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Required practical 2: antiseptics and antibiotics

Cell biology · Cell structure · note 9 of 9

Spec 4.1.1.6
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Required practical 2: antiseptics and antibioticsSpec 4.1.1.6

In short

Required practical 2 tests the effect of antiseptics or antibiotics on bacterial growth. Paper discs soaked in each substance are placed on a bacterial lawn on an agar plate, with a sterile water control, and incubated at 25 °C. The larger the clear zone of inhibition around a disc, the more effective the substance is against that bacterium.

Required practical:

Investigate the effect of antiseptics or antibiotics on bacterial growth using agar plates and measuring zones of inhibition.

  1. Using aseptic technique, spread a sample of bacteria evenly over the surface of a sterile agar plate to make a bacterial 'lawn'.
  2. Using sterile forceps, place sterile paper discs soaked in different antiseptics or antibiotics (or different concentrations of one) onto the agar. Include a control disc soaked in sterile water.
  3. Secure the lid with adhesive tape, label the base, store the plate upside down and incubate at 25 °C for about two days.
  4. Measure the diameter of the clear area (zone of inhibition) around each disc and calculate its area using πr².

Where the antiseptic or antibiotic has killed or stopped the growth of the bacteria, there is a clear zone around the disc. The larger the zone of inhibition, the more effective the substance is against that bacterium. The control disc should have no clear zone, which shows that the water itself does not kill the bacteria.

  • Independent variable: the type or concentration of antiseptic or antibiotic.
  • Dependent variable: the size (diameter or area) of the zone of inhibition.
  • Control variables: type and amount of bacteria, type and depth of agar, size of disc, volume of solution on the disc, temperature, incubation time.

Safety: the cultures may contain harmful microorganisms. Use aseptic technique, wash hands, keep the plate taped shut, never open it after incubation, and disinfect the work area and dispose of plates safely.

Quick check

  1. How is the genetic material of a bacterial cell arranged?

    Show answer

    As a single DNA loop that is not enclosed in a nucleus. There may also be plasmids.

  2. Which sub-cellular structure is the site of aerobic respiration?

    Show answer

    Mitochondria.

  3. What is the formula for magnification?

    Show answer

    Magnification = size of image ÷ size of real object.

  4. Why does an electron microscope show more detail than a light microscope?

    Show answer

    It has much higher magnification and resolving power.

  5. Triple only: why are inoculating loops passed through a flame?

    Show answer

    To sterilise them and kill unwanted microorganisms.

Written and checked against the AQA GCSE Biology (8461) specification · Updated October 2026

Frequently asked questions

What is the difference between prokaryotic and eukaryotic cells?

Eukaryotic cells, such as plant and animal cells, have their genetic material enclosed in a nucleus. Prokaryotic cells, such as bacteria, are much smaller and their genetic material is not in a nucleus. It is a single DNA loop, and there may be plasmids. Both types of cell have a cell membrane and cytoplasm.

Why do plant cells have a cell wall?

Plant cells have a cell wall made of cellulose because it strengthens the cell. Animal cells do not have a cell wall. Plant cells also often have a permanent vacuole filled with cell sap, which helps keep the cell firm, and chloroplasts, which contain chlorophyll to absorb light energy for photosynthesis.

How do you calculate magnification in biology?

Magnification is calculated with magnification = size of image ÷ size of real object. Make sure both measurements are in the same units before you divide. Rearranged, real size = image size ÷ magnification. For a light microscope, the total magnification is the eyepiece lens magnification multiplied by the objective lens magnification.

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