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CIE A-Level Biology 9700 Paper 5 Practice Paper (Planning, Analysis and Evaluation) – 30 marks, 1 hour 15 minutes, answers and mark scheme
TutorZone original practice paper – CIE A-Level Biology 9700 Paper 5 (Planning, Analysis and Evaluation)
2 questions, 10 parts – 30 marks – 1 hour 15 minutes – calculator allowed
Paper comparison
| Paper | Time | Marks | Role |
|---|---|---|---|
| Paper 5 (planning, analysis and evaluation) | 1 h 15 min | 30 | this paper |
| Paper 4 (A Level structured) | 2 hours | 100 | other A Level paper |
| Paper 3 (AS practical skills) | 2 hours | 40 | AS paper |
| Paper 2 (AS structured) | 1 h 15 min | 60 | AS paper |
| Paper 1 (multiple choice) | 1 hour | 40 | AS paper |
Data and formulae
| Quantity | Value |
|---|---|
| Acceleration of free fall g | 9.81 m s-2 |
| Molar volume at room temperature | 24.0 dm3 mol-1 |
| Molar mass of oxygen | 32.0 g mol-1 |
| Mr of glucose | 180 |
| Standard atmospheric pressure | 101 kPa |
| Vapour pressure of water at 20 °C | 2.3 kPa |
Questions
Total: 30 marks · 1 h 15 min · calculator allowed
Q 1 · Planning: effect of temperature on catalase activity [15]
(a) A student investigates the effect of temperature on catalase activity in potato, using the rate of oxygen release from hydrogen peroxide as the measurement. State the independent variable, the dependent variable and two control variables. [2]
(b) Describe how the reaction mixture is kept at each chosen temperature between 10 °C and 50 °C and how the temperature of the mixture is determined accurately. [3]
(c) Describe how the rate of oxygen release is measured accurately. [4]
(d) State which graph should be plotted to show the relationship between temperature and rate, describe its expected shape, and explain how Q10 is calculated from the rates. [3]
(e) State two precautions that make the experiment more reliable or safer and explain the reason for each. [3]
Q 2 · Analysis and evaluation: light intensity and the rate of photosynthesis [15]
(a) A student investigates the effect of light intensity on photosynthesis in pondweed (Elodea). Oxygen released in 5 minutes is collected in a gas-tight measuring tube, and the distance d between the lamp and the weed is changed; the light intensity at the weed is assumed to be proportional to 1/d². The results are shown in the table. Complete the table by calculating the rate of oxygen release in cm3 min-1 for each distance, giving each value to 2 decimal places. [3]
(b) Plot a graph of rate (cm3 min-1) on the y-axis against 1/d² (cm-2) on the x-axis, describe the relationship obtained and state what the gradient represents. [4]
(c) Use the gradient of your graph to predict the rate at d = 25 cm, and compare it with the value obtained from the raw data. [3]
(d) Calculate the percentage uncertainty in the rate at d = 50 cm (volume reading ±0.05 cm3, timing ±1 s). [2]
(e) State two limitations or sources of error that affect the results and suggest an improvement for each. [3]
Answers and mark scheme
| Q | Part | Answer | Mark points | marks |
|---|---|---|---|---|
| 1 | (a) | Independent variable: temperature (10, 20, 30, 40 and 50 °C). Dependent variable: the rate of oxygen release (cm3 s-1). Controls: the concentration and volume of hydrogen peroxide (20 cm3 of 2.0% w/v), the number and total surface area of the potato discs (5 discs, 2 mm thick, cut with the same cork borer), the pH (same buffer) and the length of time over which gas is collected. | 1 mark: temperature independent, rate of oxygen release dependent 1 mark: two controls (hydrogen peroxide concentration / volume, disc number and size, pH, collection time) |
2 |
