A student wants to enrich ammonia-oxidising bacteria from garden soil. Which medium and incubation will most selectively favour them over the far more numerous heterotrophs?

Mineral salts with ammonium as the only energy source and bicarbonate as the only carbon, aerobic, in the dark

Mineral salts with glucose and ammonium, aerobic, in the dark, so that ammonium is in plentiful supply

Mineral salts with mannitol and no combined nitrogen, aerobic, so that ammonium is made only by the cells

Mineral salts with lactate, sulphate and ammonium, anaerobic, so that oxygen cannot oxidise ammonium

Which of the following statements about culture media are correct?

MacConkey agar is both selective and differential

Mannitol salt agar uses 7.5% NaCl to select for staphylococci

MacConkey agar is both selective and differential

A medium containing peptone and yeast extract is chemically defined

Mannitol salt agar uses 7.5% NaCl to select for staphylococci

An auxotroph grows on a minimal medium lacking the growth factor it requires

Which biological indicator is routinely used to confirm that a steam autoclave cycle has achieved sterilisation?

Spores of Geobacillus stearothermophilus in a sealed vial

Vegetative cells of Staphylococcus aureus dried on a paper strip

Spores of Bacillus atrophaeus dried on a paper strip

Spores of Geobacillus stearothermophilus in a sealed vial

A suspension of Pseudomonas aeruginosa in nutrient broth

For a spore suspension, D at 121 °C is 1.5 min and D at 111 °C is 15 min. Assuming a straight-line relation between log D and temperature, which pair gives the z-value and the expected D at 131 °C?

z = 10 °C; D at 131 °C = 0.15 min

z = 10 °C; D at 131 °C = 0.75 min

z = 13.5 °C; D at 131 °C = 0.15 min

z = 1 °C; D at 131 °C = 0.015 min

z = 10 °C; D at 131 °C = 0.15 min

Which of the following statements about sterilisation and disinfection are correct?

Filtration through a 0.22 µm membrane is suitable for a heat-labile antibiotic solution

70% ethanol is a better disinfectant than absolute ethanol because water aids protein denaturation

Filtration through a 0.22 µm membrane is suitable for a heat-labile antibiotic solution

70% ethanol is a better disinfectant than absolute ethanol because water aids protein denaturation

UV light at 254 nm sterilises the contents of closed opaque plastic containers

Dry heat at 121 °C for 15 min is as effective as autoclaving at 121 °C for 15 min

A phase-contrast microscope lets a student see the endospores and inclusions of living, unstained bacteria. What makes this possible?

Small phase shifts caused by refractive-index differences are converted into differences in brightness

A shorter-wavelength light source raises the resolving power well beyond that of bright-field optics

Only light scattered by the cell edges enters the objective, so cells glow against a black field

Cell components absorb ultraviolet excitation light and emit visible light of a longer wavelength

Which of the following statements about fluorescence and electron microscopy are correct?

A fluorochrome emits light of longer wavelength than the light that excites it

TEM images are formed by electrons passing through ultrathin sections stained with heavy metals

A fluorochrome emits light of longer wavelength than the light that excites it

TEM images are formed by electrons passing through ultrathin sections stained with heavy metals

SEM resolves internal structures of a cell better than TEM does

In negative staining of viruses the particles appear dark against a light background

In a tube agglutination test, doubling dilutions of a patient's serum from 1:10 to 1:1280 are mixed with a killed bacterial suspension. Clumping is absent at 1:10 to 1:40, present at 1:80 to 1:640 and absent at 1:1280. Which interpretation is correct?

A prozone of antibody excess masks the low dilutions, and the titre is 640

The serum contains no specific antibody, and clumping in the middle tubes is non-specific

The titre is 1280, because the last tube marks the end point of antibody activity

A prozone of antibody excess masks the low dilutions, and the titre is 640

Antigen excess in the low dilutions blocks clumping, and the titre is 80

Why does a TaqMan hydrolysis-probe assay not need a melt-curve check for primer dimers, whereas a SYBR Green assay does?

The reporter is freed only when the polymerase cleaves a probe hybridised to the target sequence

SYBR Green fluoresces only with single-stranded DNA, so it reports the amplified product unreliably

The TaqMan reporter fluoresces whenever any double-stranded DNA is present in the reaction tube

The polymerase used with TaqMan probes proofreads and so destroys primer dimers as they form

The reporter is freed only when the polymerase cleaves a probe hybridised to the target sequence

A bacterium carries a CRISPR spacer that exactly matches a sequence in phage P. Why does the crRNA transcribed from that spacer not direct Cas9 to cut the bacterium's own CRISPR array, which contains the identical sequence?

