Prokaryotic Cells: Walls, Membranes, Transport, Appendages, Inclusions and Chemotaxis

This chapter answers Section XL3.4 of XL3 Microbiology, one of the five optional sections of the Life Sciences (XL) paper: “Cell walls, cell membranes and their biosynthesis; Mechanisms of solute transport across membranes; Flagella and Pili, Capsules, Cell inclusions like endospores and gas vesicles; Bacterial locomotion, including positive and negative chemotaxis.” It works from the outside in and back out: why small cells pay, how peptidoglycan and the two kinds of envelope are built, how membrane lipids are made, how solutes cross membranes and at what energetic price, how flagella, pili and capsules are built and used, what endospores, gas vesicles and storage granules do, and how a bacterium swims towards attractants and away from repellents. The numericals are surface-to-volume ratios, peptidoglycan cross-link arithmetic, accumulation ratios from the proton motive force, and flagellar speeds.

1. Small cells and the surface-to-volume ratio

Most bacteria are 0.5–5 µm long. Everything a cell takes in or gets rid of crosses its surface, while its demand scales with its volume, so a high surface-to-volume (S/V) ratio supports fast uptake and fast growth. For a sphere S/V = 4πr²/(4/3 πr³) = 3/r: halving the radius doubles the ratio. For a rod of radius r with hemispherical ends and a cylindrical part of length l, S = 2πrl + 4πr² and V = πr²l + (4/3)πr³. Worked example: a rod 1 µm wide and 2 µm long (r = 0.5 µm, l = 1 µm) has S = π + π = 2π µm² and V = 0.25π + 0.167π = 0.417π µm³, so S/V = 4.8 µm⁻¹, against 6 µm⁻¹ for a coccus 1 µm across.

2. Cell walls: peptidoglycan and the two envelopes

Peptidoglycan (murein) is one bag-shaped molecule, the sacculus, that resists the cell's turgor pressure. Its glycan strands alternate N-acetylglucosamine (NAG) and N-acetylmuramic acid (NAM) joined β(1→4). Each NAM carries a short peptide, L-Ala–D-Glu–X–D-Ala, where X is meso-diaminopimelic acid (DAP) in E. coli and most Gram-negatives and L-lysine in most Gram-positives. Peptides on neighbouring strands are cross-linked from the D-Ala at position 4 of one to the amino group of residue 3 of the other, directly in E. coli and through a pentaglycine bridge in Staphylococcus aureus. D-amino acids protect the wall from ordinary proteases. Lysozyme cuts the NAM–NAG β(1→4) bond: in an isotonic sucrose solution this leaves a protoplast (Gram-positive) or spheroplast (Gram-negative, outer membrane remnants attached), which bursts if diluted into water.

Gram-positive and Gram-negative envelopes
FeatureGram-positiveGram-negative
PeptidoglycanThick, many layers (about 20–80 nm)Thin, one or a few layers, in the periplasm
Other wall polymersTeichoic acids (polyol-phosphate chains) on peptidoglycan; lipoteichoic acids anchored in the membraneBraun's lipoprotein ties peptidoglycan to the outer membrane
Outer membraneAbsentInner leaflet phospholipid, outer leaflet lipopolysaccharide (lipid A, core, O antigen); porins pass small hydrophilic solutes
Gram stainKeeps crystal violet–iodine after alcohol: purpleDecolourised, counterstained by safranin: pink
Notable propertyMore open to lysozyme and penicillinLipid A is endotoxin; outer membrane excludes many drugs and detergents
ℹ️ Walls that break the pattern
Mycobacteria are acid-fast: mycolic acids esterified to an arabinogalactan layer make the envelope waxy and impermeable. Mycoplasmas have no wall and stiffen their membranes with sterols taken from the host. Archaea lack peptidoglycan; some methanogens have pseudomurein (N-acetyltalosaminuronic acid, β(1→3) links and L-amino acids, so lysozyme and penicillin do not act), and many archaea have only a protein S-layer.

