Microbial Growth: Growth Curve and Kinetics, Measurement and Yields, Continuous Culture, Environment, Sporulation, Quorum Sensing and Biofilms

Practice questions (17)

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

1. A flask of E. coli in mid-exponential phase in glucose minimal medium at 37 °C is split. Portion P is diluted into fresh glucose minimal medium at 37 °C; portion Q is held until it reaches stationary phase and is then diluted into the same fresh medium. What is expected?

   1. Both show the same lag, because the lag phase depends only on the composition of the fresh medium
   2. P continues growing with little or no lag, whereas Q shows a lag while its cells rebuild ribosomes and repair damage
   3. P shows a longer lag than Q, because exponential cells must first pass through stationary phase before they can divide again in fresh medium
   4. Neither shows a lag, because both are transferred into a medium that contains the same carbon source
   Show answer

   Answer: B — P continues growing with little or no lag, whereas Q shows a lag while its cells rebuild ribosomes and repair damage

   Exponential cells moved into fresh medium of the same composition already have the ribosomes and enzymes they need, so they keep growing with essentially no lag. Stationary-phase cells have run down their ribosome content and accumulated damage, so they need a lag before dividing, even in the same medium. The tempting answer that neither shows a lag ignores the physiological state of the inoculum, which matters as much as the medium.

2. In a batch culture kept for several days, the optical density at 600 nm stays almost constant while the plate count falls from 10⁹ to 10⁷ CFU mL⁻¹. The best explanation is that

   1. the culture is in the death phase, and the dead cells remain intact and still scatter light although they no longer form colonies
   2. the culture is still in exponential phase, and the plate count is wrong because colonies merge when cells divide rapidly
   3. the cells have entered a second lag phase, during which optical density is constant and the viable count always falls
   4. the cells have lysed completely, and the optical density now comes from dissolved proteins absorbing at 600 nm
   Show answer

   Answer: A — the culture is in the death phase, and the dead cells remain intact and still scatter light although they no longer form colonies

   Turbidity measures light scattered by all cells, living or dead, while a plate count detects only cells able to form colonies. A falling viable count with unchanged turbidity is the signature of the death phase, in which most dead cells stay intact. Complete lysis would lower the turbidity, and dissolved proteins do not absorb appreciably at 600 nm, so that distractor fails on both counts.

3. A broth is inoculated with 5.0 × 10⁴ cells mL⁻¹. The culture shows a lag of exactly 1 h and then grows exponentially without interruption. Four hours after inoculation the count is 3.2 × 10⁶ cells mL⁻¹. The generation time during exponential growth is ____ min.

   Numerical answer — type the value.

   Show answer

   Answer: 30

   N/N₀ = 3.2 × 10⁶/5.0 × 10⁴ = 64 = 2⁶, so n = 6 generations. These occurred only in the exponential period, 4 − 1 = 3 h = 180 min. g = 180/6 = 30 min. Using the whole 4 h gives the wrong value of 40 min.

4. A bacterium growing exponentially has a specific growth rate μ = 1.2 h⁻¹ (defined by dN/dt = μN). How many generations does the population complete in 2.5 h? Take ln 2 = 0.693 and give the answer to two decimal places.

   Numerical answer — type the value.

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   Answer: 4.33

   μ is a natural-log rate, so ln(N/N₀) = μt = 1.2 × 2.5 = 3.0. The number of doublings is n = ln(N/N₀)/ln 2 = 3.0/0.693 = 4.329, which is 4.33. Equivalently g = 0.693/1.2 = 0.5775 h and n = 2.5/0.5775 = 4.33. Taking μt = 3 as the number of generations is the trap.

5. Which method of measuring a bacterial population counts only cells that are able to multiply?

   1. Direct microscopic counting of an unstained sample in a Petroff–Hausser chamber
   2. Measuring optical density at 600 nm and reading cell number from a standard curve
   3. Drying a known volume of culture and weighing the cell mass that is left behind
   4. Spreading serial dilutions on agar plates and counting the colonies that appear
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   Answer: D — Spreading serial dilutions on agar plates and counting the colonies that appear

   A colony arises only from a cell (or clump) that can divide repeatedly, so the plate count gives viable cells as CFU. A counting chamber counts every cell it sees, dead or alive, unless a viability stain is used; turbidity and dry weight measure mass that includes dead cells.

