature, involves no enzymes and releases all the energy as heat and light. Respiration is enzyme-controlled, happens in many small steps at body temperature, and traps a large share of the energy in ATP.)

Curiosity Corner

Q4. Why does a cell break glucose down through roughly thirty small enzyme-controlled steps instead of oxidising it in one quick burst, the way a flame does?

A. For three reasons. Small packets of energy can actually be captured in ATP, whereas one large burst would be lost as heat and would damage the cell. Many steps mean many enzymes, and every enzyme is a point at which the pathway can be speeded up or slowed down to match the cell’s needs. And each intermediate formed along the way can be drawn off as a building block for other biomolecules.

Hydrogen, not oxygen, is the primary concern

Side Note

Removal of hydrogen from respiratory materials is the primary process in respiration. The fact that during respiration oxygen is taken in and carbon dioxide is given out may give a false impression that respiratory materials directly unite with oxygen. It must be remembered that oxygen does not play such a primary role in the process of respiration. The primary process in respiration consists of the removal of hydrogen from the respiratory materials. The reactions in which hydrogen is removed are catalysed by enzymes called dehydrogenases, as free hydrogen cannot exist in the cell. As soon as it is removed from the respiratory material it is picked up by substances known as acceptors. In aerobic respiration this hydrogen is ultimately handed over to oxygen. These two combine with each other and form water.

This explains the design of the whole pathway. The acceptors are the coenzymes \(NAD^+\) and \(FAD^+\); oxygen simply waits at the end of the chain as the final acceptor. It also explains why an anaerobic cell can still run glycolysis: it needs an acceptor for hydrogen, not oxygen in particular, and pyruvic acid or acetaldehyde will serve.

Curiosity Corner

Q3. If oxygen is so essential to us, why do biologists say that the primary event in respiration is the removal of hydrogen rather than the addition of oxygen?

A. Because the respiratory substrate never combines with oxygen directly. Dehydrogenase enzymes strip hydrogen from the substrate and pass it to acceptors such as \(NAD^+\) and \(FAD^+\). Oxygen enters only at the very last step of the electron transport chain, where it accepts the electrons and combines with protons to form water. Oxygen is indispensable, but it is the final acceptor rather than the starting attacker.

Respiration and photosynthesis compared

Comparison of the overall equations of photosynthesis and respiration shows that, to some extent, the two processes are the reverse of each other. Photosynthesis involves the reduction of \(CO_2\) to glucose, and respiration involves the oxidation of glucose.

Compare Table

The word “to some extent” is important. The two are not simply the same reaction run backwards — they use different enzymes, different coenzymes and different organelles.

Respiratory Quotient

The ratio of the volume of \(CO_2\) released to the volume of \(O_2\) consumed in respiration is called the respiratory quotient (RQ) or respiratory ratio. It depends on the type of respiratory substrate.

\[RQ = \frac{\text{Volume of } CO_2 \text{ released}}{\text{Volume of } O_2 \text{ consumed}}\]

Both volumes are measured under the same conditions of temperature and pressure, so RQ has no unit — it is a pure ratio.

Definition

Respiratory quotient (RQ) is the ratio of the volume of carbon dioxide released to the volume of oxygen consumed during respiration, measured under the same conditions; its value depends on the respiratory substrate being oxidised.

When carbohydrates are used as the respiratory substrate and are completely oxidised, the RQ is 1, because the volume of \(CO_2\) released is equal to the volume of \(O_2\) consumed, as shown in the equation.

\[C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O\]

Six volumes of oxygen are taken in and six volumes of carbon dioxide come out, so \(RQ = 6/6 = 1\).

When fats or proteins are used as a substrate, the RQ is less than 1, as the volume of \(CO_2\) released is always less than the volume of \(O_2\) consumed. The reason lies in the composition of these substrates: fats in particular are poor in oxygen and rich in hydrogen, so a great deal of atmospheric oxygen must be spent on oxidising that hydrogen to water rather than on producing carbon dioxide. Mostly for fats, RQ is about 0.7, and for proteins it is about 0.9.

In the case of anaerobic respiration, RQ is always infinity, as \(CO_2\) is evolved without taking \(O_2\).

Compare Table

Memory Aid

the order of RQ values (Carbohydrate > Protein > Fat, that is 1 > 0.9 > 0.7 — the more oxygen a substrate already contains, the higher its RQ.)

