AQA AS Level Biology Paper 1, June 2024: Question 8
9 marks · Medium difficulty · Practical Techniques & Data Analysis
Explain how a non-competitive inhibitor decreases the rate of an enzyme-controlled reaction, identify variables in an enzyme-substrate investigation, calculate reaction rate from a tangent on a casein concentration-time graph, and sketch a predicted curve at an optimum temperature.
Practise this questionQuestion
Question text
08.1 A non-competitive inhibitor decreases the rate of an enzyme-controlled reaction.
Explain how.
[3 marks]
08.2 A scientist investigated the hydrolysis of the protein casein.
The scientist:
• mixed a solution of a protease enzyme with a solution of casein
• then measured the casein concentration in the mixture at intervals
• controlled all relevant variables appropriately.
For this investigation, identify:
[2 marks]
the independent variable
the dependent variable
08.3 Figure 10 shows the scientist’s results.
Figure 10
Use Figure 10 to determine the rate of casein hydrolysis at 2 minutes.
Show how you obtained your answer.
[2 marks]
Answer mg dm–3 minute–1
08.4 The scientist repeated the investigation but increased the temperature to the optimum
temperature for this protease.
Sketch a line on Figure 10 showing the results you predict if the investigation is
repeated at the optimum temperature for the protease.
[2 marks]
Mark scheme
Show the mark scheme
Question Marking Guidance Mark Comments
1. (Inhibitor) binds (to enzyme) away from active Ignore inhibitor not
site similar or not same
shape as substrate
OR
1. Accept binds to
(Inhibitor) does not bind to active site;
allosteric (binding) site
2. Changing (enzyme) tertiary structure for does not bind to
OR active site
3 3. Accept a
08.1 Changing active site (shape); (3 x
description of
AO1) enzyme-substrate
3. No/fewer enzyme-substrate complexes (form)
complex, eg so
no/fewer substrate
OR molecules
fit/bind/enter (the
Enzyme-substrate not complementary; active site)
3. Accept no/fewer E-
S complexes
1. (Independent) Time (of measurement/test 2
08.2 /sample taken); (2 x
2. (Dependent) Casein concentration; AO2)
1. and 2. Correct answer of 20–30 = 2 marks;;
Tangent drawn touching (2, 35) with incorrect
calculation = 1 mark 2
08.3 OR (2 x
AO2)
17.5 (for incorrect method with correct reading at 2
minutes (35) and division by correct time, (2))
= 1 mark;
1. Curve drawn entirely to the left of the curve
given; 2
08.4 (2 x
2. Similar shaped curve from (0, 180) to intersect x AO3)
axis between (0, 0) and (6, 0);
How to answer it
Enzyme Action, Inhibition and Rate Calculations Study Guide
What this question tests
This question assesses your understanding of enzyme kinetics, the specific mechanism of non-competitive inhibition, experimental design identification (independent and dependent variables), graphical analysis (calculating gradients/rates from curves), and predicting the effects of temperature on enzyme-catalysed reactions.
Mechanism of Non-Competitive Inhibition
✅ Correct Answer Structure
- Point 1: Inhibitor binds to the enzyme away from its active site (e.g., at an allosteric site).
- Point 2: This binding alters/changes the tertiary structure (and therefore the active site shape) of the enzyme.
- Point 3: Consequently, fewer/no enzyme-substrate complexes can form because substrates are no longer complementary to the active site.
❌ Common Errors
- Stating that the inhibitor "blocks" or "competes for" the active site (this describes competitive inhibition).
- Vague phrasing like "changes the shape of the enzyme" without specifying the tertiary structure or the active site.
- Failing to link the structural change directly to the inability of substrate molecules to bind.
Identifying Experimental Variables
✅ Correct Answer
- Independent variable: Time (of measurement / when sample taken).
- Dependent variable: Casein concentration.
🧠 Exam Technique
Remember the definitions: The independent variable is what the investigator deliberately changes or selects intervals for (time). The dependent variable is what is measured as a result (concentration of casein remaining).
Determining Rate from a Curve
📐 Step-by-Step Calculation
- Step 1: Locate Time = 2 minutes on the x-axis of Figure 10 and move up to the curve.
- Step 2: Draw a precise tangent touching the curve strictly at the point where time equals 2 minutes.
- Step 3: Calculate the gradient of your tangent using change in y divided by change in x (Delta y / Delta x).
- Step 4: Expected valid range for full marks: 20 to 30 mg dm⁻³ minute⁻¹ (based on accurate tangent construction).
❌ Common Errors & Partial Marks
- Reading the concentration value directly at 2 minutes (approx 35) and dividing by 2 to get 17.5. Examiners award 1 mark for this incorrect method if the reading (35) and division by 2 are shown, but you lose the second mark for not drawing a tangent.
- Drawing a chord instead of a tangent (connecting two points on the curve rather than touching at a single instant).
Predicting Temperature Effects on Rate
💡 Key Knowledge
Increasing temperature to the optimum increases kinetic energy, leading to more frequent successful collisions and a higher rate of enzyme-substrate complex formation.
Therefore, the reaction finishes quicker, meaning casein concentration drops to zero faster.
✅ Sketch Requirements
- Mark 1: The entire new curve must lie to the left of the original curve provided in Figure 10 (indicating a faster reaction/steeper initial drop).
- Mark 2: The curve must start at the exact same initial concentration ( 180 mg dm⁻³ at time 0) and level off to intersect the x-axis earlier (between 0, 0 and 6, 0 minutes).
Topics
Biology · Practical skills · 3.1 Biological molecules · Experimental design · Data analysis
Question and mark scheme from the AQA AS Level Biology examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.