IGCSE Biology 0610 Paper 3 Practice Paper (Core Theory) — 80 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 3 format: Core level, 6 structured questions, 33 parts, 80 marks and 1 hour 15 minutes. All questions are original and are not past paper questions or an official mark scheme.

Total 80 marks · 1 hour 15 minutes · 33 parts

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

Level Core (same content as Paper 1 and Paper 3)

Paper at a glance

Paper	Level	Time	Marks
Paper 3	Core	1 hour 15 minutes	80 marks

Questions

Total: 80 marks · 1 hour 15 minutes · calculator allowed

Question 1 · Cells and movement in and out of cells [14]

(a) Name the part of a cell that controls the movement of substances into and out of the cell. [1]

(b) A photomicrograph of a cell structure has an image length of 30 mm. The actual length of the structure is 0.06 mm. Calculate the magnification. Show your working. [2]

(c) Two cubes of plant tissue have sides of 3 mm and 6 mm. Calculate the surface area to volume ratio of each cube, and explain which cube takes in oxygen faster by diffusion. [3]

(d) Explain how the long extension and the many mitochondria of a root hair cell help it to absorb water and mineral ions. [3]

(e) Define diffusion. [2]

(f) State one difference between diffusion and osmosis. [3]

Question 2 · Enzymes and digestion [13]

(a) Define the term enzyme and state its chemical nature. [2]

(b) In a catalase experiment, 24 cm3 of oxygen was collected in 30 s. Calculate the mean rate of reaction in cm3 per second. [2]

(c) Catalase works fastest at pH 7. Explain why the rate is very low at pH 2. [3]

(d) State two ways in which bile helps in the digestion of fats. [3]

(e) An enzyme solution is boiled for 5 minutes and then cooled to its optimum temperature. The rate stays at zero. Explain why. [3]

Question 3 · Plant nutrition and transport [14]

(a) Write the word equation for photosynthesis. [2]

(b) Pondweed 20 cm from a lamp produces 36 bubbles in 3 minutes. Calculate the bubbles per minute, predict the value at 40 cm where the light intensity falls to one quarter, and explain. [3]

(c) State two mineral ions needed by plants and their use. [2]

(d) A plant transpires 18 g of water in 4 hours. Calculate the mean rate of transpiration in g per hour, and state one environmental condition that increases it. [3]

(e) Explain how water moves up the xylem to the leaves. [2]

(f) State two advantages of having more stomata on the lower surface of a leaf. [2]

Question 4 · Human transport and gas exchange [14]

(a) Name the blood vessel that carries blood away from the heart, and state which type of blood vessel contains valves. [2]

(b) A person has a heart rate of 70 beats per minute and a stroke volume of 65 cm3. Calculate the cardiac output per minute in cm3 and in dm3. [3]

(c) A person breathes 12 times per minute, taking in 0.45 dm3 of air each breath. Calculate the ventilation rate per minute. [2]

(d) State three features of alveoli that make gas exchange efficient, and explain each. [3]

(e) Explain why the wall of the left ventricle is thicker than that of the right ventricle. [2]

(f) State two components of blood and their functions. [2]

Question 5 · Coordination and homeostasis [13]

(a) Define homeostasis. [2]

(b) Write the order of the nervous pathway in the withdrawal reflex, from stimulus to response, and explain why a reflex is faster than a conscious action. [3]

(c) Body temperature is kept near 37 °C. When it rises, state two responses of the skin and explain how they cool the body. [3]

(d) After a meal raises blood glucose, name the hormone, its gland and its action. [3]

(e) Explain why negative feedback is important in keeping the internal environment stable. [2]

Question 6 · Inheritance and ecology [12]

(a) A plant of genotype Tt is crossed with a plant of genotype tt (T = tall, dominant). Give the genotype ratio of the offspring and the percentage that are tall. [3]

(b) Two Tt plants are crossed and produce 400 seeds. Using a 3 : 1 ratio, predict how many seedlings are tall and how many are short. [2]

(c) A quadrat of 0.5 m2 contains 12 plants of one species. Estimate the number per square metre and the total in a 500 m2 field. [3]

(d) Explain why food chains usually have only four or five trophic levels. [2]

