Energy Metabolism: Glycolysis, TCA Cycle, Pentose Phosphate Pathway, Oxidative Phosphorylation, Gluconeogenesis and Fatty Acids

This chapter opens Section XL1.3 of XL1 Biochemistry, one of the five optional sections of the Life Sciences (XL) paper, of which a candidate answers any two. The heading reads: “Glycolysis, TCA cycle, pentose phosphate pathway, oxidative phosphorylation, gluconeogenesis, and fatty acid metabolism; Metabolism of nitrogen-containing compounds: nitrogen fixation, amino acids, and nucleotides. Photosynthesis, photophosphorylation, Calvin cycle. Metabolic disorders (Sugars, amino acids, and lipids).” This chapter takes the first clause: the carbon and energy pathways, each with its control points, its poisons and its energy yield. The next chapter takes nitrogen, photosynthesis and the disorders. The numericals are ATP yields under a stated P/O convention, the cost of gluconeogenesis, NADPH from the pentose phosphate pathway, P/O ratios from oxygen-electrode data and carbon tracing.

1. Glycolysis: ten steps, three control points

Glycolysis runs in the cytosol of every cell and splits one glucose into two pyruvate. The preparatory phase spends 2 ATP: hexokinase makes glucose-6-phosphate, an isomerase makes fructose-6-phosphate, phosphofructokinase-1 (PFK-1) makes fructose-1,6-bisphosphate, and aldolase cleaves it into dihydroxyacetone phosphate and glyceraldehyde-3-phosphate, which triose phosphate isomerase interconverts. The payoff phase runs twice per glucose: glyceraldehyde-3-phosphate dehydrogenase oxidises the aldehyde with NAD⁺ and Pi to 1,3-bisphosphoglycerate; phosphoglycerate kinase and, after a mutase and enolase, pyruvate kinase each make ATP by substrate-level phosphorylation. Net: glucose + 2 NAD⁺ + 2 ADP + 2 Pi → 2 pyruvate + 2 NADH + 2 ATP.

The irreversible steps are the control points
EnzymeActivated byInhibited by, and notes
Hexokinase (most tissues)Low Km for glucoseGlucose-6-phosphate, its product
Glucokinase (liver, pancreatic β cells)High Km (about 10 mM), so it tracks blood glucoseNot inhibited by glucose-6-phosphate; acts as the glucose sensor
PFK-1 (committed step)AMP, ADP, fructose-2,6-bisphosphateATP, citrate, H⁺; the main valve of the pathway
Pyruvate kinaseFructose-1,6-bisphosphate (feed-forward)ATP, alanine; the liver isozyme is inactivated by protein kinase A phosphorylation

Glycolysis needs a steady supply of NAD⁺. With oxygen, cytosolic NADH is reoxidised by the mitochondria through a shuttle. Without it, lactate dehydrogenase reduces pyruvate to lactate (sprinting muscle, and red cells, which have no mitochondria), and yeast decarboxylates pyruvate to acetaldehyde and reduces it to ethanol. Either way the net gain is only 2 ATP per glucose.

⚠️ Poisons and bypasses that change the yield
Arsenate replaces Pi at glyceraldehyde-3-phosphate dehydrogenase; the 1-arseno-3-phosphoglycerate formed hydrolyses spontaneously, so the phosphoglycerate kinase ATP is lost and the net yield falls to zero although glycolysis keeps running. Fluoride inhibits enolase (hence fluoride tubes for blood glucose samples), and iodoacetate inhibits glyceraldehyde-3-phosphate dehydrogenase. Red cells divert some 1,3-bisphosphoglycerate to 2,3-bisphosphoglycerate, the haemoglobin effector, and forgo that ATP.

2. Pyruvate dehydrogenase and the TCA cycle

The pyruvate dehydrogenase (PDH) complex in the mitochondrial matrix converts pyruvate + CoA + NAD⁺ to acetyl-CoA + CO₂ + NADH. The reaction is irreversible, which is why acetyl-CoA can never be turned back into glucose. It uses three enzymes and five cofactors: E1 (thiamine pyrophosphate, TPP), E2 (lipoamide, CoA) and E3 (FAD, NAD⁺); α-ketoglutarate dehydrogenase has the same design. PDH is switched off by phosphorylation by PDH kinase, which its products acetyl-CoA and NADH and also ATP stimulate; PDH phosphatase, stimulated by Ca²⁺ and, in adipose tissue, by insulin, switches it back on. Thiamine deficiency (beriberi; Wernicke's encephalopathy in alcoholism) slows PDH and α-ketoglutarate dehydrogenase, so pyruvate and lactate rise; arsenite poisons the same two complexes by binding lipoamide.