In the array the sequence is flanked by repeats, not by the PAM that Cas9 needs beside its target

The array is methylated at every adenine, and methylation blocks Cas9 from binding host DNA

A crRNA pairs only with RNA, so Cas9 is guided to phage transcripts and never to DNA at all

Cas9 is made only after phage infection and is degraded before the array is next replicated

Which of the following statements about CRISPR-Cas systems are correct?

New spacers are captured from invading DNA by the Cas1–Cas2 complex and added at the leader end of the array

Cas13 targets RNA, and its collateral cleavage of reporter RNA is used in diagnostics

A single-guide RNA used for editing fuses the crRNA and the tracrRNA into one molecule

New spacers are captured from invading DNA by the Cas1–Cas2 complex and added at the leader end of the array

Cas9 cuts both DNA strands to leave staggered ends with four-nucleotide 5′ overhangs

Cas13 targets RNA, and its collateral cleavage of reporter RNA is used in diagnostics

A single-guide RNA used for editing fuses the crRNA and the tracrRNA into one molecule

Methods in Microbiology: Pure Culture, Sterilisation, Nutrition, Diagnosis, Microscopy, PCR and CRISPR-Cas

This chapter answers Section XL3.2 of XL3 Microbiology, one of the five optional sections of the Life Sciences (XL) paper: “Pure culture techniques; Disinfection and Sterilization: principles, methods and assessment of efficacy; Principles of microbial nutrition; Enrichment culture techniques for isolation of microorganisms; Antigen and antibody detection methods for microbial diagnosis; Light-, phase contrast-, fluorescence- and electron-microscopy; PCR, Real-time PCR for quantitation of microbes; CRISPR-Cas.” It teaches how a single kind of microbe is isolated, fed and counted, how microbes are killed and how the killing is measured, how they are seen, and how they are detected by their antigens, by the antibodies they raise and by their DNA. The numericals are plate counts from serial dilutions, log reductions and D- and z-values, the Abbe resolution limit, and copy numbers from real-time PCR cycle thresholds.

1. Pure culture techniques

A pure (axenic) culture descends from a single cell. It is obtained by aseptic technique (flamed loops, sterile media, work beside a flame or in a biosafety cabinet) and by spreading cells so thinly on a solid medium that each grows into a separate colony. Agar, a red-algal polysaccharide, is the solidifying agent of choice: few microbes digest it, it melts only at about 85 °C or above, yet stays liquid on cooling to about 45 °C and sets near 40 °C. A streak plate dilutes cells mechanically across sectors of the plate and is for isolation only. A spread plate spreads 0.1 mL of a dilution over the surface. A pour plate mixes 1 mL with molten agar cooled to 45–50 °C, so colonies grow both within and on the agar; brief heating can harm heat-sensitive cells. Pure cultures are kept by subculture, freezing in glycerol at −80 °C or freeze-drying.

🧠 Worked example: viable count from a serial dilution

CFU per mL = colonies ÷ (volume plated in mL × dilution). A culture is diluted tenfold in series; 0.1 mL of the 10⁻⁵ dilution gives 214 colonies and 0.1 mL of the 10⁻⁶ dilution gives 23. Count only the plate in the conventional 30–300 range (some standards use 25–250): 214 ÷ (0.1 × 10⁻⁵) = 2.14 × 10⁸ CFU mL⁻¹. Forgetting the 0.1 mL plated gives an answer ten times too low.

⚠️ A colony is not always one cell

Chains and clumps each give one colony, and dead or viable-but-non-culturable cells give none, so a plate count reports colony-forming units and is usually lower than a microscopic count, which includes dead cells.

2. Microbial nutrition and enrichment culture

Every cell needs a source of energy, of electrons (reducing power) and of carbon. Light-users are photo-, chemical-users chemo-; inorganic electron donors make an organism a litho-, organic ones an organo-troph; CO₂ as carbon makes it an auto-, organic carbon a heterotroph. Macronutrients are C, H, O, N, S, P, K, Mg, Ca and Fe; trace elements such as Mn, Zn, Co, Cu, Mo (in nitrogenase) and Ni (in urease and hydrogenases) are needed in tiny amounts. Growth factors (amino acids, purines and pyrimidines, vitamins) are required by auxotrophs but not by prototrophs. Media are defined (every component known) or complex (peptone, yeast extract); selective media suppress unwanted organisms, differential media make groups look different, and enriched media (blood, chocolate agar) support fastidious ones. MacConkey agar is both selective (bile salts and crystal violet inhibit Gram-positives) and differential (lactose fermenters turn pink with neutral red); mannitol salt agar selects staphylococci with 7.5% NaCl.