Real-World Application

Seeds differ in what they store, and their RQ shows it. A germinating pea or wheat grain, living on stored starch, respires with an RQ close to 1, while a germinating castor or groundnut seed, which lives on stored fat, shows an RQ well below 1. The same principle is used in hospitals, where the ratio of carbon dioxide produced to oxygen consumed by a patient is measured to judge whether the body is burning mainly carbohydrate or mainly fat, and the patient’s diet is adjusted accordingly.

Numerical 13.3

In a respirometer experiment, a germinating seed released 18 mL of \(CO_2\) while consuming 25 mL of \(O_2\) under identical conditions of temperature and pressure. Calculate its RQ and state which type of respiratory substrate the seed is most likely using.

Try Yourself

Calculate the RQ for different respiratory substrates using the appropriate formula.

Significance of Respiration

Respiration provides energy for the synthesis of biomolecules.
It is also a source of energy for cell division, growth, repair and replacement of worn-out parts, movements, locomotion and so on.
Various intermediates of Krebs cycle are used as building blocks for the synthesis of other complex compounds.
Coupled with photosynthesis, it helps to maintain the balance between \(CO_2\) and \(O_2\) in the atmosphere.
Anaerobic respiration (fermentation) is used in various industries such as dairies, bakeries, distilleries, leather industries, paper industries and others. It is used in the commercial production of alcohol, organic acids, vitamins, antibiotics and so on.

RQ is one of the few quantitative items in this chapter, so it appears regularly as a short numerical or as a match-the-following item pairing substrates with their RQ values. The utility of stepwise oxidation is usually asked as a reasoning question or an assertion–reason item.

