All subjects · Biology · WBI16 Practical Skills in Biology II

Skills applied: CP13 microbial growth, CP14 antibiotics

Edexcel IAL Biology · WBI16 Practical Skills in Biology II

Microbiology questions combine aseptic technique, serial dilutions, counting methods, logarithmic growth and t-tests on zones of inhibition.

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CP13: growth in liquid culture

Method recap (see The bacterial growth curve, Unit 4): inoculate sterile broth with yeast or bacteria, incubate, and sample at intervals.

Measuring population size:

Cells per cm³ = colonies ÷ volume plated x dilution factor.

CP14: effect of antibiotics

Method recap (see Core practical 14, Unit 4): spread bacteria evenly on agar (a lawn), add antibiotic discs (or wells) with a control disc, incubate at 25 °C, measure the clear zone.

Aseptic technique

Flame loops and bottle necks; work near a Bunsen; minimal lid opening; sterile pipettes and spreaders (dip in ethanol, flame, let cool); tape lids; incubate at 25 °C; autoclave everything afterwards.

Worked example

0.1 cm³ of a 10⁻⁵ dilution produces 34 colonies. Calculate the number of bacteria per cm³ in the original culture.

  1. Colonies per cm³ of the diluted sample = 34 ÷ 0.1 = 340
  2. Multiply by the dilution factor 10⁵
  3. = 3.4 x 10⁷

Answer: 3.4 x 10⁷ bacteria per cm³

Key idea: CP13: turbidity, haemocytometer or dilution plating; cells per cm³ = colonies ÷ volume x dilution factor; log graphs. CP14: lawn plates, control disc, zone area πr², t-test for two antibiotics; strict aseptic technique.

Check yourself

  1. 0.1 cm³ of a 10⁻⁴ dilution gives 56 colonies. The original culture has a x 10⁶ bacteria per cm³. What is a?
    Show answer

    5.6. 56 / 0.1 = 560; x 10⁴ = 5.6 x 10⁶.

  2. A haemocytometer square holds 0.004 mm³ and contains on average 12 yeast cells. Calculate cells per mm³.
    Show answer

    3000 cells per mm³. 12 / 0.004 = 3000 cells per mm³.

  3. Zones (mm) for antibiotic X: 21, 23, 22, 24, 20; antibiotic Y: 17, 19, 18, 16, 18. Calculate t to 2 decimal places.
    Show answer

    5.05. Means 22.0 and 17.6; s = 1.581 and 1.140; t = 4.4 / √(0.500 + 0.260) = 5.05.

  4. For the zone data, t = 5.05 with 8 degrees of freedom (critical value 2.31). What is the conclusion?
    • Accept the null hypothesis: no significant difference
    • Reject the null hypothesis: antibiotic X gives significantly larger zones than Y
    • There is a significant correlation
    • The antibiotics are identical
    Show answer

    Reject the null hypothesis: antibiotic X gives significantly larger zones than Y. 5.05 > 2.31.

  5. Why are plates with about 30 to 300 colonies used for counting?
    • Only these plates grow bacteria
    • These plates are sterile
    • Colonies must be over 300 to see them
    • Fewer gives large sampling error; more means colonies overlap and cannot be counted
    Show answer

    Fewer gives large sampling error; more means colonies overlap and cannot be counted. A countable range balances accuracy and practicality.

  6. What is a disadvantage of measuring culture growth by turbidity?
    • It takes many days
    • It cannot distinguish living from dead cells
    • It needs no calibration
    • It can only measure fungi
    Show answer

    It cannot distinguish living from dead cells. Dead cells still scatter light.

  7. What is a bacteriostatic antibiotic?
    • One that kills bacteria
    • One that only works on viruses
    • One that makes bacteria grow faster
    • One that stops bacteria from growing and dividing without killing them
    Show answer

    One that stops bacteria from growing and dividing without killing them. Bactericidal antibiotics kill; bacteriostatic ones stop growth.

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Last updated 8 November 2026. Spotted a mistake? Tell us.