3. Building the wall and the membrane

Peptidoglycan synthesis runs in three compartments. (1) Cytoplasm: UDP-NAG is converted to UDP-NAM (MurA, the target of fosfomycin, then MurB), and amino acids are added one at a time by ATP-using ligases (MurC to MurF) to give UDP-NAM-pentapeptide. The final D-Ala–D-Ala dipeptide is made by alanine racemase and D-Ala–D-Ala ligase, both blocked by D-cycloserine. (2) Membrane: MraY transfers phospho-NAM-pentapeptide to the C₅₅ lipid carrier undecaprenyl phosphate (bactoprenol), giving lipid I; MurG adds NAG to give lipid II, which is flipped to the outer face. (3) Outside the membrane: transglycosylases add the disaccharide-pentapeptide to a growing glycan strand, and transpeptidases (penicillin-binding proteins) form cross-links. Transpeptidation releases the terminal D-Ala of the donor peptide and uses the energy of the D-Ala–D-Ala bond, because no ATP is available outside the cell. The carrier leaves as undecaprenyl pyrophosphate and must be dephosphorylated to be reused, the step blocked by bacitracin. β-lactams acylate the transpeptidases as D-Ala–D-Ala mimics, and vancomycin binds the D-Ala–D-Ala end of lipid II and blocks both polymerisation steps. Autolysins open the existing sacculus so new material can be inserted, along the side wall of rods and at the septum during division.

Membranes are phospholipid bilayers carrying proteins (the fluid mosaic). Bacteria generally lack sterols; many use hopanoids to similar effect. In E. coli the main lipid is phosphatidylethanolamine, with phosphatidylglycerol and cardiolipin. Fatty acids are made by a dissociated (type II) synthase working on acyl carrier protein; glycerol-3-phosphate is acylated twice to phosphatidic acid, activated with CTP to CDP-diacylglycerol, and converted to phosphatidylserine (then decarboxylated to phosphatidylethanolamine) or to phosphatidylglycerol (and cardiolipin). Cells keep their membranes fluid by homeoviscous adaptation: at lower temperature they make more unsaturated (and shorter) fatty acids. Archaea build isoprenoid ether lipids instead, and some thermophiles span the membrane with tetraether lipids that form a monolayer.

🧠 Worked example: counting cross-links
Digesting a sacculus with a muramidase gives monomers, dimers (two peptide stems, one cross-link) and trimers (three stems, two cross-links). If 1,000 stems are 610 in monomers, 300 in dimers and 90 in trimers, there are 150 + 60 = 210 cross-links, 21 per 100 stems, and each one released one D-Ala when it was formed.

4. Mechanisms of solute transport

Transport mechanisms
MechanismEnergy and directionExample
Passive diffusionNone; down the gradient; not saturableO₂, CO₂, water (aided by aquaporin)
Facilitated diffusionNone; down the gradient through a specific protein; saturableGlycerol facilitator GlpF of E. coli
Primary active: ABC transporterATP hydrolysis; against the gradient; Gram-negative importers use a periplasmic substrate-binding proteinMaltose (MalE with MalFGK₂), histidine
Secondary active: symport, antiport, uniportAn ion gradient (H⁺ or Na⁺) pays; solute can be concentratedLactose permease LacY (H⁺ symport); Na⁺/H⁺ antiporters
Group translocation (PTS)Phosphoryl from PEP via enzyme I, HPr, EIIA and EIIB; the sugar enters through EIIC as sugar phosphateGlucose, mannitol in E. coli
Iron uptakeSecreted siderophores bind Fe³⁺; TonB–ExbB–ExbD transmit inner-membrane proton motive force to outer-membrane receptorsEnterobactin of E. coli

Porins of the outer membrane are open water-filled channels that pass small hydrophilic molecules (up to about 600 Da) by diffusion and cannot concentrate anything. The PTS costs one PEP per sugar, equal to one ATP, and the phosphorylated sugar cannot leak out; its EIIA component also signals glucose availability for catabolite repression. Energetics: for an uncharged solute, the free energy of moving it in is ΔG = 2.303RT log(C_in/C_out); a charged solute also feels the membrane potential. The proton motive force has two parts, Δp = Δψ − (2.303RT/F)ΔpH, with 2.303RT/F ≈ 59 mV at 25 °C; in respiring E. coli Δψ is negative inside and the inside is alkaline, so both parts drive H⁺ inward. A 1:1 H⁺ symporter of a neutral solute reaches equilibrium when log(C_in/C_out) = |Δp|/59 mV. Worked example: |Δp| = 118 mV allows at most a 10² = 100-fold accumulation; doubling the protons per solute doubles the logarithm.

