An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V1) Cambridge IGCSE Biology (0610) paper. Not affiliated with or reproduced from Cambridge.
Paper 21 (選擇題)
Answer all forty questions. Choose the single best answer for each question.
40 题目 · 40 分
题目 1 · multiple_choice
1 分
Four identical cylinders of fresh turnip tissue, each with an initial mass of 5.0 g, are placed into four different concentrations of sucrose solution. After 60 minutes, the cylinders are reweighed. In which solution was the water potential lower than the water potential of the turnip cells?
A.final mass of cylinder = 5.6 g
B.final mass of cylinder = 5.3 g
C.final mass of cylinder = 5.0 g
D.final mass of cylinder = 4.4 g
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解题
When plant tissue is placed in a solution with a lower water potential (a hypertonic solution), water moves out of the cells by osmosis down the water potential gradient. This loss of water causes a decrease in the mass of the tissue below its starting mass of 5.0 g, resulting in a final mass of 4.4 g.
评分标准
D [1]
题目 2 · multiple_choice
1 分
A sample of food is tested with four different reagents. The Biuret test gives a purple colour, the ethanol emulsion test forms a milky-white emulsion, the iodine test remains yellow-brown, and the Benedict's test remains blue after heating. Which biological molecules are present in the food sample?
A.protein and lipid only
B.protein and reducing sugar only
C.starch and lipid only
D.starch and reducing sugar only
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解题
A purple colour with the Biuret test indicates the presence of protein. A milky-white emulsion with ethanol indicates the presence of fats/lipids. The yellow-brown iodine test indicates starch is absent, and the blue Benedict's test indicates reducing sugars are absent.
评分标准
A [1]
题目 3 · multiple_choice
1 分
Which row correctly identifies the site of urea production and the main organ of urea excretion in the human body?
A.site of production: liver; organ of excretion: kidney
B.site of production: liver; organ of excretion: bladder
C.site of production: kidney; organ of excretion: liver
D.site of production: kidney; organ of excretion: bladder
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解题
Urea is produced in the liver by the deamination of excess amino acids. It is then transported in blood plasma to the kidneys, which excrete it in urine. The bladder stores urine but does not produce or filter urea.
评分标准
A [1]
题目 4 · multiple_choice
1 分
A person is looking at an aeroplane high in the sky and then looks down to read a message on a mobile phone. Which changes occur in the eye to focus on the nearby mobile phone?
A.ciliary muscles contract, suspensory ligaments slacken, lens becomes more convex
D.ciliary muscles relax, suspensory ligaments tighten, lens becomes more convex
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解题
To focus on a near object (accommodation), the ciliary muscles contract, releasing tension on the suspensory ligaments (they slacken). This allows the lens to become more convex (fatter/more rounded), increasing its refractive power.
评分标准
A [1]
题目 5 · multiple_choice
1 分
A food chain in a grassland ecosystem is shown: grass -> grasshopper -> frog -> snake. Why is only a small percentage of the total energy contained in the grass transferred to the tissues of the grasshopper?
A.Energy is lost as heat from cellular respiration in the grass and not all plant material is eaten or digested.
B.All light energy absorbed by the grass is permanently stored as starch and cannot be released.
C.The grasshopper converts all consumed plant matter directly into kinetic energy without producing waste.
D.Energy cannot be transferred across trophic levels because it is destroyed during consumption.
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解题
Energy transfer between trophic levels is inefficient because producers lose energy as heat through cellular respiration, and consumers do not ingest or digest all parts of the plant (e.g. roots or indigestible fibre).
评分标准
A [1]
题目 6 · 選擇題
1 分
Plant cells with a high water potential were placed into a concentrated sucrose solution. Which row correctly describes the direction of net water movement and the resulting condition of the plant cells?
A.Net water movement: into the cells; Condition of cells: turgid
B.Net water movement: into the cells; Condition of cells: lysed
C.Net water movement: out of the cells; Condition of cells: plasmolysed
D.Net water movement: out of the cells; Condition of cells: lysed
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解题
Water moves by osmosis down a water potential gradient from a region of higher water potential (inside the plant cell) to a region of lower water potential (the concentrated sucrose solution). As water leaves the plant cell, the cytoplasm and cell membrane pull away from the cell wall, causing the cell to become plasmolysed.
评分标准
C is correct [1]; A and B are incorrect as water moves out of the cells down the water potential gradient; D is incorrect because plant cells do not lyse when losing water (lysis occurs in animal cells in pure water).
题目 7 · 選擇題
1 分
A sample of food was tested using four different chemical tests. The observations are recorded below: - Biuret test: purple - Benedict's test (after heating): blue - Ethanol emulsion test: milky-white emulsion - Iodine solution test: yellow-brown
Which nutrients are present in this food sample?
A.protein and fat only
B.reducing sugar and protein only
C.starch and fat only
D.reducing sugar and starch only
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解题
A purple colour in the Biuret test indicates the presence of protein. A milky-white emulsion in the ethanol emulsion test indicates the presence of fats/lipids. The Benedict's test remained blue (negative for reducing sugars) and the iodine test remained yellow-brown (negative for starch). Therefore, the sample contains protein and fat only.
评分标准
A is correct [1]; B, C, and D are incorrect because Benedict's remaining blue shows no reducing sugars, and iodine remaining yellow-brown shows no starch.
