IGCSE Biology 0610 Paper 5 Practice Paper (Practical Test) — 40 marks, 1 hour 15 minutes, answers and marking scheme

2026-10-30

IGCSE Biology practice – Core

An original practice paper written to the Cambridge IGCSE Biology 0610 Paper 5 format: Core level, 6 structured questions, 20 parts, 40 marks and 1 hour 15 minutes. All questions are original and are not past paper questions or an official mark scheme.

Total 40 marks · 1 hour 15 minutes · 20 parts

Mark distribution 14 + 13 + 14 + 14 + 13 + 12 = 80

Level Core (same content as Paper 1 and Paper 5)

Paper at a glance

Paper	Level	Time	Marks
Paper 5	Core	1 hour 15 minutes	40 marks

Questions

Total: 40 marks · 1 hour 15 minutes · calculator allowed

Question 1 · Food tests and practical skills [10]

(a) State the reagent used to test for starch and the colour change seen in a positive result. [2]

(b) Describe how to carry out the Benedict’s test and state the positive result for a reducing sugar. [3]

(c) A 2.0 g sample of food contains 0.50 g of protein. Calculate the percentage of protein in the sample. Show your working. [2]

(d) Explain why the Benedict’s test is heated in a water bath instead of holding the tube in a Bunsen flame. [2]

(e) State one safety precaution to follow when carrying out food tests. [1]

Question 2 · Osmosis and water potential [10]

(a) A potato strip has an initial mass of 4.8 g. After 30 minutes in distilled water its mass is 5.4 g. Calculate the percentage change in mass. Show your working. [2]

(b) Another potato strip changes from 5.0 g to 4.4 g in 0.5 mol/dm3 sucrose solution. Calculate the percentage decrease in mass. [2]

(c) Use the movement of water molecules to explain why the mass of the potato strip increased in (a). [3]

(d) State why the potato strips are blotted dry with paper towel before being weighed. [1]

(e) One concentration is repeated three times and gives mass changes of 12.5%, 12.1% and 15.0%. Identify the anomalous result and calculate the mean of the other two values. [2]

Question 3 · Respiration and enzyme activity [10]

(a) The table shows bubbles per minute from yeast at different temperatures: 20 °C 18, 30 °C 42, 40 °C 30, 50 °C 6. Calculate the percentage increase in bubble count from 20 °C to 30 °C. [2]

(b) Explain why the number of bubbles rises between 20 °C and 30 °C. [2]

(c) Explain why the number of bubbles falls greatly at 50 °C. [3]

(d) Apart from temperature, state one variable that must be controlled and describe how to control it. [2]

(e) State one limitation of measuring the rate of respiration by counting bubbles. [1]

Question 4 · Microscopy and plant tissue observation [10]

(a) A cell in a photomicrograph has an image length of 45 mm and an actual length of 0.15 mm. Calculate the magnification. Show your working. [2]

(b) State two features of a good biological drawing. [2]

(c) Twelve stomata are counted in a field of view of area 0.25 mm2. Calculate the number of stomata per mm2. [2]

(d) Explain why the lower epidermis of a leaf usually has more stomata than the upper epidermis. [2]

(e) State one technique to use when making a peel slide and explain why it matters. [2]

