An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V1) Cambridge IGCSE 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
An experiment is set up to measure the rate of diffusion of a colored dye into four agar blocks of different sizes: block P (\(1\text{ cm} \times 1\text{ cm} \times 1\text{ cm}\)), block Q (\(2\text{ cm} \times 2\text{ cm} \times 2\text{ cm}\)), block R (\(3\text{ cm} \times 3\text{ cm} \times 3\text{ cm}\)), and block S (\(4\text{ cm} \times 4\text{ cm} \times 4\text{ cm}\)). Which block will have the highest percentage of its total volume penetrated by the dye after 5 minutes?
A.block P
B.block Q
C.block R
D.block S
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Worked solution
Block P has the smallest dimensions, which gives it the largest surface area to volume ratio (6:1). A larger surface area to volume ratio allows a substance to diffuse and penetrate a greater percentage of the total volume in a given period of time compared to larger blocks with smaller surface area to volume ratios.
Marking scheme
1 mark for identifying that block P has the highest surface area to volume ratio, leading to the greatest percentage penetration of its volume.
Question 2 · multiple-choice
1 marks
A student monitors the rate of an enzyme-controlled reaction. At very high substrate concentrations, the rate of reaction remains constant and does not increase further. What is the explanation for this plateau?
A.All of the active sites on the enzyme molecules are fully occupied.
B.The enzyme molecules have been completely denatured.
C.The activation energy of the reaction has increased.
D.The substrate molecules have run out of kinetic energy.
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Worked solution
At extremely high substrate concentrations, all of the active sites of the enzyme molecules are fully occupied (saturated) with substrate molecules. Therefore, any additional substrate cannot be processed until an active site becomes free, and the rate of reaction reaches its maximum limit.
Marking scheme
1 mark for the correct explanation that all active sites are occupied at high substrate concentrations, making enzyme concentration the limiting factor.
Question 3 · multiple-choice
1 marks
The table shows some features of three types of blood vessels: X, Y, and Z.
| Feature | Vessel X | Vessel Y | Vessel Z | | :--- | :--- | :--- | :--- | | Wall thickness | Thick | Very thin (one cell thick) | Thin | | Lumen diameter | Narrow | Very narrow | Wide | | Presence of valves | Absent | Absent | Present |
Which row correctly identifies vessels X, Y, and Z?
A.X = artery, Y = capillary, Z = vein
B.X = artery, Y = vein, Z = capillary
C.X = vein, Y = capillary, Z = artery
D.X = capillary, Y = artery, Z = vein
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Worked solution
Arteries (X) have thick muscular walls and a narrow lumen to withstand high blood pressure. Capillaries (Y) have walls that are only one cell thick to allow rapid exchange of substances. Veins (Z) have thin walls, a wide lumen, and valves to prevent backflow of blood under low pressure.
Marking scheme
1 mark for correctly matching all three vessel types based on their structural features.
Question 4 · multiple-choice
1 marks
Which scenario describes how passive, artificial immunity is acquired?
A.An infant receiving antibodies naturally through breast milk.
B.A person receiving an injection of anti-venom antibodies after a snakebite.
C.A child producing antibodies in response to a weakened virus vaccine.
D.An adult developing antibodies after recovering from an active virus infection.
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Worked solution
Passive immunity involves receiving antibodies from an external source rather than producing them. Artificial passive immunity occurs when antibodies are medically administered (e.g., an injection of anti-venom containing pre-made antibodies to neutralize snake venom).
Marking scheme
1 mark for identifying the correct scenario that represents passive, artificial immunity.
Question 5 · multiple-choice
1 marks
Under which combination of environmental conditions will the rate of transpiration from a leafy shoot be the lowest?
A.high humidity, low temperature, low wind speed
B.low humidity, high temperature, high wind speed
C.high humidity, high temperature, low wind speed
D.low humidity, low temperature, high wind speed
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Worked solution
Transpiration is lowest when the concentration gradient of water vapor between the inside of the leaf and the outside air is minimal (high humidity), when the kinetic energy of water molecules is low (low temperature), and when water vapor is not blown away from the stomatal pores (low wind speed).
Marking scheme
1 mark for selecting the combination of high humidity, low temperature, and low wind speed.
Question 6 · multiple-choice
1 marks
A person looks up from reading a book to focus on a distant ship on the horizon. What changes occur in their eyes during this process of accommodation?
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Worked solution
When focusing on a distant object, the ciliary muscles relax, which pulls the suspensory ligaments tight. This tension pulls on the lens, causing it to become thinner (less convex) to focus the light from the distant object onto the retina.
Marking scheme
1 mark for identifying the correct state of ciliary muscles (relaxed), suspensory ligaments (tightened), and lens shape (thinner).
Question 7 · multiple-choice
1 marks
In genetic engineering, the human insulin gene is inserted into a bacterial plasmid. What are the specific functions of restriction enzymes and DNA ligase in this process?
A.Restriction enzymes cut DNA at specific sites to form sticky ends; DNA ligase joins the insulin gene to the plasmid.
B.Restriction enzymes join the insulin gene to the plasmid; DNA ligase cuts the bacterial chromosome.
C.Restriction enzymes replicate the human insulin gene; DNA ligase inserts the plasmid into the host bacterium.
D.Restriction enzymes isolate the plasmid; DNA ligase transcribes the insulin gene into mRNA.
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Worked solution
Restriction enzymes are used to cut the DNA at specific sequences, leaving complementary sticky ends. DNA ligase is then used to join the sticky ends of the human gene and the cut plasmid DNA together, sealing the sugar-phosphate backbone.
Marking scheme
1 mark for correctly matching the cutting role to restriction enzymes and the joining role to DNA ligase.
Question 8 · multiple-choice
1 marks
Untreated sewage is released into a river, triggering eutrophication. What is the correct sequence of events that leads directly to the death of fish in the river?
1. Decomposers (bacteria) multiply rapidly. 2. Dissolved oxygen levels in the water drop. 3. Decomposers respire aerobically. 4. Fish die from a lack of oxygen.
A.1 → 3 → 2 → 4
B.3 → 1 → 2 → 4
C.2 → 1 → 3 → 4
D.1 → 2 → 3 → 4
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Worked solution
The correct sequence of events is: 1 (sewage provides nutrients, so decomposers multiply rapidly) -> 3 (the large population of decomposers respires aerobically) -> 2 (aerobic respiration depletes the dissolved oxygen) -> 4 (fish suffocate and die due to lack of oxygen).
Marking scheme
1 mark for correctly ordering the sequence: bacterial growth -> aerobic respiration -> oxygen depletion -> fish mortality.
Question 9 · multiple-choice
1 marks
Which statement describes a feature of active transport but not diffusion?
A.It involves the net movement of particles down a concentration gradient.
