Which pair of enzymes is unique to the Entner–Doudoroff pathway and distinguishes it from both glycolysis and the pentose-phosphate pathway?

6-Phosphogluconate dehydratase and 2-keto-3-deoxy-6-phosphogluconate aldolase

Glucose-6-phosphate dehydrogenase and 6-phosphogluconate dehydrogenase

Phosphofructokinase and fructose-1,6-bisphosphate aldolase of the hexose-splitting steps

Transketolase and transaldolase of the reversible sugar-rearranging reactions

An *E. coli* mutant lacks isocitrate lyase but is otherwise normal. On which sole carbon source would it fail to grow?

Acetate, because without the bypass every acetyl-CoA carbon is lost as CO₂ and no oxaloacetate accumulates

Glucose, because isocitrate lyase is needed to convert pyruvate into acetyl-CoA

Succinate, because succinate can enter central metabolism only through the glyoxylate pathway

Acetate, because without the bypass every acetyl-CoA carbon is lost as CO₂ and no oxaloacetate accumulates

Glycerol, because glycerol is oxidised to glyoxylate before it can enter the citric acid cycle

A protonophore that makes the cytoplasmic membrane freely permeable to H⁺ is added to an aerobic, flagellated *E. coli* culture growing on succinate. Which combination of effects is expected?

O₂ consumption continues or rises, while oxidative ATP synthesis and flagellar rotation both stop

O₂ consumption stops at once, because electron transport cannot run once protons leak back across the membrane

O₂ consumption continues or rises, while oxidative ATP synthesis and flagellar rotation both stop

ATP synthesis continues normally, because bacterial ATP synthase is driven by electron flow rather than by protons

Flagella keep rotating, because bacterial flagellar motors are powered directly by the hydrolysis of ATP

Which of the following statements about bacterial respiration are correct?

In *E. coli* the respiratory chain lies in the cytoplasmic membrane, and the proton motive force it generates also drives flagellar rotation

With the same electron donor, respiration with nitrate releases more energy than respiration with sulfate, because the nitrate couple has the more positive E°′

In *E. coli* the respiratory chain lies in the cytoplasmic membrane, and the proton motive force it generates also drives flagellar rotation

In anaerobic respiration the terminal electron acceptor is an organic intermediate produced within the same catabolic pathway

With the same electron donor, respiration with nitrate releases more energy than respiration with sulfate, because the nitrate couple has the more positive E°′

Cytochrome bd of *E. coli* has a low affinity for O₂ and is the oxidase used mainly when oxygen is plentiful

A Gram-negative enteric isolate ferments glucose, gives a negative methyl red test and a positive Voges–Proskauer test, and produces much 2,3-butanediol. Which fermentation does it carry out?

Butanediol fermentation, in which pyruvate is condensed to acetoin and then reduced to the diol

Mixed-acid fermentation of the *E. coli* type, which drives the pH of the medium down to a low value

Homolactic fermentation, in which pyruvate is reduced to lactate by lactate dehydrogenase alone

Propionic acid fermentation, in which pyruvate is carboxylated and converted to propionate

Butanediol fermentation, in which pyruvate is condensed to acetoin and then reduced to the diol

Which of the following statements about microbial fermentations are correct?

In fermentation ATP is formed almost entirely by substrate-level phosphorylation

Heterolactic fermentation by the phosphoketolase pathway yields 2 ATP per glucose, the same as homolactic fermentation

*Zymomonas mobilis* obtains 2 ATP per glucose in its ethanol fermentation because it uses glycolysis

Four autotrophic chemolithotrophs fix CO₂ by the Calvin cycle and need NADH for it. Take E°′(NAD⁺/NADH) = −0.32 V. Which of them can reduce NAD⁺ directly from its energy source, without reverse electron transport?

*Cupriavidus necator*, oxidising H₂ with a soluble hydrogenase (E°′ of 2H⁺/H₂ = −0.42 V)

*Nitrobacter*, oxidising nitrite to nitrate (E°′ of NO₃⁻/NO₂⁻ about +0.42 V)

*Cupriavidus necator*, oxidising H₂ with a soluble hydrogenase (E°′ of 2H⁺/H₂ = −0.42 V)

*Acidithiobacillus ferrooxidans*, oxidising Fe²⁺ to Fe³⁺ at pH 2 (E°′ of Fe³⁺/Fe²⁺ about +0.77 V)

*Nitrosomonas*, oxidising ammonia to nitrite (E°′ of NO₂⁻/NH₄⁺ about +0.34 V)

A phototroph grows anaerobically in light with H₂S as electron donor and does not evolve O₂. It deposits sulfur globules outside its cells, has chlorosomes containing bacteriochlorophyll c, and fixes CO₂ without RuBisCO. It is best placed among the

green sulfur bacteria, which use an iron–sulfur-type reaction centre and the reverse citric acid cycle

purple sulfur bacteria, which use a quinone-type reaction centre and fix CO₂ by the Calvin cycle

cyanobacteria, which use two photosystems and fix CO₂ inside carboxysomes by the Calvin cycle

green sulfur bacteria, which use an iron–sulfur-type reaction centre and the reverse citric acid cycle

heliobacteria, which contain bacteriochlorophyll g and grow as photoheterotrophs on organic carbon compounds

Triclosan and the antitubercular drug isoniazid (through its activated form) both block bacterial fatty-acid synthesis. Which step of the type II elongation cycle do they inhibit?

Reduction of the trans-2-enoyl-ACP intermediate by enoyl-ACP reductase (FabI or InhA)

Carboxylation of acetyl-CoA to malonyl-CoA by the biotin-containing acetyl-CoA carboxylase

Condensation of the growing acyl chain with malonyl-ACP, with release of carbon dioxide

Reduction of the trans-2-enoyl-ACP intermediate by enoyl-ACP reductase (FabI or InhA)

Dehydration of the 3-hydroxyacyl-ACP intermediate to form the trans double bond

In the *trp* operon of *E. coli*, a point mutation destroys the start codon of the leader peptide so that ribosomes can no longer translate the leader. Regions 1–4 of the leader RNA are otherwise intact. What happens to transcription past the attenuator?

