Cambridge IGCSE · Thinka-original Practice Paper

2024 Cambridge IGCSE Biology (0610) Practice Paper with Answers

Thinka Jun 2024 (V1) Cambridge International A Level-Style Mock — Biology (0610)

160 marks180 mins2024
AnalysisSample paper
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V1) Cambridge International A Level Biology (0610) paper. Not affiliated with or reproduced from Cambridge.

Paper 2 (Multiple Choice - Extended)

Answer all forty questions on the multiple choice answer sheet. Each correct answer scores one mark. You may use a calculator.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
Which statement correctly explains the rate of reaction at temperatures higher than the optimum temperature of an enzyme?
  1. A.The kinetic energy of the substrate and enzyme molecules decreases, reducing the rate of collisions.
  2. B.The active site of the enzyme changes shape permanently, so the substrate can no longer fit.
  3. C.The activation energy of the reaction increases, preventing the substrate from being converted into product.
  4. D.The enzyme molecules are completely broken down into individual amino acids.
Show answer & marking scheme

Worked solution

At temperatures higher than the optimum, the increased kinetic energy causes bonds holding the enzyme's tertiary structure to vibrate and break. This permanently changes the shape of the active site, denaturing the enzyme. Consequently, the substrate can no longer fit into the active site, stopping the reaction.

Marking scheme

1 mark for identifying that denaturation involves a permanent change in the shape of the active site preventing substrate binding.
Question 2 · multiple-choice
1 marks
A cylinder of fresh potato tissue is placed in a concentrated sucrose solution. Which row correctly describes the net movement of water, the water potential of the sucrose solution compared to the cell sap, and the change in mass of the potato cylinder?
  1. A.Net movement of water: into the cells | Water potential of solution: higher than cell sap | Change in mass: increases
  2. B.Net movement of water: into the cells | Water potential of solution: lower than cell sap | Change in mass: decreases
  3. C.Net movement of water: out of the cells | Water potential of solution: higher than cell sap | Change in mass: increases
  4. D.Net movement of water: out of the cells | Water potential of solution: lower than cell sap | Change in mass: decreases
Show answer & marking scheme

Worked solution

A concentrated sucrose solution has a lower water potential than the sap of the potato cells. Water therefore moves out of the cells by osmosis down a water potential gradient, causing the cells to become flaccid and the overall mass of the cylinder to decrease.

Marking scheme

1 mark for correctly matching water movement out of cells, a lower water potential in the external solution, and a decrease in tissue mass.
Question 3 · multiple-choice
1 marks
Which row correctly describes the structural features of an artery and a vein?
  1. A.Artery wall: thick | Artery lumen: narrow | Vein wall: thin | Vein lumen: wide
  2. B.Artery wall: thin | Artery lumen: wide | Vein wall: thick | Vein lumen: narrow
  3. C.Artery wall: thick | Artery lumen: wide | Vein wall: thin | Vein lumen: narrow
  4. D.Artery wall: thin | Artery lumen: narrow | Vein wall: thick | Vein lumen: wide
Show answer & marking scheme

Worked solution

Arteries have thick walls containing muscle and elastic fibres to withstand and maintain high blood pressure, resulting in a narrow lumen. Veins have much thinner walls because blood flows under low pressure, which allows for a wider lumen to reduce resistance to flow.

Marking scheme

1 mark for correctly describing the combination of wall thickness and lumen diameter for both arteries and veins.
Question 4 · multiple-choice
1 marks
Which statement about active immunity is correct?
  1. A.It is acquired immediately after receiving an injection of pre-formed antibodies.
  2. B.It results in the long-term production of memory cells by the body.
  3. C.It only occurs following a natural infection and cannot be stimulated by vaccination.
  4. D.It provides short-term protection because the antibodies are rapidly broken down.
Show answer & marking scheme

Worked solution

Active immunity is the defense against a pathogen by antibody production in the body. It is long-term because memory cells are produced and remain in the body to rapidly respond to subsequent infections. It can be stimulated either by natural infection or artificially by vaccination.

Marking scheme

1 mark for identifying that active immunity leads to the production of memory cells, providing long-term protection.
Question 5 · multiple-choice
1 marks
A photomicrograph of a red blood cell is magnified \(\times 2500\). The image size of the red blood cell in the photograph is \(17.5\text{ mm}\). What is the actual size of the red blood cell in micrometres (\(\mu\text{m}\))?
  1. A.70.0 \(\mu\text{m}\)
  2. B.7.0 \(\mu\text{m}\)
  3. C.0.7 \(\mu\text{m}\)
  4. D.0.07 \(\mu\text{m}\)
Show answer & marking scheme

Worked solution

First, convert the image size from millimetres to micrometres: \(17.5\text{ mm} \times 1000 = 17500\ \mu\text{m}\). Next, use the formula Actual size = Image size / Magnification: \(17500\ \mu\text{m} / 2500 = 7.0\ \mu\text{m}\).

Marking scheme

1 mark for executing the conversion and calculating the actual size to be 7.0 micrometres.
Question 6 · multiple-choice
1 marks
The statements describe some events that occur during eutrophication in a river: 1. Decomposers respire aerobically and deplete oxygen levels; 2. Algae grow rapidly on the water surface, blocking light; 3. Aquatic plants die due to lack of light; 4. Run-off of fertilisers into the river; 5. Fish and other active aquatic organisms die. What is the correct sequence of these events?
  1. A.4 \(\rightarrow\) 2 \(\rightarrow\) 3 \(\rightarrow\) 1 \(\rightarrow\) 5
  2. B.4 \(\rightarrow\) 3 \(\rightarrow\) 2 \(\rightarrow\) 5 \(\rightarrow\) 1
  3. C.2 \(\rightarrow\) 4 \(\rightarrow\) 3 \(\rightarrow\) 1 \(\rightarrow\) 5
  4. D.4 \(\rightarrow\) 1 \(\rightarrow\) 2 \(\rightarrow\) 3 \(\rightarrow\) 5
Show answer & marking scheme

Worked solution

Eutrophication begins with the run-off of fertilisers (4), which causes rapid algal growth (an algal bloom) blocking light (2). Deprived of light, submerged aquatic plants die (3). Bacteria/decomposers feed on the dead plant matter, respiring aerobically and exhausting the dissolved oxygen (1). This lack of oxygen causes fish and other organisms to suffocate and die (5).

Marking scheme

1 mark for correctly ordering the sequence of events of eutrophication as 4, 2, 3, 1, 5.
Question 7 · multiple-choice
1 marks
During the genetic modification of bacteria to produce human insulin, which enzymes are used to cut the DNA and to join the DNA fragments together?
  1. A.Enzyme used to cut DNA: protease | Enzyme used to join DNA: DNA ligase
  2. B.Enzyme used to cut DNA: restriction enzyme | Enzyme used to join DNA: DNA ligase
  3. C.Enzyme used to cut DNA: restriction enzyme | Enzyme used to join DNA: amylase
  4. D.Enzyme used to cut DNA: DNA ligase | Enzyme used to join DNA: restriction enzyme
Show answer & marking scheme

Worked solution

Restriction enzymes (restriction endonucleases) are used to cut open plasmids and isolate the human insulin gene, leaving sticky ends. DNA ligase is used to join the sticky ends of the gene and plasmid together, forming recombinant DNA.

Marking scheme

1 mark for identifying restriction enzymes as the cutting tool and DNA ligase as the joining tool.
Question 8 · multiple-choice
1 marks
A plant is placed in four different environmental conditions. In which set of conditions will the rate of transpiration be the highest?
  1. A.Humidity: low | Temperature: high | Wind speed: high
  2. B.Humidity: high | Temperature: high | Wind speed: low
  3. C.Humidity: low | Temperature: low | Wind speed: low
  4. D.Humidity: high | Temperature: low | Wind speed: high
Show answer & marking scheme

Worked solution

Transpiration rate increases when: 1. Humidity is low (increases the water potential gradient between the leaf interior and the atmosphere); 2. Temperature is high (increases kinetic energy and rate of evaporation); 3. Wind speed is high (blows away water vapour near stomata, maintaining a steep concentration gradient).

