OCR A Level Biology A: Planning, implementing, analysing and evaluating practicals (H420) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 1.1 of OCR A Level Biology A (H420, Version 4.1, April 2026, first assessment in 2025): planning, implementing, analysing and evaluating practicals. Use the study guide and revision notes first. Also see the OCR A Level Biology hub and the printable checklist.

Questions

1. A student investigates how the nitrate concentration of the water supplied affects the growth of cress seedlings.

(a) Identify the independent variable and the dependent variable, and state how the dependent variable could be measured. [2] (b) Give two variables that must be controlled, and say how each would be controlled. [2]

2. A student tests the effect of light colour on photosynthesis. Pondweed is placed 10 cm from a filament lamp covered with a coloured filter, and bubbles are counted for 1 minute. One test is done per colour. Suggest three reasons why this method is not appropriate. [3]

3. A scale bar on a micrograph represents 25 µm and is 5 mm long on the image. A cell on the same image is 38 mm long.

(a) Calculate the magnification. [1] (b) Calculate the actual length of the cell in µm. [1] (c) Give this length in mm. [1]

4. (a) 8.25 cm³ of gas is collected in 1.5 minutes. Calculate the rate in cm³ min⁻¹, quoting the answer to an appropriate number of significant figures. [2] (b) Three masses are 2.31 g, 2.35 g and 2.30 g. Calculate the mean. [1]

5. State the most suitable way to display each set of data, and give a reason for each.

(a) Mean shoot length of seedlings grown in four types of soil. [1] (b) The lengths of 80 leaves, grouped into 5 mm classes. [1] (c) Stomatal density and leaf area for 25 leaves, to look for a relationship. [1]

6. A calibration line for a dye in a colorimeter passes through (0.00, 0.05) and (0.80, 0.65), where x is dye concentration in mol dm⁻³ and y is absorbance (no units).

(a) Calculate the gradient, with its unit. [2] (b) State what the y-intercept represents. [1] (c) A sample has absorbance 0.41. Calculate its concentration. [2]

7. Volume of product from an enzyme reaction:

Time / s	0	10	20	30	40	50	60
Volume / cm³	0.0	11.8	19.0	23.3	25.9	27.5	28.5

(a) Describe how to find the rate of reaction at 20 s from a graph of these data. [2] (b) A tangent drawn at 20 s passes through (0, 7.9) and (40, 30.0). Calculate the rate, with its unit. [2]

8. Times for leaf discs to rise in a solution (s): 54, 57, 55, 56, 71, 55.

(a) Identify the anomaly and justify your choice. [1] (b) Calculate the mean without the anomaly. [1] (c) State the range of the remaining values. [1] (d) Calculate the rate as 1 ÷ mean time, to 3 significant figures, with its unit. [1] (e) Suggest one cause of the anomaly. [1]

9. A potato disc has mass 2.40 g before and 2.15 g after soaking. The balance reads to 2 decimal places, so each reading has an uncertainty of ±0.005 g.

(a) Calculate the change in mass. [1] (b) State the absolute uncertainty in the change. [1] (c) Calculate the percentage uncertainty in the change. [1] (d) A thermometer reading 20 °C has an uncertainty of ±0.5 °C. Calculate the percentage uncertainty. [1] (e) Suggest one way to reduce the percentage uncertainty in the mass change. [1]

10. (a) A student estimates a population as 148 individuals. A full count gives 160. Calculate the percentage error. [2] (b) A rate rises from 0.082 s⁻¹ to 0.131 s⁻¹. Calculate the percentage increase. [2]

11. Agar cubes contain sodium hydroxide and an indicator that is pink in alkali. Dilute hydrochloric acid diffuses in and turns the agar colourless. Plan an investigation into how the size of the cube affects the time for the colour to disappear. [8]

Answers

1. (a) Independent: nitrate concentration; dependent: shoot length [1], measured with a ruler (mm) after a set number of days, or mass [1]. (b) Light intensity, with plants the same distance from the same lamp [1]; volume of solution given daily, using a measuring cylinder (or temperature, or seed batch) [1]. Examiner insight: A control variable earns credit when the method of control is given, not when it is merely named.

2. Any three, each with its reason. The lamp heats the water, so temperature changes along with the filter [1]. Filters transmit different light intensities, so colour is not the only change [1]. Bubbles vary in size, so counting them does not measure gas volume [1]. Other valid points: one test per colour gives no repeats or spread; bubbles are counted by eye. [3] Examiner insight: “Not a fair test” is too vague to credit; name the variable that changes and its effect.