It is mostly terminated even when tryptophan is scarce, because 1:2 pairing forms and leaves 3:4 free to make the terminator

It becomes constitutively high, because without a ribosome the 2:3 antiterminator always forms in the leader RNA

It responds to tryptophan exactly as in the wild type, because attenuation depends only on the Trp repressor

It stops entirely, because RNA polymerase cannot initiate at the *trp* promoter unless the leader peptide is being translated

It is mostly terminated even when tryptophan is scarce, because 1:2 pairing forms and leaves 3:4 free to make the terminator

*E. coli* is grown in minimal glucose medium supplemented with excess phenylalanine and tyrosine but no tryptophan. The first step of aromatic amino-acid synthesis (DAHP synthase) is catalysed by three isoenzymes, each feedback-inhibited by a different aromatic amino acid. Which outcome is expected?

DAHP synthesis continues through the tryptophan-sensitive isoenzyme, so the cell can still make tryptophan

All DAHP synthesis stops, because two end products together inhibit every isoenzyme in a concerted manner

DAHP synthesis continues through the tryptophan-sensitive isoenzyme, so the cell can still make tryptophan

The cell becomes a tryptophan auxotroph until tryptophan is added, because the common pathway is shut off

DAHP synthesis rises above normal, because phenylalanine and tyrosine act as allosteric activators of DAHP synthase

The first enzyme of a branched biosynthetic pathway is barely affected by an excess of either end product alone, but is strongly inhibited when both end products are in excess together. This pattern of control is called

concerted feedback inhibition, in which significant inhibition needs all the end products together

cumulative feedback inhibition, in which each end product causes a partial inhibition and the effects add

sequential feedback inhibition, in which a branch-point intermediate inhibits the first common enzyme

concerted feedback inhibition, in which significant inhibition needs all the end products together

isoenzyme control, in which separate enzymes for the same step respond to different end products

Which of the following statements about the regulation of central metabolism in *E. coli* are correct?

Phosphofructokinase is allosterically activated by ADP and inhibited by phosphoenolpyruvate

Glucose taken up by the phosphotransferase system lowers cAMP, reducing cAMP–CRP activation of operons for other sugars

During growth on acetate, phosphorylation inactivates isocitrate dehydrogenase and diverts isocitrate into the glyoxylate bypass

Phosphofructokinase is allosterically activated by ADP and inhibited by phosphoenolpyruvate

Glucose taken up by the phosphotransferase system lowers cAMP, reducing cAMP–CRP activation of operons for other sugars

During growth on acetate, phosphorylation inactivates isocitrate dehydrogenase and diverts isocitrate into the glyoxylate bypass

End products of biosynthetic pathways usually inhibit the last enzyme of their pathway, so that no intermediates are left unused

Microbial Metabolism: Energetics, Central Pathways, Respiration and Fermentation, Chemolithotrophy, Photosynthesis, Biosynthesis and Regulation

This chapter answers Section XL3.6, “Microbial Metabolism”, of XL3 Microbiology, one of the five optional sections of the Life Sciences (XL) paper: “Energetics: redox reactions and electron carriers; Electron transport and oxidative phosphorylation; An overview of metabolism; Glycolysis; Pentose-phosphate pathway; Entner-Doudoroff pathway; Glyoxalate pathway; The citric acid cycle; Fermentation; Aerobic and anaerobic respiration; Chemolithotrophy; Photosynthesis; Calvin cycle; Biosynthetic pathway for fatty acids synthesis; Common regulatory mechanisms in synthesis of amino acids; Regulation of major metabolic pathways.” It teaches the pathways from the microbe's point of view: how redox potentials set the energy available, how bacteria choose among glycolysis, the pentose-phosphate and Entner–Doudoroff routes, how the glyoxylate bypass lets them live on acetate, how respiration, fermentation, chemolithotrophy and phototrophy differ, and how biosynthesis is controlled. The numericals are ΔG°′ from ΔE°′, ATP yields under a stated P/O convention, and the cofactor cost of the Calvin cycle and of fatty-acid synthesis.

1. Energetics, redox reactions and electron carriers

Metabolism is catabolism, which releases energy and reducing power, plus anabolism, which spends them. The central (amphibolic) pathways serve both: glycolysis, the pentose-phosphate pathway and the citric acid cycle supply about a dozen precursor metabolites (glucose-6-phosphate, ribose-5-phosphate, erythrose-4-phosphate, 3-phosphoglycerate, phosphoenolpyruvate, pyruvate, acetyl-CoA, 2-oxoglutarate, oxaloacetate and a few others) from which every monomer is built. Microbes differ in their energy source (light or chemicals), electron donor (organic or inorganic) and carbon source (organic or CO₂), giving categories such as chemoorganoheterotroph and photolithoautotroph.

Every energy-yielding reaction is a redox pair: a donor with a more negative standard reduction potential (E°′) passes electrons to an acceptor with a more positive one, and ΔG°′ = −nFΔE°′, where ΔE°′ = E°′(acceptor) − E°′(donor), n is the number of electrons and F = 96.485 kJ V⁻¹ mol⁻¹. Worked example: NADH (E°′ of NAD⁺/NADH = −0.32 V) to O₂ (O₂/H₂O = +0.82 V): ΔE°′ = 1.14 V and ΔG°′ = −2 × 96.485 × 1.14 = −220 kJ mol⁻¹. The farther apart donor and acceptor sit on this redox tower, the more energy, which is why respiration with O₂ beats respiration with sulfate.

Electron carriers are either freely diffusible coenzymes, NAD⁺ (catabolic) and NADP⁺ (biosynthetic), each carrying two electrons and a proton as a hydride, or membrane-bound: flavoproteins (FMN, FAD; 2e⁻ + 2H⁺), iron–sulfur proteins (1e⁻), quinones (lipid-soluble, 2e⁻ + 2H⁺: ubiquinone in aerobic E. coli, menaquinone anaerobically) and cytochromes (haem iron, 1e⁻). Alternation between H-carriers and pure e⁻-carriers in the membrane is what moves protons. ATP itself is made by substrate-level phosphorylation (from a high-energy intermediate such as 1,3-bisphosphoglycerate or phosphoenolpyruvate), oxidative phosphorylation or photophosphorylation.

