Environmental Microbiology: Microbial Groups, Cell Biology, Metabolism, Growth Kinetics and Water-Quality Microbiology

Section 3 of the GATE Environmental Science and Engineering (ES) paper, taken in the order of its five sub-headings and kept at the depth the syllabus words them. First the organisms: prokaryotes and eukaryotes, how they are classified, and what each group does in wastewater treatment, bioremediation and the carbon, nitrogen, sulphur and phosphorus cycles. Then the cell: the four classes of biomolecule, the membrane, the wall, the outer membrane and the glycocalyx, and two behaviours an engineer now has to know — antimicrobial resistance and quorum sensing. Then metabolism: phosphorylation, glycolysis, the TCA cycle, the electron transport chain, fermentation and anaerobic respiration, energy balances and enzyme kinetics. Then growth: specific growth rate and doubling time, Monod’s model, culture media, batch and continuous culture, environmental factors and control. Last, health: pathogens and how they travel, indicator organisms, and counting coliforms by MPN and membrane filtration.

1. Prokaryotic and eukaryotic microorganisms, and what they do in the environment

Prokaryotes — bacteria and archaea — have no nucleus or membrane-bound organelles; their DNA lies in a nucleoid, often with extra plasmids, and their ribosomes are 70S. Eukaryotes — fungi, algae and protozoa, and the helminths (worms) that are multicellular animals — have a nucleus, mitochondria (and chloroplasts in algae) and 80S ribosomes. Viruses are not cells: they are nucleic acid in a protein coat that replicates only inside a host. The modern three-domain classification (Bacteria, Archaea, Eukarya) rests on 16S/18S ribosomal RNA sequences, and the same sequencing reveals the microbial diversity of a sludge floc or a soil — thousands of taxa, most never grown in culture.

Classifying microbes by how they live
BasisClassesEnvironmental example
Carbon sourceautotroph (CO₂) or heterotroph (organic carbon)nitrifiers are autotrophs; the BOD-removing bacteria of activated sludge are heterotrophs
Energy sourcephototroph (light) or chemotroph (chemical oxidation); lithotroph (inorganic donor) or organotrophalgae in stabilisation ponds are photoautotrophs; Nitrosomonas is a chemolithoautotroph
Oxygenobligate aerobe, facultative, microaerophile, aerotolerant, obligate anaerobemethanogens are obligate anaerobes; most activated-sludge bacteria are facultative
Temperaturepsychrophile (below about 20 °C), mesophile (about 20–45 °C), thermophile (above about 45 °C)compost piles pass through a thermophilic stage that kills pathogens
  • In wastewater treatment: heterotrophic bacteria oxidise BOD and build flocs held together by extracellular polymers; nitrifiers (Nitrosomonas: NH₄⁺ → NO₂⁻; Nitrobacter: NO₂⁻ → NO₃⁻) and denitrifiers (NO₃⁻ → N₂ in the absence of oxygen) remove nitrogen; phosphate-accumulating organisms take up excess phosphorus after an anaerobic stage; methanogenic archaea finish anaerobic digestion; protozoa and rotifers graze free bacteria and signal a healthy sludge; filamentous bacteria in excess cause bulking.
  • In bioremediation: microbes degrade or immobilise pollutants in place (in situ) or after excavation (ex situ). Biostimulation adds nutrients or oxygen to the native population; bioaugmentation adds selected organisms; bioventing, land farming, composting and biopiles are its engineered forms, and some bacteria reduce toxic Cr(VI) to Cr(III) or U(VI) to insoluble U(IV).
  • In biogeochemical cycles: nitrogen fixation (N₂ → NH₃, by Rhizobium, Azotobacter and cyanobacteria), ammonification, nitrification, denitrification and anammox (NH₄⁺ + NO₂⁻ → N₂, anaerobically); sulphate reduction to H₂S (Desulfovibrio) and sulphide oxidation back to sulphuric acid (Thiobacillus — the cause of sewer crown corrosion); carbon fixation by photosynthesis and its return by respiration and methanogenesis; and phosphorus solubilisation from mineral phosphates.

