An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V1) Cambridge IGCSE Biology (0610) paper. Not affiliated with or reproduced from Cambridge.
Theory Extended (Paper 4 Specimen Layout)
Answer all questions. Write your answers in the spaces provided on the question paper.
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Worked solution
Active transport requires two key components: 1. Movement of particles / ions / molecules across a cell membrane from a lower concentration to a higher concentration (against a concentration gradient). 2. The requirement of energy released from respiration (using protein carriers).
Marking scheme
Award 1 mark for each point (max 2): • movement of particles / ions / molecules through a cell membrane from a region of lower concentration to a region of higher concentration / against a concentration gradient [1] • using energy from respiration [1]
Reject: 'along a concentration gradient', 'creation of energy'.
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Worked solution
Transpiration consists of: 1. The loss of water vapour from leaves / aerial parts of the plant. 2. Evaporation of water at the surfaces of the mesophyll cells followed by the diffusion of water vapour through the stomata into the air.
Marking scheme
Award 1 mark for each point (max 2): • loss of water vapour from leaves / aerial parts of a plant [1] • evaporation (of water) at the surfaces of mesophyll cells followed by diffusion (of water vapour) through stomata [1]
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Worked solution
(a) Excess amino acids are broken down in the liver. (b) Deamination is specifically the removal of the nitrogen-containing amino group from amino acids, resulting in the production of urea.
Marking scheme
(a) liver [1] (b) removal of the nitrogen-containing part of amino acids to form urea [1]
Accept: 'removal of amino group / \(-\text{NH}_2\)' for 'nitrogen-containing part'. Reject: 'kidney' for part (a).
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Worked solution
(a) By syllabus definition, a pathogen is a disease-causing organism. (b) Direct transmission involves passing a pathogen directly from one host to another via direct physical contact, transfer of bodily fluids, or across the placenta.
Marking scheme
(a) disease-causing organism [1] (b) any one from: • direct contact / skin-to-skin contact [1] • contact with blood / bodily fluids / semen / vaginal secretions [1] • sexual intercourse / sexual contact [1] • across placenta / breastfeeding [1]
Reject (for b): indirect methods such as 'contaminated food', 'airborne droplets', 'vectors / mosquitoes'.
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Worked solution
A hormone is defined by: 1. Being a chemical substance secreted by an endocrine gland and transported in the blood plasma. 2. Affecting or altering the metabolic activity of specific target organs/cells.
Marking scheme
Award 1 mark for each point (max 2): • chemical substance produced by a gland and carried / transported by the blood [1] • alters the activity of one or more specific target organs [1]
Accept: 'secreted by endocrine gland' for produced by a gland.
Question 7 · data_analysis
5 marks
A student investigated the effect of temperature on the rate of anaerobic respiration in yeast (Saccharomyces cerevisiae). A suspension of yeast in 5% glucose solution was placed in a water bath at different temperatures. The volume of carbon dioxide released in 10 minutes was recorded for each temperature.
Table 1.1 shows the results of the investigation.
Table 1.1 | Temperature / °C | Volume of carbon dioxide released in 10 min / cm³ | | :--- | :--- | | 10 | 1.8 | | 20 | 4.2 | | 30 | 8.6 | | 40 | 14.0 | | 50 | 6.4 | | 60 | 0.0 |
(a) Describe the effect of temperature on the volume of carbon dioxide released between 10 °C and 60 °C. [2]
(b) Calculate the percentage increase in the volume of carbon dioxide released when the temperature increased from 20 °C to 40 °C. Show your working and give your answer to one decimal place. [2]
(c) Explain, in terms of enzymes, why no carbon dioxide was produced at 60 °C. [1]
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Worked solution
(a) - Between 10 °C and 40 °C, the volume of carbon dioxide produced increases with temperature, reaching a maximum of 14.0 cm³ at 40 °C (the optimum temperature). - Above 40 °C (from 40 °C to 60 °C), the volume of carbon dioxide released decreases rapidly, reaching 0.0 cm³ at 60 °C.