Marking scheme

1 mark for identifying low humidity, high temperature, and high wind speed as the combination maximizing transpiration rate.
Question 9 · multiple-choice
1 marks
Which statement describes the effect of increasing temperature from 20 °C to 30 °C on an enzyme-controlled reaction?
  1. A.The kinetic energy of the substrate molecules decreases, reducing successful collisions.
  2. B.The active site of the enzyme changes shape permanently, preventing substrate binding.
  3. C.More enzyme-substrate complexes are formed per unit time because molecules move faster.
  4. D.The activation energy of the reaction is increased, allowing the reaction to proceed faster.
Show answer & marking scheme

Worked solution

Increasing the temperature from 20 °C to 30 °C increases the kinetic energy of both the enzyme and substrate molecules. This increases the rate of movement of these molecules, leading to a higher frequency of successful collisions between the substrate molecules and the active site of the enzyme. Consequently, more enzyme-substrate complexes are formed per unit time, which increases the overall rate of reaction. Option A is incorrect because kinetic energy increases, not decreases. Option B describes denaturation, which typically happens at much higher temperatures (e.g., above 40 °C for human enzymes). Option D is incorrect because enzymes lower the activation energy, and changing the temperature does not increase the activation energy of the reaction.

Marking scheme

1 mark for the correct option C. Reject other options.
Question 10 · multiple-choice
1 marks
At which point in the cardiac cycle does the aortic valve open?
  1. A.When the pressure in the left ventricle becomes higher than the pressure in the left atrium.
  2. B.When the pressure in the left ventricle becomes higher than the pressure in the aorta.
  3. C.When the pressure in the left atrium becomes higher than the pressure in the left ventricle.
  4. D.When the pressure in the aorta becomes higher than the pressure in the left ventricle.
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Worked solution

The aortic valve (a semi-lunar valve) is located between the left ventricle and the aorta. It remains closed during ventricular diastole and early systole while ventricular pressure is lower than the pressure in the aorta. Once the left ventricle contracts strongly during ventricular systole and its pressure exceeds the pressure inside the aorta, the aortic valve is pushed open, allowing blood to flow into the systemic circulation. Option A describes the event that closes the bicuspid (mitral) valve. Option C describes the event that opens the bicuspid valve. Option D describes the event that closes the aortic valve.

Marking scheme

1 mark for the correct option B. Reject other options.
Question 11 · multiple-choice
1 marks
Four identical cylinders of potato tissue were placed in four sucrose solutions of different concentrations: 0.1 mol/dm³, 0.3 mol/dm³, 0.5 mol/dm³, and 0.7 mol/dm³. The initial mass of each cylinder was 5.0 g. After two hours, the masses were recorded. Which cylinder was placed in the 0.7 mol/dm³ sucrose solution?
  1. A.The cylinder with a final mass of 5.4 g.
  2. B.The cylinder with a final mass of 5.0 g.
  3. C.The cylinder with a final mass of 4.6 g.
  4. D.The cylinder with a final mass of 4.1 g.
Show answer & marking scheme

Worked solution

A sucrose solution of 0.7 mol/dm³ has the lowest water potential of all the concentrations listed. The cell sap of the potato cells will have a higher water potential than this highly concentrated external solution. As a result, water will leave the potato cells by osmosis, moving down a water potential gradient from a high water potential inside the cells to a low water potential outside the cells. This net movement of water out of the tissue causes the greatest decrease in mass. Among the options given, 4.1 g represents the greatest mass loss (from 5.0 g to 4.1 g), meaning it must correspond to the 0.7 mol/dm³ solution.

Marking scheme

1 mark for the correct option D. Reject other options.
Question 12 · multiple-choice
1 marks
Which statement correctly describes passive immunity?
  1. A.It is gained by the production of antibodies by the person's own lymphocytes.
  2. B.It provides long-term protection through the production of memory cells.
  3. C.It can be acquired by a breast-feeding infant receiving antibodies from breast milk.
  4. D.It is triggered by the introduction of a weakened pathogen during vaccination.
Show answer & marking scheme

Worked solution

Passive immunity is a short-term defense against a pathogen by antibodies acquired from another individual. An example of this is a breast-feeding infant receiving antibodies (such as IgA) through breast milk, or antibodies crossing the placenta from the mother to the fetus. Options A, B, and D describe active immunity, which involves the individual's own lymphocytes producing antibodies and memory cells in response to direct exposure to an antigen (either via natural infection or vaccination).

Marking scheme

1 mark for the correct option C. Reject other options.
Question 13 · multiple-choice
1 marks
A student accidentally touches a hot object and quickly withdraws their hand. Which sequence shows the correct pathway of the nerve impulse in this reflex action?
  1. A.receptor → motor neurone → relay neurone → sensory neurone → effector
  2. B.effector → sensory neurone → relay neurone → motor neurone → receptor
  3. C.receptor → sensory neurone → relay neurone → motor neurone → effector
  4. D.effector → motor neurone → relay neurone → sensory neurone → receptor
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Worked solution

In a simple reflex arc, a receptor (such as temperature or pain receptors in the skin) detects the stimulus and generates electrical impulses. These impulses travel along a sensory neurone to the central nervous system (spinal cord). In the spinal cord, the impulses are transmitted across synapses to a relay neurone, and then across synapses to a motor neurone. The motor neurone carries the impulse out of the spinal cord to the effector (a muscle or gland, in this case, the arm muscles), which contracts to withdraw the hand. Thus, the correct sequence is: receptor → sensory neurone → relay neurone → motor neurone → effector.

Marking scheme

1 mark for the correct option C. Reject other options.
Question 14 · multiple-choice
1 marks
Fertiliser run-off into a lake can cause a sequence of events leading to the death of fish. Which event occurs first after the fertiliser enters the water?
  1. A.Rapid growth of algae blocking light from reaching submerged plants.
  2. B.Decomposition of dead organic matter by aerobic bacteria.
  3. C.Decrease in the concentration of dissolved oxygen in the water.
  4. D.Anaerobic respiration by aquatic decomposers producing toxins.
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Worked solution

When agricultural fertilisers run off into aquatic environments, they supply a surplus of nutrients like nitrates and phosphates. This nutrient enrichment initially triggers rapid population growth of algae (an algal bloom) at the water surface, which blocks light from reaching submerged plants underneath. The death of these submerged plants is followed by the proliferation of decomposers (bacteria) that break down the dead organic matter. This aerobic decomposition consumes dissolved oxygen, creating hypoxic conditions that lead to the suffocation and death of fish. Therefore, the rapid growth of algae blocking light occurs first.

Marking scheme

1 mark for the correct option A. Reject other options.
Question 15 · multiple-choice
1 marks
Under which environmental conditions will the rate of transpiration from a leafy shoot be the highest?
  1. A.Low temperature, high humidity, low wind speed
  2. B.High temperature, low humidity, high wind speed
  3. C.High temperature, high humidity, high wind speed
  4. D.Low temperature, low humidity, low wind speed
Show answer & marking scheme

Worked solution

Transpiration is the loss of water vapour from plant leaves. Its rate depends on the water vapour concentration gradient between the internal air spaces of the leaf and the external atmosphere. A high temperature increases the rate of evaporation from the wet cell walls into the air spaces. Low humidity in the external air keeps the water potential of the outside air low, steepening the concentration gradient. High wind speed removes the layer of humid air that accumulates around stomata, maintaining a steep concentration gradient. Therefore, high temperature, low humidity, and high wind speed maximize transpiration.

Marking scheme

1 mark for the correct option B. Reject other options.
Question 16 · multiple-choice
1 marks
A plant is kept in a sealed glass container under a constant light source. Which change would most likely increase the rate of photosynthesis if carbon dioxide concentration is the limiting factor?
  1. A.Increasing the oxygen concentration in the container.
  2. B.Increasing the relative humidity inside the container.
  3. C.Adding a beaker of sodium hydrogencarbonate solution to the container.
  4. D.Placing a green plastic filter over the light source.
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Worked solution

If carbon dioxide is the limiting factor, any change that increases the availability of CO2 to the plant's chloroplasts will increase the rate of photosynthesis. Adding a beaker of sodium hydrogencarbonate solution to the sealed container is a common experimental method to increase carbon dioxide concentration, as the solution releases CO2 gas. Option A is incorrect because oxygen is a product of photosynthesis, not a reactant, and does not limit the rate of the reaction. Option B might affect transpiration and turgidity but will not address a CO2 limitation. Option D is incorrect because chlorophyll poorly absorbs green light, so a green filter would drastically decrease the rate of photosynthesis.

Marking scheme

1 mark for the correct option C. Reject other options.
Question 17 · multiple-choice
1 marks
A student investigates the effect of pH on the rate of amylase activity. The student records the time taken for starch to be completely broken down. At pH 6, the time taken was 2 minutes. At pH 8, the time taken was 8 minutes. What is the rate of reaction at pH 6 compared to the rate of reaction at pH 8?
  1. A.4 times faster
  2. B.4 times slower
  3. C.6 times faster
  4. D.2 times slower
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Worked solution

The rate of a reaction is inversely proportional to the time taken (\(\text{Rate} = 1 / \text{time}\)). At pH 6, the rate is \(1 / 2 = 0.5\) \(\text{min}^{-1}\). At pH 8, the rate is \(1 / 8 = 0.125\) \(\text{min}^{-1}\). Comparing the rates: \(0.5 / 0.125 = 4\). Therefore, the rate of reaction at pH 6 is 4 times faster than at pH 8.

