Cells and organisation · GCSE Biology
Osmosis required practical
Plan the GCSE Biology potato-chip osmosis practical: independent, dependent and control variables, percentage change in mass, blotting, and how the graph crosses zero at the isotonic point.
Independent variable: sucrose concentration. Dependent variable: percentage change in mass of the potato chips. Controls: temperature, time, volume, chip size, blotting. The zero-crossing is isotonic.
The important bits
What you need to know
- 1
Aim: investigate the effect of sugar-solution concentration on the mass of potato tissue, to show osmosis.
- 2
Independent variable: concentration of sucrose (or salt) solution, for example 0.0, 0.2, 0.4, 0.6, 0.8, 1.0 mol/dm³.
- 3
Dependent variable: change in mass of the potato chips, calculated as percentage change so different start masses can be compared.
- 4
Control variables: temperature, time in solution, volume of solution, potato variety, chip length and diameter (surface area), whether chips are peeled, and blotting method.
- 5
Method outline: cut equal chips with a cork borer, blot, measure start mass, place in labelled solutions for a fixed time (for example 30 minutes), blot, measure end mass, repeat each concentration.
- 6
Percentage change = ((end mass − start mass) ÷ start mass) × 100. Positive means water entered (outside more dilute); negative means water left (outside more concentrated).
- 7
Plot percentage change against concentration. The intercept where the line crosses 0% is the sucrose concentration equal to the potato cell contents (isotonic; no net osmosis).
- 8
Blotting removes surface solution that would fake a mass gain. Repeats allow a mean and highlight anomalies. A water bath keeps temperature constant because temperature affects the rate of osmosis.
Quotations worth analysing
Short evidence. Real method.
“Independent variable: concentration of sucrose. Dependent variable: percentage change in mass.”
State both variables in that language. “I change the sugar and see what happens to the potato” is not enough for method marks.
“percentage change in mass = (change in mass / original mass) × 100”
Show the substitution. A negative sign means the chip lost water by osmosis.
“Where the graph crosses the x-axis, there is no net movement of water.”
That concentration matches the potato’s cell sap. Do not say “osmosis stopped forever”; say no net movement.
Go deeper
How do I write a method that would actually score?
Cut chips with a cork borer so diameter is constant, then trim to the same length with a scalpel on a tile. Blot each chip with paper towel to remove surface water, then record start mass to two decimal places. Place chips in equal volumes of each sucrose concentration in labelled boiling tubes. Keep tubes in a water bath at a stated temperature, or in the same room, for the same time. Remove, blot in the same way, record end mass. Repeat the whole concentration range at least twice more and calculate means. Safety: scalpel cutting away from the hand. The independent variable is sucrose concentration; the dependent is percentage change in mass; controls are temperature, time, volume, chip size and blotting. If you leave out blotting or percentage change, expect to lose interpretation marks later.
Go deeper
Why percentage change, and what does the zero intercept mean?
Chips never start at exactly the same mass. A 0.20 g gain on a 2.00 g chip is 10%; the same gain on a 4.00 g chip is only 5%. Percentage change removes that unfairness. Plot mean percentage change (y) against concentration (x). Draw a line of best fit. Above the axis, net water entered because the external solution was more dilute than the cells. Below the axis, net water left. At the intercept, water in equals water out: the sucrose is isotonic with the potato cytoplasm and vacuole. That value is an estimate of the potato’s internal solute concentration. If all points are negative, your weakest sucrose was still more concentrated than the cells, or the chips dried in air. Do not join the dots with a ruler from point to point if the task asks for a line of best fit.
Go deeper
What goes wrong, and how do I evaluate the practical?
Poor blotting is the main error: leftover sucrose looks like mass gain. Chips left in air lose water by evaporation. Uneven chip length changes surface area. Temperature drift changes rate, so a short immersion may not reach a new equilibrium. Potato from the green end versus the centre can differ in sugar content. Improvements: more intermediate concentrations to locate the intercept more precisely, a longer timed soak, a temperature-controlled bath, electronic balance to 0.01 g, randomise the order of tubes, and use a range of at least five concentrations. Validity: you are measuring mass as a proxy for water movement by osmosis through cell membranes. You are not measuring sucrose entering the cells in this design. Repeatability: similar means across repeats. A random mass jump at one concentration is an anomaly to circle, not to force the line through.
See the idea in action
Chip in 0.8 mol/dm³ sucrose: start mass 3.50 g, end mass 3.15 g. Change = −0.35 g. Percentage change = (−0.35 ÷ 3.50) × 100 = −10%. Water left by osmosis; the outside solution was more concentrated than the cell sap. A second chip in distilled water (0.0 mol/dm³) went from 3.60 g to 3.96 g: +10%. A graph of these means would cross zero between 0.0 and 0.8 mol/dm³. Independent variable: sucrose concentration. Dependent: percentage change in mass. Controls: 30 minutes, 10 cm³ of solution, 20 °C water bath, chips 3.0 cm × 0.8 cm, blotted three times.
Exam technique
Turn knowledge into marks
State IV, DV and at least three controls. Calculate percentage change with a sign. On the graph, read the intercept as the isotonic concentration where there is no net osmosis. Mention blotting when evaluating.
Common mistakes
Do not give these marks away
- 01
Calling the independent variable “the potato” or the dependent variable “osmosis” instead of percentage change in mass.
- 02
Forgetting to blot chips, or plotting raw mass instead of percentage change.
- 03
Saying the intercept is “where osmosis stops working” rather than where there is no net water movement.
In the potato osmosis practical, what is the independent variable?
APercentage change in mass of the chips
BConcentration of the sucrose solution
CTemperature of the room
DThe length of time the chips are left
Show the answer
Concentration of the sucrose solution. You change the sucrose concentration on purpose. Percentage change in mass is the dependent variable. Temperature and time should be controlled.
Quick questions
If this is the bit you searched
What is the independent variable in the osmosis potato practical?
The concentration of sucrose (or salt) solution. The dependent variable is the percentage change in mass of the potato chips.
Why do you calculate percentage change in mass?
Start masses differ. Percentage change lets you compare chips fairly and plot a graph that can cross zero at the isotonic concentration.
Why must potato chips be blotted before weighing?
Surface solution would add extra mass and fake a gain. Blotting means you are closer to measuring water that moved into or out of the cells.
What does it mean if the graph crosses the x-axis at 0.35 mol/dm³?
At that sucrose concentration there is no net osmosis: it matches the potato’s internal solute concentration (isotonic).