Citrate synthase joins acetyl-CoA (2 C) to oxaloacetate (4 C) to give citrate (6 C); aconitase gives isocitrate; isocitrate dehydrogenase and α-ketoglutarate dehydrogenase each release CO₂ and make NADH; succinyl-CoA synthetase makes GTP (ATP in some tissues) by substrate-level phosphorylation; succinate dehydrogenase, the only membrane-bound enzyme of the cycle and also complex II, makes FADH₂; fumarase adds water, and malate dehydrogenase regenerates oxaloacetate with a third NADH. Per acetyl-CoA: 3 NADH, 1 FADH₂, 1 GTP and 2 CO₂. No O₂ is used directly, yet the cycle stops without O₂, because only the respiratory chain regenerates NAD⁺ and FAD.

  • Control: isocitrate dehydrogenase (activated by ADP and Ca²⁺, inhibited by ATP and NADH) and α-ketoglutarate dehydrogenase (inhibited by succinyl-CoA and NADH, activated by Ca²⁺); citrate synthase is limited by the supply of oxaloacetate. Ca²⁺ ties the cycle to muscle contraction.
  • Anaplerosis: pyruvate carboxylase (biotin, activated by acetyl-CoA) refills oxaloacetate when intermediates are drawn off for biosynthesis or gluconeogenesis.
  • Inhibitors: malonate competes with succinate at succinate dehydrogenase; fluoroacetate is converted to fluorocitrate, which blocks aconitase.
🧠 Carbon tracing
Glucose C-1 and C-6 end up in the methyl carbon of pyruvate, C-2 and C-5 in its carbonyl carbon, and C-3 and C-4 in its carboxyl carbon, which PDH releases as CO₂. In the first turn of the cycle the two CO₂ come from oxaloacetate carbons, not from the acetyl group just added; acetyl carbons leave only in later turns.

3. Oxidative phosphorylation and the ATP balance sheet

The respiratory chain in the inner mitochondrial membrane takes electrons from NADH at complex I and from succinate at complex II, passes them through ubiquinone to complex III (cytochrome bc₁), cytochrome c and complex IV (cytochrome c oxidase), which reduces O₂ to water. Complexes I, III and IV move about 4, 4 and 2 protons out of the matrix per electron pair; complex II moves none. So NADH drives about 10 H⁺ out and succinate about 6. ATP synthase (F₀F₁) lets them back: protons turn the F₀ c-ring, and the rotating γ subunit drives the three catalytic β subunits through loose, tight and open states (the binding-change mechanism). Counting about 4 H⁺ per ATP made and exported (including Pi and ADP/ATP transport) gives P/O ratios of 2.5 for NADH and 1.5 for FADH₂; older texts use whole numbers, 3 and 2.

Respiratory control: electrons flow only as fast as ADP is supplied, because a steep proton gradient opposes further pumping. Uncouplers break this link: 2,4-dinitrophenol ferries protons across the membrane, and thermogenin (UCP1) in brown adipose tissue does so deliberately to make heat in newborns and hibernating mammals. O₂ uptake then rises while ATP synthesis falls.

Inhibitors and uncouplers
AgentSiteEffect
Rotenone, amytalComplex INADH not oxidised; succinate still oxidised
Antimycin AComplex IIIBoth NADH and succinate oxidation blocked
Cyanide, azide, COComplex IVAll electron flow to O₂ stops
OligomycinF₀ of ATP synthaseATP synthesis stops and, by respiratory control, so does O₂ uptake
AtractylosideADP/ATP translocaseADP cannot enter the matrix
2,4-Dinitrophenol, UCP1Inner membraneUncoupling: O₂ uptake continues, little ATP
ATP per glucose with P/O = 2.5 (NADH) and 1.5 (FADH₂)
StageProductsATP
Glycolysis2 ATP + 2 cytosolic NADH2 + (3 or 5)
Pyruvate dehydrogenase (×2)2 NADH5
TCA cycle (×2)6 NADH, 2 FADH₂, 2 GTP15 + 3 + 2
TotalDepends on the shuttle30 or 32

Cytosolic NADH cannot cross the inner membrane. The malate–aspartate shuttle (liver, heart) hands its electrons to matrix NAD⁺ (2.5 ATP); the glycerol-3-phosphate shuttle (skeletal muscle, brain) hands them to ubiquinone through FAD (1.5 ATP). Hence 30 or 32 ATP per glucose; with P/O values of 3 and 2 the same arithmetic gives 36 or 38. A numerical answer is fixed only by the convention stated in the question.