Nutritional types
Type	Energy; electrons; carbon	Example
Photolithoautotroph	Light; H₂O or H₂S; CO₂	Cyanobacteria, purple sulphur bacteria
Photoorganoheterotroph	Light; organic acids; organic carbon	Purple non-sulphur bacteria
Chemolithoautotroph	Oxidation of NH₃, NO₂⁻, H₂S, Fe²⁺ or H₂; CO₂	Nitrosomonas, Acidithiobacillus
Chemoorganoheterotroph	Organic compounds for all three	Escherichia coli, fungi, protozoa

Enrichment culture (Beijerinck and Winogradsky) designs a medium and conditions in which the wanted physiology wins, transfers repeatedly, then streaks to purify. No combined nitrogen, an organic carbon source and air enrich aerobic N₂ fixers such as Azotobacter; ammonium as the only energy source with bicarbonate as carbon, in the dark, enriches ammonia oxidisers; light, H₂S and no oxygen enrich phototrophic sulphur bacteria; a pollutant as the sole carbon source enriches its degraders. The Winogradsky column (mud with cellulose, sulphate and carbonate, lit from the side) builds opposing gradients of O₂ from the top and H₂S from the bottom, and the communities stratify along them: algae and cyanobacteria at the top, purple and green bacteria in the middle, sulphate reducers below.

3. Sterilisation and disinfection: principles, methods and efficacy

Sterilisation destroys or removes every viable organism, endospores included; it is absolute. Disinfection reduces pathogens on inanimate objects, antisepsis does so on living tissue, and sanitisation lowers counts to levels judged safe; none of these need kill spores. Agents are -cidal (kill) or -static (stop growth reversibly). Resistance falls roughly in this order: prions, bacterial endospores, mycobacteria, small non-enveloped viruses, vegetative bacteria and fungi, and, easiest of all, enveloped viruses.

Common methods
Method	Typical conditions and action	Used for
Autoclave (moist heat)	121 °C, about 15 psi above atmospheric, 15–20 min for small loads; coagulates proteins; kills spores	Media, glassware, waste
Hot-air oven (dry heat)	160–170 °C for about 2 h; oxidation; needs higher temperature and longer time than moist heat	Dry glassware, powders, oils
Pasteurisation	63 °C for 30 min or 72 °C for 15 s; kills pathogens, not spores	Milk, juices
Membrane filtration	0.22 µm pores retain bacteria; HEPA filters clean air	Heat-labile solutions: antibiotics, vitamins, serum
Ionising radiation	Gamma rays or electron beams; DNA breaks via radicals; penetrates packaging	Disposable plastics, syringes
UV light (about 260 nm)	Pyrimidine (thymine) dimers; very poor penetration	Surfaces, air, cabinets
Chemicals	Ethylene oxide gas and glutaraldehyde alkylate and are sporicidal; 70% ethanol, hypochlorite, iodine, phenolics and quaternary ammonium compounds disinfect	Heat-sensitive devices; surfaces, skin

Assessing efficacy. Death under a lethal agent is first order: the same fraction dies in each equal interval, so log N falls linearly with time. The D-value is the time (or dose) at a stated temperature that kills 90%, one log cycle; log reduction = log₁₀(N₀/N); the time to go from N₀ to N is D × log₁₀(N₀/N). The z-value is the temperature rise that cuts D tenfold. Example: survivors fall from 10⁶ to 10² in 12 min at 121 °C, a 4-log reduction, so D₁₂₁ = 3 min; reaching a sterility assurance level of 10⁻⁶ (one surviving item in a million) from 10⁶ needs 12 logs, 36 min. If D is 3 min at 121 °C and 30 min at 111 °C, z = 10 °C. In practice, biological indicators (spores of Geobacillus stearothermophilus for steam, of Bacillus atrophaeus for dry heat and ethylene oxide) prove a cycle works; chemical indicator tape only shows the temperature was reached. Disinfectants are compared by the phenol coefficient (the dilution of the test agent that kills a standard organism in a set time, divided by the equivalent dilution of phenol) and by use-dilution tests on contaminated carriers.

⚠️ Why 70% ethanol beats absolute ethanol

Water is needed for alcohol to denature proteins and penetrate cells; absolute ethanol dehydrates the surface and evaporates quickly. Neither kills endospores, so alcohol is a disinfectant or antiseptic, never a sterilant.

4. Light, phase-contrast, fluorescence and electron microscopy

Total magnification is objective × eyepiece, but resolution, the smallest separation at which two points are seen as two, is what limits a microscope. Diffraction sets it: d = 0.61 λ/NA (the Rayleigh form; Abbe's own form, d = λ/2NA, is also used, so a question must say which). The numerical aperture NA = n sin θ, where n is the refractive index between specimen and lens and θ is half the angle of the cone of light the objective accepts. Air (n = 1) caps NA below 1; immersion oil (n ≈ 1.5) lifts it to about 1.25–1.4. With λ = 550 nm and NA = 1.25, d = 0.61 × 550/1.25 ≈ 270 nm, so about 0.2 µm is the practical limit of light microscopy. Magnifying beyond about 1000 × NA is empty magnification: bigger, not sharper.