13.5 Summary

Respiration is a catabolic process in which a complex organic substrate is oxidised to simple components in order to generate biological energy. Cellular respiration occurs in two ways — anaerobic and aerobic.
The energy released is stored in ATP. The formation of ATP is called phosphorylation, and it occurs in three ways — photophosphorylation (studied in photosynthesis), substrate-level phosphorylation and oxidative phosphorylation.
Substrate-level phosphorylation is the direct phosphorylation of ADP by transfer of a phosphate group from a suitable substrate. It occurs in the cytoplasm and in the mitochondrial matrix.
Oxidative phosphorylation is the phosphorylation of ADP at the cost of energy released during oxidation of substrates such as \(NADH+H^+\) and \(FADH_2\). It occurs only on the inner mitochondrial membrane.
Whenever energy is needed for a metabolic process, ATP is hydrolysed and the energy released is used for that activity.
Anaerobic respiration, also called fermentation, is cellular respiration that does not involve atmospheric oxygen. It is completed through glycolysis followed by the incomplete conversion of glycolytic products into lactic acid or ethanol.
Glycolysis is the breakdown of one glucose molecule into two molecules of pyruvic acid. It occurs in the cytoplasm and is common to anaerobic as well as aerobic respiration.
Glycolysis has ten steps grouped into two phases. In the preparatory phase (first five steps) glucose is phosphorylated twice at the cost of two ATP to form fructose 1, 6-bisphosphate, which splits into glyceraldehyde 3-phosphate and dihydroxyacetone phosphate — both 3-carbon trioses and isomers. Dihydroxyacetone phosphate is isomerised, giving two molecules of glyceraldehyde-3-phosphate.
In the pay-off phase, both molecules of glyceraldehyde-3-phosphate are converted to 1, 3-bisphosphoglycerate by oxidation and phosphorylation, the phosphate coming from inorganic phosphate and not from ATP. These are then converted to two molecules of pyruvic acid, and the energy released is used to make 4 ATP by substrate-level phosphorylation.
Overall reaction of glycolysis: \[\text{Glucose} + 2\,ATP + 2\,iP + 4\,ADP + 2\,NAD^+ \rightarrow 2\,\text{Pyruvate} + 2\,ADP + 4\,ATP + 2\,NADH+H^+ + 2\,H_2O\]
The rate of glycolysis is tightly controlled by the cell’s ATP requirement, by \(NADH_2\) regeneration, by regulation of enzymes such as hexokinase, PFK-1 and pyruvate kinase, and by hormones such as glucagon, epinephrine and insulin.
In muscles, \(NADH+H^+\) is reoxidised to \(NAD^+\) by donating one proton and two electrons to pyruvic acid, forming lactic acid. Its accumulation after vigorous exercise causes muscle fatigue; during rest the lactic acid is reconverted to pyruvic acid and channeled into the aerobic pathway.
In yeast, pyruvate is decarboxylated to acetaldehyde, which is reduced by \(NADH+H^+\) to ethanol, with carbon dioxide also released. This is alcoholic fermentation. Accumulated ethanol can stop yeast multiplication and kill the cells; in the presence of oxygen yeast respires aerobically.
Aerobic respiration involves molecular oxygen and oxidises glucose completely through glycolysis, production of acetyl CoA (connecting link reaction), Krebs cycle, electron transfer chain reaction and terminal oxidation.
In the connecting link reaction, pyruvic acid undergoes oxidative decarboxylation catalysed by the pyruvate dehydrogenase complex, which requires thiamin (vitamin \(B_1\)). Acetyl CoA (2c), \(CO_2\) and \(NADH+H^+\) are formed. The complex is present in the mitochondria of eukaryotes and the cytosol of prokaryotes.
Krebs cycle (TCA or citric acid cycle) occurs in the mitochondrial matrix. Acetyl CoA condenses with oxaloacetic acid (4c) to form citric acid, which is oxidised step-wise with the release of carbon dioxide, finally regenerating oxaloacetic acid. Acetyl CoA also arises from β-oxidation of fatty acids.
The cycle has four steps of oxidation catalysed by dehydrogenases using \(NAD^+\) or \(FAD^+\), which are reduced to \(NADH+H^+\) and \(FADH_2\). One molecule of GTP (ATP) is generated for every molecule of citric acid oxidised.
Krebs cycle is an amphibolic pathway — besides oxidising acetyl CoA, its intermediates such as acetyl CoA, α-ketoglutarate and oxaloacetate serve as precursors for fatty acids, glutamic acid and aspartic acid.
In the electron transport chain, carriers and enzymes on the inner mitochondrial membrane are arranged as complexes I to V. \(NADH+H^+\) is oxidised by complex I and \(FADH_2\) by complex II; both pass electrons to ubiquinone (CoQ), which becomes ubiquinol. Only electrons move forward; protons are released into the outer chamber.
Ubiquinol is oxidised by complex III (cytochrome \(bc_1\)), which passes electrons to cytochrome C, a small iron-containing mobile carrier. Complex IV (cytochrome C oxidase, containing cytochrome a and \(a_3\)) transfers the electrons to molecular oxygen — this is terminal oxidation — and the reduced oxygen combines with protons to form water.
Protons return from the outer chamber to the inner chamber through the \(F_0\) part of the oxysome (complex V), and this flow is coupled to \(F_1\), which synthesises ATP from ADP and inorganic phosphate. This is oxidative phosphorylation, named chemiosmosis by Peter Mitchell.
Oxidation of one \(NADH+H^+\) yields 3 ATP and of one \(FADH_2\) yields 2 ATP, though the actual number depends on physiological conditions and the respiratory substrate. Complete aerobic oxidation of one glucose molecule gives a net benefit of 38 ATP.
Significance of ETS: it produces 34 of the 38 ATP molecules, regenerates \(NAD^+\) and \(FAD^+\) for recycling, produces water, and releases energy stepwise so that cells are not damaged.
Stepwise oxidation is useful because it allows a higher proportion of the bond energy to be captured as ATP, allows the pathway’s rate and energy output to be controlled through individual enzymes, and provides intermediates for the synthesis of other biomolecules such as amino acids.
The primary process in respiration is the removal of hydrogen from respiratory materials by dehydrogenases, not direct union with oxygen. Free hydrogen cannot exist in the cell, so it is at once picked up by acceptors; in aerobic respiration it is finally handed over to oxygen to form water.
Photosynthesis and respiration are to some extent the reverse of each other — photosynthesis reduces \(CO_2\) to glucose, while respiration oxidises glucose.
Respiratory quotient (RQ) is the ratio of the volume of \(CO_2\) released to the volume of \(O_2\) consumed, and it depends on the respiratory substrate. \[RQ = \frac{\text{Volume of } CO_2 \text{ released}}{\text{Volume of } O_2 \text{ consumed}}\] It is 1 for completely oxidised carbohydrates, about 0.9 for proteins, about 0.7 for fats, and infinity for anaerobic respiration, where \(CO_2\) is evolved without \(O_2\) being taken in.
Significance of respiration: it provides energy for the synthesis of biomolecules; supplies energy for cell division, growth, repair and replacement of worn-out parts, movement and locomotion; supplies Krebs cycle intermediates as building blocks for complex compounds; helps, along with photosynthesis, to maintain the atmospheric balance of \(CO_2\) and \(O_2\); and, as fermentation, is used in dairies, bakeries, distilleries, leather and paper industries and in the commercial production of alcohol, organic acids, vitamins and antibiotics.