(e) State two features of an insect-pollinated flower that attract insects. [2]

Answers and marking scheme

Part	Answer	Marking points	Marks
1(a)	Cell surface membrane (plasma membrane).	1 mark: cell surface membrane / plasma membrane.	1
1(b)	Magnification = image length / actual length = 30 / 0.06 = x500.	1 mark: correct working (image / actual). 1 mark: answer x500.	2
1(c)	3 mm cube: surface area 54 mm2, volume 27 mm3, ratio 2.0. 6 mm cube: surface area 216 mm2, volume 216 mm3, ratio 1.0. The 3 mm cube has the larger ratio, so oxygen diffuses in faster.	1 mark: 2.0. 1 mark: 1.0. 1 mark: links the larger ratio to faster diffusion.	3
1(d)	The long extension gives a large surface area for faster osmosis and ion uptake; many mitochondria respire to release energy for active transport of mineral ions into the cell.	1 mark: long extension / large surface area. 1 mark: mitochondria / energy release. 1 mark: active transport of mineral ions.	3
1(e)	The net movement of particles from a region of higher concentration to a region of lower concentration, down a concentration gradient, without energy from respiration.	1 mark: higher to lower concentration. 1 mark: down a gradient / passive, no energy needed.	2
1(f)	Osmosis involves only water molecules moving through a partially permeable membrane; diffusion can involve any particles and needs no membrane. Osmosis moves water from the higher water potential (more dilute) side to the lower.	1 mark: osmosis is water only. 1 mark: needs a partially permeable membrane. 1 mark: higher to lower water potential.	3
2(a)	Enzymes are biological catalysts that speed up reactions and are not used up; they are proteins.	1 mark: biological catalyst / speeds up reactions. 1 mark: protein.	2
2(b)	Rate = 24 / 30 = 0.8 cm3 s-1.	1 mark: 24 / 30. 1 mark: 0.8 with unit.	2
2(c)	The acidic conditions at pH 2 change the shape of the active site (the enzyme is denatured). The substrate can no longer fit the active site, so no enzyme-substrate complexes form and the rate is close to zero; the change is not reversible.	1 mark: active site / enzyme shape changed. 1 mark: substrate cannot fit. 1 mark: denatured, rate near zero.	3
2(d)	Bile emulsifies fats into small droplets, increasing the surface area for lipase; bile also neutralises the acidic chyme from the stomach so the small intestine stays slightly alkaline for lipase and pancreatic enzymes.	1 mark: emulsifies fats / increases surface area. 1 mark: neutralises acidic chyme. 1 mark: links to faster lipase action.	3
2(e)	High temperature breaks bonds in the enzyme, so its three-dimensional shape and active site are permanently changed (denatured). Substrate can no longer bind and cooling does not restore the shape, so the rate stays at zero.	1 mark: boiling denatures the enzyme. 1 mark: active site permanently changed. 1 mark: cooling cannot reverse it.	3
3(a)	Carbon dioxide + water -> (light energy, chlorophyll) glucose + oxygen.	1 mark: correct reactants. 1 mark: correct products and conditions.	2
3(b)	At 20 cm: 36 / 3 = 12 bubbles per minute. At 40 cm: about 3 per minute, because light intensity limits the rate, so one quarter of the light gives about one quarter of the oxygen.	1 mark: 12. 1 mark: 3. 1 mark: light intensity limits the rate.	3
3(c)	Nitrate: to make amino acids and proteins. Magnesium: to make chlorophyll.	1 mark: nitrate / proteins. 1 mark: magnesium / chlorophyll.	2
3(d)	Rate = 18 / 4 = 4.5 g h-1. More wind, higher temperature, lower humidity or brighter light all increase the rate.	1 mark: 18 / 4. 1 mark: 4.5 g h-1. 1 mark: any suitable condition.	3
3(e)	Transpiration from the leaves lowers the water potential there, creating a transpiration pull; cohesion between water molecules keeps a continuous column that is pulled up the xylem.	1 mark: transpiration pull. 1 mark: cohesion / continuous column.	2