Marking scheme

Award 1 mark for the correct option A. Correctly calculates rates at both pH levels and determines the relative increase.
Question 18 · multiple-choice
1 marks
Which row correctly matches the type of blood cell to its anatomical feature and physiological function?
  1. A.Lymphocyte | Large, spherical nucleus | Produces antibodies
  2. B.Phagocyte | Lobed nucleus | Produces antibodies
  3. C.Red blood cell | Biconcave disc with no nucleus | Transports carbon dioxide only
  4. D.Platelet | Fragment of cell with lobed nucleus | Engulfs pathogens
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Worked solution

Lymphocytes are characterized by a large, spherical nucleus that fills most of the cell, and their function is to produce antibodies. Phagocytes have a lobed nucleus and engulf pathogens. Red blood cells lack a nucleus and transport oxygen, while platelets are cell fragments involved in clotting.

Marking scheme

Award 1 mark for the correct option A. Identifies correct blood cell structure and function.
Question 19 · multiple-choice
1 marks
Six identical potato cylinders were placed in sucrose solutions of different concentrations: \(0.0\text{ mol/dm}^3\), \(0.2\text{ mol/dm}^3\), \(0.4\text{ mol/dm}^3\), \(0.6\text{ mol/dm}^3\), \(0.8\text{ mol/dm}^3\), and \(1.0\text{ mol/dm}^3\). After two hours, the percentage change in mass was calculated. The potato cylinders in \(0.0\text{ mol/dm}^3\) and \(0.2\text{ mol/dm}^3\) gained mass, while those in \(0.4\text{ mol/dm}^3\) to \(1.0\text{ mol/dm}^3\) lost mass. Which concentration range contains the concentration of the cytoplasm inside the potato cells?
  1. A.Between \(0.0\text{ mol/dm}^3\) and \(0.2\text{ mol/dm}^3\)
  2. B.Between \(0.2\text{ mol/dm}^3\) and \(0.4\text{ mol/dm}^3\)
  3. C.Between \(0.4\text{ mol/dm}^3\) and \(0.6\text{ mol/dm}^3\)
  4. D.Exactly \(0.0\text{ mol/dm}^3\)
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Worked solution

The internal concentration of the potato cells corresponds to the point where there is no net movement of water by osmosis (0% change in mass). This transition point must lie between the solution where the cells gained mass (\(0.2\text{ mol/dm}^3\)) and the solution where they lost mass (\(0.4\text{ mol/dm}^3\)).

Marking scheme

Award 1 mark for the correct option B. Demonstrates understanding of water potential and net movement in hypertonic/hypotonic environments.
Question 20 · multiple-choice
1 marks
A child is bitten by a venomous snake. At the hospital, they receive an emergency injection of antivenom containing specific antibodies. Two weeks later, the child has fully recovered but is bitten by the same species of snake again. Why does the child NOT have long-term immunity against this venom from the first treatment?
  1. A.The antivenom provided active immunity, which does not produce memory cells.
  2. B.The antivenom provided passive immunity, meaning no memory cells were produced by the child's own lymphocytes.
  3. C.The snake venom destroyed the child's memory cells before they could multiply.
  4. D.Passive immunity only lasts for a few hours before the injected antibodies reproduce and cause an immune response.
Show answer & marking scheme

Worked solution

The antivenom injection contains pre-formed antibodies, which provides passive immunity. Because the child's own immune system did not active-respond by recognizing antigens and producing antibodies, no memory cells were created. Once the injected antibodies break down, protection is lost.

Marking scheme

Award 1 mark for the correct option B. Distinguishes active and passive immunity and identifies lack of memory cells in passive immunity.
Question 21 · multiple-choice
1 marks
A student accidentally touches a hot object and immediately pulls their hand away. Which sequence shows the correct pathway of the electrical impulses along the neurones in this reflex arc?
  1. A.receptor → motor neurone → relay neurone → sensory neurone → effector
  2. B.effector → sensory neurone → relay neurone → motor neurone → receptor
  3. C.receptor → sensory neurone → relay neurone → motor neurone → effector
  4. D.receptor → sensory neurone → motor neurone → relay neurone → effector
Show answer & marking scheme

Worked solution

In a spinal reflex arc, the pathway starts with a receptor detecting the stimulus. The electrical impulse travels along a sensory neurone to the spinal cord, passes across a synapse to a relay neurone, then passes across another synapse to a motor neurone, which carries the impulse to the effector (muscle) to cause the response.

Marking scheme

Award 1 mark for the correct option C. Identifies correct order of neurones and components in a reflex arc.
Question 22 · multiple-choice
1 marks
When excess nitrate fertilizers leach from farmland into a river, a sequence of events occurs leading to the death of fish. Which sequence shows the correct order of these events?
  1. A.increased growth of producers → increased aerobic respiration by decomposers → death of algae → decrease in dissolved oxygen
  2. B.increased growth of producers → death of algae → increased aerobic respiration by decomposers → decrease in dissolved oxygen
  3. C.decrease in dissolved oxygen → death of algae → increased growth of producers → increased aerobic respiration by decomposers
  4. D.increased aerobic respiration by decomposers → increased growth of producers → death of algae → decrease in dissolved oxygen
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Worked solution

Leached nitrates stimulate the rapid growth of algae (producers). Eventually, the algae die due to competition for light. Decomposers (bacteria) feed on the dead algae and multiply, undergoing aerobic respiration and using up the dissolved oxygen in the water, which leads to fish suffocation.

Marking scheme

Award 1 mark for the correct option B. Correctly orders the sequential stages of eutrophication.
Question 23 · multiple-choice
1 marks
A potometer is used to measure the rate of water uptake by a leafy shoot. Which set of environmental conditions will result in the highest rate of transpiration?
  1. A.High temperature, high humidity, high wind speed
  2. B.Low temperature, low humidity, low wind speed
  3. C.High temperature, low humidity, high wind speed
  4. D.High temperature, low humidity, still air
Show answer & marking scheme

Worked solution

Transpiration rate is highest under conditions that maximize evaporation and maintain a steep water vapor concentration gradient: high temperature increases kinetic energy of water molecules, low humidity maintains a dry atmosphere outside the stomata, and high wind speed continuously removes moist air around the leaf surface.

Marking scheme

Award 1 mark for the correct option C. Identifies the combination of high temperature, low humidity, and high wind speed as optimal for transpiration.
Question 24 · multiple-choice
1 marks
An experiment measures the rate of photosynthesis of a plant at different light intensities, under three different sets of conditions:
Curve 1: \(0.04\%\text{ CO}_2\) at \(20^\circ\text{C}\)
Curve 2: \(0.12\%\text{ CO}_2\) at \(20^\circ\text{C}\)
Curve 3: \(0.12\%\text{ CO}_2\) at \(30^\circ\text{C}\)
At high light intensities, Curve 1 plateaus at a much lower rate than Curve 2. What is the limiting factor for Curve 1 at high light intensities?
  1. A.Light intensity
  2. B.Carbon dioxide concentration
  3. C.Temperature
  4. D.Chlorophyll concentration
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Worked solution

At high light intensity, the rate of photosynthesis in Curve 1 levels off, meaning light is no longer limiting. When the carbon dioxide concentration is increased to 0.12% in Curve 2 (keeping temperature constant at 20\(^\circ\)C), the rate increases significantly. This proves that carbon dioxide concentration was the factor limiting the rate in Curve 1.

Marking scheme

Award 1 mark for the correct option B. Interprets limiting factors correctly from comparison of environmental parameters.
Question 25 · multiple-choice
1 marks
An enzyme-controlled reaction is carried out at 30 °C. The temperature is then increased to 40 °C. Why does the rate of reaction increase?
  1. A.The active site of the enzyme changes shape to fit the substrate more closely.
  2. B.The enzyme molecules vibrate so much that they denature.
  3. C.The molecules have more kinetic energy, leading to more frequent successful collisions.
  4. D.The activation energy required for the reaction is lowered by the increase in temperature..
Show answer & marking scheme

Worked solution

The rate of reaction increases because increasing the temperature increases the kinetic energy of the enzyme and substrate molecules. This leads to more rapid molecular movement and a higher frequency of successful collisions between the substrates and the active sites of the enzymes. Option A is incorrect because the active site shape is genetically determined and does not change to 'fit better' upon warming. Option B is incorrect as denaturation decreases the rate of reaction (usually above 40 °C for human enzymes). Option D is incorrect because enzymes themselves lower the activation energy, but temperature does not lower it further; it provides more energy to overcome the activation energy barrier.