6. A homolactic bacterium ferments glucose to lactate by glycolysis, gaining a net 2 ATP per glucose, and uses no other energy source. Take Y_ATP = 10.5 g dry cells per mol ATP and the molar mass of glucose as 180 g mol⁻¹. If the culture consumes 5.4 g of glucose per litre, the dry cell mass formed is ____ g L⁻¹ (to two decimal places).

   Numerical answer — type the value.

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   Answer: 0.63

   Moles of glucose = 5.4/180 = 0.030 mol L⁻¹. ATP formed = 2 × 0.030 = 0.060 mol L⁻¹. Biomass = 0.060 × 10.5 = 0.63 g L⁻¹. Equivalently the molar growth yield is 2 × 10.5 = 21 g per mol glucose, and 21 × 0.030 = 0.63.

7. In the Helmstetter–Cummings technique, cells are bound to a membrane filter, the filter is inverted and medium is pumped through it. Why do the cells collected in the effluent form a synchronous population?

   1. Because the flowing medium starves the bound cells, and starvation arrests all of them at the same stage before release
   2. Because only the largest cells, which are about to divide, are detached from the filter by the flow of medium
   3. Because the cells released are newborn daughters freed at division, so all of them begin the cycle at the same point
   4. Because a temperature shift applied to the filter induces every bound cell to divide at the same moment
   Show answer

   Answer: C — Because the cells released are newborn daughters freed at division, so all of them begin the cycle at the same point

   The bound cells keep growing on the filter; when one divides, the daughter not attached to the membrane is washed off. The effluent therefore contains only newborn cells, all at the start of the division cycle. It is a selection method, so explanations based on starvation or a temperature shift (induction methods) are wrong, and it collects the smallest newborn cells, not the largest ones.

8. A glucose-limited chemostat at steady state is moved from D = 0.2 h⁻¹ to D = 0.3 h⁻¹; both values are well below the critical dilution rate, and the feed glucose concentration is unchanged. Assuming the Monod model with a constant yield, at the new steady state

   1. the specific growth rate stays at its old value, and the extra flow simply lowers the cell concentration in proportion
   2. the specific growth rate rises to 0.3 h⁻¹, the residual glucose rises and the cell concentration falls slightly
   3. the specific growth rate rises to 0.3 h⁻¹, the residual glucose falls and the cell concentration rises
   4. the specific growth rate rises to 0.3 h⁻¹, while both the residual glucose and the cell concentration stay unchanged
   Show answer

   Answer: B — the specific growth rate rises to 0.3 h⁻¹, the residual glucose rises and the cell concentration falls slightly

   At steady state μ = D, so μ becomes 0.3 h⁻¹. A higher μ requires a higher limiting-substrate concentration, S̄ = K_sD/(μ_max − D), which rises with D. Biomass X̄ = Y(S_R − S̄) therefore falls slightly. The distractor in which glucose falls and biomass rises reverses the Monod relation; μ cannot stay at its old value because cells would then wash out faster than they grow.

9. A 5.0 L chemostat is fed with medium containing 4.5 g L⁻¹ of the limiting substrate. The organism obeys the Monod equation with μ_max = 0.80 h⁻¹ and K_s = 0.50 g L⁻¹. The largest flow rate at which a steady-state population can still be maintained (the flow rate at the critical dilution rate) is ____ L h⁻¹ (to one decimal place).

   Numerical answer — type the value.

   Show answer

   Answer: 3.6

   The highest growth rate attainable is at the feed concentration: D_c = μ_maxS_R/(K_s + S_R) = 0.80 × 4.5/(0.50 + 4.5) = 3.6/5.0 = 0.72 h⁻¹. F = D_c × V = 0.72 × 5.0 = 3.6 L h⁻¹. Using μ_max itself gives 4.0 L h⁻¹, which overestimates the limit.