4. The pentose phosphate pathway

The pentose phosphate pathway, also cytosolic, makes NADPH for reductive biosynthesis and for keeping glutathione reduced, and ribose-5-phosphate for nucleotides. Its oxidative phase is irreversible: glucose-6-phosphate dehydrogenase (G6PD, the rate-limiting enzyme, controlled by the NADP⁺/NADPH ratio) gives 6-phosphoglucono-δ-lactone and NADPH; a lactonase opens the ring, and 6-phosphogluconate dehydrogenase releases C-1 as CO₂, making a second NADPH and ribulose-5-phosphate. The non-oxidative phase is reversible: an isomerase and an epimerase make ribose-5-phosphate and xylulose-5-phosphate, and transketolase (TPP, moves two-carbon units) and transaldolase (moves three-carbon units) reshuffle pentoses into fructose-6-phosphate and glyceraldehyde-3-phosphate.

  • Ribose needed more than NADPH (dividing cells): the non-oxidative phase runs backwards from glycolytic intermediates.
  • Both needed: the oxidative phase supplies ribose-5-phosphate and NADPH together.
  • NADPH only (adipose tissue, adrenal cortex, red cells): pentoses are recycled to glucose-6-phosphate, so in the limit one glucose-6-phosphate is oxidised completely to 6 CO₂ with 12 NADPH.
🎯 Why G6PD deficiency harms red cells
Red cells have no mitochondria and depend on G6PD for NADPH, which keeps glutathione reduced so that it can destroy H₂O₂. In G6PD deficiency (X-linked, common where falciparum malaria has been endemic) oxidant drugs such as primaquine, some sulfonamides, or fava beans cause haemolysis with Heinz bodies of precipitated haemoglobin. Erythrocyte transketolase activity is used to assess thiamine status.

5. Gluconeogenesis and reciprocal control

The liver (and, in prolonged fasting, the kidney cortex) makes glucose from lactate, glycerol, propionate and glucogenic amino acids. Seven glycolytic steps run in reverse; the three irreversible ones are bypassed by four enzymes:

  • Pyruvate carboxylase (mitochondrial, biotin, activated by acetyl-CoA): pyruvate + CO₂ + ATP → oxaloacetate, which leaves the mitochondrion as malate or aspartate.
  • PEP carboxykinase: oxaloacetate + GTP → phosphoenolpyruvate + CO₂.
  • Fructose-1,6-bisphosphatase-1: inhibited by AMP and fructose-2,6-bisphosphate.
  • Glucose-6-phosphatase: in the endoplasmic reticulum of liver and kidney; absent from muscle, so muscle glycogen cannot supply blood glucose.

Cost: 2 pyruvate + 4 ATP + 2 GTP + 2 NADH → glucose, six high-energy phosphates against the two ATP that glycolysis yields; the difference is the price of making each direction irreversible. The Cori cycle returns lactate from muscle and red cells to the liver; the glucose–alanine cycle brings back carbon and amino nitrogen together. Even-chain fatty acids give no net glucose in animals: PDH is irreversible, and when two acetyl carbons enter the TCA cycle two carbons leave as CO₂, so no oxaloacetate is gained. Glycerol and the propionyl-CoA from odd-chain fatty acids are glucogenic.

Glycolysis and gluconeogenesis must not run together, or ATP is wasted in a futile cycle. The liver's switch is fructose-2,6-bisphosphate, made and destroyed by the bifunctional enzyme PFK-2/FBPase-2. In the fed state insulin keeps the enzyme dephosphorylated, fructose-2,6-bisphosphate is high, PFK-1 is activated and FBPase-1 inhibited: glycolysis runs. Glucagon raises cAMP; protein kinase A phosphorylates the bifunctional enzyme, its phosphatase activity prevails, fructose-2,6-bisphosphate falls and gluconeogenesis runs, while PKA also inactivates liver pyruvate kinase. Acetyl-CoA from fatty acid oxidation in fasting activates pyruvate carboxylase and, through PDH kinase, turns PDH off, steering pyruvate towards glucose.

6. Fatty acid oxidation and synthesis

Activation and entry. Acyl-CoA synthetase forms fatty acyl-CoA, hydrolysing ATP to AMP + PPi (two ATP equivalents). Long-chain acyl groups cross the inner membrane as acylcarnitine through carnitine palmitoyltransferase I (CPT-I), a translocase and CPT-II. CPT-I is inhibited by malonyl-CoA, the first intermediate of fatty acid synthesis, so a cell does not burn the fatty acids it is making. β-Oxidation then repeats four matrix steps: acyl-CoA dehydrogenase (FAD → FADH₂), enoyl-CoA hydratase, 3-hydroxyacyl-CoA dehydrogenase (NAD⁺ → NADH) and thiolase, which cleaves off acetyl-CoA. A saturated acid with n carbons (n even) needs n/2 − 1 rounds and gives n/2 acetyl-CoA.

