AQA GCSE Biology Biology Paper 1 (Foundation), June 2024: Question 6
14 marks · Low Demand difficulty · Short Answer
Investigate osmosis in potato tissue across varying salt concentrations by identifying control variables, calculating percentage change in mass, and determining isotonic concentration.
Practise this questionQuestion
Question text
06 A student investigated the effect of different concentrations of salt solution on the
mass of uncooked pieces of potato.
This is the method used.
1. Cut four pieces of a potato to the same size.
2. Record the mass of each piece of potato.
3. Put one of the pieces of potato into a beaker containing 100 cm3 of 0.1 mol/dm3
salt solution.
4. Repeat step 3 using the other pieces of potato, each in a different concentration of
salt solution.
5. After 20 minutes, remove the pieces of potato from the solutions.
6. Record the mass of each piece of potato.
06.1 Give two control variables the student used in the investigation.
[2 marks]
06.2 The student needed to be sure the measurements were as accurate as possible.
What should be done to each piece of potato after removing from the solution and
before measuring the mass?
[1 mark]
06.3 Name the piece of apparatus the student could use to measure the mass of the
pieces of potato.
[1 mark]
Table 4 shows the results.
*26* Table 4
Mass of piece of potato in grams
Concentration Percentage (%)
Piece of
of salt solution change in mass
potato 3 After 20
in mol/dm At start Change of piece of potato
minutes
A 0.1 6.2 6.5 + 0.3 + 4.8
B 0.3 6.8 6.5 − 0.3 − 4.4
C 0.5 6.5 5.8 − 0.7 − 10.8
D 0.7 6.0 4.9 − 1.1 X
06.4 What was the resolution of the apparatus used for measuring mass?
Use Table 4.
[1 mark]
Tick ( ) one box.
0.01 g 0.1 g 1.0 g 1.1 g
06.5 Which piece of potato had the greatest change in mass in the investigation?
*27* [1 mark]
Tick ( ) one box.
A B 29 C D
06.6 Calculate value X in Table 4.
Use the equation:
change in mass in grams
percentage change in mass = × 100
mass at start in grams
Give your answer to 1 decimal place.
[3 marks]
X (1 decimal place) = %
06.7 What is the best way to present the data in Table 4?
[1 mark]
Tick ( ) one box.
Bar chart
Line graph
Pie chart 30
06.8 Complete the sentences.
[3 marks]
Some of the pieces of potato decreased in mass because the potato cells
lost .
The decrease in mass was due to a process called .
The structure surrounding each cell in a piece of potato is
partially .
06.9 Table 4 is repeated below.
Table 4
Mass of piece of potato in grams
Concentration Percentage (%)
Piece of
of salt solution change in mass
potato 3 After 20
in mol/dm At start Change of piece of potato
minutes
A 0.1 6.2 6.5 + 0.3 + 4.8
B 0.3 6.8 6.5 − 0.3 − 4.4
C 0.5 6.5 5.8 − 0.7 − 10.8
D 0.7 6.0 4.9 − 1.1 X
Estimate the concentration of salt solution that would not cause a change in mass of
these pieces of potato.
[1 mark]
Concentration = mol/dm3
Mark scheme
Show the mark scheme
Question 6
AO /
Question Answers Extra information Mark
Spec Ref.
06.1 any two from: 2 AO1
• size of piece of potato ignore size of potato 4.1.3.2
• the (type of) potato RPA3
• volume / 100 cm3 of salt allow amount of salt solution
solution
• time (pieces of potato are allow 20 minutes (the pieces of
kept) in the solution / beaker potato are kept) in the solution /
beaker
ignore time unqualified
• the potato was uncooked
AO /
Spec Ref.
06.2 blot allow descriptions of blotting 1 AO3
or 4.1.3.2
dry (the surface) allow descriptions of drying (the RPA3
surface)
AO /
Spec Ref.
06.3 balance or weighing scale 1 AO1
4.1.3.2
RPA3
AO /
Spec Ref.
06.4 0.1 g 1 AO2
4.1.3.2
RPA3
AO /
Spec Ref.
06.5 D 1 AO3
RPA3
AO /
Spec Ref.
06.6 ignore minus sign throughout AO2
18 1.1 1 4.1.3.2
× 100 RPA3
6.0
18.333… 1
18.3(%) allow correct conversion to 1 1
decimal place from student’s
incorrect calculation using
figures from potato piece D
AO /
Spec Ref.
06.7 line graph 1 AO2
4.1.3.2
RPA3
AO /
Spec Ref.
06.8 must be in this order
water 1 AO2
osmosis allow diffusion 1 AO1
permeable (membrane) 1 AO2
4.1.3.2
RPA3
AO /
Spec Ref.
06.9 answer in the range 0.15 to 0.25 1 AO3
(mol/dm3) 4.1.3.2
RPA3
Total Question 6 14
How to answer it
AQA GCSE Biology: Required Practical 3 (Osmosis in Potato)
What this question tests
This question examines your knowledge of Required Practical 3: investigating the effect of a range of concentrations of salt or sugar solutions on the mass of plant tissue. Specifically, it tests:
- Variables: Identifying control variables to ensure a fair, valid investigation.
- Practical Accuracy: Explaining why drying potato pieces is essential prior to weighing.