2. Cell chemistry and cell biology

Four classes of macromolecule build a cell. Proteins are chains of twenty amino acids joined by peptide bonds; their primary structure (sequence) folds into secondary structures (α-helix, β-sheet), a tertiary shape and, for multi-chain proteins, a quaternary arrangement — and the shape is what makes an enzyme work, which is why heat and extreme pH, by unfolding it, kill. Nucleic acids are chains of nucleotides: DNA is a double helix of deoxyribose nucleotides with the bases A, T, G and C paired A–T and G–C; RNA is single-stranded, uses ribose and U in place of T, and comes as messenger, transfer and ribosomal RNA. Lipids include the phospholipids whose bilayer is the cell membrane (archaea use distinctive ether-linked lipids). Polysaccharides store energy (glycogen, starch), build walls (peptidoglycan in bacteria, cellulose in algae and plants, chitin in fungi) and make up most of the extracellular slime.

The bacterial envelope, from inside out
StructureCompositionFunction
Cytoplasmic membranephospholipid bilayer with embedded proteinsselective permeability; passive and active transport and group translocation; site of the electron transport chain and proton motive force in prokaryotes
Cell wallpeptidoglycan — thick with teichoic acids in Gram-positive cells, thin in Gram-negative cellsshape and resistance to osmotic lysis; the Gram stain distinguishes the two types
Outer membrane (Gram-negative only)lipopolysaccharide (LPS) outer leaflet with porin channelsa barrier to many antibiotics and detergents; LPS is the endotoxin
Glycocalyx (capsule or slime layer)extracellular polysaccharides and proteinsattachment to surfaces, biofilm formation, protection from drying and grazing; binds cells into settleable activated-sludge flocs

Antimicrobial resistance arises when bacteria survive a drug that once killed them. The mechanisms are enzymatic inactivation (β-lactamases that open penicillin’s ring), efflux pumps that expel the drug, alteration of the drug’s target, and reduced permeability. Resistance genes spread vertically to daughter cells and horizontally between cells by conjugation (plasmid transfer through a pilus), transformation (uptake of free DNA) and transduction (carried by a bacteriophage). Wastewater treatment plants, where dense mixed populations meet sub-lethal antibiotic residues from hospitals and homes, are recognised hotspots for this exchange. Quorum sensing is cell-to-cell signalling by diffusible autoinducers — acyl-homoserine lactones in Gram-negative bacteria, short peptides in Gram-positive ones: when the population is dense enough for the signal to pass a threshold, genes for biofilm formation, bioluminescence or virulence switch on together. Blocking it (quorum quenching) is studied as a way to limit membrane biofouling.

3. Microbial metabolism and enzyme kinetics

Metabolism is catabolism, which breaks substrates down and releases energy, coupled to anabolism, which uses that energy to build cell material. The energy currency is ATP, made in three ways: substrate-level phosphorylation (a phosphate transferred directly from a high-energy intermediate, as in glycolysis), oxidative phosphorylation (driven by the proton gradient an electron transport chain creates) and photophosphorylation (the same, driven by light). Glycolysis turns one glucose into two pyruvate with a net gain of 2 ATP and 2 NADH. With oxygen, pyruvate becomes acetyl-CoA and enters the TCA cycle, each turn of which releases 2 CO₂ and yields 3 NADH, 1 FADH₂ and 1 ATP (or GTP). The NADH and FADH₂ pass their electrons down the electron transport chain to a terminal acceptor; the protons pumped out return through ATP synthase. Aerobic respiration of glucose therefore yields far more ATP than glycolysis alone — up to 38 in the classical prokaryotic accounting.

Three ways to dispose of electrons
ModeTerminal electron acceptorEnergy yield and example
Aerobic respirationO₂, externalhighest; activated sludge, trickling filters
Anaerobic respirationexternal inorganic acceptors — NO₃⁻, SO₄²⁻, CO₂ — used through an electron transport chainlower than aerobic, falling down the redox ladder; denitrification, sulphate reduction, methanogenesis
Fermentationan internal organic intermediate (pyruvate or its products); no electron transport chainlowest, about 2 ATP per glucose by substrate-level phosphorylation; lactic acid, ethanol and mixed-acid fermentations, the acidogenic step of digestion

Energy balances link this chemistry to sludge production. Writing the electron donor and acceptor as half-reactions per electron equivalent, the free energy released decides how much can be spent on synthesis: aerobic heterotrophs, with the most energy per electron, convert a large fraction of substrate into new cells (a high yield Y, of the order of 0.4–0.6 g VSS/g BOD), whereas methanogens, with little energy per electron, grow slowly and produce little sludge — the practical reason anaerobic treatment makes a fraction of the sludge of aerobic treatment. Enzymes are protein catalysts that lower activation energy without being consumed; many need cofactors (metal ions) or coenzymes (NAD⁺, FAD). Their rate follows Michaelis–Menten kinetics, v = V_max S/(K_m + S), where K_m is the substrate concentration giving half the maximum rate. The double-reciprocal Lineweaver–Burk plot, 1/v = (K_m/V_max)(1/S) + 1/V_max, turns it into a straight line. A competitive inhibitor raises the apparent K_m and leaves V_max unchanged; a non-competitive inhibitor lowers V_max and leaves K_m unchanged.