B.It requires energy released from aerobic respiration.
C.It is a passive process that occurs randomly.
D.It only occurs across a fully permeable cellulose cell wall.
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Worked solution
Active transport is the movement of particles through a cell membrane from a region of lower concentration to a region of higher concentration (i.e. against a concentration gradient), which requires energy released from respiration. Diffusion is a passive process that does not require metabolic energy and occurs down a concentration gradient.
Marking scheme
1 mark for the correct choice (B).
Question 10 · multiple-choice
1 marks
What describes the effect of increasing temperature from \(20^\circ\text{C}\) to \(35^\circ\text{C}\) on an enzyme-controlled reaction?
A.The kinetic energy of molecules increases, leading to more frequent collisions.
B.The active site of the enzyme changes shape permanently.
C.The activation energy required for the reaction increases.
D.The enzymes are denatured, causing the reaction rate to fall to zero.
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Worked solution
As temperature increases up to the optimum, the kinetic energy of the enzyme and substrate molecules increases. This results in faster movement and a higher frequency of successful collisions, increasing the rate of reaction. Denaturation typically happens above the optimum temperature (usually above \(40^\circ\text{C}\) for human enzymes).
Marking scheme
1 mark for the correct choice (A).
Question 11 · multiple-choice
1 marks
Which chamber of the human heart generates the highest pressure to pump blood around the systemic circulation?
A.left atrium
B.left ventricle
C.right atrium
D.right ventricle
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Worked solution
The left ventricle has the thickest muscular wall of all the chambers because it must contract with enough force to generate high pressure, pumping oxygenated blood through the aorta to reach all the tissues of the body (systemic circulation).
Marking scheme
1 mark for the correct choice (B).
Question 12 · multiple-choice
1 marks
A plant is placed in a warm room with a strong breeze. What is the main reason for the subsequent increase in the rate of transpiration?
A.The humidity of the air surrounding the leaf increases.
B.The water molecules inside the leaf have less kinetic energy.
C.The wind removes water vapour from near the stomata, maintaining a steep concentration gradient.
D.The stomata close in response to the moving air.
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Worked solution
A strong breeze continuously blows away water vapour accumulating near the stomata on the leaf surface. This maintains a steep concentration gradient of water vapour between the inside of the leaf and the outside air, accelerating transpiration.
Marking scheme
1 mark for the correct choice (C).
Question 13 · multiple-choice
1 marks
A student measures a diagram of a mitochondrion. The length of the mitochondrion in the diagram is \(60\text{ mm}\). The actual length of the mitochondrion is \(2\ \mu\text{m}\). What is the magnification of the diagram?
A.\(\times 30\)
B.\(\times 300\)
C.\(\times 3000\)
D.\(\times 30,000\)
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Worked solution
First, convert all measurements to the same unit. \(60\text{ mm} = 60,000\ \mu\text{m}\). Using the formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\), we get \(\frac{60,000\ \mu\text{m}}{2\ \mu\text{m}} = 30,000\). Thus, the magnification is \(\times 30,000\).
Marking scheme
1 mark for the correct choice (D).
Question 14 · multiple-choice
1 marks
Which scenario describes an example of passive immunity?
A.A child receives a tuberculosis vaccine containing weakened bacteria.
B.A person produces antibodies after being infected with the influenza virus.
C.A baby receives ready-made antibodies from breast milk.
D.A teenager recovers from chickenpox and retains memory cells.
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Worked solution
Passive immunity is a short-term defense against a pathogen by antibodies acquired from another individual (such as a baby receiving antibodies from breast milk or across the placenta). It does not involve the recipient's immune system producing its own antibodies or developing memory cells.
Marking scheme
1 mark for the correct choice (C).
Question 15 · multiple-choice
1 marks
What is the correct sequence of structures through which a nerve impulse passes during a simple withdrawal reflex?
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Worked solution
The reflex arc begins with a receptor detecting a stimulus, which initiates an impulse in a sensory neurone. The impulse travels to a relay neurone in the central nervous system, which then transmits it to a motor neurone, leading to the effector (muscle or gland) to produce a response.
Marking scheme
1 mark for the correct choice (B).
Question 16 · multiple-choice
1 marks
What is the direct consequence of an overgrowth of algae (algal bloom) on the surface of a lake during eutrophication?
A.Increased penetration of sunlight to underwater plants.
B.Reduction of light reaching plants at the bottom of the lake.
C.A rapid increase in dissolved oxygen levels throughout the lake depths.
D.A decrease in the population of aerobic decomposers.
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Worked solution
An algal bloom blocks out light from reaching the plants at the bottom of the lake. These submerged plants are unable to photosynthesise without light, causing them to die.
Marking scheme
1 mark for the correct choice (B).
Question 17 · multiple-choice
1 marks
The optimal pH values of three different digestive enzymes (X, Y, and Z) found in the human alimentary canal are pH 2.0, pH 7.0, and pH 8.5 respectively. Which row correctly identifies the likely locations where these enzymes are active?
A.X = stomach; Y = mouth; Z = small intestine
B.X = mouth; Y = stomach; Z = small intestine
C.X = stomach; Y = small intestine; Z = mouth
D.X = small intestine; Y = mouth; Z = stomach
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Worked solution
Enzyme X functions best in highly acidic conditions (pH 2.0), which are found in the stomach due to hydrochloric acid. Enzyme Y functions best in neutral conditions (pH 7.0), typical of the mouth (saliva). Enzyme Z functions best in alkaline conditions (pH 8.5), typical of the small intestine (duodenum/ileum) due to bile and pancreatic juice.
Marking scheme
1 mark for the correct option A.
Question 18 · multiple-choice
1 marks
Which change would increase the rate of diffusion of oxygen from the alveoli into the blood capillaries in the lungs?
A.an increase in the thickness of the alveolus wall
B.an increase in the surface area of the alveoli
C.a decrease in the concentration gradient of oxygen
D.a decrease in the temperature of the gas in the lungs
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Worked solution
According to the principles of diffusion, an increase in surface area provides more space for gas molecules to pass through, thereby increasing the rate of diffusion. Increasing thickness, decreasing the concentration gradient, or lowering the temperature would all decrease the rate of diffusion.
Marking scheme
1 mark for the correct option B.
Question 19 · multiple-choice
1 marks
A student uses a microscope to observe a pollen grain. The image of the pollen grain has a length of 4.5 cm. The actual length of the pollen grain is 0.09 mm. What is the magnification of the image?
A.\(\times 0.05\)
B.\(\times 50\)
C.\(\times 500\)
D.\(\times 5000\)
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Worked solution
First, convert both measurements to the same unit (e.g., millimetres). Image length = 4.5 cm = 45 mm. Actual length = 0.09 mm. Magnification = Image length / Actual length = 45 / 0.09 = 500. Therefore, the magnification is \(\times 500\).