- Scientific Apparatus & Data Handling: Naming balances, identifying apparatus resolution, and choosing the appropriate graph format.
- Maths Skills: Calculating percentage change in mass to 1 decimal place and interpreting graphs/tables to determine the isotonic point (zero change in mass).
- Core Biology Concepts: Explaining water movement across a partially permeable membrane by osmosis.
Control Variables in the Potato Investigation [2 marks]
✅ Correct Answers (Choose Any Two)
- Size (or length/surface area) of potato pieces
- The type/variety or source of potato used
- Volume of salt solution (e.g. 100 cm³)
- Time the potato is left in solution (e.g. 20 minutes)
- The potato was uncooked / temperature
❌ Common Errors & Mark Scheme Guidance
- Do NOT write just "size of potato": You must say the size of the cut piece of potato, not the whole potato vegetable!
- Do NOT write just "time": "Time" alone is too vague; specify "time potato pieces were kept in the solution".
- "Amount of solution" is allowed, but volume is the scientifically precise term.
Experimental Accuracy and Measuring Apparatus [2 marks]
✅ 06.2 What to do before weighing
Blot or dry the surface of each potato piece using a paper towel.
💡 Why Blotting Matters
Liquid clinging to the outside of the potato piece contributes to the mass reading. Excess water on the surface would cause an inaccurate, falsely high mass measurement that does not reflect actual water absorbed or lost by osmosis.
✅ 06.3 Apparatus to measure mass
Balance (or weighing scale / digital top-pan balance).
🧠 Top Tip
Never write "weighing machine". Always use the correct scientific term: electronic balance or balance.
Reading Data from Table 4 [2 marks]
✅ 06.4 Resolution of Apparatus
Correct box to tick: 0.1 g
✅ 06.5 Greatest Change in Mass
Correct box to tick: D
🧠 Resolution vs. Accuracy
Resolution is the smallest change in quantity being measured that gives a perceptible change in the reading. Since the values change in steps of 0.1 g, the balance resolution is 0.1 g.
Percentage Change Calculation [3 marks]
📐 Step-by-Step Calculation for Value X (Piece D)
Use the provided formula:
percentage change in mass = (change in mass in grams ÷ mass at start in grams) × 100
- Identify the figures for Potato D from Table 4:
Mass at start = 6.0 g
Mass after 20 minutes = 4.9 g
Change in mass = 4.9 - 6.0 = -1.1 g - Substitute into formula [Mark 1]:
(-1.1 ÷ 6.0) × 100 (Examiners ignore minus sign during substitution) - Calculate raw answer [Mark 2]:
18.3333... % - Round to 1 decimal place as requested [Mark 3]:
-18.3 % (or 18.3 %)
❌ Calculation Traps to Avoid
- Dividing by the final mass: A classic error is dividing by the end mass ( 4.9 g ) instead of the initial mass ( 6.0 g ). Always divide by the start mass.
- Incorrect rounding: 18.333... rounds to 18.3 , NOT 18 or 18.33. Follow the instruction in the question prompt!
🧠 Why calculate percentage change?
Potato cylinders never start with identical masses. Calculating percentage change allows a valid comparison of mass changes between pieces with different initial masses.
Displaying Data: Graph Choice [1 mark]
✅ Best Presentation Method
Tick: Line graph
💡 Line Graph vs. Bar Chart
Both variables (concentration of solution and percentage change in mass) are continuous quantitative data. Continuous data should always be plotted on a line graph (or scatter graph with a line of best fit), whereas bar charts are used for categoric/discrete data.
Explaining the Biological Mechanism [3 marks]
✅ Completed Sentences (Must be in this exact order)
Some of the pieces of potato decreased in mass because the potato cells lost water [1 mark].
The decrease in mass was due to a process called osmosis [1 mark] (diffusion is also allowed).
The structure surrounding each cell in a piece of potato is partially permeable [1 mark].
💡 Key Knowledge: Definition of Osmosis
Osmosis is the net movement of water molecules from a dilute solution (high water concentration) to a more concentrated solution (low water concentration) across a partially permeable membrane.
❌ Common Misconceptions
- Never say "salt moved into/out of the potato". Salt ions are solutes and do not move freely in this osmosis experiment. Only water moves.
- Don't write "semi-permeable" if you can avoid it; the spec specifically requires partially permeable.
Estimating the Isotonic Point [1 mark]
✅ Correct Answer Range
Any concentration in the range:
0.15 to 0.25 mol/dm³
🧠 How to Deduce This from the Table
- At 0.1 mol/dm³: mass increased by +4.8% (water entered).
- At 0.3 mol/dm³: mass decreased by -4.4% (water left).
- No change in mass ( 0% ) must occur between 0.1 and 0.3 mol/dm³. Since +4.8 and -4.4 have roughly equal magnitudes, zero change is roughly halfway: 0.2 mol/dm³.
💡 Biological Meaning
The concentration where there is no change in mass is the point where the solution has the same concentration as the cytoplasm inside the potato cells (isotonic). Water enters and leaves the cells at the same rate, resulting in no net movement.
Topics
Biology · Required Practicals · B1: Cell Biology · Required Practicals
Question and mark scheme from the AQA GCSE Biology examination, Biology Paper 1 (Foundation), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.