4. Growth and control of microorganisms

Microbes need carbon, an energy source, the macronutrients N and P (a BOD : N : P ratio of about 100 : 5 : 1 is the usual rule for aerobic treatment), sulphur, trace metals and sometimes growth factors such as vitamins. In a closed batch culture a population passes through the lag, exponential (log), stationary and death phases. In the exponential phase dX/dt = μX, where μ is the specific growth rate, so X = X₀e^(μt) and the doubling time is t_d = ln 2/μ. Counting by generations, the number of generations is n = (log N − log N₀)/log 2 and the generation time is g = t/n. Monod’s model relates μ to the limiting substrate: μ = μ_max S/(K_s + S), where K_s, the half-saturation constant, is the substrate concentration at which μ = μ_max/2. At S ≫ K_s growth is zero order in substrate; at S ≪ K_s it is first order.

In continuous culture a chemostat is fed at flow F into volume V, giving a dilution rate D = F/V. At steady state the cells grow exactly as fast as they are washed out, so μ = D, and Monod’s equation then fixes the residual substrate: S = K_s D/(μ_max − D). If D is raised past the maximum growth rate the feed allows, D_crit = μ_max S₀/(K_s + S₀), the culture washes out. This is the same logic by which an activated-sludge plant must hold its solids retention time above the minimum the slowest organisms need — nitrifiers wash out first. Culture media are defined (synthetic) or complex, liquid or solidified with agar, and are made selective (inhibiting unwanted organisms), differential (showing a reaction by colour) or enriched: MacConkey and EMB agars are both selective and differential for lactose-fermenting Gram-negatives, which is why they appear in coliform work.

Environmental factors shift growth: rate roughly doubles for each 10 °C rise within the organism’s range and collapses above its optimum; most bacteria prefer near-neutral pH (fungi tolerate acid); low water activity and high salt dehydrate cells; oxygen is required, tolerated or lethal according to the organism’s class. Control is physical or chemical. Physical: moist heat (autoclaving at 121 °C and about 15 psi gauge for 15 minutes kills spores), dry heat, pasteurisation, filtration through membranes of about 0.22–0.45 μm, and ultraviolet radiation near 254–260 nm, which damages DNA. Chemical: chlorine and its compounds, ozone, alcohols, phenolics, aldehydes and quaternary ammonium compounds. Killing is approximately first order, N = N₀e^(−kt), so each equal interval removes the same fraction; the Chick–Watson law adds the disinfectant concentration, ln(N/N₀) = −k′Cⁿt, which is the origin of the "C × t" design of disinfection.

⚠️ μ = D only at steady state, and only below washout
With μ_max = 0.4 h⁻¹, K_s = 50 mg/L and S₀ = 450 mg/L, washout begins at D = 0.4 × 450/500 = 0.36 h⁻¹, not at 0.4 h⁻¹: the feed concentration limits the fastest growth the reactor can sustain. At D = 0.1 h⁻¹ the residual substrate is S = 50 × 0.1/(0.4 − 0.1) = 16.7 mg/L, independent of S₀.

5. Microbiology and health: pathogens, indicators, MPN and membrane filtration

Waterborne pathogens by group
GroupExamples and diseaseEngineering note
BacteriaVibrio cholerae (cholera), Salmonella Typhi (typhoid), Shigella (dysentery), pathogenic E. coli, Campylobacter; Legionella (inhaled from aerosols)readily killed by chlorine
Viruseshepatitis A and E, rotavirus, norovirus, enteroviruses including poliovirusvery small, low infectious dose, more resistant than bacteria
ProtozoaGiardia lamblia, Cryptosporidium parvum, Entamoeba histolytica (amoebic dysentery)cysts and oocysts resist chlorine (Cryptosporidium strongly) — removed by filtration, inactivated by UV
HelminthsAscaris (roundworm), hookworm, Taenia (tapeworm)eggs are large and settle — the concern in sludge and wastewater reuse on crops

Modes of transmission are classified by how water is involved: waterborne (the pathogen is drunk — cholera, typhoid, hepatitis), water-washed (spread by too little water for hygiene — trachoma, scabies, and faecal–oral diseases again), water-based (the pathogen spends part of its life cycle in an aquatic host — schistosomiasis in snails, guinea worm), and water-related insect vector (mosquitoes breeding in water — malaria, dengue, filariasis). The faecal–oral route through contaminated water, food and hands is the dominant one for enteric pathogens.