ℹ️ Worked example: palmitate (16:0)
With P/O = 2.5 and 1.5 and 10 ATP per acetyl-CoA: 7 rounds give 7 FADH₂ (10.5 ATP) and 7 NADH (17.5 ATP); 8 acetyl-CoA give 80 ATP; total 108, minus 2 for activation, is 106 ATP. With P/O = 3 and 2 (12 ATP per acetyl-CoA) the same count gives 129.
  • Unsaturated: a cis double bond at an odd-numbered carbon (oleate, Δ⁹) reaches the cis-Δ³ position, and enoyl-CoA isomerase moves it to trans-Δ²; one acyl-CoA dehydrogenase step is skipped, so one FADH₂ is lost per such bond. A bond at an even-numbered carbon also needs NADPH-dependent 2,4-dienoyl-CoA reductase.
  • Odd-chain: the last thiolysis leaves propionyl-CoA, which propionyl-CoA carboxylase (biotin) converts to methylmalonyl-CoA and methylmalonyl-CoA mutase (vitamin B₁₂) to succinyl-CoA, a glucogenic product. B₁₂ deficiency causes methylmalonic aciduria.
  • Ketone bodies: in fasting liver mitochondria surplus acetyl-CoA forms acetoacetate and β-hydroxybutyrate through HMG-CoA synthase. Brain and muscle use them, but the liver cannot, lacking the CoA transferase that activates acetoacetate. Uncontrolled insulin deficiency leads to ketoacidosis.

Synthesis is not β-oxidation reversed: it runs in the cytosol, uses NADPH and carries intermediates on the acyl carrier protein (ACP). Acetyl-CoA leaves the mitochondrion as citrate, which ATP-citrate lyase cleaves. Acetyl-CoA carboxylase (biotin) makes malonyl-CoA and is the control point: citrate activates it, palmitoyl-CoA inhibits it, phosphorylation by AMP-activated protein kinase (favoured by glucagon and adrenaline signalling) inactivates it, and insulin activates it. Fatty acid synthase adds two carbons per round (condensation releasing CO₂, reduction, dehydration, reduction) for seven rounds: 8 acetyl-CoA + 7 ATP + 14 NADPH → palmitate. NADPH comes from the pentose phosphate pathway and malic enzyme. Elongation and desaturation occur in the endoplasmic reticulum; animals cannot insert double bonds beyond C-9, so linoleic and α-linolenic acids are essential.

Key takeaways

  • Glycolysis nets 2 ATP and 2 NADH per glucose; PFK-1 is the main valve (AMP and fructose-2,6-bisphosphate on, ATP and citrate off); arsenate cuts the net yield to zero and fluoride blocks enolase.
  • PDH (TPP, lipoamide, CoA, FAD, NAD⁺) is irreversible and switched off by phosphorylation; each acetyl-CoA gives 3 NADH, 1 FADH₂, 1 GTP and 2 CO₂, and pyruvate carboxylase refills oxaloacetate.
  • NADH moves about 10 H⁺ and FADH₂ about 6; with P/O 2.5 and 1.5 a glucose gives 30 or 32 ATP (36 or 38 with 3 and 2), depending on the shuttle. Uncouplers raise O₂ uptake and stop ATP synthesis; oligomycin stops both.
  • The oxidative pentose phosphate pathway gives 2 NADPH and 1 CO₂ (from C-1) per glucose-6-phosphate, or 12 NADPH with full recycling; G6PD deficiency causes oxidant-induced haemolysis.
  • Gluconeogenesis bypasses three steps with four enzymes at 6 high-energy phosphates per glucose; fructose-2,6-bisphosphate switches the liver between glycolysis and gluconeogenesis, and even-chain fatty acids give no net glucose.
  • β-Oxidation gives 1 FADH₂, 1 NADH and 1 acetyl-CoA per round (palmitate 106 ATP at 2.5/1.5); synthesis uses malonyl-CoA, ACP and 14 NADPH per palmitate, and malonyl-CoA blocks CPT-I.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Blood samples for glucose estimation are collected in tubes containing sodium fluoride so that the cells in the tube stop consuming glucose. Which glycolytic enzyme does fluoride inhibit?

    1. Hexokinase, which traps glucose in the cell as glucose-6-phosphate
    2. Glyceraldehyde-3-phosphate dehydrogenase, the NAD⁺-dependent step
    3. Enolase, which dehydrates 2-phosphoglycerate
    4. Pyruvate kinase, the last ATP-forming step of the pathway
    Show answer

    Answer: C — Enolase, which dehydrates 2-phosphoglycerate

    Fluoride inhibits enolase, blocking the dehydration of 2-phosphoglycerate to phosphoenolpyruvate, so glycolysis in the blood cells stops and the glucose value does not fall while the sample waits. Glyceraldehyde-3-phosphate dehydrogenase, the tempting answer, is the classical target of iodoacetate (and the step arsenate subverts), not of fluoride; hexokinase and pyruvate kinase are not its targets either.

  2. Cells exposed to arsenate keep converting glucose to pyruvate at a normal rate, yet the net ATP gained from glycolysis falls to zero. What explains this?