Marking scheme

1 mark: correct option C.
Question 26 · multiple-choice
1 marks
Which statement describes the advantage of a double circulation system in mammals?
  1. A.It ensures that blood can only flow through the heart in one direction.
  2. B.It allows oxygenated and deoxygenated blood to mix to help maintain body temperature.
  3. C.It allows blood to be pumped to body tissues at a higher pressure than to the lungs.
  4. D.It allows the heart muscle to contract rhythmically without needing nerve impulses.
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Worked solution

In a double circulatory system, blood is pumped to the lungs at a lower pressure to prevent damage to the delicate lung tissues, and then returns to the heart to be pumped to the rest of the body at a much higher pressure. This ensures efficient delivery of oxygen to respiring tissues. Option A describes the function of valves, not double circulation. Option B is incorrect because oxygenated and deoxygenated blood are kept separate. Option D is incorrect because the heart is myogenic and does not require nerve impulses to beat, which is unrelated to double circulation.

Marking scheme

1 mark: correct option C.
Question 27 · multiple-choice
1 marks
A plant cell is placed in a solution with a lower water potential than its cytoplasm. Which row correctly describes the net movement of water and the final state of the cell?
  1. A.Net movement of water: out of the cell | Final state of cell: plasmolysed
  2. B.Net movement of water: out of the cell | Final state of cell: turgid
  3. C.Net movement of water: into the cell | Final state of cell: plasmolysed
  4. D.Net movement of water: into the cell | Final state of cell: turgid
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Worked solution

Because the surrounding solution has a lower water potential than the cytoplasm, there is a net movement of water out of the cell by osmosis. As water leaves the vacuole and cytoplasm, the volume decreases, causing the cell membrane to pull away from the cell wall. The cell is described as plasmolysed.

Marking scheme

1 mark: correct option A.
Question 28 · multiple-choice
1 marks
Which statement correctly describes passive immunity?
  1. A.It is gained by the production of antibodies by the person's own immune system.
  2. B.It provides long-term protection through the production of memory cells.
  3. C.It involves the transfer of antibodies from another individual, providing temporary protection.
  4. D.It is achieved only by injecting a weakened or harmless pathogen into the body.
Show answer & marking scheme

Worked solution

Passive immunity is the temporary defense against a pathogen, which is acquired by the transfer of antibodies from another individual (for example, across the placenta or through breast milk). It is temporary because no memory cells are produced by the recipient's own immune system. Options A, B, and D describe features of active immunity.

Marking scheme

1 mark: correct option C.
Question 29 · multiple-choice
1 marks
Which process occurs at a synapse to transmit an impulse from one neurone to the next?
  1. A.Active transport of neurotransmitters against a concentration gradient.
  2. B.Diffusion of neurotransmitter molecules across the synaptic cleft.
  3. C.Electrical impulses jumping directly across the gap between neurones.
  4. D.Osmosis of water containing dissolved neurotransmitter molecules.
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Worked solution

When an electrical impulse reaches the end of a neurone, it triggers the release of neurotransmitter chemicals. These molecules diffuse across the tiny gap (synaptic cleft) and bind to receptor proteins on the membrane of the next neurone, triggering a new electrical impulse. This movement occurs down a concentration gradient and does not involve active transport, osmosis, or direct electrical jumping.

Marking scheme

1 mark: correct option B.
Question 30 · multiple-choice
1 marks
Which sequence of events describes the process of eutrophication in a river after fertiliser run-off occurs?
  1. A.increased algae growth → increased aerobic respiration by decomposers → decrease in dissolved oxygen → death of fish
  2. B.death of algae → increased photosynthesis by water plants → decrease in dissolved oxygen → death of fish
  3. C.increased aerobic respiration by decomposers → increased algae growth → decrease in dissolved oxygen → death of fish
  4. D.decrease in dissolved oxygen → death of decomposers → increased algae growth → death of fish
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Worked solution

Eutrophication begins with fertiliser run-off, which causes rapid growth of algae (algae bloom). The algae block sunlight, causing submerged plants to die. Aerobic decomposers (bacteria) feed on the dead plant matter, multiplying rapidly and using up the dissolved oxygen in the water through aerobic respiration. This lack of oxygen causes fish and other aquatic organisms to suffocate and die.

Marking scheme

1 mark: correct option A.
Question 31 · multiple-choice
1 marks
A potometer is used to measure the rate of transpiration in a leafy shoot. Which set of environmental conditions would produce the slowest rate of water uptake?
  1. A.High humidity, low temperature, still air
  2. B.High humidity, high temperature, moving air
  3. C.Low humidity, low temperature, moving air
  4. D.Low humidity, high temperature, still air
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Worked solution

The rate of transpiration (and thus water uptake) is slowest when the water potential gradient between the inside of the leaf and the external air is as small as possible. High humidity keeps the external air moist, low temperature reduces the rate of water evaporation from the spongy mesophyll cell walls, and still air allows a boundary layer of water vapor to build up around the stomata. All three conditions together minimize transpiration.

Marking scheme

1 mark: correct option A.
Question 32 · multiple-choice
1 marks
A student views a micrograph of a plant cell. The image size of the cell is 48 mm. The actual size of the cell is 0.08 mm. What is the magnification of this image?
  1. A.\(\times 0.0016\)
  2. B.\(\times 1.67\)
  3. C.\(\times 600\)
  4. D.\(\times 3840\)
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Worked solution

To calculate magnification, use the formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\). Since both measurements are already in the same unit (mm), we calculate: \(M = \frac{48}{0.08} = 600\). Thus, the magnification is \(\times 600\).

Marking scheme

1 mark: correct option C.
Question 33 · multiple-choice
1 marks
An enzyme-catalysed reaction is set up at four different temperatures: \(10^\circ\text{C}\), \(20^\circ\text{C}\), \(30^\circ\text{C}\), and \(40^\circ\text{C}\). At \(40^\circ\text{C}\), the rate of product formation is initially very high but quickly drops to zero before all of the substrate is used up. What explains this observation?
  1. A.The activation energy of the reaction increased.
  2. B.The enzyme was denatured because the kinetic energy of the substrate molecules was too low.
  3. C.The high temperature caused the active sites of the enzyme molecules to change shape permanently.
  4. D.The substrate molecules were denatured by the high temperature.
Show answer & marking scheme

Worked solution

At higher temperatures such as \(40^\circ\text{C}\), enzymes have high kinetic energy, leading to a high initial rate of reaction. However, because the temperature is above the optimum, the active site of the enzyme quickly denatures (loses its specific shape permanently). As a result, substrates can no longer bind to the active site, and the reaction stops abruptly even if substrate molecules are still available.

Marking scheme

1 mark for the correct answer C. [Reject: A, because activation energy is not increased; B, because kinetic energy is high at high temperatures; D, because substrate molecules do not denature.]
Question 34 · multiple-choice
1 marks
Which sequence correctly identifies the vessels and heart chambers through which a red blood cell passes as it travels from the left kidney to the left lung?
  1. A.renal vein \(\rightarrow\) vena cava \(\rightarrow\) right atrium \(\rightarrow\) right ventricle \(\rightarrow\) pulmonary artery
  2. B.renal artery \(\rightarrow\) vena cava \(\rightarrow\) left atrium \(\rightarrow\) left ventricle \(\rightarrow\) pulmonary artery
  3. C.renal vein \(\rightarrow\) vena cava \(\rightarrow\) left atrium \(\rightarrow\) left ventricle \(\rightarrow\) pulmonary vein
  4. D.renal artery \(\rightarrow\) aorta \(\rightarrow\) right atrium \(\rightarrow\) right ventricle \(\rightarrow\) pulmonary vein
Show answer & marking scheme

Worked solution

Blood leaving the left kidney travels through the renal vein into the vena cava, which delivers it to the right atrium of the heart. From there, it passes into the right ventricle and is pumped through the pulmonary artery to reach the lungs.

Marking scheme

1 mark for the correct sequence starting with the renal vein and ending with the pulmonary artery (A). [Reject: B, C, D which contain incorrect anatomical pathways or wrong vessels/chambers.]
Question 35 · multiple-choice
1 marks
Four cylinders of potato tissue of equal initial length were placed into four different sucrose solutions (P, Q, R, and S). After 24 hours, the change in length of each cylinder was measured: Solution P: \(-1.2\text{ mm}\), Solution Q: \(+2.4\text{ mm}\), Solution R: \(0.0\text{ mm}\), Solution S: \(-0.5\text{ mm}\). Which solution had the highest water potential?
  1. A.Solution P
  2. B.Solution Q
  3. C.Solution R
  4. D.Solution S
Show answer & marking scheme

Worked solution

A positive change in length indicates that water has entered the potato cells by osmosis, meaning the surrounding solution had a higher water potential than the cell sap. Solution Q resulted in the greatest increase in length (\(+2.4\text{ mm}\)), indicating it has the highest water potential among all tested solutions.

