Free Pearson Edexcel International A-level Biology · Unit 3

Membrane permeability and changes in tissue mass

What you’ll learn

  • Interpret pigment leakage as evidence about membrane permeability.
  • Control tissue preparation and calculate percentage mass changes in osmosis.

Understand the idea

Core practical 3 investigates how temperature and alcohol affect membrane permeability. In a suitable pigmented tissue, release of pigment into the surrounding liquid provides an indirect measure of membrane disruption. A colorimeter can measure the resulting absorbance when its calibration and useful range are respected. Greater absorbance usually means more pigment in the sample, but it does not directly count damaged cells or identify exactly which membrane component changed.

Make comparisons using comparable tissue dimensions, exposed surface area, number of pieces, solution volume and exposure time. Cutting can release pigment independently of the treatment, so consistent preparation and the approved removal of initial surface pigment matter. Use a suitable blank to correct background absorbance. Temperature can affect membrane organisation and proteins; alcohol can disturb membrane structure. Neither effect should be inferred from unmatched pieces of tissue.

An associated osmosis investigation measures tissue mass before and after exposure to solutions of known concentration. Calculate percentage mass change to account for different initial masses, and blot consistently before weighing to avoid including surface liquid. A concentration giving approximately zero mass change estimates where there is no net water movement under those conditions; it does not mean that water molecules have stopped moving.

Key formulae

Acorrected=Asample−AblankA_{\mathrm{corrected}}=A_{\mathrm{sample}}-A_{\mathrm{blank}}

mass change (%)=mfinal−minitialminitial×100\mathrm{mass\ change}\ (\%)=\frac{m_{\mathrm{final}}-m_{\mathrm{initial}}}{m_{\mathrm{initial}}}\times100

The zero crossing estimates no net tissue mass change

Conditions. Invented local linear response, Δmass% = 15 − 50c, shown only over c = 0–0.6 mol dm⁻³. Same tissue type, exposure and handling are assumed; this is not an observed universal osmotic law.

External solute concentration / mol dm⁻³; Tissue mass change / %.

The line crosses zero at 0.30 mol dm⁻³. Positive values mean net mass gain and negative values net loss; zero net change still permits water movement in both directions.

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

Worked example

Two membrane-treatment samples have absorbances 0.170.17 and 0.530.53. The appropriate blank is 0.050.05 for both. Equal tissue quantities and solution volumes were used, and these readings are within the linear measurement range. Compare the corrected pigment signals.

  1. The first corrected absorbance is 0.17−0.05=0.120.17-0.05=0.12.
  2. The second is 0.53−0.05=0.480.53-0.05=0.48.
  3. The corrected signal ratio is 0.48/0.12=40.48/0.12=4. This supports four times the pigment signal, not a claim that exactly four times as many cells were damaged.

Answer: The second corrected pigment signal is four times the first.

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