Answers and marking scheme

Part	Answer	Marking points	Marks
1(a)	Iodine solution; the colour changes from yellow-brown to blue-black.	1 mark: iodine solution. 1 mark: turns blue-black.	2
1(b)	Put about 2 cm3 of the food solution in a test tube, add an equal volume of Benedict’s solution and heat it in a water bath at about 75 °C for a few minutes. A positive result is a change from blue to green, yellow or a brick-red precipitate.	1 mark: add Benedict’s solution. 1 mark: heat in a water bath. 1 mark: positive = green / yellow / brick-red.	3
1(c)	0.50 / 2.0 x 100 = 25%.	1 mark: correct working (0.50 / 2.0 x 100). 1 mark: 25%.	2
1(d)	A water bath heats the tube evenly at a controlled temperature (about 75 °C), so the solution does not suddenly boil over. Direct heating makes one part of the tube much hotter, so the solution can spit out and scald the student, and the tube may crack.	1 mark: even heating / controlled temperature. 1 mark: avoids boiling over / scalding or cracking the tube.	2
1(e)	Wear eye protection and hold the test tube with a test-tube holder so that hot water and reagents are not touched directly.	1 mark: eye protection / test-tube holder / no direct contact with hot water or reagents.	1
2(a)	Increase in mass = 5.4 – 4.8 = 0.6 g; 0.6 / 4.8 x 100 = a 12.5% increase.	1 mark: 0.6 g change / correct working. 1 mark: +12.5% (increase).	2
2(b)	5.0 – 4.4 = 0.6 g; 0.6 / 5.0 x 100 = a 12.0% decrease.	1 mark: 0.6 / 5.0 x 100. 1 mark: 12.0%.	2
2(c)	Distilled water is pure and has the highest water potential; the potato cell sap contains solutes and has a lower water potential. Water moves by osmosis through the partially permeable membranes from the higher water potential outside into the cells, so the cells become turgid and the mass increases.	1 mark: higher water potential outside. 1 mark: water enters through partially permeable membranes (osmosis). 1 mark: cells become turgid / mass increases.	3
2(d)	So that surface water is not included in the mass and the change in mass only shows water that has entered or left the cells.	1 mark: stops surface water affecting the mass / makes the result accurate.	1
2(e)	15.0% is much further from the other two, so it is anomalous. Mean of the other two = (12.5 + 12.1) / 2 = 12.3%.	1 mark: 15.0% identified as anomalous. 1 mark: 12.3%.	2
3(a)	Percentage increase = (42 – 18) / 18 x 100 = about 133%.	1 mark: (42 – 18) / 18 x 100. 1 mark: about 133%.	2
3(b)	A higher temperature gives the enzyme and substrate molecules more kinetic energy, so they collide more often. More enzyme-substrate complexes form, respiration speeds up and more carbon dioxide is released.	1 mark: more kinetic energy / more collisions. 1 mark: faster respiration / more carbon dioxide released.	2
3(c)	The high temperature denatures the enzyme, which is a protein. The active site changes shape so the substrate can no longer bind, no enzyme-substrate complexes form and the rate of respiration falls greatly. This denaturation is irreversible.	1 mark: enzyme denatured / active site changed shape. 1 mark: substrate cannot bind. 1 mark: irreversible / rate falls greatly.	3
3(d)	The concentration and volume of the yeast suspension: use the same concentration and the same volume every time (for example 10 cm3 of 2% suspension). Alternatively keep the time between placing the tube in the bath and starting the count the same (for example 5 minutes).	1 mark: suitable variable (yeast concentration / volume / time / substrate amount). 1 mark: a specific way of controlling it.	2
3(e)	The bubbles are not all the same size and some stick to the tube, so counting them does not measure the gas volume accurately; a gas syringe or a pressure sensor should be used instead.	1 mark: uneven bubble size / counting inaccurate (use a gas syringe or sensor).	1
4(a)	Magnification = image length / actual length = 45 / 0.15 = x300.	1 mark: 45 / 0.15. 1 mark: x300.	2
4(b)	It is drawn with a sharp pencil using clear unbroken lines and no shading or colour; the proportions are accurate and the structures are the right size; labelling lines are straight and do not cross, and labels, a title and the magnification are given.	1 mark: pencil, clear lines, no shading or colour. 1 mark: accurate proportions / straight label lines with labels and magnification.	2
4(c)	12 / 0.25 = 48 per mm2.	1 mark: 12 / 0.25. 1 mark: 48 per mm2 with unit.	2
4(d)	The lower surface is shaded and cooler, so water is lost more slowly. Stomata there still let carbon dioxide in and oxygen out while reducing transpiration, and they are less likely to be blocked by dust or rainwater.	1 mark: shaded / less water lost. 1 mark: gas exchange still possible (CO2 in, O2 out).	2
4(e)	Peel a very thin layer (one cell thick) and lay it flat on the slide, then lower the cover slip at about 45 degrees to avoid air bubbles. Bubbles appear as dark circles and hide the cells, so the observation is unreliable.	1 mark: thin peel / laid flat without folding. 1 mark: cover slip lowered at an angle to avoid bubbles (bubbles hide cells).	2

Timing plan

Section	Suggested time	Marks
Questions 1-2	25 min	27 marks
Questions 3-4	26 min	28 marks
Questions 5-6	24 min	25 marks
Checking	remaining time	–

Common pitfalls

Calculations written without working lose every method mark.
Explanations that list key words without a cause-and-effect sentence.
Missing units: cm3 s-1, g h-1, dm3 min-1, plants m-2.
Core answers may be shorter, but you still need one point per mark available.

How to use this paper Sit the paper closed-book in 75 minutes, then mark point by point with the answer booklet and redo every lost mark.

Syllabus reference Format follows Cambridge IGCSE Biology 0610 (2026-2028) Paper 5 Core: 1 hour 15 minutes, 40 marks.

Frequently asked questions

Who is this paper for?
Students preparing for IGCSE Biology 0610 Core papers (Paper 1 and Paper 5), especially those who need practice in writing structured answers.

Does the answer booklet include a mark scheme?
Yes. Every part lists its marking points so you can mark yourself or a tutor can mark point by point.

Can I print it?
Yes, both the question paper and the answer booklet are A4 PDFs ready to print.

Print question paper (PDF)Print answer booklet (PDF)

Originality statement: every question, data set and context in this paper was written by TutorZone. It is not a Cambridge IGCSE Biology 0610 past paper or an official mark scheme.

This paper is not an IGCSE Biology 0610 past paper; only the structure and mark distribution follow the official Paper 5 (Core) format.