Pathogens are too many, too sparse and too hard to culture to test for routinely, so water is tested for indicator organisms instead. A good indicator is present whenever faecal contamination (and so possibly pathogens) is present and absent otherwise; is more numerous than the pathogens; survives at least as long in water and resists treatment at least as well; is not itself harmful; and is quick and cheap to count. Total coliforms (Gram-negative, non-spore-forming rods that ferment lactose with gas at 35 °C) include soil organisms; faecal (thermotolerant) coliforms, which do so at 44.5 °C, and above all E. coli are specific to faecal origin. Faecal streptococci (enterococci), spores of Clostridium perfringens (persistent, marking old or intermittent pollution) and coliphages (surrogates for viruses) complete the set. For drinking water the accepted criterion is that E. coli or thermotolerant coliforms are not detectable in any 100 mL sample.

Counting coliforms
MethodProcedureResult
Multiple-tube fermentation (MPN)series of tubes (e.g. five each of 10, 1 and 0.1 mL); presumptive test in lactose or lauryl tryptose broth at 35 °C for 24–48 h (gas = positive); confirmed test in brilliant green lactose bile broth; completed test on agara statistical estimate, MPN/100 mL, read from tables of the pattern of positive tubes, or by Thomas’s formula: MPN/100 mL = 100P/√(N T), where P = positive tubes, N = mL of sample in negative tubes, T = mL of sample in all tubes
Membrane filtrationa measured volume filtered through a 0.45 μm membrane, which is incubated on m-Endo (coliform colonies with a metallic sheen) or on m-FC medium at 44.5 °C for faecal coliformsa direct count: colonies/100 mL = colonies counted × 100/mL filtered; choose the volume so that the plate carries a countable number (about 20–80 coliform colonies)
ℹ️ MPN is an estimate, MF is a count
MPN handles turbid samples and gives confidence limits but takes two to four days; membrane filtration gives a direct count within a day and uses large volumes of clean water, but turbidity clogs the filter and suspended solids hide colonies. That is why MF suits drinking water and MPN suits wastewater and sediment.

Key takeaways

  • Prokaryotes lack a nucleus and have 70S ribosomes; classify microbes by carbon source, energy source, oxygen and temperature, and know who does what: heterotrophs remove BOD, nitrifiers are autotrophs, methanogens are archaea and obligate anaerobes.
  • Gram-positive walls are thick peptidoglycan; Gram-negative cells add an LPS outer membrane. Resistance genes spread horizontally by conjugation, transformation and transduction; quorum sensing switches on density-dependent genes through autoinducers.
  • Glycolysis nets 2 ATP; respiration uses an external acceptor through an electron transport chain, fermentation an internal organic one. Michaelis–Menten v = V_max S/(K_m + S); competitive inhibition raises K_m, non-competitive lowers V_max.
  • t_d = ln 2/μ; Monod μ = μ_max S/(K_s + S); in a chemostat μ = D and S = K_s D/(μ_max − D), with washout above μ_max S₀/(K_s + S₀). Killing is first order in time, and Chick–Watson adds concentration.
  • Indicators stand in for pathogens: faecal coliforms grow at 44.5 °C and E. coli is the specific faecal marker. MPN by Thomas’s formula is 100P/√(NT); membrane filtration counts colonies × 100/mL filtered.

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. Which of the following is a prokaryote?

    1. a methanogenic archaeon
    2. Giardia lamblia
    3. a green alga
    4. a filamentous fungus
    Show answer

    Answer: A — a methanogenic archaeon

    Archaea, like bacteria, have no nucleus and 70S ribosomes, so they are prokaryotes, though a separate domain from bacteria. Giardia is a protozoan, algae and fungi are eukaryotes with nuclei and organelles.
  2. Which of the following statements about microbial nitrogen transformations are correct?