10. A chemostat is run at D = 0.40 h⁻¹ with a feed containing 10 g L⁻¹ of the limiting substrate. The organism follows the Monod equation with μ_max = 0.60 h⁻¹ and K_s = 0.10 g L⁻¹, and its yield is Y_X/S = 0.45 g g⁻¹. The steady-state biomass productivity (D × X̄) is ____ g L⁻¹ h⁻¹ (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 1.76

    Residual substrate S̄ = K_sD/(μ_max − D) = 0.10 × 0.40/0.20 = 0.20 g L⁻¹. Biomass X̄ = Y(S_R − S̄) = 0.45 × (10 − 0.20) = 4.41 g L⁻¹. Productivity = D × X̄ = 0.40 × 4.41 = 1.764, which is 1.76 g L⁻¹ h⁻¹. Ignoring S̄ gives 0.40 × 4.5 = 1.80.

11. Which of the following statements about continuous culture are correct?

    1. In a chemostat at steady state, the specific growth rate of the cells equals the dilution rate
    2. At a fixed dilution rate, doubling the feed concentration of the limiting nutrient doubles its residual concentration in the vessel at steady state
    3. A turbidostat keeps one nutrient limiting and works best at dilution rates far below the maximum growth rate
    4. If the dilution rate of a chemostat is held above μ_max, the cell population is progressively washed out of the vessel
    Show answer

    Answer: A — In a chemostat at steady state, the specific growth rate of the cells equals the dilution rate; D — If the dilution rate of a chemostat is held above μ_{max}, the cell population is progressively washed out of the vessel

    At steady state μ = D, and when D exceeds μ_max cells leave faster than they can possibly grow, so they wash out. The residual substrate S̄ = K_sD/(μ_max − D) depends only on D and the organism's constants, not on the feed concentration; doubling S_R raises the biomass X̄ = Y(S_R − S̄), not S̄. A turbidostat is the opposite of a chemostat: it uses a photocell to hold turbidity constant with no nutrient limiting and operates near μ_max.

12. A bacterium inoculated into a tube of thioglycollate broth grows only as a narrow band a short distance below the surface, with no growth at the very top or in the depths. It is best described as

    1. an aerotolerant anaerobe that ferments and is unaffected by the oxygen gradient
    2. a facultative anaerobe that respires near the surface and ferments in the depths
    3. a microaerophile that needs oxygen but is harmed by its concentration in air
    4. an obligate anaerobe that is killed wherever any oxygen has diffused into the tube
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    Answer: C — a microaerophile that needs oxygen but is harmed by its concentration in air

    A band below the surface marks the zone of reduced oxygen tension that a microaerophile needs. A facultative anaerobe grows throughout the tube, densest at the top; an aerotolerant anaerobe grows evenly throughout; an obligate anaerobe grows only at the bottom.

13. Which of the following statements about the effect of environmental factors on microbial growth are correct?

    1. Psychrotrophs can grow at refrigeration temperatures, but their optimum temperature is well above 15 °C
    2. Thermophiles keep their membranes functional at high temperature by using lipids richer in unsaturated fatty acids than those of mesophiles
    3. Many bacteria tolerate low water activity by accumulating compatible solutes such as glycine betaine or proline
    4. Obligate anaerobes generally lack catalase and superoxide dismutase, so reactive oxygen species formed in air harm them
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    Answer: A — Psychrotrophs can grow at refrigeration temperatures, but their optimum temperature is well above 15 °C; C — Many bacteria tolerate low water activity by accumulating compatible solutes such as glycine betaine or proline; D — Obligate anaerobes generally lack catalase and superoxide dismutase, so reactive oxygen species formed in air harm them

    Psychrotrophs grow in the cold but have optima of about 20–30 °C; compatible solutes balance external osmotic pressure without disturbing enzymes; and the lack of superoxide dismutase and catalase is why oxygen is toxic to strict anaerobes. Thermophiles do the reverse of the false statement: they use more saturated (and, in archaea, ether-linked) lipids, because unsaturated chains would make the membrane too fluid at high temperature.

14. During endospore formation in Bacillus subtilis, in which compartment, and in what order, do the sigma factors σF and σG act?