⚠️ Concentrating needs energy; modifying is not concentrating
Only active systems can raise a solute above its outside concentration. The PTS is classed separately because the solute that arrives (glucose 6-phosphate) is not the one that left (glucose); a protonophore that collapses Δp stops LacY at once but leaves ATP-driven ABC import and PEP-driven PTS running as long as ATP and PEP last.

5. Flagella, pili and capsules

A bacterial flagellum is a rotary propeller, not a whip. Its filament is a hollow helical polymer of flagellin (FliC) about 20 nm thick and several µm long, which grows at its tip as new subunits travel up the central channel, exported by a type III secretion-like apparatus at the base. A flexible hook joins it to the basal body, a rod with rings: in Gram-negatives an L ring in the outer membrane, a P ring in the peptidoglycan, the MS ring in the cytoplasmic membrane and the cytoplasmic C ring (FliG, FliM, FliN), which is the rotor and the switch; Gram-positives lack the L and P rings. Around the rotor sit stator units of MotA and MotB that conduct protons; each proton's passage down the proton motive force pushes the rotor. Some motors, such as the polar flagellum of Vibrio, run on Na⁺ instead. Flagella are monotrichous (Vibrio cholerae, Pseudomonas aeruginosa), lophotrichous (a polar tuft, Helicobacter pylori), amphitrichous (one at each pole, Campylobacter jejuni) or peritrichous (E. coli, Proteus); spirochaetes carry periplasmic (endo)flagella beneath an outer sheath. Archaeal flagella (archaella) are related to type IV pili, are driven by ATP and grow from the base.

Pili (fimbriae) are thinner, straighter protein appendages that do not rotate. Fimbriae attach cells to surfaces and host tissues: the FimH tip adhesin of type 1 fimbriae binds mannose, and P pili of uropathogenic E. coli bind kidney glycolipids. The conjugative (sex) pilus of F⁺ cells contacts a recipient and retracts to bring the cells together for DNA transfer. Type IV pili extend, attach and retract under the PilT ATPase, pulling the cell in twitching motility (Pseudomonas aeruginosa, Neisseria) and helping take up DNA. The glycocalyx is a capsule when organised and firmly attached and a slime layer when loose. Most capsules are polysaccharide, but that of Bacillus anthracis is poly-D-glutamic acid; capsules resist phagocytosis and desiccation and help form biofilms (unencapsulated rough pneumococci are avirulent), are seen as clear halos by negative staining with India ink.

6. Inclusions: storage granules, gas vesicles and endospores

Storage and special inclusions. Poly-β-hydroxybutyrate (PHB) granules store carbon and energy when carbon is plentiful but another nutrient limits growth; glycogen is the other carbon store. Polyphosphate (volutin) granules store phosphate and stain metachromatically (red with blue dyes), as in Corynebacterium diphtheriae. Sulphur oxidisers such as Beggiatoa store sulphur globules. Carboxysomes are protein-shelled compartments packed with RuBisCO and carbonic anhydrase that concentrate CO₂ for fixation. Magnetosomes are membrane-bound crystals of magnetite (Fe₃O₄) or greigite in a chain, which align magnetotactic bacteria such as Magnetospirillum with the Earth's field lines and so guide them up or down towards low-oxygen layers.

Gas vesicles are hollow, spindle-shaped structures whose wall is a single layer of protein (mainly GvpA, strengthened by GvpC). The inner surface is hydrophobic, so gases diffuse in freely but liquid water cannot enter or condense; the vesicles lower the cell's density and let aquatic cyanobacteria (Anabaena, bloom-forming Microcystis) and halophilic archaea float to the light. They collapse irreversibly above a critical pressure, and cells adjust their depth by making new vesicles or by adding dense carbohydrate ballast.