The whole chapter can be held together as one sequence, shown in FC 13.6.

Flowchart

FC 13.6 – Overall pathway of respiration: Glucose → glycolysis in cytoplasm → 2 Pyruvic acid + 2 ATP (net) + 2 NADH+H+; if oxygen absent → lactic acid (muscle) or ethanol + CO2 (yeast), NAD+ regenerated, no further ATP; if oxygen present → pyruvic acid enters mitochondrion → connecting link reaction (pyruvate dehydrogenase complex) → Acetyl CoA + CO2 + NADH+H+ → Krebs cycle in matrix → CO2 + NADH+H+ + FADH2 + GTP(ATP) → electron transport chain on inner membrane (complex I/II → CoQ → complex III → cytochrome C → complex IV) → terminal oxidation, O2 + protons → H2O; protons return through F0–F1 (complex V) → oxidative phosphorylation → ATP; net yield 38 ATP per glucose

13.6 Points to Ponder

Glycolysis occurs in the cytoplasm of almost every living cell — bacterial, plant and animal — and follows the same ten steps everywhere. What might this universality tell us about the age of the pathway and about the ancestry of living organisms?
Some reactions of glycolysis can run in either direction, while a few can run only one way. Why would a cell find it useful to place its control points at the irreversible steps rather than the reversible ones?
Magnesium and zinc ions appear beside several glycolytic enzymes, yet neither becomes part of the product. What role could such ions be playing, and what would happen to the pathway if the diet were deficient in them?
Fermentation gives a cell only a small fraction of the energy available in glucose, and yet it has been retained through the whole of evolution. Under what conditions would a wasteful pathway still be worth keeping?
Yeast can be killed by the very ethanol it produces. What does this suggest about the limits of any metabolic strategy that leaves a partly oxidised waste product behind?
Oxygen never touches the respiratory substrate directly; it waits at the end of the electron transport chain. Why might it be safer for a cell to hand electrons to oxygen only after passing them through a long series of carriers?
Carbon monoxide and cyanide act on the cytochromes of complex IV. If only the last step of a thirty-step pathway is blocked, why should the effect on the organism be so rapid and so severe?
Krebs cycle both breaks molecules down and supplies material for building them up. If a cell were to withdraw its intermediates too freely for synthesis, what would happen to the cycle itself, and how might a cell guard against this?
The overall equations of photosynthesis and respiration look like mirror images of each other. In what important ways are the two processes not simply the same reaction run backwards?
Do plants breathe in the way animals do? If they do, how is it managed without lungs, gills or blood, and why is it needed at all in a plant that also photosynthesises?
The respiratory quotient of a germinating fatty seed differs from that of a starchy one. If you were given only the RQ value of an unknown tissue, how far could you go in working out what it is living on — and what would you still not be able to say?
A cell cannot store ATP in the way it stores starch or fat. What does this constraint imply about how tightly the rate of respiration must be matched to the rate at which energy is spent?