Marking scheme
1 mark for correct calculation leading to option C.
Question 20 · multiple-choice
1 marks
Why is a double circulatory system more advantageous to highly active mammals than a single circulatory system?
A.Blood pressure to the body tissues can be maintained at a higher level.
B.Oxygenated and deoxygenated blood are completely mixed in the heart.
C.The blood travels slower through the body tissues, allowing more gas exchange.
D.It requires less metabolic energy from the heart to pump blood through two circuits.
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Worked solution
In a double circulatory system, blood is pumped to the lungs at a lower pressure to prevent damage, returns to the heart to be repressurized, and is then pumped to the rest of the body at a much higher pressure. This maintains a rapid flow of oxygenated blood to active tissues, unlike a single circulatory system where pressure is lost in the gill capillaries before reaching the rest of the body.
Marking scheme
1 mark for selecting option A.
Question 21 · multiple-choice
1 marks
A person is bitten by a venomous snake and is immediately injected with an antivenom containing ready-made antibodies. Which type of immunity does this injection provide, and are memory cells produced in the recipient's body?
A.active immunity; memory cells are produced
B.active immunity; memory cells are not produced
C.passive immunity; memory cells are produced
D.passive immunity; memory cells are not produced
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Worked solution
Injecting ready-made antibodies provides passive immunity because the recipient's body did not produce the antibodies itself. Since the recipient's own lymphocytes are not activated to clonal selection and division, no memory cells are produced.
Marking scheme
1 mark for identifying passive immunity with no memory cell production (option D).
Question 22 · multiple-choice
1 marks
Under which set of environmental conditions will a well-watered plant have the lowest rate of transpiration?
A.high temperature, high humidity, high wind speed
B.low temperature, high humidity, low wind speed
C.high temperature, low humidity, low wind speed
D.low temperature, low humidity, high wind speed
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Worked solution
Transpiration rate decreases when: 1. Temperature is low (less kinetic energy for evaporation). 2. Humidity is high (reduces the water potential gradient between the inside of the leaf and the external air). 3. Wind speed is low (water vapour accumulates around the stomata, further reducing the water potential gradient). Therefore, option B results in the lowest rate of transpiration.
Marking scheme
1 mark for selecting option B.
Question 23 · multiple-choice
1 marks
A plant is grown under a constant light intensity at 0.04% carbon dioxide concentration. When the light intensity is high, the rate of photosynthesis remains constant. When the carbon dioxide concentration is increased to 0.15%, the rate of photosynthesis increases significantly. What was the limiting factor at 0.04% carbon dioxide when the light intensity was high?
A.carbon dioxide concentration
B.light intensity
C.temperature
D.water availability
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Worked solution
At high light intensity, the rate of photosynthesis is constant (plateaued) because some other factor is limiting. Since increasing the carbon dioxide concentration from 0.04% to 0.15% increases the rate, carbon dioxide concentration was the factor limiting the rate at 0.04%.
Marking scheme
1 mark for correct selection of option A.
Question 24 · multiple-choice
1 marks
Untreated sewage is accidentally discharged into a freshwater river. What are the immediate effects on the concentration of dissolved oxygen and the population of aerobic bacteria in the river?
A.dissolved oxygen decreases; aerobic bacteria population decreases
B.dissolved oxygen decreases; aerobic bacteria population increases
C.dissolved oxygen increases; aerobic bacteria population decreases
D.dissolved oxygen increases; aerobic bacteria population increases
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Worked solution
Sewage contains organic nutrients that act as food for decomposers. Consequently, the population of aerobic bacteria increases rapidly. As these bacteria multiply and respire aerobically, they consume large quantities of dissolved oxygen, causing its concentration in the river water to decrease.
Marking scheme
1 mark for selecting option B.
Question 25 · multiple-choice
1 marks
A dialysis tubing bag containing a 10% starch solution and a 5% glucose solution is placed in a beaker of distilled water. After 30 minutes, samples of water from the beaker are tested with iodine solution and heated Benedict's solution.
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Worked solution
Starch molecules are too large to pass through the pores of the semi-permeable dialysis tubing, so no starch enters the beaker. Thus, the iodine test remains brown. Glucose molecules are small enough to diffuse out into the beaker, resulting in a positive (orange-red) Benedict's test.
Marking scheme
1 mark for the correct option.
Question 26 · multiple-choice
1 marks
The rate of an enzyme-controlled reaction was measured at different pH values:
- pH 4.0: 10 arbitrary units - pH 5.0: 25 arbitrary units - pH 6.0: 48 arbitrary units - pH 7.0: 30 arbitrary units - pH 8.0: 5 arbitrary units
Which statement is correct?
A.The enzyme is completely denatured at pH 5.0.
B.The optimum pH of this enzyme is between pH 5.0 and pH 7.0.
C.The enzyme activity increases continuously as pH increases.
D.The enzyme works best in strongly alkaline conditions.
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Worked solution
The rate of reaction peaks at pH 6.0 (48 units), indicating that the optimum pH lies around 6.0, which is between pH 5.0 and pH 7.0.
Marking scheme
1 mark for the correct option.
Question 27 · multiple-choice
1 marks
Which chamber of the human heart has the thickest muscular wall, and into which major blood vessel does it pump blood?
A.Left ventricle, pumping into the pulmonary artery
B.Left ventricle, pumping into the aorta
C.Right ventricle, pumping into the pulmonary artery
D.Right ventricle, pumping into the aorta
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Worked solution
The left ventricle has the thickest muscular wall because it must generate enough pressure to pump blood to all parts of the body (systemic circulation) through the aorta.
Marking scheme
1 mark for the correct option.
Question 28 · multiple-choice
1 marks
What is the correct role of lymphocytes and phagocytes in defending the human body against pathogens?
A.Lymphocytes: perform phagocytosis; Phagocytes: produce antibodies
B.Lymphocytes: produce antibodies; Phagocytes: perform phagocytosis
C.Lymphocytes: produce fibrinogen; Phagocytes: assist in blood clotting
D.Lymphocytes: assist in blood clotting; Phagocytes: produce antibodies
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Worked solution
Lymphocytes produce specific proteins called antibodies that bind to antigens on pathogens. Phagocytes engulf and digest pathogens through phagocytosis.
Marking scheme
1 mark for the correct option.
Question 29 · multiple-choice
1 marks
Which changes occur in the skin to help maintain body temperature when the external temperature falls below normal?
A.Vasodilation of skin arterioles and shivering
B.Vasoconstriction of skin arterioles and sweating
C.Vasodilation of skin arterioles and sweating
D.Vasoconstriction of skin arterioles and shivering
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Worked solution
When body temperature falls, vasoconstriction of skin arterioles reduces blood flow to skin capillaries to minimise heat loss by radiation. Shivering involves rapid muscle contractions to generate metabolic heat.