Microscopes compared
Microscope	How contrast or the image arises	Best for
Bright-field	Absorption by stained cells against a bright background	Gram, acid-fast, endospore stains; negative stains (India ink) for capsules
Dark-field	A hollow cone of light; only light scattered by the object enters the objective	Thin live spirochaetes such as Treponema
Phase-contrast (Zernike)	An annular condenser diaphragm and a phase plate turn phase shifts from refractive-index differences into brightness differences	Live, unstained cells, endospores, inclusions, motility
Fluorescence (epifluorescence, confocal)	A fluorochrome absorbs short-wavelength light and emits longer-wavelength light (Stokes shift); filters and a dichroic mirror pass only the emission; a confocal pinhole rejects out-of-focus light	DAPI for DNA, auramine for mycobacteria, labelled antibodies, rRNA probes (FISH), GFP fusions
Transmission electron (TEM)	Electrons pass through ultrathin sections stained with heavy metals, in vacuum, focused by electromagnetic lenses; resolution well below 1 nm	Internal structure, viruses by negative staining, cryo-EM of macromolecules
Scanning electron (SEM)	A focused beam scans a metal-coated surface; secondary electrons build the image	Three-dimensional surface views; resolution of a few nm, coarser than TEM

⚠️ What each method does not do

Phase contrast adds contrast, not resolution. Fluorescence emission is always at a longer wavelength than excitation. Electrons resolve far more because their wavelength is far shorter, but the vacuum means TEM and SEM look at dead, fixed specimens (cryo-EM preserves them frozen-hydrated). In negative staining the background is dark and the object stays light.

5. Antigen and antibody detection for diagnosis

Finding a microbial antigen shows the microbe or its product is present now; finding antibody shows the host has responded. IgM marks a recent or current infection; IgG persists after recovery or vaccination, so a single IgG titre cannot date an infection, but a fourfold or greater rise between acute and convalescent sera can. The titre is the reciprocal of the highest serum dilution that still gives a positive reaction.

Agglutination: antibody cross-links particles into visible clumps. The Widal test measures antibody to Salmonella Typhi O and H antigens; latex beads coated with antibody detect capsular antigens in cerebrospinal fluid. In the prozone, excess antibody coats every particle singly, so no lattice forms and undiluted serum can read falsely negative.
Precipitation of soluble antigen in gels (Ouchterlony double diffusion, radial immunodiffusion) is maximal at the zone of equivalence.
ELISA: an enzyme (horseradish peroxidase or alkaline phosphatase) on an antibody turns a substrate coloured. Indirect ELISA coats antigen and detects the patient's antibody with enzyme-labelled anti-human immunoglobulin (antibody screening). Sandwich ELISA captures antigen between two antibodies to different epitopes (antigen detection). In competitive ELISA the signal falls as analyte rises.
Lateral-flow rapid tests run the same chemistry on a strip with gold-labelled antibody, a test line and a control line; immunofluorescence uses fluorochrome-labelled antibody directly on the specimen or indirectly via a labelled second antibody; a Western blot separates proteins by size before probing, which makes it a confirmatory test.

6. PCR and real-time PCR for quantitation

PCR copies a stretch of DNA between two primers through repeated cycles of denaturation (about 94–95 °C), annealing of primers (typically 50–65 °C, a few degrees below their melting temperature) and extension at 72 °C by a thermostable DNA polymerase: Taq from Thermus aquaticus, which lacks 3′→5′ proofreading, or a proofreading enzyme such as Pfu from Pyrococcus furiosus. The reaction also needs dNTPs and Mg²⁺. Ideally the target doubles each cycle, N = N₀ × 2ⁿ; at efficiency E, N = N₀(1 + E)ⁿ. Thirty perfect cycles multiply one copy by 2³⁰ ≈ 1.07 × 10⁹ before reagents run out and the reaction plateaus. RNA viruses are detected after reverse transcription (RT-PCR). PCR finds unculturable and slow-growing pathogens such as Mycobacterium tuberculosis within hours, but it also detects DNA from dead cells.