2. Glycolysis, the pentose-phosphate pathway and the Entner–Doudoroff pathway

In many bacteria glucose enters by the phosphotransferase system (PTS), which phosphorylates it during uptake using phosphoenolpyruvate rather than ATP. Glycolysis (the Embden–Meyerhof–Parnas pathway) then splits fructose-1,6-bisphosphate (made by phosphofructokinase, the key control point) into two trioses and yields, per glucose, 2 pyruvate, a net 2 ATP and 2 NADH. The pentose-phosphate pathway oxidises glucose-6-phosphate (glucose-6-phosphate dehydrogenase, then 6-phosphogluconate dehydrogenase) to ribulose-5-phosphate + CO₂, making 2 NADPH; its non-oxidative branch (transketolase, transaldolase) interconverts sugars to supply ribose-5-phosphate for nucleotides and erythrose-4-phosphate for aromatic amino acids. The Entner–Doudoroff (ED) pathway, used by Pseudomonas, Zymomonas and many other Gram-negatives, dehydrates 6-phosphogluconate to 2-keto-3-deoxy-6-phosphogluconate (KDPG) and cleaves it to pyruvate + glyceraldehyde-3-phosphate.

The three routes of glucose breakdown compared (per glucose)
Pathway	Key enzymes	Net yield	Main role
Glycolysis (EMP)	Phosphofructokinase, fructose-bisphosphate aldolase	2 ATP, 2 NADH, 2 pyruvate	Energy and precursors
Entner–Doudoroff	6-Phosphogluconate dehydratase, KDPG aldolase	1 ATP, 1 NADH, 1 NADPH, 2 pyruvate	Energy in organisms lacking phosphofructokinase; sugar acids
Pentose phosphate (oxidative branch)	Glucose-6-phosphate dehydrogenase, 6-phosphogluconate dehydrogenase	2 NADPH and 1 CO₂ per glucose-6-phosphate	NADPH, pentoses, tetroses

⚠️ Exam trap

ED gives only half the ATP of glycolysis because only one of its two products (glyceraldehyde-3-phosphate) passes through the ATP-forming triose steps; the other is pyruvate straight away. Glucose-6-phosphate dehydrogenase and 6-phosphogluconate are shared with the pentose-phosphate pathway, so only the dehydratase and KDPG aldolase identify ED.

3. The citric acid cycle and the glyoxylate pathway

Pyruvate dehydrogenase converts pyruvate to acetyl-CoA + CO₂ + NADH. In the citric acid cycle acetyl-CoA joins oxaloacetate (citrate synthase); two decarboxylations (isocitrate dehydrogenase, 2-oxoglutarate dehydrogenase) release 2 CO₂, and each turn yields 3 NADH, 1 FADH₂ (as reduced quinone, via succinate dehydrogenase, which sits in the membrane) and 1 ATP or GTP (succinyl-CoA synthetase). The cycle also feeds biosynthesis, so intermediates drained for amino acids must be replaced by anaplerotic reactions such as PEP carboxylase or pyruvate carboxylase.

A cell growing on acetate or fatty acids has a problem: each acetyl-CoA that enters the cycle loses two carbons as CO₂, so there is no net gain of oxaloacetate to make sugars or amino acids. The glyoxylate pathway solves it with two enzymes: isocitrate lyase splits isocitrate to succinate + glyoxylate, and malate synthase joins glyoxylate with a second acetyl-CoA to give malate. The two decarboxylations are bypassed, and net 2 acetyl-CoA → 1 C₄ compound (succinate), which feeds gluconeogenesis. In E. coli the branch point is controlled by phosphorylation: on acetate, a kinase (AceK) inactivates isocitrate dehydrogenase, diverting isocitrate into the bypass.

4. Electron transport, oxidative phosphorylation, and aerobic and anaerobic respiration

Bacteria have no mitochondria: the chain sits in the cytoplasmic membrane and pumps protons out to the periplasm. In E. coli, NADH dehydrogenase I pumps protons and NADH dehydrogenase II does not; electrons pass through the quinone pool to a terminal oxidase, either cytochrome bo₃ (lower O₂ affinity, pumps protons, used when O₂ is plentiful) or cytochrome bd (high O₂ affinity, no pumping, used at low O₂). Chains are thus branched and modular, and the ATP yield depends on which branches are in use. By Mitchell's chemiosmotic theory, the proton motive force (membrane potential, inside negative, plus pH gradient, inside alkaline) drives protons back through the F₀F₁ ATP synthase; the same force powers flagellar rotation and many transporters. Uncouplers such as dinitrophenol carry protons across the membrane, so electron transport and O₂ consumption continue while ATP synthesis stops.

ATP yield per glucose depends on the convention. Taking 2.5 ATP per NADH and 1.5 per FADH₂, aerobic oxidation via glycolysis and the cycle gives 4 ATP by substrate-level phosphorylation (2 in glycolysis, 2 in the cycle) + 10 NADH × 2.5 + 2 FADH₂ × 1.5 = 32 ATP; with the older 3 and 2 it is 38. Bacteria have no mitochondrial shuttle to pay for, but their real yields are usually lower and vary with the chain branches in use. Glucose through ED instead of glycolysis gives one ATP less.

Terminal electron acceptors (approximate E°′ of the acceptor couple)
Acceptor → product	E°′ (V)	Process and example organisms
O₂ → H₂O	+0.82	Aerobic respiration; aerobes and facultative anaerobes
NO₃⁻ → NO₂⁻ → NO → N₂O → N₂	+0.42 (NO₃⁻/NO₂⁻)	Nitrate respiration and denitrification; Pseudomonas, Paracoccus
Fumarate → succinate	+0.03	Fumarate respiration; E. coli anaerobically
SO₄²⁻ → H₂S	−0.22	Sulfate reduction, after activation of sulfate to APS with ATP; Desulfovibrio
CO₂ → CH₄	−0.24	Methanogenesis by archaea, usually with H₂ as donor

Anaerobic respiration still uses a membrane chain and ATP synthase, but with an acceptor other than O₂; since these acceptors have less positive potentials, less energy is released. Facultative E. coli uses the best acceptor available, O₂ before nitrate before fumarate, switching through the oxygen-sensing regulators FNR and ArcB–ArcA and the nitrate-sensing NarX–NarL system.