(b) - Initial volume at 20 °C = 4.2 cm³ - Final volume at 40 °C = 14.0 cm³ - Increase = \(14.0 - 4.2 = 9.8\text{ cm}^3\) - Percentage increase = \(\frac{9.8}{4.2} \times 100 = 233.333...\% \approx 233.3\%
(c) - At 60 °C, the enzymes involved in anaerobic respiration are denatured; high kinetic energy alters the shape of their active sites, preventing the substrate from binding to form enzyme-substrate complexes.
Marking scheme
(a) [max 2] 1. Increase in volume of \(\text{CO}_2\) from 10 °C to 40 °C / maximum at 40 °C (14.0 cm³); [1] 2. Decrease in volume of \(\text{CO}_2\) above 40 °C / from 40 °C to 60 °C / reaches 0 cm³ at 60 °C; [1]
(c) [1] 1. Enzymes (controlling respiration) are denatured / active site changed shape / substrate no longer fits/binds; [1] Reject: enzymes killed / died.
Question 8 · data_analysis
5 marks
A potometer was used to investigate the effect of wind speed on the rate of water uptake in a leafy shoot of a dicotyledonous plant. The temperature, light intensity, and humidity were kept constant.
(a) Describe the relationship between wind speed and the rate of water uptake shown in Table 2.1. [1]
(b) Calculate the difference in water uptake per hour between a wind speed of 0.0 m s⁻¹ and 4.5 m s⁻¹. Give your answer in mm³ h⁻¹. [2]
(c) Explain why an increase in wind speed increases the rate of transpiration from the leaves. [2]
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Worked solution
(a) - As wind speed increases, the rate of water uptake increases (from 12 mm³ min⁻¹ at 0.0 m s⁻¹ to 48 mm³ min⁻¹ at 6.0 m s⁻¹).
(b) - Difference in rate per minute = \(42 - 12 = 30\text{ mm}^3\text{ min}^{-1}\) - Difference in rate per hour = \(30 \times 60 = 1800\text{ mm}^3\text{ h}^{-1}\)
(c) - Moving air / wind blows away the layer of water vapour accumulating near the stomata / outside of the leaf surface. - This maintains a steep concentration gradient / diffusion gradient of water vapour between the air spaces inside the leaf (spongy mesophyll) and the external atmosphere, resulting in a faster rate of diffusion / transpiration.
Marking scheme
(a) [1] 1. As wind speed increases, the rate of water uptake / transpiration increases (positive correlation); [1]
(b) [2] 1. Correct calculation of rate difference per minute: \(42 - 12 = 30\) [1] 2. Correct conversion to hourly rate: \(30 \times 60 = 1800\text{ (mm}^3\text{ h}^{-1}\text{)}\); [1] (Award 2 marks for correct answer 1800 without working shown)
(c) [2] 1. Wind blows/removes water vapour away from the leaf surface / stomata; [1] 2. Maintains / steepens concentration gradient of water vapour (between leaf interior and atmosphere); [1]
Question 9 · data_analysis
5 marks
A healthy adult consumed a meal rich in starch at time = 0 minutes. Blood samples were taken every 30 minutes for two hours to measure blood glucose concentration and blood insulin concentration.
(a) State the organ that secretes insulin in response to increased blood glucose concentration. [1]
(b) Describe the changes in blood glucose concentration between 0 and 120 minutes. [2]
(c) Explain how insulin causes the blood glucose concentration to decrease between 60 and 120 minutes. [2]
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Worked solution
(a) - The pancreas (specifically beta cells in the islets of Langerhans) detects high blood glucose and secretes insulin.
(b) - From 0 to 60 minutes, blood glucose concentration increases from 4.8 mmol dm⁻³ to reach a maximum / peak of 8.4 mmol dm⁻³. - From 60 to 120 minutes, blood glucose concentration decreases steadily from 8.4 mmol dm⁻³ to 5.0 mmol dm⁻³.
(c) - Insulin travels in the blood to target organs, predominantly the liver and muscle cells. - It causes cells to increase their uptake of glucose from the blood and stimulates the conversion of glucose into glycogen (glycogenesis) for storage, as well as increasing the rate of cellular respiration.
