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IGCSE Biology Past Papers & Drilling (0610 Paper 6): 6 Questions with Answers
TUTORZONE ORIGINAL DRILLING MOCK · CAMBRIDGE IGCSE BIOLOGY (0610) PAPER 6
Written to the Cambridge 0610 syllabus (2026-2028 sessions): Paper 6 Alternative to Practical is 1 hour, 40 marks and 20% of the qualification, assessing experimental skills (AO3) entirely on paper, so candidates sit no experiment; the other route is Paper 5, the Practical Test (1 h 15 min, 40 marks, 20%). This is a Paper 6 drilling mock: 6 questions for 40 marks in 60 minutes on food tests, osmosis, enzyme rate, photosynthesis, transpiration and magnification, with answers.
How is IGCSE Biology (0610) Paper 6 assessed, and what does it carry?
Direct answer: Paper 6 Alternative to Practical is 1 hour, 40 marks and 20% of the qualification, assessing experimental skills (AO3) entirely on paper with no experiment performed. The alternative is Paper 5, the Practical Test (1 h 15 min, 40 marks, 20%); a candidate sits one of the two, graded A* to G. The other two papers are Paper 1/2 multiple choice and Paper 3/4 theory.
Assessment at a glance (2027)
| Component | Content | Weighting | Duration |
|---|---|---|---|
| Paper 6 (Alternative to Practical) | Experimental skills, 40 marks, no experiment performed | 20% | 1 hour |
| Paper 5 (Practical Test) | Experimental skills, 40 marks, experiments in a laboratory | 20% | 1 h 15 min |
Paper 6 assesses experimental skills only: 40 marks in 1 hour, about 1.5 minutes a mark. Three rules: write the column headings and units of a results table before any numbers; show the working for a mean or a percentage change with the unit and decimal place the question asks for; and when asked why or how to improve, give a cause and a consequence and name the variable or the source of error involved. On a graph question, even described on paper, name both axes, their units and a sensible scale.
How this paper was built
This is a TutorZone original drilling mock: the questions follow the Cambridge 0610 structure (Paper 6 Alternative to Practical, 40 marks, 1 hour, 20% of the qualification, all AO3 experimental skills) and the answers are our own, not Cambridge past papers or marking references.
Every numerical answer was recomputed programmatically (mean length, percentage change, reaction rate, bubble rate, transpiration rate and percentage increase, magnification, surface area to volume ratio).
No laboratory is needed: all readings are fictional results written for this paper, used to train recording, calculation and error analysis.
Pacing
| Section | Marks | Suggested time | What to do first |
|---|---|---|---|
| Paper 6 (Alternative to Practical) | 20% | about 1.5 min a mark, 6 min to check units | data table and means first, then variables and errors |
| Paper 5 (Practical Test) | 20% | read the whole instruction first, record readings at once | apparatus handling, precision of readings, safety |
How many questions and marks does this mock have, and how long should it take?
Direct answer: 6 questions for 40 marks covering food tests, osmosis, enzyme rate, photosynthesis, transpiration, magnification and surface area to volume, each with a full answer. Allow 60 minutes (the official Paper 6 is 40 marks in 1 hour, about 1.5 minutes a mark).
Paper 6 Alternative to Practical drilling (6 questions, 40 marks)
Food tests 7, osmosis 7, enzyme rate 7, photosynthesis 6, transpiration 7, magnification and surface area to volume 6. Allow 60 minutes (the official Paper 6 is 40 marks in 1 hour).
Question 1 (7 marks)
A student tests one food sample with four reagents. The results are:
– Benedict’s solution turns from blue to green-yellow after heating;
– iodine solution turns from yellow-brown to blue-black;
– Biuret reagent turns from blue to purple;
– the ethanol emulsion test gives a cloudy white layer.
(a) Which reagent must be heated in a water bath, and why? (2)
(b) Deduce which three types of food molecule the sample contains, giving one result as evidence for each. (3)
(c) Why is the iodine put on a spotting tile rather than dropped into the test tube? (1)
(d) State one safety precaution for these tests. (1)
Question 2 (7 marks)
Potato strips cut to the same size, each 40.0 mm long, are put into sucrose solutions of different concentrations and measured again after 30 minutes. Two strips are used at each concentration. Final lengths (mm):
– 0.0 mol/dm3: 42.8, 43.2
– 0.2 mol/dm3: 41.4, 41.8
– 0.4 mol/dm3: 40.1, 39.9
– 0.6 mol/dm3: 38.2, 37.8
– 0.8 mol/dm3: 36.3, 36.5
(a) Calculate the mean final length of the two strips in 0.8 mol/dm3. (1)
(b) Using that mean, calculate the percentage change in length to one decimal place. (2)
(c) Deduce which concentration is closest to the cell sap of the potato, and explain. (1)
(d) Use water potential to explain why the strips shorten in the stronger sucrose solutions. (2)
(e) State one other variable that must be controlled. (1)
Question 3 (7 marks)
Amylase is added to starch solution and iodine is used every 30 seconds to find the time taken for the starch to disappear, repeated at different temperatures. Times (s): 10 C 480; 20 C 240; 30 C 120; 40 C 80; 50 C 120; 60 C 600.