5. Fermentation

Fermentation is energy metabolism without an external electron acceptor: the NADH made in glycolysis is reoxidised by reducing pyruvate or a compound derived from it, and ATP comes almost entirely from substrate-level phosphorylation. Yields are therefore small (1–2 ATP per glucose in the common types) and the products are partly reduced compounds excreted in bulk, which is why fermentations are both diagnostic and industrially useful.

Major microbial fermentations of glucose
Type	Products and ATP per glucose	Organisms and notes
Homolactic	2 lactate; 2 ATP (glycolysis)	Lactococcus, Streptococcus, many Lactobacillus; curd and yoghurt
Heterolactic	Lactate + ethanol + CO₂; 1 ATP (phosphoketolase pathway)	Leuconostoc, some Lactobacillus
Alcoholic	2 ethanol + 2 CO₂ (pyruvate decarboxylase, alcohol dehydrogenase)	Saccharomyces via glycolysis, 2 ATP; Zymomonas via ED, 1 ATP
Mixed acid	Lactate, acetate, succinate, ethanol, formate (split to H₂ + CO₂ by formate hydrogenlyase)	E. coli, Salmonella; strong acid, methyl red positive
Butanediol	2,3-Butanediol, via acetoin, with less acid	Enterobacter, Serratia; Voges–Proskauer positive (detects acetoin)
Butyric, acetone–butanol	Butyrate, butanol, acetone, H₂, CO₂	Clostridium
Propionic	Propionate, acetate, CO₂	Propionibacterium; CO₂ forms the holes of Swiss cheese

⚠️ Exam trap

Do not confuse fermentation with anaerobic respiration. Both run without O₂, but anaerobic respiration uses an external inorganic (or organic) acceptor and a membrane chain, and yields more ATP; fermentation uses an internal organic acceptor and no chain.

6. Chemolithotrophy

Chemolithotrophs oxidise inorganic donors and pass the electrons, usually to O₂, through a membrane chain; most are also autotrophs that fix CO₂ by the Calvin cycle. Examples: hydrogen bacteria (H₂ → H⁺; Cupriavidus necator, formerly Ralstonia eutropha); sulfur oxidisers (H₂S, S⁰, thiosulfate → sulfate; Thiobacillus, Acidithiobacillus, Beggiatoa); iron oxidisers (Fe²⁺ → Fe³⁺ at pH about 2; Acidithiobacillus ferrooxidans, important in bioleaching and acid mine drainage); and the nitrifiers, Nitrosomonas (NH₃ → NO₂⁻, starting with ammonia monooxygenase) and Nitrobacter (NO₂⁻ → NO₃⁻, nitrite oxidoreductase). Anammox bacteria oxidise NH₄⁺ with NO₂⁻ anaerobically to N₂ inside a membrane compartment, the anammoxosome.

Two consequences follow from the redox tower. First, donors such as nitrite (NO₃⁻/NO₂⁻ about +0.42 V) or Fe²⁺ sit close to O₂, so little energy is released per electron and these organisms must oxidise large amounts of substrate per gram of cells. Second, their electrons are too positive to reduce NAD⁺ (−0.32 V), so they make NADH for CO₂ fixation by reverse electron transport, spending proton motive force to push electrons uphill. Hydrogen (2H⁺/H₂ = −0.42 V) is the exception: a soluble hydrogenase can reduce NAD⁺ directly.

7. Microbial photosynthesis and the Calvin cycle

Groups of phototrophic bacteria
Group	Pigment and reaction centre	Electron donor; O₂ evolved?	CO₂ fixation
Cyanobacteria	Chlorophyll a; photosystems I and II	H₂O; yes (oxygenic)	Calvin cycle (carboxysomes)
Purple sulfur (Chromatium)	Bacteriochlorophyll a or b; quinone-type centre	H₂S, S⁰ (sulfur stored inside cells); no	Calvin cycle
Purple non-sulfur (Rhodobacter)	Bacteriochlorophyll a or b; quinone-type centre	H₂ or organic compounds, often photoheterotrophic; no	Calvin cycle
Green sulfur (Chlorobium)	Bacteriochlorophylls c, d, e in chlorosomes; iron–sulfur-type centre	H₂S, S⁰ (sulfur deposited outside cells); no	Reverse citric acid cycle
Filamentous anoxygenic (Chloroflexus)	Chlorosomes plus bacteriochlorophyll a; quinone-type centre	Organic compounds or H₂S; no	3-Hydroxypropionate pathway
Heliobacteria	Bacteriochlorophyll g; iron–sulfur-type centre	Organic compounds; no	None (photoheterotrophs)

Anoxygenic phototrophs have a single photosystem and cannot split water. Those with a quinone-type centre (purple bacteria) run cyclic electron flow to make proton motive force and must use reverse electron transport to make NAD(P)H; those with an iron–sulfur-type centre (green sulfur bacteria, heliobacteria) reduce ferredoxin directly. Halobacterium makes ATP from light with bacteriorhodopsin, a retinal-containing proton pump, without chlorophyll or electron transport.

The Calvin cycle has three stages. Carboxylation: RuBisCO adds CO₂ to ribulose-1,5-bisphosphate, giving two 3-phosphoglycerate. Reduction: 3-phosphoglycerate is phosphorylated and reduced to glyceraldehyde-3-phosphate. Regeneration: sugar rearrangements and phosphoribulokinase remake ribulose-1,5-bisphosphate. Per CO₂ the cost is 3 ATP and 2 NADPH; a net triose phosphate (3 CO₂) costs 9 ATP and 6 NADPH, and a hexose 18 ATP and 12 NADPH. RuBisCO and phosphoribulokinase are unique to the cycle, and cyanobacteria and many chemolithotrophs pack RuBisCO with carbonic anhydrase in carboxysomes to concentrate CO₂.