Marking scheme
(a) [1] 1. Pancreas; [1]
(b) [2] 1. (Glucose) rises from 0 to 60 min / peaks at 60 min (at 8.4 mmol dm⁻³); [1] 2. (Glucose) falls / decreases from 60 to 120 min (to 5.0 mmol dm⁻³); [1] Note: Award max 1 mark if no data/times are referenced.
(c) [2] 1. Stimulates liver / muscle cells to take up / absorb glucose from blood; [1] 2. Glucose is converted to glycogen (for storage) / increases rate of respiration using glucose; [1]
Question 10 · data_analysis
5 marks
A population of the single-celled protoctist Paramecium caudatum was grown in a closed culture vessel with a fixed supply of nutrients. The number of individuals per cm³ was measured over a 12-day period.
(a) Identify the growth phase of the population between Day 2 and Day 8. [1]
(b) Calculate the mean rate of population increase between Day 2 and Day 6. Give your answer in individuals per cm³ per day. [2]
(c) Explain two factors that cause the population growth to slow down and reach the stationary phase between Day 8 and Day 12. [2]
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Worked solution
(a) - Between Day 2 and Day 8, the population undergoes rapid growth known as the exponential phase (or log phase).
(b) - Population at Day 2 = 25 individuals per cm³ - Population at Day 6 = 310 individuals per cm³ - Total increase in population = \(310 - 25 = 285\text{ individuals per cm}^3\) - Time interval = \(6 - 2 = 4\text{ days}\) - Mean rate of increase = \(\frac{285}{4} = 71.25\text{ individuals per cm}^3\text{ per day}\)
(c) - As the population density reaches carrying capacity, growth levels off due to limiting factors: 1. Depletion of nutrients / food source in the closed container. 2. Build-up / accumulation of toxic metabolic waste products (e.g. nitrogenous waste, excretory products) inhibiting growth/reproduction, or increased intraspecific competition.
Marking scheme
(a) [1] 1. Exponential (phase) / log (phase); [1]
(b) [2] 1. Correct calculation of change in population: \(310 - 25 = 285\); [1] 2. Correct division by time (4 days): \(71.25\) (or \(71.3\)); [1] (Award 2 marks for correct final answer without working)
(c) [2] Any two from: 1. Running out of nutrients / food supply / exhaustion of resources; [1] 2. Accumulation of toxic / metabolic wastes; [1] 3. (Intraspecific) competition for resources / space; [1] 4. Death rate equals reproduction / birth rate; [1]
Question 11 · extended
6 marks
Explain how transpiration produces a tension that pulls water up the xylem from the roots to the leaves of a plant.
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Worked solution
1. Water evaporates from the moist surfaces of mesophyll cells into the intercellular air spaces within the leaf. 2. Water vapour diffuses down its concentration gradient out of the leaf into the atmosphere through open stomata. 3. The loss of water reduces the water potential of mesophyll cells compared to neighboring cells and xylem vessels. 4. Water moves out of the leaf xylem into the mesophyll cells by osmosis down a water potential gradient. 5. This creates a negative pressure / tension / suction force called the transpiration pull at the top of the xylem column. 6. Water molecules are attracted to one another by cohesion (forming an unbroken, continuous column) and to the xylem vessel walls by adhesion, pulling water upward from roots to leaves.
Marking scheme
Award 1 mark for each valid point up to a maximum of 6 marks: 1. Evaporation of water from surface of mesophyll cell walls into (intercellular) air spaces; 2. Diffusion of water vapour out of leaf through stomata (down a concentration gradient); 3. Loss of water lowers water potential in mesophyll cells / creates water potential gradient between leaf cells and xylem; 4. Water moves from leaf xylem to mesophyll cells (by osmosis); 5. Transpiration pull / suction force / tension generated in xylem vessels; 6. Cohesion between water molecules maintains an unbroken / continuous column of water; 7. Adhesion of water molecules to walls of xylem vessels helps sustain the upward movement.
Question 12 · extended
6 marks
Describe and explain the mechanism of inspiration (breathing in) in a human.