Marking scheme
1 mark for the correct option.
Question 30 · multiple-choice
1 marks
An excess of nitrate ions from agricultural fertilisers enters a freshwater lake. Which of the following shows the correct sequence of events leading to eutrophication?
A.increased growth of algae → decomposition of dead algae by bacteria → decrease in dissolved oxygen → death of fish
B.decrease in dissolved oxygen → death of fish → increased growth of algae → decomposition by bacteria
C.decomposition by bacteria → increased growth of algae → decrease in dissolved oxygen → death of fish
D.death of fish → decrease in dissolved oxygen → decomposition by bacteria → increased growth of algae
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Worked solution
Nitrate runoff causes rapid growth of algae (algal bloom). When the algae die, bacterial decomposers multiply rapidly, using up dissolved oxygen for aerobic respiration. This lack of oxygen leads to the suffocation and death of fish.
Marking scheme
1 mark for the correct option.
Question 31 · multiple-choice
1 marks
Under which combination of environmental conditions will a well-watered plant transpire most rapidly?
A.High humidity, high temperature, high wind speed
B.Low humidity, low temperature, low wind speed
C.Low humidity, high temperature, high wind speed
D.High humidity, low temperature, low wind speed
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Worked solution
Transpiration occurs most rapidly when the water vapour concentration gradient between the leaf interior and the air is greatest. This gradient is maximised by low humidity (dry air), high temperature (increases rate of evaporation), and high wind speed (removes water vapour from the leaf surface).
Marking scheme
1 mark for the correct option.
Question 32 · multiple-choice
1 marks
A micrograph of a leaf cell shows a chloroplast with a measured length of 24 mm. If the actual length of this chloroplast is 8 µm, what is the magnification of the image?
A.×3
B.×300
C.×3000
D.×30000
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Worked solution
Using the formula: \( \text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} \). First convert units to be the same: \( 24\text{ mm} = 24,000\text{ µm} \). Therefore, \( \text{Magnification} = \frac{24,000\text{ µm}}{8\text{ µm}} = \times 3000 \).
Marking scheme
1 mark for the correct option.
Question 33 · multiple-choice
1 marks
The initial mass of four potato pieces was 5.0 g. Each piece was placed in a different sucrose solution. After two hours, the final masses were measured. Which solution has the same water potential as the potato cells?
A.Solution A (final mass 5.0 g)
B.Solution B (final mass 5.8 g)
C.Solution C (final mass 4.2 g)
D.Solution D (final mass 3.5 g)
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Worked solution
When a plant cell is placed in an isotonic solution (a solution with the same water potential as the cell cytoplasm), there is no net movement of water by osmosis. Consequently, the mass of the tissue remains unchanged. In Solution A, the mass of the potato piece remained at 5.0 g, indicating no net water movement and therefore equal water potential.
Marking scheme
1 mark for selecting the correct solution (A) where no change in mass occurs.
Question 34 · multiple-choice
1 marks
Which statement correctly describes what happens to salivary amylase when it is placed in a highly acidic environment of pH 2.0?
A.The active site changes shape permanently, preventing substrate binding.
B.The kinetic energy of the amylase molecules decreases, reducing the rate of collisions.
C.The starch substrate molecules are denatured, preventing them from fitting into the active site.
D.The activation energy required for the starch breakdown reaction increases.
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Worked solution
Extremely low pH levels disrupt the ionic and hydrogen bonds that maintain the specific three-dimensional shape of the amylase protein. This causes the enzyme to denature, which permanently alters the shape of its active site so that starch molecules can no longer bind to it.
Marking scheme
1 mark for identifying that the active site permanently changes shape due to denaturation at extreme pH.
Question 35 · multiple-choice
1 marks
Which sequence correctly identifies the pathway of a red blood cell returning from the kidneys as it passes through the heart to the lungs?
A.renal vein \(\rightarrow\) vena cava \(\rightarrow\) right atrium \(\rightarrow\) right ventricle \(\rightarrow\) pulmonary artery
B.renal artery \(\rightarrow\) vena cava \(\rightarrow\) left atrium \(\rightarrow\) left ventricle \(\rightarrow\) pulmonary vein
C.renal vein \(\rightarrow\) aorta \(\rightarrow\) right atrium \(\rightarrow\) right ventricle \(\rightarrow\) pulmonary vein
D.renal artery \(\rightarrow\) aorta \(\rightarrow\) left atrium \(\rightarrow\) left ventricle \(\rightarrow\) pulmonary artery
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Worked solution
Deoxygenated blood from the kidneys leaves via the renal vein, enters the vena cava, and flows into the right side of the heart (right atrium followed by right ventricle). From the right ventricle, it is pumped to the lungs via the pulmonary artery.
Marking scheme
1 mark for identifying the correct vascular and cardiac sequence from kidneys to lungs.
Question 36 · multiple-choice
1 marks
Which statement correctly describes a type of immunity, how it is acquired, and whether memory cells are produced?
A.Active immunity: acquired by injection of antibodies; memory cells are produced.
B.Active immunity: acquired by vaccination with a weakened pathogen; memory cells are produced.
C.Passive immunity: acquired by infection with a live pathogen; memory cells are not produced.
D.Passive immunity: acquired by vaccination with a weakened pathogen; memory cells are not produced.
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Worked solution
Active immunity is stimulated by exposing the body to antigens, which can occur via a vaccine containing a weakened pathogen. This stimulates the production of antibodies and memory cells, providing long-term immunity. Passive immunity involves the transfer of pre-made antibodies (e.g., across the placenta or via injection), which does not produce memory cells.
Marking scheme
1 mark for identifying active immunity via vaccination producing memory cells.
Question 37 · multiple-choice
1 marks
Under which set of environmental conditions would a plant show the lowest rate of transpiration?
A.High humidity, low temperature, and low wind speed
B.High humidity, high temperature, and high wind speed
C.Low humidity, low temperature, and high wind speed
D.Low humidity, high temperature, and low wind speed
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Worked solution
Transpiration is the loss of water vapour from plant leaves. A high relative humidity reduces the water potential gradient between the inside of the leaf and the outside air. Low temperatures reduce the kinetic energy of water molecules, slowing evaporation. Low wind speed allows water vapour to accumulate near the leaf surface, further reducing the concentration gradient.
Marking scheme
1 mark for selecting the combination of high humidity, low temperature, and low wind speed.
Question 38 · multiple-choice
1 marks
What is the correct sequence of events that occurs in a river immediately after untreated sewage is discharged into it?