8. Fatty-acid synthesis

The committed step is acetyl-CoA carboxylase, a biotin enzyme: acetyl-CoA + HCO₃⁻ + ATP → malonyl-CoA. Bacteria then use a type II (dissociated) synthase, separate enzymes acting on substrates attached to acyl carrier protein (ACP); yeast and animals use a type I multifunctional synthase. Each elongation cycle has four steps: condensation of the acyl chain with malonyl-ACP (releasing CO₂, which drives the reaction), reduction of the 3-ketoacyl group, dehydration, and reduction of the enoyl group (enoyl-ACP reductase, FabI). Each cycle adds two carbons and uses two reduced nicotinamide cofactors. Palmitate (C16) needs 1 acetyl-CoA + 7 malonyl-CoA, hence 7 cycles, 7 ATP and 14 NAD(P)H.

E. coli makes unsaturated fatty acids without O₂: at the C10 stage the dehydratase–isomerase FabA converts trans-2-decenoyl-ACP to cis-3-decenoyl-ACP, which the enoyl reductase cannot reduce; the condensing enzyme FabB elongates it, so the cis double bond is carried into palmitoleate and cis-vaccenate and one enoyl reduction is skipped. Because FabI is essential, it is a drug target: triclosan inhibits it, and isoniazid inhibits the corresponding mycobacterial enzyme, InhA.

9. Regulation of amino-acid synthesis and of the major pathways

Biosynthetic pathways are controlled at two levels. Enzyme activity responds within seconds: in feedback (end-product) inhibition the end product binds an allosteric site on the first committed enzyme (isoleucine inhibits threonine deaminase, the first enzyme of its own branch). Gene expression responds more slowly: the end product acts as a co-repressor (tryptophan activates the Trp repressor) or controls attenuation. Branched pathways need finer patterns, listed below.

Isoenzymes: separate enzymes for one step, each inhibited by a different end product. E. coli has three aspartokinases (inhibited by threonine or lysine, or repressed by methionine) and three DAHP synthases (sensitive to phenylalanine, tyrosine and tryptophan respectively), so excess of one product does not starve the cell of the others.
Concerted (multivalent) inhibition: the shared first enzyme is strongly inhibited only when all end products are in excess together.
Cumulative inhibition: each end product inhibits partially and the effects add; E. coli glutamine synthetase, whose product feeds many pathways, is the classic case (it is also controlled by reversible adenylylation).
Sequential inhibition: each end product inhibits the first enzyme after its own branch point; the branch-point intermediate then accumulates and inhibits the first enzyme of the common pathway.

Attenuation of the trp operon. The leader transcript encodes a 14-residue peptide with two adjacent tryptophan codons, and its four regions can pair as 1:2, 2:3 or 3:4; the 3:4 hairpin is a terminator. With plenty of tryptophan the ribosome translates the leader quickly and covers region 2, so 3:4 forms and transcription stops. When tryptophan is scarce the ribosome stalls at the Trp codons over region 1, region 2 pairs with region 3 (antiterminator), 3:4 cannot form and RNA polymerase continues into the structural genes. The his operon leader has seven consecutive histidine codons and works the same way. Major pathways are also tuned allosterically (in E. coli, phosphofructokinase is activated by ADP and inhibited by phosphoenolpyruvate), by covalent modification (isocitrate dehydrogenase, above), and globally by catabolite repression: glucose taken up by the PTS lowers cAMP, so the cAMP–CRP activator is unavailable for operons of other sugars.

Key takeaways

ΔG°′ = −nFΔE°′ with ΔE°′ = E°′(acceptor) − E°′(donor); NADH → O₂ releases about 220 kJ mol⁻¹. Carriers are NAD(P)⁺, flavins, Fe–S proteins, quinones and cytochromes; ATP comes from substrate-level, oxidative or photophosphorylation.
Glycolysis gives 2 ATP + 2 NADH; ED (6-phosphogluconate dehydratase, KDPG aldolase) gives 1 ATP + 1 NADH + 1 NADPH; the pentose-phosphate pathway gives NADPH, ribose-5-phosphate and erythrose-4-phosphate. The glyoxylate bypass (isocitrate lyase, malate synthase) lets acetate support growth.
Bacterial chains are branched and sit in the cytoplasmic membrane; proton motive force drives ATP synthase, flagella and transport. Aerobic glucose oxidation gives 32 ATP at 2.5/1.5 or 38 at 3/2; anaerobic respiration uses acceptors such as nitrate, fumarate, sulfate or CO₂ and yields less.
Fermentation regenerates NAD⁺ with an internal organic acceptor and makes ATP only by substrate-level phosphorylation; mixed-acid (methyl red positive) and butanediol (Voges–Proskauer positive) fermentations separate enteric bacteria.
Chemolithotrophs oxidise H₂, reduced sulfur, Fe²⁺, NH₃ or NO₂⁻; high-potential donors need reverse electron transport to make NADH. Anoxygenic phototrophs use one photosystem and donors other than water; the Calvin cycle (RuBisCO, phosphoribulokinase) costs 3 ATP + 2 NADPH per CO₂.
Type II fatty-acid synthesis adds two carbons per cycle from malonyl-ACP at a cost of two NAD(P)H; FabA–FabB make unsaturated chains anaerobically. Amino-acid synthesis is controlled by feedback inhibition of the first committed enzyme (isoenzyme, concerted, cumulative and sequential patterns), repression and attenuation.

Practice questions (17)

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.