Marking scheme

1 mark for selecting B. [Reject: A, C, D because negative or zero change in length indicates a lower or equal water potential relative to the potato cells.]
Question 36 · multiple-choice
1 marks
A patient is bitten by a venomous snake and is immediately injected with an antivenom containing specific antibodies. Which statement correctly describes the type of immunity provided and its characteristics?
  1. A.Active immunity, because the patient's body produces memory cells in response to the injection.
  2. B.Active immunity, because the antibodies act immediately to destroy the snake toxin.
  3. C.Passive immunity, because the antibodies are introduced from an external source and no memory cells are produced.
  4. D.Passive immunity, because the treatment triggers the rapid production of antibodies by the patient's lymphocytes.
Show answer & marking scheme

Worked solution

An injection of ready-made antibodies provides passive immunity. It provides immediate protection, but because the patient's own lymphocytes did not produce the antibodies, no memory cells are created, and the protection is only temporary.

Marking scheme

1 mark for identifying passive immunity and lack of memory cell production (C). [Reject: A and B because injecting pre-formed antibodies is passive, not active. Reject D because passive immunity does not trigger the body's own lymphocytes to produce antibodies.]
Question 37 · multiple-choice
1 marks
What is the correct sequence of events that occurs at a synapse when an impulse arrives at the sensory neurone terminal?
  1. A.Neurotransmitter is released into the synaptic cleft \(\rightarrow\) neurotransmitter diffuses across the cleft \(\rightarrow\) neurotransmitter binds to receptor proteins on the relay neurone membrane
  2. B.Electrical impulses jump across the synaptic cleft \(\rightarrow\) neurotransmitter is released from receptor proteins on the relay neurone
  3. C.Neurotransmitter is released from the relay neurone \(\rightarrow\) neurotransmitter diffuses across the cleft \(\rightarrow\) an electrical impulse is generated in the sensory neurone
  4. D.Receptor proteins release neurotransmitter into the synaptic cleft \(\rightarrow\) neurotransmitter binds to the sensory neurone membrane
Show answer & marking scheme

Worked solution

When an impulse reaches the end of the sensory neurone (presynaptic membrane), neurotransmitters are released from vesicles into the synaptic cleft. They diffuse across the gap and bind to specific receptor proteins on the membrane of the next neurone (e.g., relay neurone), triggering a new electrical impulse.

Marking scheme

1 mark for the correct physiological sequence (A). [Reject: B, because electrical impulses cannot jump across the cleft directly; C and D, because the sequence of transmission is from presynaptic (sensory) to postsynaptic (relay) neurone.]
Question 38 · multiple-choice
1 marks
Untreated sewage is accidentally discharged into a river ecosystem. Which sequence represents the correct order of events leading to the death of fish in the river?
  1. A.increased growth of algae \(\rightarrow\) algae block light \(\rightarrow\) underwater plants die \(\rightarrow\) decomposers multiply and use up dissolved oxygen \(\rightarrow\) fish suffocate
  2. B.increased growth of decomposers \(\rightarrow\) pH of water increases \(\rightarrow\) algae die \(\rightarrow\) toxic chemicals are released \(\rightarrow\) fish are poisoned
  3. C.decreased concentration of nitrate ions \(\rightarrow\) death of algae \(\rightarrow\) decomposers run out of food \(\rightarrow\) oxygen levels rise \(\rightarrow\) fish die from oxygen toxicity
  4. D.sewage acts as a poison \(\rightarrow\) decomposers die \(\rightarrow\) algae populations increase \(\rightarrow\) algae use up all the dissolved oxygen \(\rightarrow\) fish die
Show answer & marking scheme

Worked solution

The addition of sewage (which contains organic matter and nutrients like nitrates) causes rapid growth of algae (algal bloom). This blocks light to deeper water, causing submerged plants to die. Decomposers (bacteria) feed on the dead plants and multiply rapidly, consuming dissolved oxygen during aerobic respiration, causing fish to die from suffocation.

Marking scheme

1 mark for the correct sequence of eutrophication events (A). [Reject: B, C, D which describe incorrect ecological pathways or biological processes.]
Question 39 · multiple-choice
1 marks
How does a high concentration of water vapour in the surrounding atmosphere affect the rate of transpiration in a healthy plant, and what is the reason for this?
  1. A.It decreases the rate because the concentration gradient of water vapour between the inside and outside of the leaf is reduced.
  2. B.It decreases the rate because the stomata close completely to prevent water from entering the leaf.
  3. C.It increases the rate because water molecules diffuse more rapidly into the leaf.
  4. D.It increases the rate because the water potential gradient between the leaf cells and the atmosphere becomes steeper.
Show answer & marking scheme

Worked solution

A high atmospheric concentration of water vapour reduces the difference in water vapour concentration between the air spaces inside the leaf and the external air. This decreases the diffusion gradient (concentration gradient), thereby reducing the rate of transpiration.

Marking scheme

1 mark for correctly matching the effect (decrease) with the scientific reason (reduced concentration gradient) (A). [Reject: B, because stomata do not close completely to prevent water entry; C and D, because high humidity decreases rather than increases the rate of transpiration.]
Question 40 · multiple-choice
1 marks
A plant is kept at a constant temperature of \(20^\circ\text{C}\). At high light intensity, the rate of photosynthesis is found to be significantly higher at a carbon dioxide concentration of \(0.15\%\) than at \(0.04\%\). What is the limiting factor for photosynthesis at high light intensity when the carbon dioxide concentration is \(0.04\%\)?
  1. A.Light intensity
  2. B.Carbon dioxide concentration
  3. C.Temperature
  4. D.Chlorophyll concentration
Show answer & marking scheme

Worked solution

Since increasing the concentration of carbon dioxide from \(0.04\%\) to \(0.15\%\) increases the rate of photosynthesis, the carbon dioxide concentration must have been the limiting factor when it was at \(0.04\%\).

Marking scheme

1 mark for B. [Reject: A, because light intensity is already high; C, because temperature is kept constant and is not altered; D, because chlorophyll content is constant within the same plant and not the factor being manipulated to increase the rate.]

Paper 4 (Theory - Extended)

Answer all questions. Show your working in the spaces provided. Write in dark blue or black pen.
7 Question · 79.80000000000001 marks
Question 1 · structured
11.4 marks
An investigation is performed to study the rate of a reaction catalyzed by the enzyme lactase. (a) State the four chemical elements that are present in all enzymes. [2] (b) Describe the mechanism of enzyme action using the lock-and-key hypothesis. [4] (c) Predict and explain the effect on the rate of reaction when the temperature is increased from 35 degrees C to 75 degrees C. [6]
Show answer & marking scheme

Worked solution

(a) Carbon, hydrogen, oxygen, and nitrogen. (b) The enzyme is the lock and the substrate is the key; they have complementary shapes; they collide and the substrate binds to the active site; an enzyme-substrate complex is formed; the reaction occurs and products are released, leaving the enzyme unchanged. (c) At 35 degrees C, the enzyme is highly active with a high rate due to sufficient kinetic energy. At 75 degrees C, the rate drops to zero because high thermal energy breaks hydrogen bonds, altering the three-dimensional shape of the protein. The active site is denatured and no longer fits the substrate.

Marking scheme

(a) 2 marks: 1 mark for any two of C, H, O; 2 marks for all four (C, H, O, N). (b) Max 4 marks: 1 mark for lock-and-key analogy; 1 mark for complementary shapes; 1 mark for substrate binding to active site; 1 mark for enzyme-substrate complex; 1 mark for products leaving active site unchanged. (c) Max 6 marks: 1 mark for identification that 75 degrees C causes denaturation; 1 mark for kinetic energy increases initially; 1 mark for breaking of bonds in enzyme structure; 1 mark for alteration of active site shape; 1 mark for substrate no longer fitting; 1 mark for reaction rate dropping to zero/stopping.
Question 2 · structured
11.4 marks
Vaccines are used to provide artificial active immunity. (a) Define the term pathogen. [2] (b) Explain how vaccination leads to long-term active immunity against a specific pathogen. [6] (c) Contrast active immunity with passive immunity, identifying one scenario where passive immunity would be preferred. [4]
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Worked solution

(a) A pathogen is a disease-causing organism (such as a bacterium or virus). (b) Vaccine contains harmless, weakened, or dead pathogens or antigens; lymphocytes recognize antigens; lymphocytes divide by mitosis; lymphocytes produce specific antibodies; memory cells are formed and remain in the blood; upon second exposure, memory cells quickly produce large quantities of antibodies to destroy the pathogen. (c) Active immunity involves the body making its own antibodies and memory cells (long-term protection). Passive immunity involves receiving ready-made antibodies from another source (short-term, no memory cells). Passive immunity is preferred for immediate protection, such as treating rabies or a snake bite.