A.bacterial population increases \(\rightarrow\) dissolved oxygen decreases \(\rightarrow\) fish die
B.algae population decreases \(\rightarrow\) dissolved oxygen increases \(\rightarrow\) fish die
C.bacterial population decreases \(\rightarrow\) dissolved oxygen decreases \(\rightarrow\) fish survive
D.algae population increases \(\rightarrow\) dissolved oxygen increases \(\rightarrow\) fish survive
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Worked solution
Untreated sewage contains organic waste which acts as a food source for decomposers. The bacterial population increases rapidly and respires aerobically, consuming dissolved oxygen. This rapid reduction in dissolved oxygen levels leads to the suffocation and death of fish.
Marking scheme
1 mark for the correct sequence of bacterial growth, oxygen depletion, and fish mortality.
Question 39 · multiple-choice
1 marks
The rate of photosynthesis of a plant was measured at different light intensities, at a constant temperature of 20 °C and a constant carbon dioxide concentration of 0.04%. At high light intensities, the rate of photosynthesis became constant. What could be done to increase the rate of photosynthesis at these high light intensities?
A.Increase the carbon dioxide concentration.
B.Decrease the light intensity.
C.Decrease the temperature to 10 °C.
D.Remove oxygen from the surrounding air.
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Worked solution
At high light intensities, light is no longer the limiting factor. Since the atmospheric concentration of carbon dioxide is low (0.04%), increasing the carbon dioxide concentration will increase the rate of photosynthesis as carbon dioxide becomes the new limiting factor.
Marking scheme
1 mark for identifying carbon dioxide concentration as the limiting factor at high light intensity.
Question 40 · multiple-choice
1 marks
An image of a plant cell nucleus has a diameter of 24 mm. If the magnification of the image is \(\times 3000\), what is the actual diameter of the nucleus?
A.0.008 \(\mu\)m
B.8 \(\mu\)m
C.0.08 mm
D.72 \(\mu\)m
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Worked solution
Using the formula Actual Size (A) = Image Size (I) / Magnification (M):
Answer all questions. Show your working in the spaces provided. Write in dark blue or black pen.
7 Question · 80 marks
Question 1 · structured
11 marks
Pancreatic secretory cells have a very high density of rough endoplasmic reticulum (rER) and mitochondria. (a) State the function of the rough endoplasmic reticulum and explain why secretory cells require a high number of mitochondria. (b) A student measures the diameter of a secretory vesicle in an electron micrograph of this cell. The image size of the vesicle is 18 mm. The actual size of the vesicle is 0.6 micrometres. Calculate the magnification of this image. Show your working and state the formula used. (c) Secretory vesicles fuse with the cell membrane to release proteins. Describe how the diffusion of small molecules across a membrane differs from active transport in terms of energy requirements and concentration gradients. (d) State the name of the cell structure that controls the entry and exit of substances in all cells.
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Worked solution
(a) The rough endoplasmic reticulum is responsible for protein synthesis due to the presence of ribosomes on its surface. Mitochondria are the sites of aerobic respiration, which releases energy in the form of ATP. Secretory cells require a continuous supply of energy to transport vesicles through the cytoplasm and to fuse them with the cell membrane during exocytosis. (b) Magnification = Image size / Actual size. Convert 18 mm to micrometres: 18 * 1000 = 18,000 micrometres. Magnification = 18,000 / 0.6 = x30,000. (c) Diffusion is a passive process that does not require metabolic energy (ATP) from respiration, whereas active transport is an active process that requires ATP. Diffusion involves the net movement of particles down a concentration gradient (from high to low concentration), while active transport moves particles against a concentration gradient (from low to high concentration). (d) The cell membrane.
Marking scheme
Part (a): [3 marks] rER synthesizes proteins / has ribosomes (1 mark); mitochondria release energy / ATP through respiration (1 mark); energy is needed for active processes such as vesicle transport / exocytosis (1 mark). Part (b): [3 marks] Magnification = Image size / Actual size (1 mark); correct conversion of mm to micrometres (18 mm = 18,000 micrometres) (1 mark); correct calculation of magnification as x30,000 (1 mark). Part (c): [3 marks] Diffusion is passive / does not require energy AND active transport requires energy / ATP (1 mark); diffusion is down a concentration gradient AND active transport is against / up a concentration gradient (1 mark); active transport requires protein carriers (1 mark). Part (d): [2 marks] Cell membrane / plasma membrane (2 marks).
Question 2 · structured
12 marks
(a) Define the term enzyme. (b) Table 2.1 shows the volume of oxygen gas produced in 2 minutes when potato catalase was mixed with different concentrations of hydrogen peroxide substrate. [1% substrate: 4.2 cm3; 2% substrate: 8.5 cm3; 3% substrate: 12.0 cm3; 4% substrate: 14.5 cm3; 5% substrate: 15.0 cm3; 6% substrate: 15.0 cm3]. Describe and explain the trend shown in Table 2.1. (c) Explain why the rate of reaction levels off at substrate concentrations of 5% and above. (d) Suggest how the temperature of the reaction mixture was kept constant during this investigation. (e) State the type of biological molecule that enzymes are made of.
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Worked solution
(a) An enzyme is a protein that functions as a biological catalyst, speeding up metabolic reactions while remaining chemically unchanged at the end of the process. (b) As the substrate concentration increases from 1% to 5%, the rate of reaction increases, represented by the rise in oxygen volume from 4.2 cm3 to 15.0 cm3. This occurs because more substrate molecules are available, increasing the frequency of successful collisions with active sites and forming more enzyme-substrate complexes. At concentrations above 5%, the volume of oxygen remains constant at 15.0 cm3. (c) The reaction rate levels off because the enzyme's active sites are fully saturated / occupied at any given moment. Adding more substrate does not increase the rate because the enzymes are already working at their maximum capacity. The concentration of the enzyme has become the limiting factor. (d) The temperature can be kept constant by placing the reaction test tubes inside a thermostatically-controlled water bath. (e) Enzymes are proteins, which are made up of long chains of amino acids.
Marking scheme
Part (a): [2 marks] Biological catalyst / speeds up chemical reactions (1 mark); remains chemically unchanged at the end of the reaction / is not used up (1 mark). Part (b): [4 marks] Increasing substrate concentration from 1% to 5% increases volume of oxygen produced (1 mark); more substrate molecules collide with enzyme active sites (1 mark); more enzyme-substrate complexes are formed per unit time (1 mark); volume of oxygen levels off / remains constant at 5% and 6% (1 mark). Part (c): [3 marks] Active sites are fully saturated / occupied (1 mark); substrate is in excess / all enzymes are working at maximum rate (1 mark); enzyme concentration is the limiting factor (1 mark). Part (d): [1 mark] Thermostatically-controlled water bath (1 mark). Part (e): [2 marks] Protein (2 marks).