Which pair of enzymes is unique to the Entner–Doudoroff pathway and distinguishes it from both glycolysis and the pentose-phosphate pathway?
1. 6-Phosphogluconate dehydratase and 2-keto-3-deoxy-6-phosphogluconate aldolase
2. Glucose-6-phosphate dehydrogenase and 6-phosphogluconate dehydrogenase
3. Phosphofructokinase and fructose-1,6-bisphosphate aldolase of the hexose-splitting steps
4. Transketolase and transaldolase of the reversible sugar-rearranging reactions
Show answer
Answer: A — 6-Phosphogluconate dehydratase and 2-keto-3-deoxy-6-phosphogluconate aldolase
ED dehydrates 6-phosphogluconate to KDPG and cleaves KDPG to pyruvate and glyceraldehyde-3-phosphate; neither enzyme occurs in the other two pathways. Glucose-6-phosphate dehydrogenase is shared by ED and the pentose-phosphate pathway, and 6-phosphogluconate dehydrogenase belongs to the pentose-phosphate pathway, so that pair is the tempting but wrong choice. Phosphofructokinase and aldolase define glycolysis; transketolase and transaldolase define the non-oxidative pentose-phosphate branch.
An E. coli mutant lacks isocitrate lyase but is otherwise normal. On which sole carbon source would it fail to grow?
1. Glucose, because isocitrate lyase is needed to convert pyruvate into acetyl-CoA
2. Succinate, because succinate can enter central metabolism only through the glyoxylate pathway
3. Acetate, because without the bypass every acetyl-CoA carbon is lost as CO₂ and no oxaloacetate accumulates
4. Glycerol, because glycerol is oxidised to glyoxylate before it can enter the citric acid cycle
Show answer
Answer: C — Acetate, because without the bypass every acetyl-CoA carbon is lost as CO₂ and no oxaloacetate accumulates
On acetate the only route to a net C₄ compound is the glyoxylate bypass: isocitrate lyase and malate synthase skip the two decarboxylations. Without isocitrate lyase the cycle can oxidise acetate for energy but cannot make the oxaloacetate needed for gluconeogenesis and amino acids. Succinate is itself a cycle intermediate and is converted to oxaloacetate directly, so that distractor is wrong; glucose and glycerol enter through glycolysis.
A Pseudomonas strain phosphorylates glucose with glucokinase (1 ATP), degrades it exclusively by the Entner–Doudoroff pathway and oxidises both pyruvates completely through pyruvate dehydrogenase and the citric acid cycle. Assume 2.5 ATP per NADH, count each NADPH as equivalent to one NADH, 1.5 ATP per FADH₂, and 1 ATP from each succinyl-CoA synthetase step. The total ATP formed per glucose is ____.
Numerical answer — type the value.
Show answer
Answer: 31
ED: −1 ATP (glucokinase) + 2 ATP (from the one glyceraldehyde-3-phosphate) = net 1 ATP, plus 1 NADPH and 1 NADH. Pyruvate dehydrogenase: 2 NADH. Cycle, two turns: 6 NADH, 2 FADH₂, 2 ATP. Reduced nicotinamides = 1 + 1 + 2 + 6 = 10, giving 25 ATP; FADH₂ gives 3 ATP; substrate-level ATP = 1 + 2 = 3. Total = 25 + 3 + 3 = 31, one fewer than the 32 obtained via glycolysis.
A protonophore that makes the cytoplasmic membrane freely permeable to H⁺ is added to an aerobic, flagellated E. coli culture growing on succinate. Which combination of effects is expected?
1. O₂ consumption stops at once, because electron transport cannot run once protons leak back across the membrane
2. O₂ consumption continues or rises, while oxidative ATP synthesis and flagellar rotation both stop
3. ATP synthesis continues normally, because bacterial ATP synthase is driven by electron flow rather than by protons
4. Flagella keep rotating, because bacterial flagellar motors are powered directly by the hydrolysis of ATP
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Answer: B — O₂ consumption continues or rises, while oxidative ATP synthesis and flagellar rotation both stop
A protonophore dissipates the proton motive force. Electron transport is no longer held back by a back-pressure of protons, so O₂ consumption continues or increases, but ATP synthase and the flagellar motor, both driven by proton flow, stop. On succinate the cell cannot fall back on glycolytic substrate-level ATP. Respiration stopping at once is the tempting error: uncouplers separate electron flow from phosphorylation rather than blocking it.
A sulfate-reducing bacterium oxidises H₂ with sulfate: 4 H₂ + SO₄²⁻ + H⁺ → HS⁻ + 4 H₂O. Take E°′(2H⁺/H₂) = −0.42 V, E°′(SO₄²⁻/HS⁻) = −0.22 V and F = 96.485 kJ V⁻¹ mol⁻¹. The magnitude of ΔG°′ per mole of sulfate reduced is ____ kJ (to one decimal place).
Numerical answer — type the value.
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Answer: 154.4
ΔE°′ = E°′(acceptor) − E°′(donor) = −0.22 − (−0.42) = 0.20 V. Sulfate (S at +6) to HS⁻ (S at −2) takes 8 electrons, supplied by 4 H₂. ΔG°′ = −nFΔE°′ = −8 × 96.485 × 0.20 = −154.38 kJ per mol sulfate, magnitude 154.4. Using n = 2 (as if for one H₂) gives 38.6 kJ, which is per mole of H₂, not per mole of sulfate.
Which of the following statements about bacterial respiration are correct?
1. In E. coli the respiratory chain lies in the cytoplasmic membrane, and the proton motive force it generates also drives flagellar rotation
2. In anaerobic respiration the terminal electron acceptor is an organic intermediate produced within the same catabolic pathway
3. With the same electron donor, respiration with nitrate releases more energy than respiration with sulfate, because the nitrate couple has the more positive E°′
4. Cytochrome bd of E. coli has a low affinity for O₂ and is the oxidase used mainly when oxygen is plentiful
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Answer: A — In *E. coli* the respiratory chain lies in the cytoplasmic membrane, and the proton motive force it generates also drives flagellar rotation; C — With the same electron donor, respiration with nitrate releases more energy than respiration with sulfate, because the nitrate couple has the more positive E°′
Bacterial chains sit in the cytoplasmic membrane and the proton motive force powers flagella; nitrate (about +0.42 V) lies far above sulfate (about −0.22 V), so ΔE°′ and the energy released are larger. An internal organic acceptor defines fermentation, not anaerobic respiration, which uses an external acceptor and a membrane chain. Cytochrome bd is the high-affinity oxidase used at low O₂; bo₃ is the low-affinity one used when O₂ is plentiful.