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Worked solution
During inspiration (inhalation): 1. The external intercostal muscles contract. 2. The internal intercostal muscles relax. 3. This pulls the ribcage upwards and outwards. 4. At the same time, the diaphragm muscles contract, causing the diaphragm to flatten and move downwards. 5. These coordinated movements increase the volume of the thorax (chest cavity) and lungs. 6. The increase in volume leads to a decrease in air pressure inside the thorax below atmospheric pressure, forcing air down the trachea into the lungs until pressures equalize.
Marking scheme
Award 1 mark for each valid point up to a maximum of 6 marks: 1. External intercostal muscles contract; 2. Internal intercostal muscles relax; 3. Ribs / ribcage move upwards and outwards; 4. Diaphragm (muscles) contract; 5. Diaphragm flattens / moves downwards; 6. Volume of thorax / thoracic cavity / chest increases; 7. Pressure inside thorax / lungs decreases / becomes lower than atmospheric pressure; 8. Air enters / is drawn into the lungs (down a pressure gradient).
Question 13 · extended
6 marks
Explain how the human body returns blood glucose concentration to its normal set point after a meal rich in carbohydrates.
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Worked solution
1. Following carbohydrate digestion, elevated blood glucose concentration is detected by receptor cells in the pancreas. 2. The pancreas responds by secreting the hormone insulin into the bloodstream. 3. Insulin travels through the circulation to target organs, predominantly the liver and skeletal muscles. 4. Insulin increases cell membrane permeability to glucose, stimulating target cells to take up more glucose from the blood. 5. Inside liver and muscle cells, insulin activates enzymes that convert excess soluble glucose into insoluble glycogen for storage. 6. Insulin also increases the rate of cellular respiration using glucose; blood glucose concentration decreases back to the normal set point, illustrating homeostatic negative feedback.
Marking scheme
Award 1 mark for each valid point up to a maximum of 6 marks: 1. Receptors in pancreas detect increase in blood glucose concentration; 2. Pancreas releases / secretes insulin; 3. Insulin is transported in blood / bloodstream (to target organs); 4. Target cells / liver cells / muscle cells absorb / take up more glucose from blood; 5. Glucose is converted to glycogen (in liver / muscle cells); 6. Increased rate of cellular respiration / breakdown of glucose in cells; 7. Blood glucose concentration decreases back to set point / normal level; 8. Correct reference to negative feedback.
Question 14 · extended
6 marks
Describe and explain the sequence of events that occurs in a freshwater lake when excess nitrate fertiliser enters the water from nearby agricultural fields.
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Worked solution
1. Nitrate ions leach / runoff from agricultural land into the lake, causing nutrient enrichment (eutrophication). 2. Increased nitrate availability triggers rapid growth and reproduction of algae at the water surface, forming an algal bloom. 3. The thick layer of algae blocks sunlight from penetrating deeper water layers. 4. Submerged aquatic plants can no longer photosynthesise due to lack of light and die. 5. Decomposing bacteria feed on the abundant dead plant and algal material and multiply rapidly. 6. These decomposers carry out aerobic respiration, using up the dissolved oxygen in the water, which leads to hypoxia / anoxia, causing fish and other aquatic fauna to die.
Marking scheme
Award 1 mark for each valid point up to a maximum of 6 marks: 1. Leaching / runoff of fertiliser / nitrates / minerals into lake / water body; 2. Rapid growth / proliferation of algae / algal bloom on surface; 3. Sunlight is blocked / prevented from reaching submerged plants; 4. Submerged plants cannot photosynthesise and die; 5. Bacteria / decomposers feed on / break down dead organic matter and multiply / population increases; 6. Decomposers respire aerobically / consume dissolved oxygen in the water; 7. Dissolved oxygen concentration drops / water becomes deoxygenated / hypoxic; 8. Aquatic animals / fish die due to lack of oxygen / suffocate.
Question 15 · extended
6 marks
Describe the sequence of events that occurs at a synapse to transmit an electrical impulse from a presynaptic neurone to a postsynaptic neurone.