Question 3 · structured
11 marks
(a) Mammals possess a double circulatory system, whereas fish possess a single circulatory system. State two advantages of a double circulatory system compared to a single circulatory system. (b) Describe the role of the coronary arteries in the mammalian heart. (c) Explain how the structure of a vein is adapted to its function of returning blood to the heart at low pressure.
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Worked solution
(a) A double circulatory system maintains higher blood pressure to body tissues. This allows for faster, more efficient delivery of oxygen and glucose, supporting a high metabolic rate. It also separates oxygenated and deoxygenated blood. (b) Coronary arteries branch off the aorta and run across the surface of the heart. They supply the cardiac muscle tissue of the heart wall with oxygenated blood, containing oxygen and glucose. These substances are essential for continuous aerobic respiration, releasing the energy required for constant heart contraction. (c) Veins carry blood at low pressure back to the heart. They have thin walls containing fewer muscle and elastic fibers than arteries because the pressure is low. They have a wide lumen, which reduces resistance to blood flow. Most importantly, veins contain valves at intervals to prevent the backflow of blood, ensuring it flows in one direction back towards the heart.
Marking scheme
Part (a): [4 marks] Maintains high blood pressure to body tissues (1 mark); allows faster / more efficient transport of oxygen / glucose / nutrients (1 mark); keeps oxygenated and deoxygenated blood separate (1 mark); supports a higher metabolic rate / homeothermic body temperature regulation (1 mark). Part (b): [3 marks] Supplies oxygenated blood / oxygen / glucose to cardiac muscle cells (1 mark); for aerobic respiration (1 mark); to provide energy / ATP for heart muscle contraction (1 mark). Part (c): [4 marks] Thin walls due to low blood pressure (1 mark); wide lumen to minimize resistance to blood flow (1 mark); valves present to prevent the backflow of blood (1 mark); skeletal muscles squeeze veins to help return blood (1 mark).
Question 4 · structured
12 marks
(a) Distinguish between active immunity and passive immunity. (b) Explain the difference between the primary immune response (upon first exposure to a pathogen) and the secondary immune response (upon second exposure to the same pathogen) with reference to the concentration of antibodies and the role of memory cells. (c) Explain how a vaccine stimulates active immunity without causing the disease. (d) State two chemical barriers the human body uses to defend itself against pathogens.
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Worked solution
(a) Active immunity is the defense against a pathogen by antibody production in the body, which is long-term and produces memory cells. Passive immunity is short-term, temporary immunity acquired by receiving antibodies from another individual (e.g., across the placenta or via breast milk), which does not produce memory cells. (b) During the primary response, antibody production is slow to rise and reaches a relatively low peak because the body must first identify the foreign antigens. In the secondary response, antibody production is extremely rapid and reaches a much higher peak concentration. This is because memory cells from the first exposure recognize the antigen immediately, clonal selection occurs rapidly, and large populations of plasma cells are quickly formed to secrete antibodies. (c) Vaccines contain dead, weakened, or fragments of pathogens containing the specific antigen. When injected, the immune system recognizes these antigens as foreign. Lymphocytes are stimulated to undergo clonal expansion, producing specific antibodies and memory cells. Because the pathogen is altered/harmless, no disease symptoms occur. (d) Stomach acid (hydrochloric acid) destroys ingested pathogens, and tears/saliva contain the enzyme lysozyme which destroys bacterial cell walls.
Marking scheme
Part (a): [3 marks] Active immunity involves antibody production by the host (1 mark); active immunity is long-term whereas passive is short-term (1 mark); active immunity produces memory cells whereas passive does not (1 mark). Part (b): [4 marks] Secondary response has a faster rate of antibody production (1 mark); secondary response reaches a higher peak concentration of antibodies (1 mark); primary response has no pre-existing memory cells (1 mark); secondary response is triggered by the rapid division/activation of memory cells (1 mark). Part (c): [3 marks] Vaccine contains dead / weakened / harmless pathogen / antigen (1 mark); stimulates specific lymphocytes to produce antibodies (1 mark); produces memory cells for long-term protection (1 mark). Part (d): [2 marks] Stomach acid / hydrochloric acid (1 mark); tears / saliva / mucus containing lysozyme (1 mark).
Question 5 · structured
11 marks
(a) Describe how the human body detects and responds to a decrease in external environmental temperature to maintain a constant core body temperature. (b) State the term used to describe the maintenance of a constant internal environment. (c) Explain the role of insulin in regulating blood glucose levels when they are too high.
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Worked solution
(a) Thermoreceptors in the skin and hypothalamus detect a decrease in temperature. The hypothalamus sends nerve impulses to effectors. Vasoconstriction occurs, where arterioles supplying skin surface capillaries constrict/narrow, reducing blood flow to the skin and minimizing heat loss by radiation. Shivering is triggered, causing rapid involuntary muscle contractions that release heat energy through aerobic respiration. Hair erector muscles contract, raising hairs to trap a layer of warm air. (b) Homeostasis. (c) When blood glucose levels rise, the pancreas detects the change and secretes insulin into the bloodstream. Insulin travels to target organs, primarily the liver and muscle cells. It stimulates these cells to absorb glucose from the blood and convert it into the insoluble storage carbohydrate glycogen. This reduces the concentration of glucose in the blood back to normal levels.
Marking scheme
Part (a): [5 marks] Thermoreceptors detect temperature drop and send nerve impulses to the brain / hypothalamus (1 mark); vasoconstriction occurs: arterioles supplying skin capillaries constrict / narrow (1 mark); less blood flows through skin surface capillaries to reduce heat loss (1 mark); shivering occurs: involuntary muscle contractions release heat through respiration (1 mark); hair erector muscles contract to trap a layer of insulating air (1 mark). Part (b): [1 mark] Homeostasis (1 mark). Part (c): [5 marks] Pancreas detects high blood glucose levels and secretes insulin (1 mark); insulin is transported in the blood (1 mark); insulin binds to target cells in the liver and muscles (1 mark); stimulates increased glucose uptake by cells (1 mark); stimulates conversion of glucose to insoluble glycogen (1 mark).
Question 6 · structured
12 marks
(a) Outline how sulfur dioxide pollution leads to the formation of acid rain and describe its damaging effects on plants. (b) Explain how a high humidity environment affects the rate of transpiration in plants. (c) State two structural features of leaves that reduce the rate of transpiration in xerophytic plants and explain how each feature functions.