    1. Nitrosomonas oxidises ammonium to nitrite
    2. Denitrification proceeds fastest at high dissolved oxygen
    3. Anammox converts ammonium and nitrite to N₂ under anaerobic conditions
    4. Nitrogen fixation reduces N₂ to ammonia
    Show answer

    Answer: A — Nitrosomonas oxidises ammonium to nitrite; C — Anammox converts ammonium and nitrite to N₂ under anaerobic conditions; D — Nitrogen fixation reduces N₂ to ammonia

    (a) is the first step of nitrification. (c) is anaerobic ammonium oxidation, NH₄⁺ + NO₂⁻ → N₂. (d) is correct: nitrogenase reduces N₂ to NH₃. (b) is false — denitrifiers use nitrate as their electron acceptor only when oxygen is absent, so denitrification needs anoxic conditions.
  3. Lipopolysaccharide is a characteristic component of:

    1. the outer membrane of Gram-negative bacteria
    2. the thick peptidoglycan wall of Gram-positive bacteria
    3. the nuclear envelope of fungi
    4. the capsid of a virus
    Show answer

    Answer: A — the outer membrane of Gram-negative bacteria

    LPS forms the outer leaflet of the Gram-negative outer membrane and acts as endotoxin. Gram-positive walls carry teichoic acids instead; fungi have no LPS; a viral capsid is protein.
  4. Which of the following statements about antimicrobial resistance and quorum sensing are correct?

    1. Conjugation can transfer a resistance plasmid between bacteria of different species
    2. Acyl-homoserine lactones are autoinducers used by Gram-negative bacteria
    3. Quorum-sensing responses switch on only when the cell density is high enough
    4. Resistance genes cannot move between cells inside a wastewater treatment plant
    Show answer

    Answer: A — Conjugation can transfer a resistance plasmid between bacteria of different species; B — Acyl-homoserine lactones are autoinducers used by Gram-negative bacteria; C — Quorum-sensing responses switch on only when the cell density is high enough

    (a) plasmids cross species boundaries by conjugation, which is why resistance spreads so fast. (b) and (c) describe quorum sensing: the signal accumulates with population density and triggers genes at a threshold. (d) is false — dense mixed populations and antibiotic residues make treatment plants hotspots of horizontal gene transfer.
  5. What is the net number of ATP molecules produced by glycolysis from one molecule of glucose?

    Numerical answer — type the value.

    Show answer

    Answer: 2

    Glycolysis invests 2 ATP in its first half and makes 4 by substrate-level phosphorylation in its second, a net gain of 2 ATP (with 2 NADH). Quoting 4 forgets the investment; quoting 36–38 is complete aerobic respiration.
  6. The essential difference between fermentation and anaerobic respiration is that fermentation:

    1. uses nitrate as the terminal electron acceptor
    2. requires traces of oxygen to proceed
    3. passes electrons to an internal organic intermediate without an electron transport chain
    4. yields more ATP per glucose than aerobic respiration
    Show answer

    Answer: C — passes electrons to an internal organic intermediate without an electron transport chain

    In fermentation the electrons from glucose end up on an organic product of its own breakdown (pyruvate becoming lactate or ethanol), and ATP comes only by substrate-level phosphorylation. Anaerobic respiration uses an external inorganic acceptor (NO₃⁻, SO₄²⁻, CO₂) through an electron transport chain. Option A describes denitrification, a form of anaerobic respiration; option D is the reverse of the truth.
  7. A competitive inhibitor of an enzyme:

    1. lowers V_max and leaves K_m unchanged
    2. raises the apparent K_m and leaves V_max unchanged
    3. lowers both K_m and V_max in proportion
    4. has no effect at any substrate concentration
    Show answer

    Answer: B — raises the apparent K_m and leaves V_max unchanged

    A competitive inhibitor competes for the active site, so enough substrate outcompetes it: the maximum rate is still reached, but a higher S is needed for half of it — K_m rises, V_max is unchanged, and on a Lineweaver–Burk plot the lines meet on the 1/v axis. Option A is non-competitive inhibition; option C is uncompetitive.
  8. An enzyme follows Michaelis–Menten kinetics with K_m = 5 mg/L. At what substrate concentration, in mg/L, does the rate reach 80% of V_max?

    Numerical answer — type the value.

    Show answer

    Answer: 20

    0.8 = S/(K_m + S) gives 0.8K_m + 0.8S = S, so S = 4K_m = 20 mg/L. The rate approaches V_max only slowly: 50% needs S = K_m, 80% needs 4K_m, and 90% needs 9K_m.
  9. A bacterial culture in exponential growth has a specific growth rate of 0.35 h⁻¹. What is its doubling time, in hours, to two decimal places?

    Numerical answer — type the value.