    1. Arsenate inhibits phosphoglycerate kinase directly, so 1,3-bisphosphoglycerate accumulates and is not used
    2. Arsenate replaces phosphate at glyceraldehyde-3-phosphate dehydrogenase, and the product hydrolyses before ATP can be made
    3. Arsenate inhibits pyruvate kinase, so the final substrate-level phosphorylation of the pathway cannot take place
    4. Arsenate inhibits hexokinase, so each glucose enters glycolysis only after an extra ATP has been spent
    Show answer

    Answer: B — Arsenate replaces phosphate at glyceraldehyde-3-phosphate dehydrogenase, and the product hydrolyses before ATP can be made

    Glyceraldehyde-3-phosphate dehydrogenase accepts arsenate in place of Pi and forms 1-arseno-3-phosphoglycerate, which hydrolyses spontaneously to 3-phosphoglycerate. The pathway continues, but the phosphoglycerate kinase step that normally makes 2 ATP per glucose is bypassed, leaving the 2 ATP from pyruvate kinase only to repay the 2 ATP invested: net zero. If phosphoglycerate kinase or pyruvate kinase were blocked, flux to pyruvate would stop, which contradicts the stem.

  3. Glucose labelled with ¹⁴C only at C-4 is given to liver cells with active pyruvate dehydrogenase. After glycolysis and the pyruvate dehydrogenase reaction, where is the label found?

    1. In the methyl carbon of acetyl-CoA, which then enters the TCA cycle
    2. In the carbonyl (thioester) carbon of acetyl-CoA, which enters the TCA cycle
    3. Split equally between the methyl carbon of acetyl-CoA and the released CO₂
    4. In the CO₂ released by pyruvate dehydrogenase, so none enters the cycle
    Show answer

    Answer: D — In the CO₂ released by pyruvate dehydrogenase, so none enters the cycle

    Aldolase splits fructose-1,6-bisphosphate between C-3 and C-4. C-4 becomes the aldehyde carbon (C-1) of glyceraldehyde-3-phosphate, which is oxidised to a carboxyl group and ends as the carboxyl carbon (C-1) of pyruvate. Pyruvate dehydrogenase removes exactly that carbon as CO₂. Glucose C-1 or C-6 would label the methyl carbon of acetyl-CoA, and C-2 or C-5 the carbonyl carbon, which is why those options tempt.

  4. A man with long-standing alcohol misuse and a poor diet has raised blood pyruvate and lactate, which fall after thiamine is given. Which pair of enzymes was limiting?

    1. Pyruvate dehydrogenase and α-ketoglutarate dehydrogenase, which both need thiamine pyrophosphate
    2. Pyruvate carboxylase and acetyl-CoA carboxylase, which both need biotin to fix CO₂
    3. Lactate dehydrogenase and malate dehydrogenase, which both need NAD⁺ for their reactions
    4. Citrate synthase and isocitrate dehydrogenase, which together set the rate of the TCA cycle
    Show answer

    Answer: A — Pyruvate dehydrogenase and α-ketoglutarate dehydrogenase, which both need thiamine pyrophosphate

    Thiamine pyrophosphate is the cofactor of E1 of the pyruvate dehydrogenase complex and of α-ketoglutarate dehydrogenase (and of transketolase). When it is lacking, pyruvate cannot be oxidised in the mitochondrion and is reduced to lactate instead. Biotin carboxylases do not depend on thiamine, and lactate and malate dehydrogenases use NAD⁺, not TPP, so their activity would not respond to thiamine.

  5. Which of the following statements about the pyruvate dehydrogenase complex and the TCA cycle are correct?

    1. Malonate inhibits succinate dehydrogenase by competing with succinate for the active site
    2. Phosphorylation of the pyruvate dehydrogenase complex by its specific kinase activates the complex
    3. Fluoroacetate is toxic because it is converted to fluorocitrate, which inhibits aconitase
    4. In the first turn, both CO₂ molecules released come from the acetyl group that has just entered the cycle
    Show answer

    Answer: A — Malonate inhibits succinate dehydrogenase by competing with succinate for the active site; C — Fluoroacetate is toxic because it is converted to fluorocitrate, which inhibits aconitase

    Malonate, a dicarboxylate analogue of succinate, is the textbook competitive inhibitor of succinate dehydrogenase. Fluoroacetate is condensed (as fluoroacetyl-CoA) with oxaloacetate to fluorocitrate, which blocks aconitase so citrate accumulates. PDH kinase phosphorylates E1 and inactivates the complex; the phosphatase reactivates it. Both CO₂ of the first turn come from oxaloacetate carbons; the acetyl carbons are lost only in later turns.

  6. Isolated mitochondria oxidise succinate rapidly when ADP and Pi are present. Oligomycin is added and O₂ consumption nearly stops; 2,4-dinitrophenol is then added and O₂ consumption becomes rapid again. Which conclusion follows?