Endospores are dormant, highly resistant cells formed inside a mother cell by Bacillus, Clostridium and their relatives when nutrients run out. From the inside: a dehydrated core holding DNA protected by small acid-soluble spore proteins (SASPs) and rich in calcium dipicolinate (up to about 10% of dry weight); an inner membrane; a germ-cell wall; a thick cortex of loosely cross-linked peptidoglycan; protein coats; and in some species an exosporium. Low core water, dipicolinate and SASPs give heat and UV resistance; the coats keep out chemicals and lytic enzymes. The spore's position (central, subterminal or terminal, as in the drumstick of Clostridium tetani) helps identification, and it is stained green with malachite green applied with heat. Germination follows activation (for example by mild heat) when germinants such as L-alanine are sensed: dipicolinate is released, the cortex is hydrolysed, the core rehydrates and the cell grows out.

⚠️ An endospore is not a means of multiplication
One vegetative cell makes one endospore, which gives back one cell: endospores are for survival, not reproduction, unlike the many conidia of Streptomyces or of fungi.

7. Locomotion and positive and negative chemotaxis

A swimming E. coli covers some tens of micrometres a second, roughly ten body lengths. At this scale viscosity dominates and inertia is negligible, so a cell stops almost instantly when its motors stop; it cannot coast. When the motors turn counter-clockwise (viewed from the filament tip towards the cell) the peritrichous filaments wind into one bundle and the cell runs smoothly for about a second; when one or more turn clockwise the bundle flies apart and the cell tumbles for about a tenth of a second, then sets off in a random new direction. Polar flagellates reverse instead of tumbling. A cell is too small to compare concentrations across its own length, so it compares in time: if conditions are improving it suppresses tumbles and extends the run. In a gradient of attractant the result is a biased random walk up the gradient (positive chemotaxis); with a repellent runs that lead away are extended (negative chemotaxis).

The signalling circuit. Transmembrane methyl-accepting chemotaxis proteins (MCPs), such as Tar (aspartate, and maltose via its binding protein) and Tsr (serine), sit in polar arrays with the adaptor CheW and the histidine kinase CheA. CheA autophosphorylates and passes the phosphoryl to CheY; CheY-P binds FliM in the switch and promotes clockwise rotation, hence tumbling. CheZ dephosphorylates CheY-P. Attractant binding inhibits CheA, CheY-P falls and runs lengthen; repellent binding stimulates CheA and tumbling rises. Adaptation gives memory: CheR steadily methylates MCPs, which raises CheA activity, while CheB, activated when CheA phosphorylates it, removes methyl groups. During a sustained attractant stimulus methylation creeps up until tumbling returns to its baseline, so the cell responds to changes, not to absolute levels, over a very wide concentration range.

Mutant phenotypes, all non-chemotactic
Mutant lackingSwimmingReason
CheA, CheW or CheYSmooth: runs onlyNo CheY-P can be made
CheZTumblyCheY-P is not removed
CheRSmoothUnmethylated receptors keep CheA activity low
CheBTumblyOver-methylated receptors keep CheA activity high
MotA or MotBParalysed, though flagella are presentNo proton-driven torque

Chemotaxis is measured with Adler's capillary assay: a capillary holding attractant collects far more cells than one holding buffer, while one holding repellent collects fewer. Related responses are aerotaxis, phototaxis and magnetotaxis. Bacteria also move without flagella: gliding over surfaces (Myxococcus, Flavobacterium, many cyanobacteria), twitching by type IV pilus retraction, swarming of hyperflagellated cells across moist surfaces (Proteus), and the corkscrew motion of spirochaetes, which bores through viscous media.