    1. Both act in the mother cell, σG first and σF later, to make the cortex and the coat
    2. σF acts in the mother cell and σG in the forespore, both before the asymmetric septum forms
    3. Both act in the forespore, σF early after the asymmetric division and σG after engulfment
    4. σF acts in the forespore and σG in the mother cell, both only after the spore is released
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    Answer: C — Both act in the forespore, σF early after the asymmetric division and σG after engulfment

    The forespore programme runs σF (activated just after asymmetric division) and then σG (after engulfment); the mother cell runs σE and then σK, which direct cortex and coat synthesis. The tempting answer that places both in the mother cell confuses σF/σG with σE/σK. The released spore is dormant, so nothing in the cascade waits until after release.

15. Two non-luminous mutants of Vibrio fischeri are grown to high density: one lacks a functional LuxI, the other a functional LuxR. Cell-free spent medium from a dense wild-type culture is added to each. What result is expected?

    1. Only the LuxR mutant luminesces, because the medium supplies the receptor protein that it is missing
    2. Only the LuxI mutant luminesces, because the medium supplies the diffusible acyl-homoserine lactone it cannot make
    3. Both mutants luminesce, because any mutant in the lux system is rescued once the population is dense enough
    4. Neither mutant luminesces, because autoinducers act only inside the cell that made them and cannot be supplied from outside
    Show answer

    Answer: B — Only the LuxI mutant luminesces, because the medium supplies the diffusible acyl-homoserine lactone it cannot make

    LuxI is the synthase of the AHL autoinducer, which diffuses freely across membranes and accumulates in spent medium, so adding wild-type medium supplies the signal and the LuxI mutant lights up. The LuxR mutant lacks the cytoplasmic receptor-activator; a protein in the medium cannot enter the cell to replace it, and extra AHL has nothing to bind, so it stays dark.

16. A Pseudomonas aeruginosa strain is engineered to overproduce a diguanylate cyclase, raising its intracellular cyclic di-GMP. Which change in behaviour is expected?

    1. Higher swimming motility and dispersal of cells from existing biofilms into the surrounding liquid
    2. Loss of quorum sensing, because cyclic di-GMP degrades the acyl-homoserine lactone signals
    3. Faster planktonic growth with no change in surface attachment or matrix production
    4. Increased matrix polysaccharide production, stronger surface attachment and reduced motility
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    Answer: D — Increased matrix polysaccharide production, stronger surface attachment and reduced motility

    Cyclic di-GMP is the intracellular switch between lifestyles: high levels promote EPS synthesis, adhesion and the sessile biofilm state while repressing flagellar motility; low levels (made by phosphodiesterases) favour swimming and dispersal. The motility-and-dispersal distractor describes the low-c-di-GMP state, the opposite of what was engineered. c-di-GMP does not destroy AHLs.

17. Which of the following statements about bacterial biofilms and biofouling are correct?

    1. Biofilm cells are more tolerant of antibiotics partly because inner cells grow slowly or are dormant, and many antibiotics act on actively growing cells
    2. The biofilm matrix consists of capsular polysaccharide alone, which is why biofilms form only from encapsulated species
    3. Biofouling of a heat exchanger reduces heat transfer, and fouling of a ship's hull increases drag and fuel consumption
    4. Biofilms of sulfate-reducing bacteria on steel surfaces can accelerate corrosion of pipelines
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    Answer: A — Biofilm cells are more tolerant of antibiotics partly because inner cells grow slowly or are dormant, and many antibiotics act on actively growing cells; C — Biofouling of a heat exchanger reduces heat transfer, and fouling of a ship's hull increases drag and fuel consumption; D — Biofilms of sulfate-reducing bacteria on steel surfaces can accelerate corrosion of pipelines

    Slow growth and dormancy inside the biofilm, together with reduced penetration, explain antibiotic tolerance; fouling layers insulate heat exchangers and roughen hulls; and sulfate reducers in biofilms drive microbially influenced corrosion. The matrix is a mixture of polysaccharides, proteins and extracellular DNA made by many species, encapsulated or not, so the statement restricting it to capsular polysaccharide is false.