    1. Oligomycin blocks complex IV, and dinitrophenol restores O₂ uptake by reactivating cytochrome c oxidase
    2. Dinitrophenol restores ATP synthesis by opening an alternative proton path through the F₁ head of the synthase
    3. Electron flow is coupled to ATP synthesis through the proton gradient, which dinitrophenol dissipates
    4. Succinate oxidation pumps no protons, so oligomycin must have inhibited succinate dehydrogenase itself
    Show answer

    Answer: C — Electron flow is coupled to ATP synthesis through the proton gradient, which dinitrophenol dissipates

    Oligomycin blocks the F₀ proton channel, so protons cannot re-enter; the gradient builds until it halts further pumping, and O₂ uptake falls (respiratory control). Dinitrophenol carries protons back across the membrane, the gradient collapses and electrons flow to O₂ again, but no ATP is made. Oligomycin does not act on complex IV, dinitrophenol restores respiration rather than phosphorylation, and succinate oxidation does pump protons at complexes III and IV.

  7. Which of the following statements about inhibitors and uncouplers of oxidative phosphorylation are correct?

    1. Mitochondria treated with rotenone can still oxidise succinate and make ATP from it
    2. Cyanide blocks the F₀ channel of ATP synthase, leaving electron transport to O₂ intact
    3. Thermogenin (UCP1) lets protons return to the matrix without making ATP, so the energy appears as heat
    4. Antimycin A blocks the oxidation of both NADH and succinate by the respiratory chain
    Show answer

    Answer: A — Mitochondria treated with rotenone can still oxidise succinate and make ATP from it; C — Thermogenin (UCP1) lets protons return to the matrix without making ATP, so the energy appears as heat; D — Antimycin A blocks the oxidation of both NADH and succinate by the respiratory chain

    Rotenone blocks complex I, but succinate feeds electrons in at ubiquinone through complex II, downstream of the block, so its oxidation and phosphorylation continue. Antimycin A blocks complex III, through which electrons from both NADH and succinate must pass. UCP1 is a physiological uncoupler that dissipates the gradient as heat. Cyanide binds the haem a₃ centre of complex IV and stops electron flow to O₂; the F₀ channel is the site of oligomycin.

  8. During prolonged fasting, why can the liver make glucose from glycerol and from propionyl-CoA but not, in net terms, from the acetyl-CoA of even-chain fatty acids?

    1. Pyruvate dehydrogenase is irreversible, and the two carbons entering the cycle are matched by two leaving as CO₂
    2. Acetyl-CoA inhibits pyruvate carboxylase, so no oxaloacetate can be made while fatty acids are oxidised
    3. Fatty acids are oxidised only in muscle, so their acetyl-CoA never reaches the gluconeogenic enzymes of liver
    4. Acetyl-CoA leaves mitochondria only as citrate, and citrate strongly inhibits phosphoenolpyruvate carboxykinase
    Show answer

    Answer: A — Pyruvate dehydrogenase is irreversible, and the two carbons entering the cycle are matched by two leaving as CO₂

    Glycerol enters at dihydroxyacetone phosphate, and propionyl-CoA becomes succinyl-CoA, which adds net carbon to the cycle and can leave as oxaloacetate. Acetyl-CoA cannot be converted back to pyruvate because PDH is irreversible, and for every two carbons it brings into the TCA cycle two leave as CO₂ before oxaloacetate is regenerated, so there is no net gain. (Plants and bacteria escape this with the glyoxylate cycle, which animals lack.) Acetyl-CoA actually activates pyruvate carboxylase, and the fasting liver itself oxidises fatty acids extensively.

  9. Glucagon raises cAMP in hepatocytes. Acting through the bifunctional enzyme PFK-2/FBPase-2, what is the immediate effect on the opposing enzymes of the fructose-6-phosphate ⇌ fructose-1,6-bisphosphate step?

    1. Fructose-2,6-bisphosphate rises, activating PFK-1 and inhibiting fructose-1,6-bisphosphatase, so glycolysis speeds up
    2. Fructose-2,6-bisphosphate falls, so PFK-1 loses an activator and fructose-1,6-bisphosphatase loses an inhibitor
    3. Fructose-2,6-bisphosphate falls, so PFK-1 and fructose-1,6-bisphosphatase both become inactive and the step stops
    4. Fructose-2,6-bisphosphate rises, inhibiting both PFK-1 and fructose-1,6-bisphosphatase so neither direction runs
    Show answer

    Answer: B — Fructose-2,6-bisphosphate falls, so PFK-1 loses an activator and fructose-1,6-bisphosphatase loses an inhibitor

    Protein kinase A phosphorylates the liver bifunctional enzyme, inhibiting its kinase (PFK-2) and activating its phosphatase (FBPase-2), so fructose-2,6-bisphosphate falls. Because it activates PFK-1 and inhibits fructose-1,6-bisphosphatase, its loss slows glycolysis and releases gluconeogenesis, as the fasting liver needs. A rise in fructose-2,6-bisphosphate is the insulin (fed) state, and it never inhibits PFK-1; nor does its fall inactivate the phosphatase.