3(f)	The lower surface is shaded, so less water is lost by transpiration, while gas exchange of carbon dioxide and oxygen can still take place.	1 mark: reduces transpiration / water loss. 1 mark: gas exchange still occurs.	2
4(a)	Arteries carry blood away from the heart; veins contain valves.	1 mark: artery. 1 mark: vein.	2
4(b)	Cardiac output = 70 x 65 = 4,550 cm3 min-1 = 4.55 dm3 min-1.	1 mark: 70 x 65. 1 mark: 4,550 cm3 min-1. 1 mark: 4.55 dm3 min-1.	3
4(c)	Ventilation = 12 x 0.45 = 5.4 dm3 min-1.	1 mark: 12 x 0.45. 1 mark: 5.4 dm3 min-1.	2
4(d)	Very many alveoli give a large surface area for faster diffusion; the wall is one cell thick so the diffusion distance is short; a moist lining and a dense capillary network keep a steep concentration gradient.	1 mark: large surface area. 1 mark: thin wall / short diffusion distance. 1 mark: moist lining / rich blood supply maintains gradient.	3
4(e)	The left ventricle pumps blood around the whole body, which needs a higher pressure; the right ventricle only pumps blood to the lungs, a shorter distance needing lower pressure.	1 mark: body-wide pumping needs higher pressure. 1 mark: right ventricle only pumps to the lungs.	2
4(f)	Red blood cells contain haemoglobin and carry oxygen; plasma transports glucose, carbon dioxide, hormones and waste.	1 mark: red blood cells / oxygen transport. 1 mark: plasma / transport of dissolved substances.	2
5(a)	The maintenance of a stable internal environment, such as body temperature, blood glucose and water content, despite changes inside and outside the body.	1 mark: keeps the internal environment stable. 1 mark: despite external changes.	2
5(b)	Stimulus -> receptor -> sensory neurone -> spinal cord (coordinator) -> motor neurone -> effector (muscle) -> response. The pathway does not pass through the brain for a decision, so the response is faster and injury is reduced.	1 mark: receptor -> sensory neurone. 1 mark: spinal cord -> motor neurone -> effector. 1 mark: no brain decision, so faster.	3
5(c)	More sweat is produced: evaporation of sweat removes latent heat. Blood vessels near the skin surface dilate (vasodilation): more blood flows near the surface, so more heat is lost.	1 mark: more sweating. 1 mark: evaporation removes heat. 1 mark: vasodilation / more heat lost.	3
5(d)	Insulin, secreted by the pancreas, makes the liver and muscles convert glucose to glycogen and increases glucose uptake by cells, so blood glucose falls.	1 mark: insulin. 1 mark: pancreas. 1 mark: glucose to glycogen / more uptake.	3
5(e)	Negative feedback triggers the opposite response whenever a value moves away from normal (for example, high glucose triggers lowering), bringing it back into range and preventing harmful extremes.	1 mark: opposite response when values drift. 1 mark: returns values to the normal range.	2
6(a)	Gametes: Tt gives T and t; tt gives only t. Offspring genotypes Tt : tt = 2 : 2, i.e. 1 : 1; 50% are tall (Tt).	1 mark: Tt : tt = 1 : 1. 1 mark: 50% tall. 1 mark: supported by gametes or a Punnett square.	3
6(b)	Tall = 400 x 3/4 = 300; short = 400 x 1/4 = 100.	1 mark: 300. 1 mark: 100.	2
6(c)	Per square metre = 12 / 0.5 = 24 plants; in 500 m2 = 24 x 500 = 12,000 plants.	1 mark: 12 / 0.5. 1 mark: 24 per m2. 1 mark: 12,000 plants.	3
6(d)	Only about one tenth of the energy passes to the next level, the rest being lost in respiration, excretion and undigested material, so there is too little energy to support more levels.	1 mark: energy is lost at each level. 1 mark: too little energy for more levels.	2
6(e)	Brightly coloured petals attract insects; nectar and scent attract insects to feed.	1 mark: brightly coloured petals. 1 mark: nectar / scent.	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 3 Core: 1 hour 15 minutes, 80 marks.

Frequently asked questions

Who is this paper for?
Students preparing for IGCSE Biology 0610 Core papers (Paper 1 and Paper 3), 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 3 (Core) format.