3. (a) 5 mm = 5000 µm; 5000 ÷ 25 = ×200 [1]. (b) 38 ÷ 200 = 0.19 mm = 190 µm [1]. (c) 0.19 mm [1]. Examiner insight: Convert to one unit before dividing, and write the unit with the answer.

4. (a) 8.25 ÷ 1.5 = 5.5 [1]; 5.5 cm³ min⁻¹ to 2 significant figures, the precision of the time [1]. (b) (2.31 + 2.35 + 2.30) ÷ 3 = 2.32 g [1]. Examiner insight: The significant figures of the least precise measurement set the number of figures in the answer.

5. (a) Bar chart, because soil types are categories [1]. (b) Histogram, because the data are continuous and grouped in classes [1]. (c) Scattergram with a line of best fit, because two variables are compared for a relationship [1]. Examiner insight: Give the reason as well as the name; the reason shows you know what type of data you have.

6. (a) (0.65 − 0.05) ÷ (0.80 − 0.00) = 0.75 [1] dm³ mol⁻¹ [1]. (b) The absorbance with no dye present (the blank or solvent) [1]. (c) (0.41 − 0.05) ÷ 0.75 [1] = 0.48 mol dm⁻³ [1]. Examiner insight: Subtract the intercept before dividing, and give the unit of concentration.

7. (a) Draw a smooth curve through the points [1]. Place a ruler touching the curve at 20 s only, draw the tangent, and calculate its gradient (change in volume ÷ change in time) [1]. (b) (30.0 − 7.9) ÷ (40 − 0) [1] = 0.5525, so 0.55 cm³ s⁻¹ [1]. Examiner insight: Take both points from the tangent line, not from the data table.

8. (a) 71 s, because it is far from the other five, which lie within 54 to 57 s [1]. (b) (54 + 57 + 55 + 56 + 55) ÷ 5 = 55.4 s [1]. (c) 57 − 54 = 3 s [1]. (d) 1 ÷ 55.4 = 0.0181 s⁻¹ [1]. (e) A disc stuck to the side of the beaker or a late stopwatch start [1]. Examiner insight: Justify the anomaly by comparing it with the other repeats, and give a specific cause.

9. (a) 2.40 − 2.15 = 0.25 g (a decrease) [1]. (b) Add the two uncertainties: ±0.01 g [1]. (c) 0.01 ÷ 0.25 × 100 = 4.0 % [1]. (d) 0.5 ÷ 20 × 100 = 2.5 % [1]. (e) Use larger discs so the change is bigger, or a balance reading to 3 decimal places [1]. Examiner insight: A difference of two readings carries the uncertainties of both, so small changes give large percentage uncertainties.

10. (a) (160 − 148) ÷ 160 × 100 [1] = 7.5 % [1]. (b) (0.131 − 0.082) ÷ 0.082 × 100 [1] = 59.8 % [1]. Examiner insight: Divide by the accepted value for error and by the starting value for change.

11. Any eight points from:

• Independent variable: cubes of at least four side lengths, such as 5, 10, 15 and 20 mm, cut with the same blade [1].
• Dependent variable: time for the pink colour to disappear completely, measured with a stopwatch to the nearest second [1].
• Control: same concentration and volume of acid, stated and measured with a measuring cylinder [1].
• Control: constant temperature, such as a water bath checked with a thermometer, and the same agar batch [1].
• Method: drop one cube into the acid, start timing, keeping each cube fully covered by the same volume of acid [1].
• Repeat at least three times per size and calculate a mean, excluding any anomaly [1].
• Prediction: smaller cubes have a larger surface area to volume ratio and a shorter diffusion distance, so the colour disappears faster [1].
• Processing: calculate surface area to volume ratio for each size and plot time against it [1]. [8] Examiner insight: A plan gains credit for detail that someone else could follow, with quantities and a stated way to read the end point.

Where marks are usually lost

• Naming a control variable without saying how to keep it constant.
• Describing a method as an “unfair test” without naming the variable.
• Giving answers with too many significant figures.
• Omitting units on gradients, rates and concentrations.
• Taking gradient points from the table rather than from the line.
• Forgetting to subtract the intercept when using a calibration line.
• Using one reading’s uncertainty for a difference of two readings.
• Dividing by the wrong value in percentage error and percentage change.
• Writing “human error” without naming a cause.

Next steps

Revisit the revision notes and study guide, then try the all free 10-minute diagnostics. Use the course hub and checklist to track coverage. Book a free trial class.

Official syllabus

OCR Level 3 Advanced GCE in Biology A (H420), Version 4.1, April 2026, for first assessment in 2025, published by OCR. Section 1.1 and the mathematical requirements in section 5d.