Key takeaways

  • Small cells have high surface-to-volume ratios (3/r for a sphere), which favour uptake and fast growth; a 1 × 2 µm rod has S/V of 4.8 µm⁻¹.
  • Peptidoglycan is NAG–NAM β(1→4) strands cross-linked 4→3 (DAP in E. coli, lysine with a pentaglycine bridge in S. aureus). Gram-positives have thick peptidoglycan with teichoic acids; Gram-negatives a thin layer plus an LPS outer membrane with porins.
  • Wall synthesis: UDP-NAM-pentapeptide in the cytoplasm, lipid I and II on bactoprenol in the membrane, transglycosylation and transpeptidation outside (one D-Ala released per cross-link). Fosfomycin, D-cycloserine, bacitracin, vancomycin and β-lactams block successive steps. Membrane phospholipids come from phosphatidic acid via CDP-diacylglycerol.
  • Transport: passive and facilitated diffusion go downhill; ABC transporters use ATP, symporters and antiporters use ion gradients, the PTS phosphorylates its sugar using PEP. A 1:1 H⁺ symporter can accumulate up to 10^(|Δp|/59 mV)-fold.
  • Flagella are proton- (or Na⁺-) driven rotary motors with a flagellin filament, hook and ringed basal body; pili adhere, conjugate or twitch; capsules resist phagocytosis. PHB, polyphosphate, carboxysomes, magnetosomes and gas vesicles are inclusions, and endospores survive heat through a dehydrated, dipicolinate-rich core.
  • Chemotaxis is a biased random walk: counter-clockwise rotation gives runs, CheY-P drives clockwise tumbles; attractants inhibit CheA, repellents stimulate it, and CheR/CheB methylation adapts the receptors. Loss of CheA, CheW, CheY or CheR gives smooth swimmers; loss of CheZ or CheB gives tumblers.

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. A sacculus is digested with a muramidase and its 2,000 peptide stems are analysed: 1,310 are in monomers, 600 in dimers and 90 in trimers. Each dimer contains one 4→3 cross-link and each trimer two. The number of terminal D-alanine residues released by the transpeptidation reactions that formed these cross-links is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 360

    600 stems in dimers = 300 dimers = 300 cross-links; 90 stems in trimers = 30 trimers = 60 cross-links; total 360. Each transpeptidation removes the terminal D-Ala of the donor pentapeptide and uses that bond's energy to link D-Ala-4 to residue 3 of the acceptor, so 360 D-Ala are released. Counting stems instead of cross-links (690) is the usual slip.
  2. Lysozyme lyses Gram-positive bacteria in water by hydrolysing

    1. the β(1→4) bond between N-acetylmuramic acid and N-acetylglucosamine
    2. the peptide cross-link between D-alanine and diaminopimelate or lysine
    3. the ester bonds that attach fatty acids to glycerol in the cell membrane
    4. the phosphodiester bonds that link the polyol units of wall teichoic acid
    Show answer

    Answer: A — the β(1→4) bond between N-acetylmuramic acid and N-acetylglucosamine

    Lysozyme is a muramidase: it cuts the glycan backbone between NAM and NAG, the sacculus loses its strength and the cell bursts under turgor. The peptide cross-link is the bond penicillin prevents from forming, the tempting distractor, but lysozyme does not cut it; membrane esters and teichoic acid phosphodiesters are not its substrates.
  3. Which of the following statements about peptidoglycan biosynthesis are correct?

    1. MraY transfers phospho-NAM-pentapeptide onto undecaprenyl phosphate, forming lipid I
    2. Transpeptidation outside the membrane is powered by ATP hydrolysed in the periplasm
    3. D-cycloserine blocks the production of the D-Ala–D-Ala dipeptide
    4. Bacitracin prevents recycling of the lipid carrier by blocking dephosphorylation of undecaprenyl pyrophosphate
    Show answer

    Answer: A — MraY transfers phospho-NAM-pentapeptide onto undecaprenyl phosphate, forming lipid I; C — D-cycloserine blocks the production of the D-Ala–D-Ala dipeptide; D — Bacitracin prevents recycling of the lipid carrier by blocking dephosphorylation of undecaprenyl pyrophosphate

    MraY makes lipid I on bactoprenol, D-cycloserine inhibits alanine racemase and D-Ala–D-Ala ligase, and bacitracin traps the carrier as its pyrophosphate. There is no ATP outside the membrane: transpeptidation is driven by the energy of the D-Ala–D-Ala bond broken as the terminal D-Ala is released.
  4. E. coli growing at 37 °C is shifted to 15 °C and allowed to adapt. Which change in its cytoplasmic membrane is expected?

