Free Pearson Edexcel International A-level Biology · Unit 4

Microbial growth, aseptic principles and measurement

What you’ll learn

  • Interpret phases of a microbial growth curve.
  • Compare direct counts, viable counts, mass and turbidity.
  • Calculate growth from model data and explain safe measurement principles.

Understand the idea

A closed culture commonly shows a lag phase while cells adapt, an exponential phase with rapid division, a stationary phase when division and death balance, and a death phase when viable numbers decline. Nutrient depletion and accumulation of inhibitory products limit growth. A straight section on a logarithmic population graph indicates exponential growth; a stationary population does not mean every cell has stopped all metabolism.

A direct microscopic count can include dead cells unless viability is assessed. A viable colony count estimates colony-forming units, which may represent individual cells or clumps. Dry mass estimates biomass; turbidity measures light scattering and requires calibration because cell size, clumps and high concentrations affect the relationship. Choose a method that matches the question and maintain consistent sampling and instrument settings.

Aseptic principles prevent unwanted organisms entering a culture and reduce exposure of people and the environment: clean equipment, protected containers, approved organisms and supervised disposal are essential. Classroom growth investigations should use approved non-pathogenic material and local safety procedures. The model below uses supplied data rather than instructions for growing pathogens. Distinguish the number of generations per time from the natural-log specific growth rate.

Key formulae

Nt=N0 2nN_t=N_0\,2^n

k=log⁡2(Nt/N0)tgenerations per unit timek=\frac{\log_2(N_t/N_0)}{t}\quad\text{generations per unit time}

μ=ln⁡Nt−ln⁡N0t,td=ln⁡2μ\mu=\frac{\ln N_t-\ln N_0}{t},\qquad t_d=\frac{\ln2}{\mu}

Exponential growth becomes a straight line after taking logs

Conditions. Invented unconstrained exponential phase with doubling time 30 min. ln(N/N₀) = (ln 2/30)t, N₀ > 0; no culturing procedure or organism-specific rate is implied.

Time / min; ln(N/N₀) / dimensionless.

At 90 min the log ratio is ln 8 ≈ 2.079, representing eight times the initial count. Slope μ ≈ 0.0231 min⁻¹ is not the same numerical value as generations per minute (1/30).

Open the interactive lesson to view plotted curves and coordinate tables.

Worked example

In supplied model data, a population increases from 400 to 3200 cells in 90 minutes. Find the number of generations and doubling time, assuming exponential growth.

  1. The population ratio is 3200/400=83200/400=8.
  2. Since 8=238=2^3, three generations have occurred.
  3. Doubling time is 90/3=30 min90/3=30\,\mathrm{min}.

Answer: Three generations and a doubling time of 30 minutes.

Open the interactive lesson to try original questions and recall. Build a revision plan.