13.7 Check Your Concepts

Why does the cell spend two molecules of ATP in the preparatory phase of glycolysis when the whole purpose of the pathway is to produce ATP?
Dihydroxyacetone phosphate is not carried directly into the next reaction of glycolysis. Explain what happens to it and why this conversion is necessary for the energy yield of the pathway.
In the pay-off phase, glyceraldehyde-3-phosphate is phosphorylated using inorganic phosphate and not ATP. Why is this detail important when calculating the net ATP gain of glycolysis?
What would happen to glycolysis in a muscle cell if the conversion of pyruvic acid to lactic acid were completely blocked during vigorous exercise? Explain your reasoning in terms of coenzyme recycling.
A yeast culture kept in a sealed vessel stops multiplying after some time even though plenty of glucose remains. Explain why, and predict what would happen if the vessel were opened and aerated.
Arrange the following intermediates of glycolysis in the correct sequence: 2-phosphoglycerate, fructose-6-phosphate, pyruvic acid, glucose-6-phosphate, 1,3-bisphosphoglycerate, phosphoenolpyruvate, fructose-1,6-diphosphate, 3-phosphoglycerate, glyceraldehyde-3-phosphate.
Why is the conversion of pyruvic acid to acetyl CoA called the ‘connecting link’ reaction? Name the enzyme complex involved and state exactly where it is located in a eukaryotic cell and in a prokaryotic cell.
A person’s diet is severely deficient in thiamin. Explain, step by step, why this leads to accumulation of pyruvic acid and lactic acid in the body.
Krebs cycle would stop if oxaloacetic acid were not regenerated at the end of each turn. Explain why, and state where in the cycle oxaloacetate is reformed.
Draw an outline of Krebs cycle and label the following: the condensation step, the two steps at which \(CO_2\) is released, the single step of substrate-level phosphorylation, and the one step at which \(FADH_2\) is formed.
Explain how α-ketoglutarate and oxaloacetate can leave Krebs cycle without the cycle coming to a halt, and name the biomolecules formed from each.
During electron transport, both electrons and protons are released from the reduced coenzymes, but the two take different routes. Describe these routes and explain how this separation makes ATP synthesis possible.
Distinguish clearly between terminal oxidation and oxidative phosphorylation, naming the complex at which each occurs.
What would be the immediate effect on the proton gradient and on ATP synthesis if the inner mitochondrial membrane suddenly became freely permeable to protons? Give reasons.
Cytochrome C is described as a mobile carrier. Explain what this means, where it is located, and why the electron transport chain needs such a carrier at all.
Trace the path of a single molecule of glucose from the cytoplasm to the point at which its last hydrogen atom becomes part of a water molecule, naming the compartment in which each stage occurs.
State the four points of significance of the electron transport system, and explain why the stepwise release of energy prevents damage to the cell.
Respiration and combustion oxidise glucose to the same end products and release the same total energy, yet only one of them is useful to a cell. Explain the difference in terms of control, temperature and energy capture.
“Oxygen is essential to aerobic respiration, but it is not the primary participant.” Justify this statement with reference to dehydrogenases and hydrogen acceptors.
Two germinating seeds are placed in separate respirometers. One shows an RQ close to 1 and the other close to 0.7. What can you conclude about the food reserve stored in each seed, and why is the second value lower?
Explain why the RQ of anaerobic respiration is taken as infinity, using alcoholic fermentation as an example.
Explain why the rate of glycolysis must be regulated, and name the enzymes and hormones through which this control is exercised.
Why can a cell not stockpile ATP for later use in the way it stockpiles starch or fat? Relate your answer to the need for continuous respiration.
When and where does anaerobic respiration occur in man and yeast?
Why is less energy produced during anaerobic respiration than in aerobic respiration?
Which is the site of ETS in mitochondrial respiration?
Which is the terminal electron acceptor in aerobic respiration?
What is RQ? What is its value for fats?
What are respiratory substrates? Name the most common respiratory substrate.
Write explanatory notes on:
1. Glycolysis
2. Fermentation by yeast
3. Electron transport chain
How are glycolysis, TCA cycle and electron transport chain linked? Explain.
How would you demonstrate that yeast can respire both aerobically and anaerobically?
What is the advantage of stepwise energy release in respiration?
Explain ETS.
Discuss: “The respiratory pathway is an amphibolic pathway.”
Why is Krebs cycle referred to as an amphibolic pathway?
Which of the following steps of aerobic respiration would be omitted when fatty acids are used as respiratory substrate?
1. Glycolysis
2. Krebs cycle
3. Electron transfer chain reaction
4. Terminal oxidation
Compare.
1. Photosynthesis and Respiration.
2. Aerobic and Anaerobic respiration.
Differentiate between.
1. Respiration and combustion
2. Glycolysis and Krebs cycle
3. Aerobic respiration and fermentation
Identify the cycle given below. Correct it, fill in the blanks and write a description of it in your own words.

Diagram

Fig. 13.7 – Incomplete and partly incorrect Krebs cycle for the student to identify and correct: Acetyl-CoA → (citrate synthase, condensation, CoA-SH and H2O) blank (6c) → (aconitase, dehydration, H2O) blank (6c) → (aconitase, labelled “dehydration”) blank (6c) → (isocitrate dehydrogenase) blank (6c) → (isocitrate dehydrogenase, CO2 released) blank (5c) → (α-ketoglutarate dehydrogenase complex, CoA-SH, NAD+ → NADH+H+, CO2 released) blank (4c) → (succinyl-CoA synthetase, GDP(ADP)+Pi → GTP(ATP), CoA-SH) blank (4c) → (succinate dehydrogenase, labelled “oxidative decarboxylation”) blank (4c) → (fumarase, hydration, H2O) blank (4c) → (malate dehydrogenase, dehydrogenation) blank (4c) → back to the condensation step