    1. A higher proportion of saturated fatty acids, to stiffen the membrane against the cold
    2. Uptake of cholesterol from the medium, to raise the fluidity of the bilayer
    3. A higher proportion of unsaturated fatty acids, to keep the membrane fluid
    4. Conversion of ester-linked lipids into ether-linked isoprenoid lipids
    Show answer

    Answer: C — A higher proportion of unsaturated fatty acids, to keep the membrane fluid

    Cold makes membranes more ordered; cis double bonds put kinks in acyl chains and restore fluidity, so cells raise the unsaturated fraction (homeoviscous adaptation). More saturated chains would make the problem worse. E. coli does not take up sterols (mycoplasmas do), and ether-linked isoprenoid lipids are an archaeal feature, not something a bacterium switches to.
  5. A respiring bacterium has a membrane potential of −118 mV (inside negative), an internal pH of 7.8 and an external pH of 6.8. Take 2.303RT/F = 59 mV. For a permease that carries one H⁺ with each uncharged lactose molecule, the maximum ratio of internal to external lactose concentration at equilibrium is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1000

    Δp = Δψ − 59 ΔpH = −118 − 59(7.8 − 6.8) = −177 mV. At equilibrium the free energy of lactose accumulation equals that released by one H⁺ entering: 59 log(C_in/C_out) = 177, so log(C_in/C_out) = 3 and the ratio is 10³ = 1000. Using Δψ alone gives 100; forgetting that both parts add gives the wrong exponent.
  6. Which feature distinguishes group translocation by the phosphotransferase system (PTS) from every other transport mechanism?

    1. It hydrolyses ATP on a cytoplasmic subunit to drive a conformational change in the carrier
    2. The sugar is phosphorylated during passage, using phosphoryl groups that come from PEP
    3. It concentrates the unmodified sugar inside the cell by co-transporting protons
    4. It depends on a periplasmic binding protein that captures the sugar at low concentration
    Show answer

    Answer: B — The sugar is phosphorylated during passage, using phosphoryl groups that come from PEP

    In the PTS a phosphoryl group passes from PEP through enzyme I, HPr, EIIA and EIIB to the sugar as it crosses EIIC, so what arrives is sugar phosphate. ATP hydrolysis and periplasmic binding proteins describe ABC transporters, the tempting distractor, and proton co-transport of an unmodified sugar describes symporters such as LacY.
  7. E. coli cells fermenting glucose anaerobically, so that ATP and PEP come from substrate-level phosphorylation, are treated with a protonophore that collapses the proton motive force. Which uptake activity is abolished most directly?

    1. Uptake of glucose by the PEP-dependent phosphotransferase system
    2. Uptake of maltose by its binding-protein-dependent ABC transporter
    3. Accumulation of lactose against its gradient by the permease LacY
    4. Entry of glycerol down its gradient through the facilitator GlpF
    Show answer

    Answer: C — Accumulation of lactose against its gradient by the permease LacY

    LacY is an H⁺ symporter, so its driving force disappears the moment Δp collapses. The PTS uses PEP and the ABC transporter uses ATP, both still made by glycolysis, so they keep working, at least until futile ATP hydrolysis by the ATP synthase drains the cell. GlpF is facilitated diffusion and needs no energy at all.
  8. Which of the following statements about solute transport in bacteria are correct?

    1. Facilitated diffusion is saturable but cannot raise a solute above its external concentration
    2. Many ABC importers of Gram-negative bacteria depend on a periplasmic substrate-binding protein
    3. Outer-membrane porins pump solutes against their gradient using ATP
    4. TonB-dependent receptors for siderophores are energised by the proton motive force of the cytoplasmic membrane
    Show answer

    Answer: A — Facilitated diffusion is saturable but cannot raise a solute above its external concentration; B — Many ABC importers of Gram-negative bacteria depend on a periplasmic substrate-binding protein; D — TonB-dependent receptors for siderophores are energised by the proton motive force of the cytoplasmic membrane