| 1 | (b) | Set up five water baths at 10, 20, 30, 40 and 50 °C and check each with a thermometer until the reading is steady. Stand the tube containing 20 cm3 of 2.0% hydrogen peroxide in the bath for 5 minutes, and pre-warm the potato discs in the same bath for 5 minutes, so that both reach the bath temperature. Just before mixing, read the temperature of the liquid in the tube with a thermometer and record it instead of assuming it equals the bath setting; it should be within ±0.5 °C of the target. | 1 mark: five water baths set at 10-50 °C 1 mark: peroxide and discs pre-warmed in the bath for 5 minutes 1 mark: temperature of the liquid measured, not assumed to equal the bath setting |
3 |
| 1 | (c) | Add the discs, immediately fit a gas-tight syringe (or an inverted gas measuring tube over water) and start the stopwatch at the same moment. Record the gas volume every 10 s for 120 s. Take the initial rate (cm3 s-1) from the steepest tangent to the volume-time graph over the first 30 s, because the substrate concentration falls and the rate slows later. Repeat each temperature three times and average, using fresh discs and fresh hydrogen peroxide each time, and blot the discs dry with filter paper. | 1 mark: gas-tight apparatus (syringe / inverted tube) with no leak 1 mark: volume read every 10 s for 120 s 1 mark: initial rate from the steepest tangent over the first 30 s, not a whole-run average 1 mark: three repeats averaged / fresh discs and fresh reagent |
4 |
| 1 | (d) | Plot the initial rate (cm3 s-1) on the y-axis against temperature (°C) on the x-axis. The rate should rise steadily from 10 °C to about 40 °C, roughly doubling for each 10 °C rise, reach a maximum near 40 °C and then fall sharply because the enzyme denatures and the active site loses its shape. Q10 equals the rate at (T + 10) °C divided by the rate at T °C, for example the rate at 30 °C divided by the rate at 20 °C. | 1 mark: rate plotted against temperature, both axes labelled with units 1 mark: expected rise to a maximum near 40 °C then a sharp fall due to denaturation 1 mark: Q10 = rate at (T + 10) divided by rate at T |
3 |
| 1 | (e) | (1) Repeat every temperature three times and average, which reduces random error. (2) Cut all discs with the same cork borer, keep the thickness the same and blot them dry, so that surface area and the amount of enzyme are controlled. (3) Hydrogen peroxide irritates skin and eyes, so wear goggles and gloves and work in a well-ventilated place; use tongs for tubes in the 50 °C bath because of the risk of scalding. | 1 mark: three repeats averaged (reduces random error) 1 mark: same cork borer / equal thickness / blotted dry (controls surface area and enzyme amount) 1 mark: goggles and gloves, ventilation, tongs for hot tubes (safety reason) |
3 |
| 2 | (a) | Rate = volume in 5 min ÷ 5: d = 10 cm gives 20.0 ÷ 5 = 4.00; 20 cm gives 5.00 ÷ 5 = 1.00; 30 cm gives 2.22 ÷ 5 = 0.44; 40 cm gives 1.25 ÷ 5 = 0.25; and 50 cm gives 0.80 ÷ 5 = 0.16 cm3 min-1. | 1 mark: all five values divided by 5 minutes (4.00, 1.00, 0.44, 0.25, 0.16) 1 mark: unit cm3 min-1 1 mark: all values given consistently to 2 decimal places |
3 |
| 2 | (b) | The five points lie close to a straight line through the origin: 1.00 × 10-2 gives 4.00; 2.50 × 10-3 gives 1.00; 1.11 × 10-3 gives 0.44; 6.25 × 10-4 gives 0.25; 4.00 × 10-4 gives 0.16. The least-squares gradient is about 4.0 × 102 cm3 min-1 cm2. A straight line through the origin shows that the rate is directly proportional to light intensity, so light is the limiting factor over this range, and the gradient is the increase in rate per unit increase in 1/d². | 1 mark: both axes labelled with variable and unit 1 mark: all five points plotted correctly 1 mark: straight line through the origin 1 mark: gradient about 4.0 × 102 cm3 min-1 cm2, with rate proportional to intensity / light limiting |
4 |