Marking scheme

(a) 2 marks: 1 mark for 'organism', 1 mark for 'causing disease'. (b) Max 6 marks: 1 mark for weakened/dead pathogen or antigen in vaccine; 1 mark for clonal selection/lymphocytes recognizing antigen; 1 mark for mitosis/division of lymphocytes; 1 mark for production of specific antibodies; 1 mark for production of memory cells; 1 mark for rapid response on reinfection. (c) Max 4 marks: 1 mark for active producing own antibodies/memory cells; 1 mark for passive receiving external antibodies/no memory cells; 1 mark for active being long-term and passive being temporary; 1 mark for correct example of passive preference (e.g., snake antivenom/tetanus shot).
Question 3 · structured
11.4 marks
The mammalian circulatory system contains different types of blood vessels. (a) Describe two structural differences between an artery and a vein. [4] (b) Explain how the structure of a capillary is adapted to its function of exchange of materials. [4] (c) Explain the advantages of a double circulatory system in mammals. [4]
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Worked solution

(a) Arteries have thicker muscular and elastic walls compared to veins; arteries have a narrower lumen; veins contain valves to prevent backflow of blood, while arteries do not. (b) Capillary walls are only one cell thick, which provides a very short diffusion pathway for gases and nutrients; they are narrow, slowing blood flow to allow time for exchange; they form extensive networks providing a very large surface area. (c) A double circulatory system pumps blood through the heart twice in one complete circuit. This keeps oxygenated and deoxygenated blood separate, ensuring high oxygen delivery to tissues. It also allows blood to be pumped to body tissues at a much higher pressure than blood going to the lungs, increasing efficiency.

Marking scheme

(a) Max 4 marks: 2 marks for structure differences (e.g., wall thickness, lumen size, valves), 2 marks for comparison details. (b) Max 4 marks: 1 mark for wall is one cell thick; 1 mark for short diffusion distance; 1 mark for highly branched/large surface area; 1 mark for narrow lumen slowing blood flow. (c) Max 4 marks: 1 mark for separating oxygenated and deoxygenated blood; 1 mark for maintaining high pressure to body tissues; 1 mark for lower pressure to lungs protecting delicate lung tissue; 1 mark for faster/more efficient supply of oxygen/nutrients.
Question 4 · structured
11.4 marks
An experiment was set up to investigate osmosis in plant tissues. Cylinders of fresh potato of equal mass were placed in different concentrations of sucrose solution for 2 hours. (a) Define the term osmosis. [3] (b) Explain why potato cylinders placed in distilled water increased in mass. [5] (c) Describe the appearance of plant cells from potato cylinders placed in a very concentrated sucrose solution, and state the term used to describe this state. [4]
Show answer & marking scheme

Worked solution

(a) Osmosis is the net movement of water molecules from a region of higher water potential (dilute solution) to a region of lower water potential (concentrated solution) through a partially permeable membrane. (b) Distilled water has a higher water potential than the cytoplasm of the potato cells; water moves into the cells by osmosis; down a water potential gradient; across the partially permeable cell membrane; causing the cells to expand/become turgid and increasing overall mass. (c) The cells will appear wilted or flaccid under a microscope; the cytoplasm and cell membrane will shrink and pull away from the cell wall; the space between the membrane and cell wall will be filled with sucrose solution; this state is described as plasmolysed.

Marking scheme

(a) 3 marks: 1 mark for net movement of water molecules; 1 mark for higher to lower water potential; 1 mark for partially permeable membrane. (b) Max 5 marks: 1 mark for distilled water having higher water potential than cells; 1 mark for water moving in by osmosis; 1 mark for movement down water potential gradient; 1 mark for crossing partially permeable membrane; 1 mark for cells becoming turgid/increasing mass. (c) Max 4 marks: 1 mark for flaccid/membrane pulling away from cell wall; 1 mark for cytoplasm shrinking; 1 mark for term 'plasmolysed'; 1 mark for cell wall remaining intact/retaining outer shape.
Question 5 · structured
11.4 marks
Agricultural fertilizers can run off into nearby rivers and lakes, causing environmental problems. (a) Describe how an excess of nitrates in a water body leads to eutrophication. [6] (b) Explain how eutrophication results in the death of aquatic animals like fish. [4] (c) Suggest two methods farmers can use to reduce the risk of eutrophication. [2]
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Worked solution

(a) Nitrate runoff causes rapid growth of algae on the surface (algal bloom); this blocks sunlight from reaching aquatic plants at the bottom; these submerged plants cannot photosynthesize and die; bacteria/decomposers feed on the dead plant matter; bacteria multiply rapidly. (b) The large population of decomposers/bacteria respires aerobically; this uses up a high proportion of the dissolved oxygen in the water; the water becomes anoxic/oxygen-depleted; aquatic animals like fish cannot respire and die of suffocation. (c) Do not apply fertilizers close to water bodies; do not apply fertilizers when heavy rain is forecast; use organic manure instead of highly soluble chemical fertilizers.

Marking scheme

(a) Max 6 marks: 1 mark for algal bloom; 1 mark for blocking sunlight; 1 mark for death of bottom-dwelling plants; 1 mark for lack of photosynthesis; 1 mark for increase in decomposers/bacteria; 1 mark for bacteria feeding on dead organic matter. (b) Max 4 marks: 1 mark for aerobic respiration by bacteria; 1 mark for depletion of dissolved oxygen; 1 mark for anoxic/hypoxic water; 1 mark for fish dying due to lack of respiration. (c) Max 2 marks: 1 mark for each sensible method (e.g., avoiding rainy days, buffer zones near rivers, controlled dosage).
Question 6 · structured
11.4 marks
Transpiration is the loss of water vapour from plant leaves. (a) Explain how water moves up the xylem vessels in a continuous column. [4] (b) Describe and explain how an increase in ambient temperature and an increase in relative humidity affect the rate of transpiration. [6] (c) State the role of stomata in controlling water loss from a leaf. [2]
Show answer & marking scheme

Worked solution

(a) Water evaporates from mesophyll cells into air spaces and diffuses out of stomata, creating a transpiration pull; this pulls water upwards through the xylem; water molecules are held together by cohesive forces (hydrogen bonding); they adhere to the cellulose walls of the xylem vessels (adhesion), forming a continuous unbroken column. (b) Temperature: an increase in temperature increases kinetic energy of water molecules, increasing the rate of evaporation and diffusion, thus increasing transpiration rate. Humidity: an increase in relative humidity increases water vapour concentration in the air around the leaf, reducing the water potential/concentration gradient between the inside and outside of the leaf, thus decreasing transpiration rate. (c) Stomata are surrounded by guard cells; when water is scarce, guard cells lose turgidity and close the stomatal pore, preventing water vapour from escaping.

Marking scheme

(a) Max 4 marks: 1 mark for transpiration pull; 1 mark for evaporation from mesophyll cells; 1 mark for cohesion between water molecules; 1 mark for adhesion to xylem walls; 1 mark for continuous column. (b) Max 6 marks: 1 mark for temperature increase increases transpiration; 1 mark for more kinetic energy/faster evaporation; 1 mark for humidity increase decreases transpiration; 1 mark for reduced concentration gradient; 2 marks for detailed physical explanation of both gradients. (c) Max 2 marks: 1 mark for guard cells controlling opening/closing; 1 mark for closure when flaccid to prevent water loss.
Question 7 · structured
11.4 marks
The nervous system coordinates rapid responses to stimuli. (a) Describe the pathway of a spinal reflex arc when a person touches a hot object. [5] (b) Explain how the structure of a synapse ensures that nerve impulses travel in only one direction. [4] (c) Contrast the nervous system and the endocrine system in terms of speed of transmission and duration of effect. [3]
Show answer & marking scheme

Worked solution

(a) Stimulus (heat) is detected by temperature/pain receptors in the skin; an impulse is generated and travels along the sensory neurone; to the spinal cord (CNS); the impulse crosses a synapse to a relay neurone; the impulse crosses another synapse to a motor neurone; the motor neurone carries the impulse to the effector (biceps muscle); the muscle contracts, pulling the hand away. (b) Synapses have a presynaptic neurone and a postsynaptic neurone separated by a synaptic cleft; neurotransmitters are only stored in vesicles within the presynaptic bulb; they are released and diffuse across the cleft; receptors for the neurotransmitter are only present on the postsynaptic membrane, ensuring the impulse can only be initiated on that side. (c) Nervous system: speed of transmission is very rapid (electrical impulses along neurones), duration of effect is very short-lived (immediate response). Endocrine system: speed of transmission is slower (hormones transported in blood), duration of effect is longer-lasting (can persist for hours/days).