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Worked solution
1. An electrical impulse travels along the axon and reaches the presynaptic terminal / synaptic knob. 2. The impulse stimulates vesicles containing neurotransmitter molecules to move toward the presynaptic membrane. 3. The vesicles fuse with the presynaptic membrane, releasing neurotransmitter into the synaptic cleft. 4. Neurotransmitter molecules diffuse across the narrow synaptic gap / cleft down a concentration gradient. 5. The neurotransmitters bind to specific, complementary receptor proteins located on the postsynaptic membrane. 6. This binding alters membrane permeability and generates a new electrical impulse in the postsynaptic neurone, ensuring unidirectional transmission.
Marking scheme
Award 1 mark for each valid point up to a maximum of 6 marks: 1. (Electrical) impulse / action potential arrives at presynaptic neurone / synaptic knob; 2. Vesicles containing neurotransmitter move towards presynaptic membrane; 3. Vesicles fuse with presynaptic membrane; 4. Neurotransmitter is released into the synaptic cleft / gap; 5. Neurotransmitter diffuses across the synaptic cleft / down a concentration gradient; 6. Neurotransmitter binds to specific / complementary receptor proteins on postsynaptic membrane; 7. Triggers / generates a new (electrical) impulse in postsynaptic neurone; 8. Synapses ensure impulses travel in one direction only.
Question 16 · extended
6 marks
Explain, using the lock and key hypothesis and kinetic theory, why the rate of an enzyme-catalysed reaction decreases rapidly when the temperature is raised significantly above the optimum temperature.
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Worked solution
1. High thermal energy increases the kinetic energy and molecular vibrations of enzyme protein molecules. 2. Intense vibrations break internal chemical bonds (e.g. hydrogen and ionic bonds) that maintain the precise three-dimensional tertiary structure of the enzyme. 3. The specific shape of the active site is altered / deformed permanently (the enzyme is denatured). 4. Under the lock and key model, the substrate shape is no longer complementary to the deformed active site. 5. Substrate molecules can no longer fit into the active site, preventing the formation of enzyme-substrate (E-S) complexes. 6. Consequently, no catalytic reaction occurs and the rate of reaction rapidly declines to zero.
Marking scheme
Award 1 mark for each valid point up to a maximum of 6 marks: 1. Increased temperature increases kinetic energy / molecular vibrations of enzyme molecule; 2. Bonds holding the 3D / tertiary shape of the enzyme break; 3. Active site changes shape / is deformed / altered; 4. Enzyme is denatured / denaturation occurs; 5. Substrate is no longer complementary to the active site / cannot fit (into active site); 6. Lock and key fit is lost / enzyme-substrate (E-S) complexes cannot form; 7. Fewer successful collisions / no catalysis takes place; 8. Rate of reaction decreases rapidly / drops to zero.
Question 17 · theory
3 marks
A student investigated osmosis in plant tissues by immersing potato cylinders in sucrose solutions of different concentrations.
Table 1.1 shows the initial and final mass of a potato cylinder placed in a \(0.6\text{ mol dm}^{-3}\) sucrose solution for \(60\text{ minutes}\).
Table 1.1
| initial mass / g | final mass / g | | :--- | :--- | | 5.60 | 4.76 |
Calculate the percentage change in mass of this potato cylinder.
Award marks as follows: - [1 mark] Correct subtraction to find mass change: \(4.76 - 5.60 = -0.84\text{ g}\) (or \(0.84\text{ g}\)) - [1 mark] Correct formula used with division by initial mass (5.60): \(\frac{-0.84}{5.60} \times 100\) (allow error carried forward from incorrect subtraction) - [1 mark] Correct final answer given to 1 decimal place: \(-15.0\%\) (accept \(15.0\%\) decrease / loss; award 3 marks for correct answer without working)
Question 18 · theory
3 marks
A student investigated anaerobic respiration in yeast by measuring the volume of carbon dioxide gas collected in a gas syringe over a period of \(15\text{ minutes}\).
At \(30\ ^\circ\text{C}\), the volume of carbon dioxide collected was \(18.6\text{ cm}^3\). At \(40\ ^\circ\text{C}\), the volume of carbon dioxide collected was \(42.3\text{ cm}^3\).