| 2 | (c) | At d = 25 cm, 1/d² = 1 ÷ 625 = 1.60 × 10-3 cm-2, so the rate is 4.0 × 102 × 1.60 × 10-3 = 0.64 cm3 min-1. From the raw data the volume in 5 minutes is 2000 ÷ 625 = 3.20 cm3, giving 3.20 ÷ 5 = 0.64 cm3 min-1. The two values agree to within 1%, which supports the conclusion that the rate is proportional to light intensity. | 1 mark: 1/d² = 1.60 × 10-3 cm-2 1 mark: gradient used to obtain 0.64 cm3 min-1 1 mark: compared with the value from the data and a conclusion drawn |
3 |
| 2 | (d) | Volume: 0.05 ÷ 0.80 × 100% = 6.25%; time: 1 ÷ 300 × 100% = 0.33%; total 6.58%, which is about 6.6% (or 7% to one significant figure). The volume is smallest at 50 cm, so the relative uncertainty is largest there and that point is the least reliable on the graph. | 1 mark: 6.25% + 0.33% = 6.6% 1 mark: notes that the smallest volume at 50 cm gives the largest percentage uncertainty |
2 |
| 2 | (e) | (1) The distance is measured with a ruler between the lamp and the weed, but the position of the filament is not well defined, and the lamp heats up and raises the water temperature, changing enzyme activity. Use a light box or LED, fix the lamp-weed distance, and keep the water at 25 °C in a water bath. (2) Some oxygen dissolves in the water or clings to the weed and never enters the tube. Let the weed acclimatise for 5 minutes, tap the tube so bubbles rise, or measure with a dissolved-oxygen electrode. (3) Only the carbon dioxide already dissolved in the water is available and it may run out within 5 minutes and become the limiting factor. Supply carbon dioxide using a solution of sodium hydrogencarbonate of fixed concentration. | 1 mark: first limitation (distance / filament position, lamp heating) with an improvement (light box, fixed distance, water bath) 1 mark: second limitation (oxygen dissolving / clinging, CO2 running out) with an improvement (acclimatise, tap the tube, hydrogencarbonate solution) 1 mark: each improvement must actually address the stated error |
3 |
Topic breakdown
| Question | Topic | Marks |
|---|---|---|
| 1 | Planning: effect of temperature on catalase activity | 15 |
| 2 | Analysis and evaluation: light intensity and the rate of photosynthesis | 15 |
Common lost marks
- Writing only “measure the oxygen” in the planning question without saying how the initial rate is found, for example from the steepest tangent over the first 30 s.
- Treating the total volume collected in 5 minutes as the rate and forgetting to divide by the time, so the unit becomes cm3 instead of cm3 min-1.
- Plotting rate against d instead of 1/d², which gives a curve and hides the proportional relationship.
- Writing “human error” as the evaluation without naming a real source (filament position, lamp heating, oxygen dissolving) and a matching improvement.
Frequently asked
What does Paper 5 test?
Paper 5 is the planning, analysis and evaluation paper: 1 hour 15 minutes, 30 marks. Question 1 asks you to design an experiment (variables, method, data handling, safety); Question 2 gives data for you to complete a table, plot a graph, find a gradient and an uncertainty, and evaluate the errors.
Can I do this paper without having done the experiment?
Yes. Paper 5 tests experimental design and data handling rather than hands-on experience, but you must be fluent with control variables, initial rates, gradients from graphs and percentage uncertainties.
Why plot 1/d² rather than d?
Light intensity is inversely proportional to the square of the distance, so 1/d² is the true independent variable. A straight line through the origin for rate against 1/d² proves that the rate is proportional to light intensity; a plot against d only gives a curve.
This paper is written by TutorZone tutors from the public syllabus; all questions and data are original and not endorsed by CIE.
The answer booklet gives mark-scheme points for study use only; please credit the source when reusing.