    A facilitator speeds equilibration but cannot concentrate; Gram-negative ABC importers use periplasmic binding proteins (MalE for maltose); and the outer membrane, having no ATP or ion gradient of its own, borrows inner-membrane Δp through TonB–ExbB–ExbD. Porins are passive diffusion channels.
  9. A rod-shaped bacterium is 1.0 µm in diameter and 3.0 µm long overall, with hemispherical ends. Its surface-to-volume ratio in µm⁻¹ is ____ (to one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 4.5

    r = 0.5 µm and the cylindrical part is 3.0 − 2(0.5) = 2.0 µm long. S = 2πrl + 4πr² = 2π(0.5)(2) + 4π(0.25) = 2π + π = 3π µm². V = πr²l + (4/3)πr³ = 0.5π + 0.1667π = 0.6667π µm³. S/V = 3π/0.6667π = 4.5 µm⁻¹. Treating the whole 3 µm as cylinder, or forgetting the end caps, gives a different value.
  10. Compared with the flagellar basal body of E. coli, that of the Gram-positive Bacillus subtilis lacks

    1. the MS ring embedded in the cytoplasmic membrane
    2. the cytoplasmic C ring that acts as the rotor and the directional switch
    3. the stator units that conduct ions to drive rotation
    4. the L and P rings of the outer membrane and peptidoglycan
    Show answer

    Answer: D — the L and P rings of the outer membrane and peptidoglycan

    The L ring sits in the outer membrane and the P ring in the thin peptidoglycan layer of Gram-negatives; a Gram-positive has no outer membrane, so it has neither. The MS ring, the C ring switch and the stators are essential motor parts present in both.
  11. A capsulated strain of a bacterium is far more resistant to being engulfed by host phagocytes than a mutant that makes no capsule. The best explanation is that the capsule

    1. secretes an A–B toxin that kills the phagocyte before it can act
    2. is a loose slime layer that dissolves away as soon as a phagocyte approaches
    3. masks surface molecules and hinders their recognition and binding by the phagocyte
    4. replaces the cell wall, so there is no rigid structure for the phagocyte to grip
    Show answer

    Answer: C — masks surface molecules and hinders their recognition and binding by the phagocyte

    A firmly attached capsule covers the cell-surface molecules that phagocytes and complement would otherwise recognise, so the cell is poorly engulfed and more virulent; this is why rough, unencapsulated pneumococci are avirulent. A capsule is not a toxin and does not dissolve on contact, and it lies outside the wall rather than replacing it.
  12. Which of the following statements about bacterial surface appendages are correct?

    1. The flagellar filament grows by adding flagellin subunits at its distal tip
    2. Type IV pili pull a cell along a surface by extending, attaching and then retracting
    3. The conjugative pilus is the channel through which ribosomes pass to a recipient cell
    4. Fimbriae such as type 1 attach bacteria to host surfaces through tip adhesins
    Show answer

    Answer: A — The flagellar filament grows by adding flagellin subunits at its distal tip; B — Type IV pili pull a cell along a surface by extending, attaching and then retracting; D — Fimbriae such as type 1 attach bacteria to host surfaces through tip adhesins

    Flagellin travels up the hollow filament and adds at the tip; type IV pili drive twitching motility by extension and retraction; and fimbrial tip adhesins (FimH on mannose) mediate attachment. The conjugative pilus brings the cells together for DNA transfer; it does not transfer ribosomes.
  13. A bacterium swims in smooth runs at a steady 25 µm per second. During one uninterrupted run that lasts 1.2 seconds, the straight-line distance it covers is ____ µm.

    Numerical answer — type the value.

    Show answer

    Answer: 30

    Distance = speed × time = 25 µm s⁻¹ × 1.2 s = 30 µm, about thirty times the cell's own length. Because inertia is negligible at this scale, the cell moves only while its motor turns and stops almost at once when it stops.
  14. Which property of the bacterial endospore is most directly responsible for its extreme resistance to moist heat?