Calculate the percentage increase in the volume of carbon dioxide gas produced at \(40\ ^\circ\text{C}\) compared to \(30\ ^\circ\text{C}\).
Step 3: Round to three significant figures. $$\text{Percentage increase} = 127\%$$
Marking scheme
Award marks as follows: - [1 mark] Correct calculation of increase in volume: \(42.3 - 18.6 = 23.7\text{ cm}^3\) - [1 mark] Correct method for percentage increase: \(\frac{23.7}{18.6} \times 100\) (allow ecf for incorrect volume difference divided by 18.6) - [1 mark] Correct final answer rounded to 3 significant figures: \(127\%\) (accept 127.4% only if 3 sig fig rule is ignored; award 3 marks for 127 with no working shown)
Question 19 · theory
3 marks
A potted plant was placed on a top-pan balance to determine the rate of transpiration under different environmental conditions. The mass of the pot and plant was recorded before and after a \(4\text{-hour}\) period.
- In still air, the mass decreased from \(520.4\text{ g}\) to \(508.8\text{ g}\). - In windy conditions, the mass decreased from \(520.4\text{ g}\) to \(486.8\text{ g}\).
Calculate the difference in the rate of water loss between windy conditions and still air.
difference in rate = ...........................................................
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Worked solution
Step 1: Calculate the rate of water loss in still air. $$\text{Mass lost in still air} = 520.4\text{ g} - 508.8\text{ g} = 11.6\text{ g}$$ $$\text{Rate in still air} = \frac{11.6\text{ g}}{4\text{ h}} = 2.9\text{ g h}^{-1}$$
Step 2: Calculate the rate of water loss in windy conditions. $$\text{Mass lost in windy conditions} = 520.4\text{ g} - 486.8\text{ g} = 33.6\text{ g}$$ $$\text{Rate in windy conditions} = \frac{33.6\text{ g}}{4\text{ h}} = 8.4\text{ g h}^{-1}$$
Step 3: Calculate the difference in rate. $$\text{Difference} = 8.4\text{ g h}^{-1} - 2.9\text{ g h}^{-1} = 5.5\text{ g h}^{-1}$$
Marking scheme
Award marks as follows: - [1 mark] Correct calculation of rate in still air (\(2.9\text{ g h}^{-1}\)) OR rate in windy conditions (\(8.4\text{ g h}^{-1}\)) - [1 mark] Subtraction of the two rates: \(8.4 - 2.9\) OR subtraction of mass losses divided by 4: \(\frac{33.6 - 11.6}{4}\) - [1 mark] Correct numerical value and appropriate unit: \(5.5\text{ g / h}\) or \(5.5\text{ g h}^{-1}\) or \(5.5\text{ g per hour}\) (accept \(0.0917\text{ g / min}\) if time converted correctly to minutes)
Question 20 · theory
3 marks
In a grassland ecosystem, the total energy fixed by producers through photosynthesis is \(18\,500\text{ kJ m}^{-2}\text{ year}^{-1}\).
The energy transferred to and incorporated into the biomass of primary consumers (herbivores) is \(2\,127.5\text{ kJ m}^{-2}\text{ year}^{-1}\).
Calculate the percentage efficiency of energy transfer from the producers to the primary consumers.
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Worked solution
Step 1: State the formula for efficiency of energy transfer. $$\text{Percentage efficiency} = \left( \frac{\text{energy incorporated by primary consumers}}{\text{energy fixed by producers}} \right) \times 100$$
Step 2: Substitute the values into the formula. $$\text{Percentage efficiency} = \left( \frac{2\,127.5}{18\,500} \right) \times 100$$ $$\text{Percentage efficiency} = 0.115 \times 100 = 11.5\%$$
Marking scheme
Award marks as follows: - [1 mark] Correct identification of numerator (\(2\,127.5\)) and denominator (\(18\,500\)) - [1 mark] Division and multiplication by 100: \(\frac{2\,127.5}{18\,500} \times 100\) - [1 mark] Correct answer: \(11.5\%\) (award 3 marks for correct answer without working shown)
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