(a) Calculate the rate at 40 C as reactions per minute (60 / time), to three significant figures. (1)
(b) State the optimum temperature and give one piece of data that supports it. (2)
(c) Explain why the rate falls above 50 C. (2)
(d) State the independent and the dependent variable. (1)
(e) Why is a sample tested at fixed intervals rather than once at the end? (1)
Question 4 (6 marks)
The same pondweed is placed at different distances from a lamp and the bubbles produced in five minutes are counted: 10 cm 60; 20 cm 40; 40 cm 22; 50 cm 12. The 30 cm group was run twice and counted 30 and 45 bubbles.
(a) Calculate the bubbles per minute at 10 cm. (1)
(b) Calculate the mean of the two 30 cm counts and its rate per minute. (2)
(c) Describe the relationship between lamp distance and the rate of bubble production, and explain it. (2)
(d) Why is a bubble count only an estimate of the rate of photosynthesis? (1)
Question 5 (7 marks)
A potometer measures the water uptake of a cut shoot. The air bubble moves: 0 min 0 mm; 2 min 14 mm; 4 min 28 mm; 6 min 42 mm; 8 min 58 mm; 10 min 71 mm. A fan is then switched on and the bubble moves 118 mm in ten minutes.
(a) Calculate the mean rate with no fan, in mm per minute to one decimal place. (1)
(b) Calculate the mean rate with the fan on, to one decimal place. (1)
(c) Calculate the percentage increase in rate with the fan. (2)
(d) Explain how moving air raises the rate of transpiration. (2)
(e) Which assumption does a potometer make about transpiration? (1)
Question 6 (6 marks)
(a) A student draws a cell 48 mm wide on paper; the real cell is 60 micrometres wide. Calculate the magnification of the drawing. (2)
(b) An agar cube of side 20 mm contains a pH indicator. Calculate its surface area, its volume and the surface area to volume ratio (SA:V) in mm-1. (2)
(c) The same indicator agar is cut into cubes of side 10 mm and the experiment repeated. Explain why the smaller cube changes colour sooner. (1)
(d) When comparing potato strips of different starting lengths, why use percentage change rather than change in length? (1)
How should the time be split inside Paper 6?
Direct answer: Take the data questions first (means, percentage changes, rates) because they cost the most time and give the surest marks; then the variable and error explanations; then the short definition and safety items. Allow about 1.5 minutes a mark and keep six minutes to check units and decimals.
Paper 6 answers (Questions 1-6)
Question 1
(a) Benedict’s solution: it is heated in a water bath at about 75-80 C for a few minutes so that reducing sugars reduce the blue copper(II) ions to a red-yellow precipitate of copper(I) oxide; heating the tube directly makes the liquid boil over (2).
(b) Any three, one mark each: starch (iodine goes blue-black); protein (Biuret goes purple); lipid (cloudy white emulsion); reducing sugar (Benedict’s goes green-yellow, so only a little) (3).
(c) The wells of a spotting tile hold a small drop of sample and iodine against a white background, so the blue-black colour is easier to see and compare and less reagent is used (1).
(d) Any one: wear eye protection; avoid skin contact with iodine and sodium hydroxide; never point a heated tube at anyone; keep ethanol away from a flame (1).
Question 2
(a) (36.3 + 36.5) / 2 = 36.4 mm (1).
(b) (36.4 – 40.0) / 40.0 x 100% = -9.0%, a decrease to one decimal place (2).
(c) 0.4 mol/dm3: the mean length there is still 40.0 mm, the same as the starting length, so net water movement is close to zero and the solution has about the same water potential as the cell sap (1).
(d) The sucrose solution has a lower water potential than the cell sap, so water leaves the cells by osmosis across the partially permeable membrane; the cells lose water, the vacuoles shrink and the tissue gets shorter (2).
(e) Any one: temperature; soaking time; the initial length and surface area of the strips; the variety and part of the potato; the number of strips per group (1).
Question 3
(a) 60 / 80 = 0.750 reactions per minute (1).
(b) The optimum is 40 C (1); there the time is shortest at 80 s and the rate highest at 0.75 per minute, while both 30 C and 50 C take 120 s (1).
(c) Above the optimum the shape of the enzyme’s active site changes (denaturation), so enzyme-substrate complexes cannot form and the rate falls sharply; at 60 C the starch takes 600 s, showing that most of the enzyme has been denatured (2).
(d) Independent variable: temperature. Dependent variable: the time taken for the starch to disappear, or the rate calculated from it (1).
(e) The experiment measures a rate, a change over time; testing only at the end cannot show when the starch disappeared, and the fall in rate would be invisible (1).
Question 4
(a) 60 / 5 = 12 bubbles per minute (1).