Real-time (quantitative) PCR reads fluorescence every cycle. SYBR Green fluoresces on binding any double-stranded DNA, so a melt curve is needed to show a single product; a TaqMan hydrolysis probe carries a reporter and a quencher and fluoresces only when the 5′→3′ exonuclease of the polymerase cleaves the probe bound to its target, which makes it sequence-specific. The threshold cycle (Ct, also Cq) is the cycle at which fluorescence crosses a set threshold; the more template, the earlier, so lower Ct means more target. At 100% efficiency one cycle is a twofold difference and a tenfold difference is log₂10 = 3.32 cycles. A standard curve of Ct against log₁₀(copies) has slope −3.32 at 100% efficiency, and efficiency E = 10^(−1/slope) − 1.

🧠 Worked example: copies from a standard curve

A standard curve reads Ct = 40.0 − 3.32 log₁₀N₀. A sample gives Ct = 26.72. Then log₁₀N₀ = (40.0 − 26.72)/3.32 = 13.28/3.32 = 4.00, so N₀ = 10⁴ copies in the reaction. Scale up by the fraction of the extract that went into the tube to get copies per mL of specimen.

7. CRISPR-Cas: a prokaryotic immune system and a tool

A CRISPR locus (clustered regularly interspaced short palindromic repeats) is an array of identical repeats separated by unique spacers, each a fragment of a phage or plasmid that once invaded, flanked by cas genes; many bacteria and most archaea carry one. Immunity works in three stages. Adaptation: the Cas1–Cas2 complex captures a short protospacer from invading DNA, chosen next to a PAM (protospacer-adjacent motif), and inserts it as a new spacer at the leader end of the array, so the array is a heritable record of past infections. Expression: the array is transcribed and processed into short crRNAs; in type II systems a tracrRNA pairs with the repeat and RNase III cuts. Interference: a crRNA guides a Cas nuclease to matching invader nucleic acid and cuts it. The spacers in the cell's own array lack an adjacent PAM, which is how the system avoids attacking itself.

Class 1 systems (types I, III, IV) use multi-protein effector complexes; Class 2 systems use one large protein. Cas9 (type II) cuts double-stranded DNA with its HNH and RuvC domains, leaving a blunt break 3 bp upstream of the PAM (5′-NGG-3′ for the Streptococcus pyogenes enzyme). Cas12a (type V) needs no tracrRNA, recognises a T-rich PAM and makes a staggered cut; Cas13 (type VI) cuts RNA. For editing, crRNA and tracrRNA are fused into one single-guide RNA with a 20-nucleotide guide; the break is repaired by error-prone non-homologous end joining (knock-outs) or by homology-directed repair from a supplied template (precise edits), and a nuclease-dead dCas9 represses genes without cutting. Once bound to target, Cas12a and Cas13 also cut nearby reporter molecules indiscriminately (collateral cleavage), the basis of CRISPR diagnostics such as DETECTR (Cas12a) and SHERLOCK (Cas13).

Key takeaways

Pure cultures come from single colonies on streak, spread or pour plates; CFU per mL = colonies ÷ (volume plated × dilution), counted on plates in the stated range (conventionally 30–300).
Nutritional type = energy (photo/chemo) + electrons (litho/organo) + carbon (auto/hetero). Media are defined or complex, selective, differential or enriched; enrichment culture lets the wanted physiology outgrow the rest before streaking.
Microbial death is first order: time = D × log₁₀(N₀/N), z is the rise in temperature that cuts D tenfold. Autoclave 121 °C, 15 psi, 15–20 min; dry heat 160–170 °C, 2 h; 0.22 µm filters for heat-labile liquids; spores of G. stearothermophilus validate steam.
Resolution d = 0.61 λ/NA (state the form); oil immersion raises NA and light microscopy bottoms out near 0.2 µm. Phase contrast adds contrast to live cells, fluorescence emits at a longer wavelength, TEM resolves below 1 nm and SEM shows surfaces.
Antigen detection shows current presence; IgM or a fourfold rise in titre shows recent infection. Indirect ELISA detects antibody, sandwich ELISA detects antigen; the prozone gives false negatives in undiluted serum.
In qPCR lower Ct means more template; at 100% efficiency a tenfold difference is 3.32 cycles and E = 10^(−1/slope) − 1. CRISPR-Cas immunity runs adaptation, expression and interference with PAM-dependent targeting; Cas9 makes blunt DNA breaks, Cas12a and Cas13 power collateral-cleavage diagnostics.

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.