A Gram-negative enteric isolate ferments glucose, gives a negative methyl red test and a positive Voges–Proskauer test, and produces much 2,3-butanediol. Which fermentation does it carry out?
1. Mixed-acid fermentation of the E. coli type, which drives the pH of the medium down to a low value
2. Homolactic fermentation, in which pyruvate is reduced to lactate by lactate dehydrogenase alone
3. Propionic acid fermentation, in which pyruvate is carboxylated and converted to propionate
4. Butanediol fermentation, in which pyruvate is condensed to acetoin and then reduced to the diol
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Answer: D — Butanediol fermentation, in which pyruvate is condensed to acetoin and then reduced to the diol
The Voges–Proskauer test detects acetoin, the precursor of 2,3-butanediol, and butanediol fermenters (Enterobacter, Serratia) make less acid, so the methyl red test is negative. Mixed-acid fermentation is the opposite pattern: strong acid, methyl red positive, Voges–Proskauer negative. Homolactic and propionic fermentations are not typical of enteric bacteria and make no acetoin.
Which of the following statements about microbial fermentations are correct?
1. In fermentation ATP is formed almost entirely by substrate-level phosphorylation
2. The NADH formed in glycolysis is reoxidised by reducing pyruvate or a compound derived from pyruvate
3. Heterolactic fermentation by the phosphoketolase pathway yields 2 ATP per glucose, the same as homolactic fermentation
4. Zymomonas mobilis obtains 2 ATP per glucose in its ethanol fermentation because it uses glycolysis
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Answer: A — In fermentation ATP is formed almost entirely by substrate-level phosphorylation; B — The NADH formed in glycolysis is reoxidised by reducing pyruvate or a compound derived from pyruvate
Fermentation has no external acceptor, so NAD⁺ is regenerated on pyruvate or its products and ATP comes from substrate-level phosphorylation. Heterolactic fermentation loses a carbon as CO₂ in the pentose route, and only one triose goes through the ATP-forming steps while the acetyl phosphate is used to reoxidise NADH as ethanol, so it yields 1 ATP, not 2. Zymomonas uses the Entner–Doudoroff pathway and gains 1 ATP per glucose, not 2.
Four autotrophic chemolithotrophs fix CO₂ by the Calvin cycle and need NADH for it. Take E°′(NAD⁺/NADH) = −0.32 V. Which of them can reduce NAD⁺ directly from its energy source, without reverse electron transport?
1. Nitrobacter, oxidising nitrite to nitrate (E°′ of NO₃⁻/NO₂⁻ about +0.42 V)
2. Cupriavidus necator, oxidising H₂ with a soluble hydrogenase (E°′ of 2H⁺/H₂ = −0.42 V)
3. Acidithiobacillus ferrooxidans, oxidising Fe²⁺ to Fe³⁺ at pH 2 (E°′ of Fe³⁺/Fe²⁺ about +0.77 V)
4. Nitrosomonas, oxidising ammonia to nitrite (E°′ of NO₂⁻/NH₄⁺ about +0.34 V)
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Answer: B — *Cupriavidus necator*, oxidising H₂ with a soluble hydrogenase (E°′ of 2H⁺/H₂ = −0.42 V)
Electrons flow spontaneously only towards a more positive potential. H₂ (−0.42 V) is more negative than NAD⁺/NADH (−0.32 V), so the soluble hydrogenase of hydrogen bacteria reduces NAD⁺ directly. Nitrite, Fe²⁺ and ammonia all have couples far more positive than −0.32 V, so these organisms must spend proton motive force to drive electrons uphill to NAD⁺ by reverse electron transport.
A phototroph grows anaerobically in light with H₂S as electron donor and does not evolve O₂. It deposits sulfur globules outside its cells, has chlorosomes containing bacteriochlorophyll c, and fixes CO₂ without RuBisCO. It is best placed among the
1. purple sulfur bacteria, which use a quinone-type reaction centre and fix CO₂ by the Calvin cycle
2. cyanobacteria, which use two photosystems and fix CO₂ inside carboxysomes by the Calvin cycle
3. green sulfur bacteria, which use an iron–sulfur-type reaction centre and the reverse citric acid cycle
4. heliobacteria, which contain bacteriochlorophyll g and grow as photoheterotrophs on organic carbon compounds
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Answer: C — green sulfur bacteria, which use an iron–sulfur-type reaction centre and the reverse citric acid cycle
Chlorosomes with bacteriochlorophyll c, sulfur deposited outside the cell and CO₂ fixation without RuBisCO (by the reverse citric acid cycle) together identify green sulfur bacteria such as Chlorobium. Purple sulfur bacteria are the tempting choice because they also oxidise H₂S anoxygenically, but they store sulfur inside the cell, lack chlorosomes and use RuBisCO. Cyanobacteria evolve O₂, and heliobacteria do not fix CO₂.
An autotrophic bacterium fixes CO₂ by the Calvin cycle. How many molecules of ATP does it spend to produce one net molecule of glyceraldehyde-3-phosphate from CO₂?
Numerical answer — type the value.
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Answer: 9
A net triose phosphate needs 3 CO₂. Each CO₂ fixed requires 2 ATP to phosphorylate the two 3-phosphoglycerates it yields and 1 ATP at phosphoribulokinase to regenerate ribulose-1,5-bisphosphate, that is 3 ATP (and 2 NADPH) per CO₂. 3 × 3 = 9 ATP (with 6 NADPH). Answering 18 confuses the triose with a hexose.
Triclosan and the antitubercular drug isoniazid (through its activated form) both block bacterial fatty-acid synthesis. Which step of the type II elongation cycle do they inhibit?
1. Carboxylation of acetyl-CoA to malonyl-CoA by the biotin-containing acetyl-CoA carboxylase
2. Condensation of the growing acyl chain with malonyl-ACP, with release of carbon dioxide
3. Reduction of the trans-2-enoyl-ACP intermediate by enoyl-ACP reductase (FabI or InhA)
4. Dehydration of the 3-hydroxyacyl-ACP intermediate to form the trans double bond
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Answer: C — Reduction of the trans-2-enoyl-ACP intermediate by enoyl-ACP reductase (FabI or InhA)
Triclosan inhibits FabI, the enoyl-ACP reductase, and activated isoniazid inhibits InhA, the enoyl-ACP reductase of Mycobacterium tuberculosis, which is needed for mycolic acid synthesis. The condensation step is the tempting distractor because it is also a well-known antibiotic target (of cerulenin and thiolactomycin), but neither triclosan nor isoniazid acts there.