(b) (30 + 45) / 2 = 37.5 bubbles (1); 37.5 / 5 = 7.5 bubbles per minute (1).
(c) The further the lamp, the lower the rate: fastest at 10 cm (12 per minute), slowest at 50 cm (2.4 per minute). A greater distance means a lower light intensity reaching the weed, so the light-dependent stage of photosynthesis slows and less oxygen is released (2).
(d) The bubbles are gas released by the weed, mostly oxygen, but some dissolves in the water or collects inside the plant, and the bubbles are not all the same size, so counting them estimates the rate rather than measuring the oxygen produced (1).
Question 5
(a) 71 / 10 = 7.1 mm per minute (1).
(b) 118 / 10 = 11.8 mm per minute (1).
(c) (11.8 – 7.1) / 7.1 x 100% = 66.2% (2).
(d) Moving air removes the humid layer next to the leaf surface, which steepens the concentration gradient of water vapour between the inside of the leaf and the air outside, so vapour diffuses out of the stomata faster (2).
(e) It assumes that the rate of water uptake equals the rate of water lost in transpiration, so any change in the shoot’s own water store is negligible (1).
Question 6
(a) First put both in the same unit: 48 mm = 48 000 micrometres. Magnification = drawing size / real size = 48 000 / 60 = x800 (2: one for the unit conversion, one for the calculation).
(b) Surface area = 6 x 20 x 20 = 2 400 mm2; volume = 20 x 20 x 20 = 8 000 mm3; SA:V = 2 400 / 8 000 = 0.30 mm-1 (2).
(c) The smaller cube has a larger SA:V (0.60 mm-1 for the 10 mm cube), so it has more surface in contact with the outside relative to its volume and the acid or alkali has a shorter distance to diffuse to the centre, so the whole cube changes colour sooner (1).
(d) With strips of different starting lengths the same change in length is a bigger relative change for the shorter strip; percentage change puts every group on the same footing and removes the bias of the starting length (1).
How do I answer the data questions for marks?
Direct answer: Three steps: record the readings in a table with headings and units; write the working before the answer (for example (36.3 + 36.5) / 2 = 36.4 mm); and give the decimal place or significant figures asked for, with percentage change divided by the starting value and its sign kept. A bare final number loses the step marks.
Where these marks are usually lost
| Mistake | What it looks like | What it costs | Instead |
|---|---|---|---|
| Units missing from the table | A column headed only ‘length’ | No marks for recording | Head it ‘length / mm’ and keep one decimal place |
| Mean with no working | Writing 36.4 on its own | All method marks lost | Write (36.3 + 36.5) / 2 = 36.4 mm |
| Percentage change made positive | Writing 9.0% for a shortening | The direction is wrong and the explanation fails with it | Divide by the starting value and write -9.0% |
| Explanation that states only the result | Writing ‘the enzyme denatures’ | One mark at most | Say the active site changes shape, the substrate cannot bind and the rate falls |
| Independent and dependent variables swapped | Calling temperature the dependent variable | The whole variable question fails | Ask what I change (independent) and what I measure (dependent) |
How to use this paper
First drill: the full 60 minutes on all 6 questions with no notes, writing each formula in the margin first.
Mark it: check every unit, decimal place and significant figure, and check that each explanation gives both a cause and a consequence.
Second drill: redo three to seven days later, aiming to finish inside 50 minutes, and copy the recurring error types into your log.
What should I do after drilling this paper?
Direct answer: Log errors in four groups (units, working, variables, error analysis); redo the paper in three to seven days aiming to finish inside 50 minutes; top up the topics you found weakest (sampling intervals for enzyme rate, the assumption behind a potometer, unit conversion in magnification); and if your school enters you for Paper 5, take the same skills into the laboratory.
FAQ
Is this a Cambridge past paper?
No. The questions and answers are TutorZone originals, not Cambridge past papers or marking references; every experimental reading in it was written for this paper.
What is the difference between Paper 6 and Paper 5?
Both are 40 marks and 20% of the qualification and both assess AO3 experimental skills, and a candidate sits only one. Paper 5 is the Practical Test (1 h 15 min), where the experiments are actually done in a laboratory; Paper 6 is the Alternative to Practical (1 hour), answered entirely on paper with no experiment.
Can I drill Paper 6 without ever doing the experiments?
Yes. Paper 6 tests exactly what you do with experimental data when there is no apparatus: record readings, take means and percentage changes, identify variables, spot errors and suggest improvements. Read every question as a real experiment, and learn the standard colour changes (iodine for starch, Benedict’s for reducing sugar, Biuret for protein).
Official references
| Resource | What to check |
|---|---|
| Cambridge IGCSE Biology 0610 programme page | Confirm the marks, times and weightings of Papers 5 and 6 |
| https://www.tutorzone.com.hk/en-igcse-biology-0610-paper-4-mock-2027/ | IGCSE Biology Paper 4 theory drilling mock (the written half of the subject) |
Related: practice area · IGCSE Biology Paper 4 drilling mock
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