    Show answer

    Answer: 1.98

    t_d = ln 2/μ = 0.693/0.35 = 1.98 h. Using 1/μ = 2.86 h gives the time for an e-fold increase, not a doubling.
  10. A culture grows from 10³ to 10⁹ cells/mL in 10 hours of exponential growth. How many generations has it passed through? Give the answer to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 19.9

    n = (log N − log N₀)/log 2 = (9 − 3)/0.301 = 19.9 generations, so the generation time is 10/19.9 = 0.50 h. Dividing the million-fold increase by 2 instead of taking its base-2 logarithm is the error to avoid.
  11. An organism has μ_max = 0.5 h⁻¹ and K_s = 20 mg/L. What is its specific growth rate, in h⁻¹, when the limiting substrate is at 30 mg/L?

    Numerical answer — type the value.

    Show answer

    Answer: 0.3

    μ = μ_max S/(K_s + S) = 0.5 × 30/(20 + 30) = 15/50 = 0.3 h⁻¹. At S = K_s = 20 mg/L it would be exactly half of μ_max, 0.25 h⁻¹.
  12. A chemostat is operated at a dilution rate of 0.1 h⁻¹ with an organism for which μ_max = 0.4 h⁻¹ and K_s = 50 mg/L. What is the steady-state substrate concentration in the reactor, in mg/L, to one decimal place?

    Numerical answer — type the value.

    Show answer

    Answer: 16.7

    At steady state μ = D = 0.1 h⁻¹. From Monod, S = K_s D/(μ_max − D) = 50 × 0.1/(0.4 − 0.1) = 5/0.3 = 16.7 mg/L. It does not depend on the feed concentration, which sets only the cell concentration.
  13. For the organism of the previous kind (μ_max = 0.4 h⁻¹, K_s = 50 mg/L) fed at a substrate concentration of 450 mg/L, above what dilution rate, in h⁻¹, does the chemostat wash out?

    Numerical answer — type the value.

    Show answer

    Answer: 0.36

    Washout occurs when D exceeds the growth rate at the feed concentration: D_crit = μ_max S₀/(K_s + S₀) = 0.4 × 450/(50 + 450) = 0.36 h⁻¹. Quoting μ_max = 0.4 h⁻¹ assumes the cells could grow at S ≫ K_s, which the feed does not provide.
  14. Disinfection kills a pathogen by first-order kinetics with k = 0.46 min⁻¹. What contact time, in minutes, gives a 3-log (99.9%) kill? Give the answer to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 15

    N/N₀ = 10⁻³ = e^(−kt), so t = ln 1000/k = 6.908/0.46 = 15.0 min. Using log₁₀ 1000 = 3 in place of ln 1000 gives 6.5 min, too short by the factor 2.303.
  15. Which of the following is NOT a requirement of an ideal indicator of faecal contamination?

    1. It is present whenever faecal pathogens are present
    2. It is more numerous than the pathogens
    3. It is itself pathogenic, so that its presence proves disease risk
    4. It survives in water at least as long as the pathogens
    Show answer

    Answer: C — It is itself pathogenic, so that its presence proves disease risk

    An indicator should be harmless, so that laboratories can handle it safely and routinely; its value is statistical association with faecal pollution, not pathogenicity. The other three are the standard criteria.
  16. Faecal (thermotolerant) coliforms are distinguished from total coliforms by their ability to ferment lactose with gas production at:

    1. 20 °C
    2. 35 °C
    3. 44.5 °C
    4. 60 °C
    Show answer

    Answer: C — 44.5 °C

    Total coliforms are incubated at 35 °C; the elevated temperature of 44.5 °C selects the thermotolerant organisms of warm-blooded gut origin, chiefly E. coli. 20 °C is the BOD incubation temperature, not a coliform test.
  17. In a membrane-filtration test, 50 mL of a water sample is filtered and 36 coliform colonies grow. What is the coliform count per 100 mL?

    Numerical answer — type the value.

    Show answer

    Answer: 72

    Count per 100 mL = colonies × 100/volume filtered = 36 × 100/50 = 72. Reporting 36 per 100 mL forgets that only half of 100 mL was filtered.
  18. Five tubes each inoculated with 10 mL of sample are tested and 3 are positive. Using Thomas’s formula, what is the MPN per 100 mL, to one decimal place?

    Numerical answer — type the value.

    Show answer

    Answer: 9.5

    P = 3 positive tubes; the 2 negative tubes hold N = 20 mL of sample; all tubes hold T = 50 mL. MPN/100 mL = 100P/√(NT) = 300/√(20 × 50) = 300/31.62 = 9.5. The simple ratio 3 × 100/50 = 6 ignores that a positive tube can hold more than one organism.