  10. A man with glucose-6-phosphate dehydrogenase deficiency develops haemolysis after taking primaquine, though his other tissues show no damage. Why are red cells the vulnerable ones?

    1. G6PD deficiency blocks glycolysis in red cells, so they cannot make the ATP that drives their ion pumps
    2. Red cells convert primaquine into a toxic quinone only when their G6PD activity is low
    3. Red cells need G6PD to make 2,3-bisphosphoglycerate, and without it their haemoglobin denatures
    4. Lacking mitochondria, red cells rely on G6PD for the NADPH that keeps glutathione reduced
    Show answer

    Answer: D — Lacking mitochondria, red cells rely on G6PD for the NADPH that keeps glutathione reduced

    NADPH from the oxidative pentose phosphate pathway keeps glutathione reduced through glutathione reductase, and reduced glutathione removes H₂O₂ and protects haemoglobin and membrane thiols. Red cells have no mitochondria and no other substantial NADPH source, so with low G6PD an oxidant drug causes Heinz bodies and haemolysis. Glycolysis does not need G6PD, so red-cell ATP is not cut off, and 2,3-bisphosphoglycerate comes from a glycolytic side branch, not from G6PD.

  11. Which of the following statements about fatty acid synthesis and its coordination with β-oxidation are correct?

    1. Malonyl-CoA inhibits carnitine palmitoyltransferase I, so newly made fatty acids are not carried into mitochondria for oxidation
    2. Fatty acid synthesis takes place in the mitochondrial matrix and uses NADH as its reductant
    3. Acetyl-CoA carboxylase is activated by citrate and inactivated by phosphorylation by AMP-activated protein kinase
    4. Each round of β-oxidation produces one NADPH, one FADH₂ and one acetyl-CoA
    Show answer

    Answer: A — Malonyl-CoA inhibits carnitine palmitoyltransferase I, so newly made fatty acids are not carried into mitochondria for oxidation; C — Acetyl-CoA carboxylase is activated by citrate and inactivated by phosphorylation by AMP-activated protein kinase

    Malonyl-CoA, the committed intermediate of synthesis, inhibits CPT-I, which prevents a futile cycle of making and burning fatty acids. Acetyl-CoA carboxylase is activated by citrate (a signal of plentiful acetyl-CoA and ATP) and switched off when AMP-activated protein kinase phosphorylates it in low-energy states. Synthesis is cytosolic and uses NADPH, and β-oxidation makes NADH, not NADPH, in each round.

  12. The propionyl-CoA left after β-oxidation of an odd-chain fatty acid is converted to succinyl-CoA. Which two vitamin-derived cofactors does this conversion require?

    1. Thiamine pyrophosphate and lipoic acid, as in the decarboxylation of α-keto acids
    2. Biotin for the carboxylase and adenosylcobalamin (vitamin B₁₂) for the mutase
    3. Pyridoxal phosphate for the transfer step and tetrahydrofolate for the methyl group
    4. FAD and NAD⁺ only, as in the two oxidative steps of every β-oxidation round
    Show answer

    Answer: B — Biotin for the carboxylase and adenosylcobalamin (vitamin B₁₂) for the mutase

    Propionyl-CoA carboxylase (biotin) gives D-methylmalonyl-CoA; an epimerase forms the L isomer, and methylmalonyl-CoA mutase, which needs 5′-deoxyadenosylcobalamin, rearranges it to succinyl-CoA. That is why vitamin B₁₂ deficiency raises methylmalonic acid in urine. TPP and lipoate belong to the α-keto acid dehydrogenases, and no folate- or PLP-dependent step is involved.

  13. Under the same P/O convention, complete oxidation of oleate (18:1, cis-Δ⁹) yields 1.5 ATP fewer than that of stearate (18:0). What is the reason?

    1. Oleate gives one acetyl-CoA fewer, because thiolase cannot cleave the unit that carried the double bond
    2. An extra NADPH is consumed by 2,4-dienoyl-CoA reductase in removing the cis double bond
    3. Oleoyl groups enter mitochondria by a separate carnitine carrier that consumes an additional ATP
    4. Isomerase makes the trans-Δ² intermediate, so one acyl-CoA dehydrogenase step and its FADH₂ are skipped
    Show answer

    Answer: D — Isomerase makes the trans-Δ² intermediate, so one acyl-CoA dehydrogenase step and its FADH₂ are skipped

    After three rounds oleoyl-CoA becomes a cis-Δ³-enoyl-CoA, which acyl-CoA dehydrogenase cannot use; enoyl-CoA isomerase converts it to trans-Δ²-enoyl-CoA, the normal substrate of the hydratase. That round therefore makes no FADH₂, and at 1.5 ATP per FADH₂ the yield falls by 1.5 (120 → 118.5). Both acids give 9 acetyl-CoA. The NADPH-dependent dienoyl-CoA reductase is needed for double bonds at even-numbered positions, such as the Δ¹² bond of linoleate, not for oleate's Δ⁹ bond.