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Worked solution
(a) Sulfur dioxide gas is released from burning fossil fuels containing sulfur impurities. It rises into the atmosphere and dissolves in water droplets in clouds, forming sulfuric acid, which falls as acid rain. Acid rain directly damages leaves, killing cells and stripping waxy cuticles. It also makes the soil acidic, which leaches vital mineral ions (like magnesium, causing chlorosis) away from plant roots. (b) High humidity means there is a high concentration of water vapor in the air surrounding the leaf. This reduces the water potential gradient between the wet cell walls inside the leaf's mesophyll and the air outside. As a result, the rate of diffusion of water vapor out of the stomata decreases, thus lowering the overall rate of transpiration. (c) Feature 1: Thick waxy cuticle. Function: Provides a barrier that significantly reduces non-stomatal evaporation of water from the upper epidermal cells. Feature 2: Sunken stomata / stomata in pits. Function: Traps a pocket of moist air close to the stomatal pore, reducing the water potential gradient between the inside of the leaf and the immediate outside environment.
Marking scheme
Part (a): [4 marks] Sulfur dioxide is released from fossil fuel combustion and dissolves in cloud water vapor to form sulfuric acid / acid rain (1 mark); damages leaf cells / waxy cuticles directly (1 mark); makes soil acidic (1 mark); leaches essential mineral ions like magnesium / releases toxic aluminum ions (1 mark). Part (b): [4 marks] High humidity means high water vapor concentration outside the leaf (1 mark); decreases the water potential gradient between inside and outside of the leaf (1 mark); reduces the rate of diffusion of water vapor (1 mark); through the stomata, thus lowering the transpiration rate (1 mark). Part (c): [4 marks] Feature 1: Thick waxy cuticle (1 mark); Function 1: Reduces non-stomatal water loss / evaporation from epidermis (1 mark). Feature 2: Sunken stomata / stomata in pits (1 mark); Function 2: Traps moist air to reduce water potential gradient (1 mark).
Question 7 · structured
11 marks
(a) Write the balanced chemical equation for photosynthesis. (b) Describe how carbon dioxide concentration acts as a limiting factor on the rate of photosynthesis. (c) Explain why bacteria are useful organisms for genetic modification to produce human proteins, such as insulin.
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Worked solution
(a) The balanced chemical equation for photosynthesis is: 6CO2 + 6H2O -> C6H12O6 + 6O2. (b) Carbon dioxide is a raw material for photosynthesis. At low carbon dioxide concentrations, the rate of photosynthesis is low because there are fewer carbon atoms available to be fixed into glucose. As the carbon dioxide concentration increases, the rate of photosynthesis increases proportionally because carbon dioxide is no longer limiting. Eventually, the rate of photosynthesis plateaus because another factor, such as light intensity or temperature, becomes the limiting factor. (c) Bacteria are highly suited for genetic engineering because: 1. They have a rapid reproduction rate, allowing massive populations to be grown quickly in fermenters. 2. They contain small circular DNA molecules called plasmids, which can easily be extracted, cut with restriction enzymes, and have foreign genes inserted. 3. They share the same universal genetic code as humans, allowing them to translate the human insulin gene into the correct protein. 4. Using bacteria avoids the ethical objections associated with using human tissue or animals to extract proteins.
Marking scheme
Part (a): [3 marks] Correct chemical formulas for reactants (CO2 + H2O) and products (C6H12O6 + O2) (1 mark); correct balancing (6, 6 -> 1, 6) (1 mark); light and chlorophyll indicated on / above arrow (1 mark). Part (b): [3 marks] Carbon dioxide is a reactant / raw material (1 mark); at low concentration, rate is low because CO2 is the limiting factor (1 mark); increasing concentration increases rate of reaction until another factor (e.g. light / temperature) becomes limiting / rate plateaus (1 mark). Part (c): [5 marks] Rapid reproduction / growth rate (1 mark); contain plasmids which are easy to manipulate / cut / insert genes (1 mark); share same universal genetic code as humans (1 mark); can produce complex human proteins / no ethical concerns compared to using animals (1 mark); cheap / easy to culture on a large scale in fermenters (1 mark).
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 · 40 marks
Question 1 · practical-alternative
13 marks
A student investigated the effect of temperature on the rate of anaerobic respiration in yeast.
The student used the following method: 1. Yeast suspension was mixed with glucose solution in a test-tube. 2. A delivery tube was attached from the test-tube containing yeast into another test-tube containing limewater. 3. The test-tube containing the yeast mixture was placed in a water-bath set at a specific temperature. 4. After 5 minutes, the number of gas bubbles escaping from the delivery tube into the limewater was counted for 3 minutes. 5. The procedure was repeated at five different temperatures: 20 °C, 30 °C, 40 °C, 50 °C and 60 °C.
(a) (i) Identify the gas produced by the yeast during this investigation and state the positive test result for this gas in limewater. (ii) State the independent variable and the dependent variable in this investigation. (iii) The student wrote down the results of the bubble counts in their notebook: - At 20 °C: 9 bubbles in 3 minutes - At 30 °C: 24 bubbles in 3 minutes - At 40 °C: 42 bubbles in 3 minutes - At 50 °C: 15 bubbles in 3 minutes - At 60 °C: 0 bubbles in 3 minutes
Prepare a table to record these results. Your table should show: - the temperature - the total number of bubbles counted in 3 minutes - the rate of gas production expressed as the number of bubbles per minute (to one decimal place).
(iv) Draw a conclusion from these results.
(b) Explain why a layer of liquid paraffin (oil) is often placed on top of the yeast and glucose mixture in anaerobic respiration experiments.
(c) Suggest why the yeast and glucose mixture was left in the water-bath for 5 minutes before counting the bubbles.
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Worked solution
(a) (i) Gas: Carbon dioxide / CO2. Positive test: Limewater turns cloudy / milky / chalky. (ii) Independent variable: Temperature. Dependent variable: Rate of gas production / rate of anaerobic respiration / number of bubbles. (iii) A suitable table of results: | Temperature / °C | Total bubbles in 3 minutes | Rate of bubble production / bubbles per minute | | 20 | 9 | 3.0 | | 30 | 24 | 8.0 | | 40 | 42 | 14.0 | | 50 | 15 | 5.0 | | 60 | 0 | 0.0 | (iv) Conclusion: As the temperature increases from 20 °C to 40 °C, the rate of anaerobic respiration in yeast increases. Above 40 °C, the rate decreases and stops completely at 60 °C. (b) Liquid paraffin forms a barrier that prevents oxygen from diffusing into the yeast suspension, ensuring anaerobic conditions are maintained. (c) To allow the yeast suspension to equilibrate and reach the exact temperature of the water-bath before bubble counting begins, ensuring temperature is kept constant and controlled.