    1. A low water content in the core, together with calcium dipicolinate and bound SASPs
    2. A thick capsule of polysaccharide that insulates the spore from the surroundings
    3. A high rate of DNA repair that corrects heat damage as fast as it occurs
    4. A set of gas vesicles that reflect heat away from the dehydrated core
    Show answer

    Answer: A — A low water content in the core, together with calcium dipicolinate and bound SASPs

    The dehydrated, dipicolinate-rich core with its DNA coated by small acid-soluble spore proteins keeps proteins and DNA from denaturing, which gives heat resistance; the coats add chemical resistance. A spore is dormant, so active DNA repair is not the mechanism, endospores are not capsules, and gas vesicles are buoyancy organelles unrelated to heat resistance.
  15. Which of the following inclusion–function pairings are correct?

    1. Poly-β-hydroxybutyrate granules: storage of carbon and energy
    2. Gas vesicles: lowering cell density so the cell can float towards light
    3. Carboxysomes: housing RuBisCO to concentrate CO₂ for fixation
    4. Magnetosomes: storing phosphate that stains metachromatically
    Show answer

    Answer: A — Poly-β-hydroxybutyrate granules: storage of carbon and energy; B — Gas vesicles: lowering cell density so the cell can float towards light; C — Carboxysomes: housing RuBisCO to concentrate CO₂ for fixation

    PHB stores carbon and energy, gas vesicles provide buoyancy, and carboxysomes package RuBisCO with carbonic anhydrase to raise local CO₂. Magnetosomes are magnetite crystals used for orientation; the phosphate store that stains metachromatically is the polyphosphate (volutin) granule.
  16. In E. coli chemotaxis, which molecular event most directly causes a cell to tumble?

    1. Phosphorylated CheY binds the FliM switch and favours clockwise flagellar rotation
    2. CheZ dephosphorylates CheY-P, which would instead lock the motor into counter-clockwise rotation
    3. An attractant binds the chemoreceptor and inhibits the kinase activity of CheA
    4. CheR methylates the receptor, which immediately relaxes the flagellar bundle
    Show answer

    Answer: A — Phosphorylated CheY binds the FliM switch and favours clockwise flagellar rotation

    CheY-P is the output of the pathway: it binds FliM in the C ring, switches rotation clockwise, the bundle flies apart and the cell tumbles. CheZ removing phosphate from CheY-P and an attractant inhibiting CheA both reduce tumbling, the opposite effect, and receptor methylation acts over seconds as adaptation rather than triggering a tumble.
  17. A mutant E. coli strain swims in continuous smooth runs and almost never tumbles, even when a repellent is added. A defect in which gene best explains this phenotype?

    1. cheZ, the phosphatase that dephosphorylates CheY-P
    2. cheA, the histidine kinase that phosphorylates CheY
    3. cheB, the methylesterase that demethylates the chemoreceptors
    4. motB, a stator protein needed to turn the flagellar motor
    Show answer

    Answer: B — cheA, the histidine kinase that phosphorylates CheY

    Without CheA there is no phosphoryl source for CheY, so CheY-P never forms, the motor stays counter-clockwise and the cell only runs, whatever the stimulus. Losing CheZ or CheB raises CheY-P or receptor methylation and makes a cell tumble more, the opposite phenotype, and a motB defect paralyses the flagella so the cell cannot swim at all.
  18. A bacterium in a uniform, unchanging concentration of attractant settles back to its normal mix of runs and tumbles after an initial burst of smooth swimming. This adaptation is brought about by

    1. saturation of the chemoreceptors, which can no longer bind any further attractant
    2. slow methylation of the receptors by CheR, which restores CheA activity
    3. exhaustion of the cell's ATP, so CheA can no longer phosphorylate CheY
    4. rapid breakdown of the attractant by periplasmic enzymes near the receptor
    Show answer

    Answer: B — slow methylation of the receptors by CheR, which restores CheA activity

    During a sustained attractant signal CheR keeps methylating the receptors, which raises CheA activity back up until tumbling returns to baseline, so the cell responds to changes rather than absolute levels. Receptor saturation would not reset behaviour, chemotaxis is not limited by ATP in this way, and the cell does not need to destroy the attractant to adapt.