  14. Glucose is completely oxidised to CO₂ in a skeletal muscle cell in which cytosolic NADH enters the mitochondria only through the glycerol-3-phosphate shuttle. Take 2.5 ATP per mitochondrial NADH and 1.5 ATP per FADH₂ (and per cytosolic NADH entering by this shuttle), and count GTP as ATP. The number of ATP formed per glucose is ____.

    Numerical answer — type the value.

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    Answer: 30

    Glycolysis: 2 ATP and 2 cytosolic NADH (2 × 1.5 = 3). Pyruvate dehydrogenase: 2 NADH = 5. TCA cycle, two turns: 6 NADH = 15, 2 FADH₂ = 3, 2 GTP = 2. Total 2 + 3 + 5 + 15 + 3 + 2 = 30. The malate–aspartate shuttle would give 32, and P/O values of 3 and 2 with this shuttle would give 36.

  15. Calculate the net number of ATP formed by the complete oxidation of one molecule of stearic acid (18:0) to CO₂ and H₂O. Assume 2.5 ATP per NADH, 1.5 ATP per FADH₂, 10 ATP per acetyl-CoA oxidised in the TCA cycle, and that activation to stearoyl-CoA costs 2 ATP equivalents.

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    Answer: 120

    An 18-carbon acid needs 18/2 − 1 = 8 rounds of β-oxidation and yields 9 acetyl-CoA. Rounds: 8 FADH₂ × 1.5 = 12 and 8 NADH × 2.5 = 20. Acetyl-CoA: 9 × 10 = 90. Gross 122; minus 2 for activation (ATP → AMP + PPi) gives 120. Counting 9 rounds (one per acetyl-CoA) gives 124, and forgetting activation gives 122.

  16. In the liver, two molecules of lactate are converted to one molecule of glucose by gluconeogenesis. Count each ATP or GTP hydrolysed as one high-energy phosphate, and note that the NADH required is supplied by the lactate dehydrogenase step. How many high-energy phosphates are consumed per glucose formed?

    Numerical answer — type the value.

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    Answer: 6

    Per pyruvate: pyruvate carboxylase 1 ATP, PEP carboxykinase 1 GTP, and phosphoglycerate kinase running in reverse 1 ATP, so 3. For two pyruvates, 6 (4 ATP + 2 GTP). Oxidising 2 lactate to 2 pyruvate makes the 2 NADH that glyceraldehyde-3-phosphate dehydrogenase consumes, so no extra reducing power is needed, and the two phosphatases release Pi without using ATP. Glycolysis returns only 2 ATP, so the Cori cycle costs the liver 4 more than the muscle gains.

  17. In adipose tissue that needs only NADPH, the pentose phosphates formed by the oxidative pentose phosphate pathway are recycled by transketolase, transaldolase and gluconeogenic enzymes back to glucose-6-phosphate, and the cycle repeats until the equivalent of one glucose-6-phosphate has been oxidised completely to 6 CO₂. How many NADPH are produced per glucose-6-phosphate oxidised in this way?

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    Answer: 12

    Each pass through the oxidative phase releases one CO₂ (from C-1) and makes 2 NADPH. Six glucose-6-phosphate passing once give 6 CO₂, 12 NADPH and 6 ribulose-5-phosphate (30 carbons), which the non-oxidative phase and gluconeogenic enzymes rebuild into 5 glucose-6-phosphate (30 carbons). Net: one glucose-6-phosphate → 6 CO₂ + 12 NADPH. Answering 2 counts a single pass, not complete oxidation.

  18. In an oxygen-electrode experiment, mitochondria oxidising succinate receive a pulse of 300 nmol ADP. During the burst of rapid respiration that ends when all of it has been phosphorylated, 100 nmol of O₂ are consumed. What is the P/O ratio (ATP formed per oxygen atom reduced), to one decimal place?

    Numerical answer — type the value.

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    Answer: 1.5

    100 nmol O₂ = 200 nmol O atoms, and 300 nmol ADP become 300 nmol ATP. P/O = 300 / 200 = 1.5, the value expected for succinate, whose electrons enter at ubiquinone and bypass complex I. Dividing by moles of O₂ instead of O atoms gives 3.0, the commonest slip.