AQA GCSE Biology Biology Paper 1 (Foundation), June 2023: Question 1

10 marks · Low Demand difficulty · Short Answer

Identify plant organisation levels, calculate a mean change in mass from an osmosis practical, identify anomalies, variables, the process of osmosis, and how to improve the investigation.

Practise this question

Question

Question 1 contains multiple sub-questions starting with 01.1 asking which part of a plant is largest (guard cell, leaf, or root hair). Then an investigation into the effect of salt solution concentration on the mass of potato pieces is described with a method and Table 1. Table 1 shows change in mass for dilute salt solution (Test 1: 1.1, Test 2: 1.1, Test 3: 1.4, Mean: X) and concentrated salt solution (Test 1: -7.2, Test 2: -6.8, Test 3: -32.4, Mean: -7.0). Questions 01.2 to 01.9 ask to calculate mean X, ring the anomaly in Table 1, explain what was done with the anomaly, name the type of variable kept the same, identify another controlled variable, state the process (osmosis) by which water moved, describe permeability, and select an improvement to the method.
Question text

01 Plants are made of cells, tissues and organs.

01.1 Which part of a plant is the largest?

[1 mark]

Tick ( ) one box.

A guard cell

A leaf

A root hair

Students investigated the effect of concentration of salt solution on the mass of pieces

of potato.

This is the method used.

1. Cut two pieces of potato to the same size.

2. Record the mass of each piece of potato.

3. Place one piece of potato into a beaker containing a dilute salt solution.

4. Place the other piece of potato into a beaker containing a concentrated salt

solution.

5. After 20 minutes, remove each piece of potato from its solution.

6. Record the change in mass of each piece of potato.

7. Repeat steps 1 to 6 two more times.

Table 1 shows the results.

Table 1

Change in mass of piece of potato in grams

Solution

Test 1 Test 2 Test 3 Mean

Dilute salt solution 1.1 1.1 1.4 X

Concentrated salt solution −37.2 −6.8 −32.4 −7.0

01.2 Calculate mean value X in Table 1.

[2 marks]

X = grams

There is an anomalous result for the concentrated salt solution in Table 1.

01.3 Draw a ring around the anomalous result in Table 1.

[1 mark]

01.4 What did the students do with the anomalous result when calculating the mean

in Table 1?

[1 mark]

01.5 What name is given to a variable that is kept the same during an investigation?

[1 mark]

Tick ( ) one box.

Control variable

Dependent variable

Independent variable 4

01.6 One variable the students kept the same during the investigation was the size of the

pieces of potato.

Which other variable did the students keep the same?

[1 mark]

*03* Tick ( ) one box.

Change in mass of pieces of potato

Concentration of salt solution

Time in the salt solution

01.7 The pieces of potato in the concentrated salt solution decreased in mass.

Complete the sentence.

Choose the answer from the box.

[1 mark]

excretion osmosis respiration

Water moved out of the potato by the process of .

01.8 The potato cells have a partially permeable membrane.

Which particles can pass through a partially permeable membrane?

[1 mark]

Tick ( ) one box.

No particles

Some particles

All particles

01.9 How could the students improve their investigation?

[1 mark]

Tick ( ) one box.

Boil the pieces of potato at the start.

Leave the skin on some pieces of potato.

Use more concentrations of salt solution.

Mark scheme

Show the mark scheme Mark scheme for Question 1 providing acceptable answers and mark allocations: 01.1: a leaf (1 mark); 01.2: (1.1 + 1.1 + 1.4)/3 or 3.6/3 (1 mark), and 1.2 grams (1 mark); 01.3: ring around -32.4 (1 mark); 01.4: did not include it / allow ignored it (1 mark); 01.5: control variable (1 mark); 01.6: time in the salt solution (1 mark); 01.7: osmosis (1 mark); 01.8: some particles (1 mark); 01.9: use more concentrations of salt solution (1 mark). Total 10 marks.

Question 1

AO /

Question Answers Extra information Mark

Spec Ref.

01.1 a leaf 1 AO2

4.2.1

4.2.3.1

AO /

Spec Ref.

1.1 + 1.1 + 1.4 if no answer given on answer

01.2 1 AO2

lines, allow an answer in Table

3 4.1.3.2

or RPA3

3.6

1.2 (grams) 1

AO /

Spec Ref.

01.3 ring around −32.4 (grams) table takes precedence 1 AO3

4.1.3.2

allow (−)32.4 (grams) written by RPA3

question

AO /

Spec Ref.

01.4 did not include it allow ignored it 1 AO2

4.1.3.2

RPA3

AO /

Spec Ref.

01.5 control variable 1 AO1

4.1.3.2

RPA3

AO /

Spec Ref.

01.6 time in the salt solution 1 AO3

4.1.3.2

RPA3

AO / 7

Spec Ref.

01.7 osmosis 1 AO2

4.1.3.2

RPA3

AO /

Spec Ref.

01.8 some particles 1 AO1

4.1.3.2

RPA3

AO /

Spec Ref.