E. coli makes palmitoleoyl-ACP (C16 with one cis double bond) by the anaerobic route: in the cycle that forms the C10 intermediate, FabA isomerises trans-2-decenoyl-ACP to cis-3-decenoyl-ACP, which escapes enoyl reduction and is elongated by the later cycles. Starting from one acetyl-CoA and malonyl units, and counting one reduced nicotinamide cofactor, NAD(P)H, for each 3-ketoacyl reduction and each enoyl reduction, how many NAD(P)H are consumed per palmitoleoyl-ACP?
Numerical answer — type the value.
Show answer
Answer: 13
A C16 chain from one acetyl starter needs 7 malonyl additions, so 7 elongation cycles. A saturated chain would use 2 NAD(P)H per cycle, 14 in all. In the C10-forming cycle the cis-3 isomer is not reduced by the enoyl reductase, so that cycle uses only its 3-ketoacyl reduction: 14 − 1 = 13 NAD(P)H.
In the trp operon of E. coli, a point mutation destroys the start codon of the leader peptide so that ribosomes can no longer translate the leader. Regions 1–4 of the leader RNA are otherwise intact. What happens to transcription past the attenuator?
1. It becomes constitutively high, because without a ribosome the 2:3 antiterminator always forms in the leader RNA
2. It responds to tryptophan exactly as in the wild type, because attenuation depends only on the Trp repressor
3. It stops entirely, because RNA polymerase cannot initiate at the trp promoter unless the leader peptide is being translated
4. It is mostly terminated even when tryptophan is scarce, because 1:2 pairing forms and leaves 3:4 free to make the terminator
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Answer: D — It is mostly terminated even when tryptophan is scarce, because 1:2 pairing forms and leaves 3:4 free to make the terminator
Antitermination needs a ribosome stalled on region 1 at the Trp codons, which frees region 2 to pair with region 3. With no translation, the default structure forms: region 1 pairs with region 2, regions 3 and 4 form the terminator hairpin, and transcription terminates at the attenuator whatever the tryptophan level. The constitutive-expression distractor is the common slip; it assumes the 2:3 structure is the default, but 2:3 forms only when a stalled ribosome sequesters region 1.
E. coli is grown in minimal glucose medium supplemented with excess phenylalanine and tyrosine but no tryptophan. The first step of aromatic amino-acid synthesis (DAHP synthase) is catalysed by three isoenzymes, each feedback-inhibited by a different aromatic amino acid. Which outcome is expected?
1. All DAHP synthesis stops, because two end products together inhibit every isoenzyme in a concerted manner
2. DAHP synthesis continues through the tryptophan-sensitive isoenzyme, so the cell can still make tryptophan
3. The cell becomes a tryptophan auxotroph until tryptophan is added, because the common pathway is shut off
4. DAHP synthesis rises above normal, because phenylalanine and tyrosine act as allosteric activators of DAHP synthase
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Answer: B — DAHP synthesis continues through the tryptophan-sensitive isoenzyme, so the cell can still make tryptophan
With isoenzymes, each end product shuts off only its own enzyme. Excess phenylalanine and tyrosine inhibit their two isoenzymes, but the tryptophan-sensitive one is uninhibited because tryptophan is absent, so enough flux continues into the common pathway for tryptophan synthesis. Shutting off the whole pathway is exactly the failure that isoenzyme control evolved to prevent; concerted inhibition describes a single shared enzyme, not this case.
The first enzyme of a branched biosynthetic pathway is barely affected by an excess of either end product alone, but is strongly inhibited when both end products are in excess together. This pattern of control is called
1. cumulative feedback inhibition, in which each end product causes a partial inhibition and the effects add
2. sequential feedback inhibition, in which a branch-point intermediate inhibits the first common enzyme
3. concerted feedback inhibition, in which significant inhibition needs all the end products together
4. isoenzyme control, in which separate enzymes for the same step respond to different end products
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Answer: C — concerted feedback inhibition, in which significant inhibition needs all the end products together
Concerted (multivalent) inhibition is defined by the requirement for all end products together. In cumulative inhibition each end product alone gives a measurable partial inhibition, which is not what the stem describes; sequential inhibition acts through the branch-point intermediate, and isoenzyme control uses separate enzymes rather than one.
Which of the following statements about the regulation of central metabolism in E. coli are correct?
1. Phosphofructokinase is allosterically activated by ADP and inhibited by phosphoenolpyruvate
2. Glucose taken up by the phosphotransferase system lowers cAMP, reducing cAMP–CRP activation of operons for other sugars
3. During growth on acetate, phosphorylation inactivates isocitrate dehydrogenase and diverts isocitrate into the glyoxylate bypass
4. End products of biosynthetic pathways usually inhibit the last enzyme of their pathway, so that no intermediates are left unused
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Answer: A — Phosphofructokinase is allosterically activated by ADP and inhibited by phosphoenolpyruvate; B — Glucose taken up by the phosphotransferase system lowers cAMP, reducing cAMP–CRP activation of operons for other sugars; C — During growth on acetate, phosphorylation inactivates isocitrate dehydrogenase and diverts isocitrate into the glyoxylate bypass
PFK senses energy charge (ADP up, activity up) and a high PEP level signals that the lower glycolytic pathway is backed up; PTS-mediated glucose uptake keeps cAMP low (catabolite repression); and AceK phosphorylates isocitrate dehydrogenase on acetate, sending isocitrate to isocitrate lyase. Feedback inhibition acts on the first committed enzyme, not the last, precisely so that intermediates do not accumulate and carbon is not wasted.