Marking scheme

(a) Max 5 marks: 1 mark for receptor detecting stimulus; 1 mark for impulse via sensory neurone; 1 mark for relay neurone in spinal cord/grey matter; 1 mark for motor neurone; 1 mark for effector/muscle contracting. (b) Max 4 marks: 1 mark for presynaptic vesicles containing neurotransmitters; 1 mark for release/diffusion across synaptic cleft; 1 mark for receptors only on postsynaptic membrane; 1 mark for neurotransmitters binding to trigger a new impulse. (c) Max 3 marks: 1 mark for nervous system being rapid and short-lived; 1 mark for endocrine system being slower and long-lasting; 1 mark for correct method of transmission (electrical impulses vs hormones in blood).

Paper 6 (Alternative to Practical)

Answer all questions on the question paper. You must use an HB pencil for any drawings or graphs. Show your calculations.
3 Question · 39.900000000000006 marks
Question 1 · practical-alternative
13.3 marks
A student investigated the effect of different concentrations of a natural plant extract, which acts as an amylase inhibitor, on the rate of starch digestion.

Amylase breaks down starch into reducing sugars.

The student used this method:
1. Label five test-tubes with the concentration of inhibitor solution: 0%, 1%, 2%, 3%, and 4%.
2. Add \(2\text{ cm}^3\) of amylase solution and \(1\text{ cm}^3\) of the respective inhibitor solution to each tube.
3. Add \(5\text{ cm}^3\) of starch suspension to the first tube and start a timer.
4. Every 30 seconds, use a clean pipette to remove a drop of the mixture and add it to a drop of iodine solution on a white spotting tile.
5. Record the time when the iodine solution no longer turns blue-black (remains yellow-brown).
6. Repeat the procedure for the remaining four inhibitor concentrations.

The results are shown in Table 1.1.

Table 1.1:
| Inhibitor concentration (%) | Time taken for starch to be completely digested / s | Rate of reaction / \(\text{s}^{-1}\) |
| :--- | :--- | :--- |
| 0 | 60 | 0.0167 |
| 1 | 90 | 0.0111 |
| 2 | 150 | 0.0067 |
| 3 | 300 | 0.0033 |
| 4 | No digestion after 600 s | 0.0000 |

**(a) (i)** Convert the time taken for starch to be completely digested at 2% inhibitor concentration into minutes and seconds. [1]

**(a) (ii)** Identify the independent variable and the dependent variable in this investigation. [2]

**(a) (iii)** State two variables that must be controlled in this investigation. [2]

**(a) (iv)** Explain why the mixture was tested with iodine solution at regular 30-second intervals. [1]

**(b) (i)** Describe how the student could test a sample of the mixture from the 0% inhibitor tube at the end of the experiment to show that starch had been broken down into reducing sugars. State the expected observations for a positive result. [4]

**(b) (ii)** State one potential hazard when carrying out the test for reducing sugars and describe a safety precaution to minimize this risk. [3.3]
Show answer & marking scheme

Worked solution

**(a) (i)**
- 150 seconds = 2 minutes and 30 seconds. (1 mark)

**(a) (ii)**
- Independent variable: Inhibitor concentration / % (1 mark)
- Dependent variable: Time taken for starch to be completely digested / s (or rate of reaction / \(\text{s}^{-1}\)) (1 mark)

**(a) (iii)**
- Any two from:
- Volume of amylase solution (\(2\text{ cm}^3\))
- Volume of starch suspension (\(5\text{ cm}^3\))
- Temperature of the reaction mixture
- Concentration of amylase solution
- Concentration of starch suspension
- pH of the solutions (2 marks)

**(a) (iv)**
- To monitor the progress of starch breakdown over time / to determine the exact point when all starch has been completely hydrolyzed. (1 mark)

**(b) (i)**
- Add Benedict's reagent/solution to the sample. (1 mark)
- Heat the mixture in a water bath / at a temperature above \(80\text{ }^\circ\text{C}\). (1 mark)
- For at least 2 to 5 minutes. (1 mark)
- Positive result: Color changes from blue to green / yellow / orange / brick-red. (1 mark)

**(b) (ii)**
- Hazard: Hot water / water bath / boiling water (could cause burns/scalds) OR corrosive/harmful chemicals (Benedict's solution). (1.3 marks)
- Safety Precaution: Use tongs or a test-tube holder to handle hot tubes / wear safety goggles / lab coat. (2 marks)

Marking scheme

**Part (a)(i)** [1 mark]
- 2 minutes 30 seconds / \(2\text{ min } 30\text{ s}\)

**Part (a)(ii)** [2 marks]
- Independent variable: Inhibitor concentration (accept: concentration of plant extract) [1]
- Dependent variable: Time taken for starch to be completely digested / rate of reaction [1]

**Part (a)(iii)** [2 marks]
- Any two from:
- Volume of amylase
- Volume of starch
- Volume of inhibitor
- Temperature
- pH
- Concentration of amylase / starch
- Reject: 'amount' of substance. Must specify 'volume' or 'concentration'.

**Part (a)(iv)** [1 mark]
- To determine when all starch is digested / to monitor the decrease in starch concentration over time / to find the endpoint of the reaction.

**Part (b)(i)** [4 marks]
- Add Benedict's solution / reagent [1]
- Heat / warm in a water bath (accept: use of water bath) [1]
- Acceptable temperature range: \(80\text{ }^\circ\text{C}\) to \(100\text{ }^\circ\text{C}\) / boiling [1]
- State positive color change: blue to green / yellow / orange / red / brick-red [1]

**Part (b)(ii)** [3.3 marks]
- Award 1.3 marks for identifying a valid hazard:
- Hot water/liquids (causes burns/scalds) OR Benedict's solution (irritant to skin/eyes) OR bunsen burner flame (causes burns/fire).
- Award 2 marks for a corresponding, valid safety precaution:
- Wear safety goggles / eye protection [1]
- Use test-tube holders/tongs when transferring hot tubes [1]
- Use a water bath instead of a direct flame / tie back long hair / use safety screen [1]
Question 2 · practical-alternative
13.3 marks
A student investigated water loss from leaves under identical ambient conditions over a 24-hour period. Four similar leaves, labeled A, B, C, and D, were taken from the same plant.

The leaves were treated as follows:
- Leaf A: Upper surface covered with a thin layer of petroleum jelly.
- Leaf B: Lower surface covered with a thin layer of petroleum jelly.
- Leaf C: Both surfaces covered with a thin layer of petroleum jelly.
- Leaf D: Untreated (control).

The mass of each leaf was measured at the start (initial mass) and after 24 hours (final mass).
The results are shown in Table 2.1.

Table 2.1:
| Leaf | Treatment | Initial mass / g | Final mass / g | Decrease in mass / g | Percentage decrease in mass (%) |
| :--- | :--- | :--- | :--- | :--- | :--- |
| A | Upper surface covered | 3.20 | 2.88 | 0.32 | 10.0 |
| B | Lower surface covered | 3.10 | 2.98 | 0.12 | **(a)(i)** |
| C | Both surfaces covered | 3.00 | 2.97 | 0.03 | 1.0 |
| D | Untreated (control) | 3.40 | 2.89 | 0.51 | 15.0 |

**(a) (i)** Calculate the percentage decrease in mass for Leaf B. Show your working. Give your answer to one decimal place. [2]

**(a) (ii)** Plot a bar chart on grid paper of the percentage decrease in mass for all four leaves (A, B, C, and D). [4]

**(a) (iii)** Using the results in Table 2.1, state what conclusion can be drawn about the distribution of stomata on the upper and lower surfaces of the leaves of this plant species. [2]

**(a) (iv)** Explain why petroleum jelly was applied to the surfaces of the leaves. [1]

**(b) (i)** Identify one potential source of error in this method that could affect the validity of the results, and suggest how to improve the investigation. [2.3]

**(b) (ii)** State one variable that should be controlled when selecting the leaves to ensure a fair comparison, and explain why this control is important. [2]
Show answer & marking scheme

Worked solution

**(a) (i)**
- Decrease in mass for Leaf B = \(3.10\text{ g} - 2.98\text{ g} = 0.12\text{ g}\)
- Percentage decrease = \(\frac{0.12}{3.10} \times 100 = 3.8709...\%\)
- Rounded to one decimal place = \(3.9\%\) (2 marks)