A bacterial culture is serially diluted tenfold. Spreading 0.1 mL of the 10⁻⁶ dilution gives 156 colonies (the 10⁻⁵ plate is too crowded to count). The viable count of the original culture, in units of 10⁸ CFU per mL, is ____ (to one decimal place).
Numerical answer — type the value.
Show answer
Answer: 15.6
CFU per mL = colonies ÷ (volume plated × dilution) = 156 ÷ (0.1 × 10⁻⁶) = 1.56 × 10⁹ CFU per mL = 15.6 × 10⁸. Ignoring the 0.1 mL plated gives 1.56, a tenfold error.
Purple non-sulphur bacteria growing anaerobically in the light, using malate as both their carbon source and their electron donor, are best described as
1. photolithoautotrophs
2. photoorganoheterotrophs
3. chemoorganoheterotrophs
4. chemolithoautotrophs
Show answer
Answer: B — photoorganoheterotrophs
Energy from light makes them photo-, an organic electron donor (malate) makes them organo-, and organic carbon makes them heterotrophs. Photolithoautotroph is the tempting distractor because these bacteria can also grow on H₂ or low sulphide with CO₂, but not under the stated conditions. The chemo- types do not use light.
A student wants to enrich ammonia-oxidising bacteria from garden soil. Which medium and incubation will most selectively favour them over the far more numerous heterotrophs?
1. Mineral salts with ammonium as the only energy source and bicarbonate as the only carbon, aerobic, in the dark
2. Mineral salts with glucose and ammonium, aerobic, in the dark, so that ammonium is in plentiful supply
3. Mineral salts with mannitol and no combined nitrogen, aerobic, so that ammonium is made only by the cells
4. Mineral salts with lactate, sulphate and ammonium, anaerobic, so that oxygen cannot oxidise ammonium
Show answer
Answer: A — Mineral salts with ammonium as the only energy source and bicarbonate as the only carbon, aerobic, in the dark
Ammonia oxidisers are aerobic chemolithoautotrophs: they need NH₄⁺ as electron donor, O₂ as acceptor and CO₂ as carbon. Leaving out organic carbon starves heterotrophs, and darkness excludes phototrophs. Adding glucose lets heterotrophs overgrow; an N-free mannitol medium enriches N₂ fixers such as Azotobacter; anaerobic lactate–sulphate medium enriches sulphate reducers, and without O₂ the nitrifiers cannot respire.
Which of the following statements about culture media are correct?
1. MacConkey agar is both selective and differential
2. A medium containing peptone and yeast extract is chemically defined
3. Mannitol salt agar uses 7.5% NaCl to select for staphylococci
4. An auxotroph grows on a minimal medium lacking the growth factor it requires
Show answer
Answer: A — MacConkey agar is both selective and differential; C — Mannitol salt agar uses 7.5% NaCl to select for staphylococci
MacConkey's bile salts and crystal violet select against Gram-positives, and lactose with neutral red differentiates fermenters (pink). Mannitol salt agar's high salt selects staphylococci. Peptone and yeast extract are of unknown exact composition, so such a medium is complex, not defined; and an auxotroph by definition cannot grow without its required growth factor.
A disinfectant applied for 10 min reduces the viable count on a contaminated carrier from 5.0 × 10⁶ CFU to 50 CFU. The log reduction achieved is ____.
Numerical answer — type the value.
Show answer
Answer: 5
Log reduction = log₁₀(N₀/N) = log₁₀(5.0 × 10⁶ / 50) = log₁₀(10⁵) = 5, a 99.999% kill. Subtracting the counts or taking the logarithm of only the starting count are the usual slips.
In a survivor-curve experiment at 121 °C, a spore count falls from 2.0 × 10⁶ to 2.0 × 10² in 10 min. Assuming first-order death, the time in minutes at 121 °C needed to reduce a load of 10⁵ spores per item to a survival probability of 10⁻⁶ per item is ____ (to one decimal place).
Numerical answer — type the value.
Show answer
Answer: 27.5
The experiment shows a 4-log fall in 10 min, so D₁₂₁ = 10/4 = 2.5 min. Going from 10⁵ to 10⁻⁶ is log₁₀(10⁵/10⁻⁶) = 11 log cycles, so t = 11 × 2.5 = 27.5 min. Counting only 5 logs (to one survivor) gives 12.5 min, the common error.
Which biological indicator is routinely used to confirm that a steam autoclave cycle has achieved sterilisation?
1. Vegetative cells of Staphylococcus aureus dried on a paper strip
2. Spores of Bacillus atrophaeus dried on a paper strip
3. Spores of Geobacillus stearothermophilus in a sealed vial
4. A suspension of Pseudomonas aeruginosa in nutrient broth
Show answer
Answer: C — Spores of Geobacillus stearothermophilus in a sealed vial
Geobacillus stearothermophilus spores are highly resistant to moist heat, so their death proves a steam cycle worked. Bacillus atrophaeus spores are the tempting distractor: they are the indicator for dry heat and ethylene oxide. Vegetative bacteria die far too easily to prove sterility.
For a spore suspension, D at 121 °C is 1.5 min and D at 111 °C is 15 min. Assuming a straight-line relation between log D and temperature, which pair gives the z-value and the expected D at 131 °C?
1. z = 10 °C; D at 131 °C = 0.75 min