Marking scheme
(a) (i) [2 marks] - Carbon dioxide (1) - Cloudy / milky (1) (a) (ii) [2 marks] - Independent: Temperature (1) - Dependent: Rate of bubble production / gas production / number of bubbles (1) (a) (iii) [4 marks] - Table drawn with clear lines, minimum two columns, and header line (1) - Correct column/row headings with units (e.g. Temperature / °C, bubbles per minute) (1) - All raw data from notebook correctly entered (1) - All calculations of bubbles per minute correct to 1 decimal place (1) (a) (iv) [2 marks] - As temperature increases from 20 °C to 40 °C, the rate of respiration increases (1) - Above 40 °C (at 50 °C and 60 °C), the rate decreases / stops (due to denaturation of enzymes) (1) (b) [1 mark] - To prevent oxygen entering the mixture / to maintain anaerobic conditions (1) (c) [2 marks] - To allow the yeast suspension to reach the temperature of the water-bath / to equilibrate (1) - To ensure that the rate measured is at that specific temperature (1)
Question 2 · practical-alternative
13 marks
(a) Plan an investigation to determine the effect of different sodium chloride (salt) concentrations on the mass of potato cylinders.
(b) A student performed a similar osmosis experiment using potato cylinders. The cylinders had an initial mass of 4.00 g. After immersion in a highly concentrated salt solution for 24 hours, the final mass of one cylinder was found to be 3.12 g. (i) Calculate the percentage change in the mass of the potato cylinder. Show your working and include whether it is an increase or a decrease. (ii) State the term used to describe cells that have lost water and their cell membrane has pulled away from the cell wall. (iii) Identify one potential source of error when preparing the potato cylinders for weighing at the end of the experiment, and suggest how to minimize this error. (iv) State why calculating the percentage change in mass is better than calculating the absolute change in mass.
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Worked solution
(a) Plan: - Use at least five different concentrations of sodium chloride solution (e.g., 0.0, 0.2, 0.4, 0.6, 0.8 mol/dm³). - Cut identical potato cylinders using a cork borer and scalpel/ruler to equal length. - Measure the initial mass of each potato cylinder using an electronic balance. - Place one potato cylinder into each of the salt solutions and leave for a set period of time (e.g., 2 hours). - Remove the cylinders, gently blot them with a paper towel to remove excess surface water, and re-weigh to find the final mass. - Keep temperature constant (using a water bath or thermostatically controlled room) and keep the volume of each solution constant. - Repeat the experiment at least three times at each concentration to identify anomalies and calculate a mean. - Maintain safety by taking care when cutting with the cork borer/scalpel. (b) (i) Change in mass = 3.12 - 4.00 = -0.88 g. Percentage change = (-0.88 / 4.00) * 100 = -22%. This is a 22% decrease. (ii) Plasmolysed / plasmolysis. (iii) Source of error: Varying amounts of excess liquid left on the surface of the potato cylinders. Minimization: Blot each potato cylinder gently with a paper towel in a standardized way before weighing. (iv) Calculating percentage change accounts for any differences in the initial masses of the potato cylinders, making comparison valid.
Marking scheme
(a) Plan [6 marks] - Use at least 5 different concentrations of sodium chloride solution (1) - Cork borer used to cut cylinders of equal diameter/length (1) - Initial mass measured using electronic balance (1) - Left in solutions for a specified duration (e.g., 1 hour / overnight) (1) - Blotting cylinders with paper towel before re-weighing (1) - Controlled variable stated (e.g., temperature, volume of solution) (1) - Repeats and calculating a mean (1) - Safety precaution mentioned (care with sharp blades/cork borer) (1) [Max 6 marks] (b) (i) Calculation [3 marks] - Working: (-0.88 / 4.00) * 100 (1) - Value: 22% (1) - Direction: decrease / negative change (1) (b) (ii) [1 mark] - Plasmolysed / plasmolysis (1) (b) (iii) [2 marks] - Error: Excess surface water on potato cylinders (1) - Minimization: Blotting gently with a paper towel before weighing (1) (b) (iv) [1 mark] - Accounts for different initial starting masses (1)
Question 3 · practical-alternative
14 marks
Ligustrum vulgare (privet) is a species of deciduous shrub. Fig. 3.1 shows a photograph of a single leaf of Ligustrum vulgare.
(a) Draw a large, clear diagram of the leaf shown in Fig. 3.1.
(b) A line XY represents the maximum width of the leaf. The length of line XY on the photograph is measured as 63 mm. The magnification of the photograph is x 1.8. Calculate the actual maximum width of the leaf in millimeters. Show your working and give your answer to two significant figures.
(c) A student investigated the rate of transpiration from both surfaces of a leaf using dry cobalt chloride paper. The cobalt chloride paper changes color from blue to pink when exposed to water vapor. The student measured the time taken (in seconds) for the paper to turn pink on the upper and lower surfaces of three separate leaves: - Leaf 1: Upper surface = 480 s; Lower surface = 90 s - Leaf 2: Upper surface = 510 s; Lower surface = 85 s - Leaf 3: Upper surface = 450 s; Lower surface = 95 s
(i) Prepare a table to record the student's results. Your table should include the results for each leaf on both surfaces, and the mean time for each surface. (ii) State and explain the difference in the rate of water loss between the upper and lower surfaces of the leaf.
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Worked solution
(a) A large, clean line drawing of the leaf including major details such as the smooth margin, midrib extending to the tip, and petiole. (b) Actual width = Image width / Magnification = 63 mm / 1.8 = 35 mm. (c) (i) Table of results: | Leaf Number | Time taken for color change on upper surface / s | Time taken for color change on lower surface / s | | 1 | 480 | 90 | | 2 | 510 | 85 | | 3 | 450 | 95 | | Mean | 480 | 90 | (ii) State: Water loss / transpiration occurs much more rapidly on the lower surface than on the upper surface (evidenced by a much shorter time taken to turn pink). Explain: - Stomata are located primarily on the lower surface of the leaf, allowing water vapor to diffuse out rapidly. - The upper surface has a thick waxy cuticle which is impermeable to water vapor, and has few or no stomata.
Marking scheme
(a) Drawing [4 marks] - Large drawing (occupies >50% of the box) with single, clear, continuous outline (no sketchy lines) (1) - Correct overall shape of leaf blade (oval/lanceolate with pointed apex) (1) - Midrib shown clearly as a single clean line extending to the apex (1) - Petiole drawn clearly and attached to leaf blade (1) (b) Calculation [3 marks] - Formula: Actual = Image / Magnification OR 63 / 1.8 (1) - Calculation: 35 (1) - Correct two significant figures: 35 (mm) (1) (c) (i) Table [4 marks] - Table drawn with ruler, fully enclosed with clear borders and header row (1) - Appropriate column headings with units (/ s) (1) - All raw data entered correctly (1) - Mean calculated correctly (Upper = 480, Lower = 90) (1) (c) (ii) State and explain [3 marks] - State: Rate of transpiration is faster from the lower surface (1) - Explain: More stomata on lower surface (1) - Explain: Waxy cuticle on upper surface / fewer stomata on upper surface (1)
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