**(a) (ii)**
- Bar chart plotting points:
- Axis labels: y-axis = "Percentage decrease in mass (%)", x-axis = "Leaf" or "Treatment" (1 mark)
- Scale: linear, using at least half the grid, values labeled on y-axis (1 mark)
- Plotting: all four bars plotted accurately to within half a small square (Leaf A = 10.0, Leaf B = 3.9, Leaf C = 1.0, Leaf D = 15.0) (1 mark)
- Presentation: bars are of equal width, with spaces between them, drawn using a ruler (1 mark)

**(a) (iii)**
- There are more stomata on the lower surface than on the upper surface of the leaves of this species. (1 mark)
- Evidence: Leaf A (lower surface exposed) lost much more mass (10.0%) than Leaf B (upper surface exposed) which lost only 3.9%. (1 mark)

**(a) (iv)**
- To block the stomata / prevent water vapor from evaporating or escaping from the leaf surface. (1 mark)

**(b) (i)**
- Source of error: The surface area or size of the leaves was not exactly the same, which means they would naturally lose water at different rates regardless of the treatment. (1.3 marks)
- Improvement: Select leaves of identical surface area, measure and calculate leaf surface area to express water loss per unit area, or repeat the experiment with multiple leaves per treatment and calculate an average. (1 mark)

**(b) (ii)**
- Controlled variable: Species of plant / age of the leaves / size (surface area) of the leaves. (1 mark)
- Explanation: Different species or ages of leaves have different densities or numbers of stomata, which would affect the rate of transpiration independently of the treatments. (1 mark)

Marking scheme

**Part (a)(i)** [2 marks]
- Correct calculation of mass loss: \(3.10 - 2.98 = 0.12\text{ g}\) [1]
- Correct percentage calculation rounded to 1 decimal place: \(\frac{0.12}{3.10} \times 100 = 3.9\%\) [1]
- Accept: \(3.9\) with no working for [2] marks.

**Part (a)(ii)** [4 marks]
- **A** (Axes): y-axis correctly labeled with units ("Percentage decrease in mass (%)") and x-axis labeled with "Leaf" or "Treatment" [1]
- **S** (Scale): linear, suitable scale on y-axis (e.g., 2% per large grid division) so that the plotted data occupies more than half the grid height [1]
- **P** (Plotting): all 4 bars plotted accurately with neat horizontal tops [1]
- **D** (Design): bars drawn with a ruler, of equal width, and clearly separated by gaps [1]

**Part (a)(iii)** [2 marks]
- Stomata are located mostly / mainly on the lower surface [1]
- Because covering the lower surface (Leaf B) reduced water loss much more than covering the upper surface (Leaf A) / Leaf A lost significantly more mass than Leaf B [1]

**Part (a)(iv)** [1 mark]
- To block/seal stomata OR to act as a barrier to water loss/evaporation.

**Part (b)(i)** [2.3 marks]
- Award 1.3 marks for identifying a valid source of error:
- Leaves have different surface areas / sizes.
- Leaves may have different initial water content.
- Difficulty in applying exactly equal layers of petroleum jelly.
- Only one leaf used per treatment (lack of repeats).
- Award 1 mark for a corresponding valid improvement:
- Use leaves of the same surface area / measure surface area.
- Use the same mass/thickness of petroleum jelly.
- Repeat the experiment with three or more leaves for each treatment (to identify anomalies / calculate average).

**Part (b)(ii)** [2 marks]
- Variable: Leaf size / surface area / species / position on stem / age [1]
- Explanation: Different sizes/ages have different numbers of stomata/cuticle thickness, which affects the rate of water loss [1]
Question 3 · practical-alternative
13.3 marks
Figure 3.1 is a line drawing of a palisade mesophyll cell as observed under a light microscope.

[The diagram shows an elongated rectangular plant cell. It has a double-lined cell wall, a large central vacuole, a nucleus positioned against the cell wall, and several chloroplasts. The vertical length of the cell in the diagram is measured using a ruler as 65 mm.]

**(a) (i)** Make a large, clean, line drawing of the palisade mesophyll cell shown in Figure 3.1. Do not label your drawing. [4]

**(a) (ii)** State two features of your drawing that show it is a high-quality biological drawing. [2]

**(b) (i)** The measured length of the palisade cell in Figure 3.1 is 65 mm. The actual length of this palisade cell is 0.05 mm.

Calculate the magnification of the palisade cell in the drawing. Show your working. [2.3]

**(b) (ii)** A student measured another palisade cell from the same tissue sample. The length of this cell in the drawing was 78 mm.

Using the magnification calculated in **(b)(i)**, calculate the actual length of this cell. Give your answer in micrometres (\(\mu\text{m}\)). Show your working.

Note: \(1\text{ mm} = 1000\text{ }\mu\text{m}\). [3]

**(c)** State the name of a chemical stain that can be used to make cell structures, such as the nucleus and cell wall, more visible when preparing a temporary slide of plant tissue for viewing under a light microscope, and state the color it stains starch grains if present. [2]
Show answer & marking scheme

Worked solution

**(a) (i)**
- Drawing criteria:
- Size: drawing occupies at least half of the available space. (1 mark)
- Detail: shows cell wall as a double line, includes nucleus, central vacuole, and several distinct chloroplasts. (1 mark)
- Outline: clean, continuous lines drawn with a sharp pencil; no shading or feathered lines. (1 mark)
- Proportion: cell is drawn vertically elongated with correct relative proportions of nucleus and vacuole. (1 mark)

**(a) (ii)**
- Any two from:
- Drawn with single, clear, continuous lines (no sketchy lines).
- No shading or coloring used.
- Occupies more than half of the blank space.
- Correct proportion of organelles to cell size. (2 marks)

**(b) (i)**
- Formula: \(\text{Magnification} = \frac{\text{Size of image}}{\text{Actual size of object}}\)
- Calculation: \(\text{Magnification} = \frac{65\text{ mm}}{0.05\text{ mm}}\)
- Magnification = \(\times 1300\) (or \(1300\) times) (2.3 marks)

**(b) (ii)**
- \(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}}\)
- \(\text{Actual size in mm} = \frac{78\text{ mm}}{1300} = 0.06\text{ mm}\)
- Convert to micrometres: \(0.06 \times 1000 = 60\text{ }\mu\text{m}\)
- (Alternatively: \(78\text{ mm} = 78000\text{ }\mu\text{m}\). \(\text{Actual size} = \frac{78000}{1300} = 60\text{ }\mu\text{m}\)) (3 marks)

**(c)**
- Stain: Iodine solution (or methylene blue). (1 mark)
- Color of starch grains: Blue-black / dark blue / purple. (1 mark)

Marking scheme

**Part (a)(i)** [4 marks]
- **S** (Size): drawing occupies at least half of the provided space [1]
- **O** (Outline): single, clear, sharp lines without hatching, shading or overlapping lines [1]
- **D** (Detail): double line for cell wall, vacuole shown, nucleus present, and multiple chloroplasts clearly drawn as individual structures [1]
- **P** (Proportions): length-to-width ratio of the cell and relative sizes of organelles are representative of Figure 3.1 [1]

**Part (a)(ii)** [2 marks]
- Any two from:
- Drawn with sharp, continuous lines (no sketching) [1]
- No shading or stippling [1]
- Large drawing (occupies more than 50% of space) [1]
- No label lines (as requested) [1]

**Part (b)(i)** [2.3 marks]
- Award 1.3 marks for correct working / formula:
- \(\text{Magnification} = \frac{\text{measured length}}{\text{actual length}}\) OR \(\frac{65}{0.05}\)
- Award 1 mark for correct calculation:
- \(\times 1300\) or \(1300\)
- Reject: units on the final magnification value (e.g. \(1300\text{ mm}\) is incorrect, must be dimensionless or preceded/followed by \(\times\) or 'times').

**Part (b)(ii)** [3 marks]
- Correct conversion of mm to \(\mu\text{m}\) for drawing length (\(78\text{ mm} = 78000\text{ }\mu\text{m}\)) OR correct conversion of calculated actual length to \(\mu\text{m}\) (\(0.06\text{ mm} \times 1000\)) [1]
- Correct substitution into formula: \(\frac{78}{1300}\) or \(\frac{78000}{1300}\) [1]
- Correct final answer: \(60\) or \(60\text{ }\mu\text{m}\) [1]
- Award [3] marks for correct final answer with units regardless of working.

**Part (c)** [2 marks]
- Iodine solution / iodine in potassium iodide [1]
- Blue-black / dark blue / black [1]

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