2. z = 13.5 °C; D at 131 °C = 0.15 min
3. z = 1 °C; D at 131 °C = 0.015 min
4. z = 10 °C; D at 131 °C = 0.15 min
Show answer
Answer: D — z = 10 °C; D at 131 °C = 0.15 min
D falls tenfold (15 to 1.5 min) for a 10 °C rise, so z = 10 °C. Another 10 °C rise cuts D tenfold again: 1.5/10 = 0.15 min. Halving D for the second step (0.75 min) treats z like a doubling rule; 13.5 comes from subtracting the D-values, which has no meaning.
Which of the following statements about sterilisation and disinfection are correct?
1. Filtration through a 0.22 µm membrane is suitable for a heat-labile antibiotic solution
2. 70% ethanol is a better disinfectant than absolute ethanol because water aids protein denaturation
3. UV light at 254 nm sterilises the contents of closed opaque plastic containers
4. Dry heat at 121 °C for 15 min is as effective as autoclaving at 121 °C for 15 min
Show answer
Answer: A — Filtration through a 0.22 µm membrane is suitable for a heat-labile antibiotic solution; B — 70% ethanol is a better disinfectant than absolute ethanol because water aids protein denaturation
Membrane filtration removes bacteria without heat, and water makes 70% ethanol penetrate and denature better than absolute ethanol. UV penetrates very poorly and cannot reach inside opaque containers. Dry heat transfers heat far less effectively than saturated steam, so it needs about 160–170 °C for about 2 h.
Using d = 0.61 λ/NA, the resolution limit in nm of an oil-immersion objective of numerical aperture 1.30 viewing with green light of wavelength 520 nm is ____.
Numerical answer — type the value.
Show answer
Answer: 244
d = 0.61 × 520 nm ÷ 1.30 = 317.2 ÷ 1.30 = 244 nm, about 0.24 µm. Using Abbe's form λ/2NA instead would give 200 nm, which is why the stem states the formula. A higher NA or a shorter wavelength lowers d, that is, improves resolution.
A phase-contrast microscope lets a student see the endospores and inclusions of living, unstained bacteria. What makes this possible?
1. Small phase shifts caused by refractive-index differences are converted into differences in brightness
2. A shorter-wavelength light source raises the resolving power well beyond that of bright-field optics
3. Only light scattered by the cell edges enters the objective, so cells glow against a black field
4. Cell components absorb ultraviolet excitation light and emit visible light of a longer wavelength
Show answer
Answer: A — Small phase shifts caused by refractive-index differences are converted into differences in brightness
Zernike's annular diaphragm and phase plate turn the tiny phase retardation of light passing through denser structures into visible differences in brightness, so no stain is needed. Phase contrast does not improve resolution. Scattered light against a black field describes dark-field microscopy, the tempting distractor, and absorption with longer-wavelength emission describes fluorescence.
Which of the following statements about fluorescence and electron microscopy are correct?
1. A fluorochrome emits light of longer wavelength than the light that excites it
2. TEM images are formed by electrons passing through ultrathin sections stained with heavy metals
3. SEM resolves internal structures of a cell better than TEM does
4. In negative staining of viruses the particles appear dark against a light background
Show answer
Answer: A — A fluorochrome emits light of longer wavelength than the light that excites it; B — TEM images are formed by electrons passing through ultrathin sections stained with heavy metals
Part of the absorbed energy is lost before emission, so fluorescence is red-shifted (Stokes shift); TEM uses transmitted electrons and heavy-metal stains for contrast. SEM images surfaces with secondary electrons and has coarser resolution than TEM. Negative stain fills the background and outlines the particle, so the particle is light against a dark field.
A dengue antigen in patient serum is captured by an antibody coated on a plate and then detected with an enzyme-linked antibody against a different epitope of the same antigen. This format is
1. indirect ELISA
2. sandwich ELISA
3. competitive ELISA
4. direct immunofluorescence
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Answer: B — sandwich ELISA
The antigen is sandwiched between a capture antibody and a detection antibody to two different epitopes, which is the standard design for detecting antigens. Indirect ELISA, the tempting distractor, coats antigen on the plate and detects the patient's antibody with a labelled anti-human immunoglobulin. In competitive ELISA the signal falls as analyte rises, and immunofluorescence uses a fluorochrome on cells, not an enzyme in a well.
In a tube agglutination test, doubling dilutions of a patient's serum from 1:10 to 1:1280 are mixed with a killed bacterial suspension. Clumping is absent at 1:10 to 1:40, present at 1:80 to 1:640 and absent at 1:1280. Which interpretation is correct?
1. The serum contains no specific antibody, and clumping in the middle tubes is non-specific