01.9 use more concentrations of salt 1 AO3

solution 4.1.3.2

RPA3

Total Question 1 10

How to answer it

Plant Organisation & Osmosis Required Practical

📌 What this question tests

This question assesses core foundation knowledge from AQA Specification 4.1.3.2 (Osmosis) and 4.2.1 / 4.2.3.1 (Plant Tissues & Organs), alongside practical experimental skills (RPA 3):

  • Levels of biological organisation (identifying organs vs specialised cells).
  • Handling experimental data: calculating a mean and identifying/treating anomalies.
  • Working Scientifically: variables (control, independent, dependent) and evaluating improvements to investigations.
  • Definition of osmosis and function of partially permeable membranes.
Question 01.1

Plant Levels of Organisation

Identifying the largest plant structure [1 mark]

✅ Correct Answer

A leaf

awarded 1 mark for ticking the box next to 'A leaf'.

💡 Key Knowledge

Remember the hierarchy of organisation in plants:

Organelles → Cells → Tissues → Organs → Organ Systems

  • Guard cell: Specialised single cell.
  • Root hair: Part of a single cell (specialised cell).
  • Leaf: A plant organ made of multiple tissues (epidermis, mesophyll, xylem, phloem). Organs are far larger than individual cells.
Question 01.2

Calculating the Mean Mass Change

Working out mean value X for dilute salt solution [2 marks]

📐 Calculation Walkthrough

  1. Identify values: In the 'Dilute salt solution' row, the three tests are 1.1 , 1.1 , and 1.4 .
  2. Sum the values: 1.1 + 1.1 + 1.4 = 3.6 [1 mark]
  3. Divide by count: 3.6 ÷ 3 = 1.2 [1 mark]

X = 1.2 grams

❌ Common Errors

  • Dividing by the wrong number (e.g. dividing by 2 instead of 3).
  • Entering values into a calculator incorrectly without brackets: typing 1.1 + 1.1 + 1.4 ÷ 3 gives 2.66 due to order of operations! Always press '=' before dividing.
  • Rounding to inappropriate significant figures (keep to 1 decimal place to match raw data).
Questions 01.3 & 01.4

Handling Anomalous Results

Identifying and processing anomalies in experimental data [2 marks]

Solution Change in mass of piece of potato in grams
Test 1 Test 2 Test 3 Mean
Concentrated salt solution -7.2 -6.8 -32.4 -7.0

✅ Correct Answers

01.3: Circle drawn clearly around -32.4 in Table 1 [1 mark].

01.4: They did not include it / ignored it when calculating the mean [1 mark].

🧠 Exam Technique

Look at the mean provided: -7.0 .

  • Notice: (-7.2 + -6.8) ÷ 2 = -7.0 .
  • This confirms that the students spotted that -32.4 was an anomaly and completely excluded it from their mean calculation.
  • Acceptable phrasing: "excluded it", "ignored it", "left it out".
Questions 01.5 & 01.6

Experimental Variables

Understanding scientific controls [2 marks]

✅ Correct Answers

  • 01.5: Control variable [1 mark]
  • 01.6: Time in the salt solution [1 mark]

💡 Key Knowledge: Variable Definitions

  • Independent variable: What you change (Concentration of salt solution).
  • Dependent variable: What you measure (Change in mass of potato).
  • Control variable: What you keep the same to ensure a fair test (Potato size, time in solution = 20 minutes, temperature).

❌ Common Trap

Do not confuse control variable with a control experiment. A control variable is simply any factor kept constant so that only the independent variable affects the dependent variable.

Questions 01.7 & 01.8

Osmosis & Membrane Permeability

Explaining movement of water across membranes [2 marks]

✅ Correct Answers

  • 01.7: osmosis [1 mark]
  • 01.8: Some particles [1 mark]

💡 Key Definition: Osmosis

Osmosis is the diffusion of water from a dilute solution to a concentrated solution through a partially permeable membrane.

  • In concentrated salt solution, water moves out of the potato cells by osmosis, causing a decrease in mass.
  • Partially permeable means it allows small molecules (e.g. water) to pass through, but prevents larger solute molecules (e.g. sucrose or complex salts) from crossing freely. Hence, only some particles can pass.
Question 01.9

Improving the Investigation

Evaluating experimental design [1 mark]

✅ Correct Answer

Use more concentrations of salt solution. [1 mark]

🧠 Exam Technique: Improving RPA 3

Why this is the best improvement:

  • The students only tested two solutions: "dilute" and "concentrated".
  • Testing a range of known concentrations (e.g. 0.0, 0.2, 0.4, 0.6, 0.8, 1.0 mol dm⁻³) allows you to plot a graph and find the exact concentration inside the potato cells (where mass change is zero).
  • Boiling the potato would destroy cell membranes and stop osmosis completely!
  • Leaving skin on would introduce an uncontrolled variable that restricts surface area for osmosis.

Topics

Biology · Required Practicals · B1: Cell Biology · B2: Organisation · Required Practicals

Question and mark scheme from the AQA GCSE Biology examination, Biology Paper 1 (Foundation), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.