Free Pearson Edexcel International A-level Biology · Unit 6

Antibiotic comparisons: inhibition and uncertainty

What you’ll learn

  • Summarise replicate inhibition-zone measurements.
  • Evaluate controls and distinguish inhibition from killing.

Understand the idea

Core practical 14 compares antibiotic effects using approved classroom work. The questions here interpret completed records rather than provide a culturing protocol. A clear region indicates reduced visible growth in the assay. The size of that region depends on the delivered amount, diffusion, the organism tested and the measurement convention, so comparisons require these factors to be specified and standardised.

Replicate results help reveal variability in the observed response. Report a mean with an appropriate measure of spread and retain the raw data so that unusual values can be checked. A no-antibiotic or carrier control checks the baseline response, while an appropriate reference can show whether the assay behaved as expected. Multiple readings of the same zone are measurement repeats and do not replace independent assay replicates.

A zone does not by itself distinguish a bactericidal effect, which kills cells, from a bacteriostatic effect, which inhibits growth. It also does not supply a clinical prescription or establish that a larger-zone treatment is more useful in a patient. Differences in diffusion and delivery limit that inference. Handling, containment and disposal belong to the institution's approved microbiology arrangements.

Key formulae

d‾=∑din\overline d=\frac{\sum d_i}{n}

range=dmax−dmin\mathrm{range}=d_{\mathrm{max}}-d_{\mathrm{min}}

Aclear=π(R2−r2)when the disc is excludedA_{\mathrm{clear}}=\pi(R^2-r^2)\quad\text{when the disc is excluded}

Worked example

Three independent supplied records give total zone diameters of 1818, 2020 and 22 mm22\ \mathrm{mm} for antibiotic A, and 1414, 1515 and 16 mm16\ \mathrm{mm} for B. The same disc size and assay conditions were used. Summarise the means and ranges, then give a justified conclusion.

  1. For A, the mean is (18+20+22)/3=20 mm(18+20+22)/3=20\ \mathrm{mm} and the range is 22−18=4 mm22-18=4\ \mathrm{mm}.
  2. For B, the mean is 15 mm15\ \mathrm{mm} and the range is 2 mm2\ \mathrm{mm}.
  3. A has the larger mean zone in these records. A formal population comparison would require an appropriate statistical analysis, and the zone difference does not establish a clinical ranking.

Answer: A: mean 20 mm20\ \mathrm{mm}, range 4 mm4\ \mathrm{mm}; B: mean 15 mm15\ \mathrm{mm}, range 2 mm2\ \mathrm{mm}.

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