2. The titre is 1280, because the last tube marks the end point of antibody activity
3. A prozone of antibody excess masks the low dilutions, and the titre is 640
4. Antigen excess in the low dilutions blocks clumping, and the titre is 80
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Answer: C — A prozone of antibody excess masks the low dilutions, and the titre is 640
In concentrated serum, antibody saturates each bacterium singly so no cross-linked lattice forms: the prozone. The titre is the reciprocal of the highest dilution still showing agglutination, 640. The 1:1280 tube is negative, so 1280 is not the titre; the antigen amount is the same in every tube, so antigen excess cannot explain failure only at low dilution, and a reaction that disappears on dilution beyond 1:640 is specific.
A real-time PCR standard curve for a pathogen gene reads Ct = 37.5 − 3.32 log₁₀N₀, where N₀ is the number of target copies in the reaction. DNA from 0.5 mL of swab transport medium is extracted into 50 µL, and 2 µL of this extract is used in a reaction that gives Ct = 27.54. Assuming complete extraction, the number of target copies per mL of transport medium, in units of 10⁴, is ____.
Numerical answer — type the value.
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Answer: 5
log₁₀N₀ = (37.5 − 27.54)/3.32 = 9.96/3.32 = 3.00, so the reaction held 10³ copies. The reaction used 2 of 50 µL, so the extract held 10³ × 25 = 2.5 × 10⁴ copies, all from 0.5 mL of medium: 2.5 × 10⁴ ÷ 0.5 = 5 × 10⁴ copies per mL. Stopping at copies per reaction (0.1 in these units) or forgetting the 0.5 mL (2.5) are the usual slips.
Why does a TaqMan hydrolysis-probe assay not need a melt-curve check for primer dimers, whereas a SYBR Green assay does?
1. SYBR Green fluoresces only with single-stranded DNA, so it reports the amplified product unreliably
2. The TaqMan reporter fluoresces whenever any double-stranded DNA is present in the reaction tube
3. The polymerase used with TaqMan probes proofreads and so destroys primer dimers as they form
4. The reporter is freed only when the polymerase cleaves a probe hybridised to the target sequence
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Answer: D — The reporter is freed only when the polymerase cleaves a probe hybridised to the target sequence
Fluorescence in a TaqMan assay requires the probe to hybridise to the specific target and be cleaved by the 5′→3′ exonuclease activity of the polymerase, which separates reporter from quencher; primer dimers lack the probe site and give no signal. SYBR Green binds any double-stranded DNA, dimers included, hence the melt curve. Taq has no proofreading, and a probe reporting any dsDNA would describe SYBR Green, not TaqMan.
A bacterium carries a CRISPR spacer that exactly matches a sequence in phage P. Why does the crRNA transcribed from that spacer not direct Cas9 to cut the bacterium's own CRISPR array, which contains the identical sequence?
1. In the array the sequence is flanked by repeats, not by the PAM that Cas9 needs beside its target
2. The array is methylated at every adenine, and methylation blocks Cas9 from binding host DNA
3. A crRNA pairs only with RNA, so Cas9 is guided to phage transcripts and never to DNA at all
4. Cas9 is made only after phage infection and is degraded before the array is next replicated
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Answer: A — In the array the sequence is flanked by repeats, not by the PAM that Cas9 needs beside its target
Cas9 cleaves only a protospacer that sits next to its PAM (5′-NGG-3′ for the Streptococcus pyogenes enzyme). Spacers in the array are flanked by repeats, which lack a PAM, so self-targeting is avoided. Methylation is how restriction–modification systems protect host DNA, the tempting analogy, but it is not how CRISPR discriminates; Cas9 targets DNA, and cas genes are commonly expressed before infection.
Which of the following statements about CRISPR-Cas systems are correct?
1. New spacers are captured from invading DNA by the Cas1–Cas2 complex and added at the leader end of the array
2. Cas9 cuts both DNA strands to leave staggered ends with four-nucleotide 5′ overhangs
3. Cas13 targets RNA, and its collateral cleavage of reporter RNA is used in diagnostics
4. A single-guide RNA used for editing fuses the crRNA and the tracrRNA into one molecule
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Answer: A — New spacers are captured from invading DNA by the Cas1–Cas2 complex and added at the leader end of the array; C — Cas13 targets RNA, and its collateral cleavage of reporter RNA is used in diagnostics; D — A single-guide RNA used for editing fuses the crRNA and the tracrRNA into one molecule
Adaptation by Cas1–Cas2 adds spacers at the leader end, Cas13 is the RNA-targeting effector behind SHERLOCK, and the sgRNA joins crRNA and tracrRNA. Cas9's HNH and RuvC domains each cut one strand at the same position, leaving a blunt end; staggered cuts are made by Cas12a.