13.8 Practice With Numericals

Using the balance sheet given in the chapter, calculate the net ATP gain when 6 molecules of glucose are completely oxidised aerobically.
How many molecules of \(CO_2\) are released when 3 molecules of glucose are completely oxidised aerobically? Show how many come from the connecting link reaction and how many from Krebs cycle.
A muscle cell oxidises 8 molecules of glucose anaerobically to lactic acid. Calculate the net number of ATP molecules gained and the number of lactic acid molecules formed.
In a respirometer, a germinating seed consumed 40 mL of \(O_2\) and released 28 mL of \(CO_2\) under identical conditions of temperature and pressure. Calculate the RQ and identify the probable respiratory substrate.
A tissue sample released 15 mL of \(CO_2\) while consuming 15 mL of \(O_2\) under the same conditions. Calculate its RQ and state what this indicates about the substrate being oxidised.
During the complete aerobic oxidation of 2 molecules of glucose, how many molecules of \(NADH+H^+\) and \(FADH_2\) are produced in total? Using the conversion values given in the chapter, calculate how many ATP molecules these reduced coenzymes yield through oxidative phosphorylation.
How many turns of Krebs cycle are completed when 5 molecules of glucose are completely oxidised, and how many GTP (ATP) molecules are generated by substrate-level phosphorylation in the cycle during this process?
A cell needs 228 ATP molecules to complete a particular activity. Using the net yield given in Table 13.3, calculate the minimum number of glucose molecules it must oxidise aerobically.
Suppose the electron transport chain of a cell is completely blocked but glycolysis continues, with pyruvate being converted to lactic acid. Calculate the net ATP gain from 10 molecules of glucose under these conditions, and state how many ATP molecules the cell has lost compared with complete aerobic oxidation.

13.9 Objective Questions

The reactions of the TCA cycle occur in
1. ribosomes
2. grana
3. mitochondria
4. endoplasmic reticulum
In eukaryotes the complete oxidation of a molecule of glucose results in the net gain of
1. 2 molecules of ATP
2. 36 molecules of ATP
3. 4 molecules of ATP
4. 38 molecules of ATP
Which step of Krebs cycle operates substrate-level phosphorylation?
1. α-ketoglutarate → succinyl CoA
2. Succinyl CoA → succinate
3. Succinate → fumarate
4. Fumarate → malate
Acetyl CoA is formed from …………….. and co-enzyme A.
In the prokaryotes …………….. molecules of ATP are formed per molecule of glucose oxidised.
Glycolysis takes place in ……………..
\(F_1\)–\(F_0\) particles participate in the synthesis of ……………..
During glycolysis …………….. molecules of \(NADH+H^+\) are formed.
Assertion (A): Glycolysis occurs in the cells of every living organism, whether oxygen is available or not. Reason (R): Glycolysis takes place in the cytoplasm and does not require molecular oxygen at any step. Choose the correct option
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Assertion (A): Oxidation of one molecule of \(FADH_2\) yields fewer ATP molecules than oxidation of one molecule of \(NADH+H^+\). Reason (R): \(FADH_2\) transfers its electrons to coenzyme Q through complex II and therefore bypasses complex I. Choose the correct option from those given in B-Q1.
Assertion (A): A mature human erythrocyte cannot carry out oxidative phosphorylation. Reason (R): Oxidative phosphorylation occurs only on the inner mitochondrial membrane. Choose the correct option from those given in B-Q1.
Match the enzyme in Column I with the reaction it catalyses in Column II.

Column I (Enzyme)	Column II (Reaction)
a. Hexokinase	i. Fructose-6-phosphate → Fructose-1,6-diphosphate
b. Aldolase	ii. 2-Phosphoglycerate → 2-Phosphoenolpyruvate
c. Phosphofructokinase	iii. Glucose → Glucose-6-phosphate
d. Enolase	iv. Fructose-1,6-diphosphate → two trioses

Match the complex of the electron transport system in Column I with its description in Column II.

Column I	Column II
a. Complex I	i. Cytochrome C oxidase, containing cytochrome a and \(a_3\)
b. Complex II	ii. NADH dehydrogenase
c. Complex III	iii. Oxysome with \(F_0\) and \(F_1\) parts
d. Complex IV	iv. Cytochrome \(bc_1\) complex
e. Complex V	v. Succinate dehydrogenase