题目 8 · 選擇題
1 分
Which row correctly identifies the organ in which urea is produced and the main organ that excretes it from the human body?
A.Organ where urea is produced: Kidney; Main organ of excretion: Bladder
B.Organ where urea is produced: Kidney; Main organ of excretion: Liver
C.Organ where urea is produced: Liver; Main organ of excretion: Kidney
D.Organ where urea is produced: Liver; Main organ of excretion: Skin
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解题
Urea is synthesised in the liver by the deamination of excess amino acids. It is then transported in the blood plasma to the kidneys, which filter it out and excrete it in urine.
评分标准
C is correct [1]; A and B are incorrect because urea is produced in the liver, not the kidneys; D is incorrect because although a tiny amount is present in sweat, the kidneys are the main organ of urea excretion.
题目 9 · 選擇題
1 分
A person shifts their focus from reading a book close to their eyes to looking at a bird flying in the distance. Which changes occur in the ciliary muscles and the suspensory ligaments of the eye?
A.Ciliary muscles contract; Suspensory ligaments become slack
B.Ciliary muscles contract; Suspensory ligaments become taut
C.Ciliary muscles relax; Suspensory ligaments become slack
D.Ciliary muscles relax; Suspensory ligaments become taut
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解题
To focus on a distant object, the ciliary muscles relax. This increases the tension on the suspensory ligaments (they become taut/tight), which pulls the lens into a thinner, less convex shape to decrease refraction.
评分标准
D is correct [1]; A is the mechanism for near vision; B and C describe impossible states as contracting muscles loosen suspensory ligaments, and relaxing muscles tighten them.
题目 10 · 選擇題
1 分
The diagram shows a food chain: maize plant → locust → lizard → snake
Which statement explains why only a small proportion of the energy captured by the maize plant reaches the snake?
A.Energy is released as heat during respiration at each trophic level.
B.Plants reflect almost all sunlight so little chemical energy is initially made.
C.The snake excretes more energy as urea than all the other organisms combined.
D.Primary consumers do not respire, so they store less energy.
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解题
At each trophic level, a significant proportion of energy is lost as heat to the environment during cellular respiration, as well as through excretion, egestion, and uneaten body parts. Therefore, only about 10% of energy is passed to the next level.
评分标准
A is correct [1]; B is biologically inaccurate (plants absorb specific wavelengths of light for photosynthesis); C is incorrect as energy loss occurs across all levels due to multiple metabolic processes, not just urea; D is false as all living organisms respire.
题目 11 · multiple_choice
1 分
Four identical plant tissue cylinders, each with an initial mass of 2.5 g, were placed into four separate test-tubes containing sucrose solutions of concentrations 0.1 mol/dm³, 0.3 mol/dm³, 0.5 mol/dm³, and 0.7 mol/dm³. The water potential of the plant cell sap is equal to that of a 0.3 mol/dm³ sucrose solution. Which final mass would be expected for the cylinder placed in the 0.7 mol/dm³ sucrose solution after 2 hours?
A.2.1 g
B.2.5 g
C.2.8 g
D.3.2 g
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解题
The 0.7 mol/dm³ sucrose solution has a lower water potential (higher solute concentration) than the plant cell sap (equivalent to 0.3 mol/dm³). As a result, water moves out of the plant cells down the water potential gradient by osmosis, causing the tissue to lose mass. Therefore, its final mass must be less than the initial 2.5 g, which corresponds to 2.1 g.
评分标准
A is correct [1]. Water moves out of cells by osmosis into the hypertonic solution, decreasing mass below 2.5 g. B is incorrect as mass remains unchanged only in isotonic solution (0.3 mol/dm³). C and D are incorrect as mass only increases in hypotonic solutions.
题目 12 · multiple_choice
1 分
A student carries out food tests on four liquid samples, P, Q, R, and S. The observations are recorded in the table:
- Sample P: Benedict's test stays blue; Biuret test turns purple; Iodine test stays yellow-brown. - Sample Q: Benedict's test turns brick-red; Biuret test stays blue; Iodine test stays yellow-brown. - Sample R: Benedict's test stays blue; Biuret test turns purple; Iodine test turns blue-black. - Sample S: Benedict's test turns brick-red; Biuret test turns purple; Iodine test turns blue-black.
Which sample contains protein and starch, but no reducing sugar?
A.Sample P
B.Sample Q
C.Sample R
D.Sample S
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解题
A blue Benedict's result indicates that reducing sugar is absent. A purple Biuret result indicates the presence of protein. A blue-black iodine result indicates the presence of starch. Sample R satisfies all three criteria.
评分标准
C is correct [1]. Sample R tests negative for reducing sugar (Benedict's blue), positive for protein (Biuret purple), and positive for starch (iodine blue-black).
题目 13 · multiple_choice
1 分
Which substance is present in the glomerular filtrate of a healthy human, but is normally absent from their urine?
A.Glucose
B.Mineral ions
C.Protein
D.Urea
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解题
Glucose molecules are small enough to pass through the capillary walls of the glomerulus during ultrafiltration and enter the Bowman's capsule as part of the glomerular filtrate. In a healthy kidney, 100% of the glucose is reabsorbed back into the bloodstream in the proximal convoluted tubule, so no glucose is present in normal urine.
评分标准
A is correct [1]. Glucose is filtered at the glomerulus and completely reabsorbed along the proximal convoluted tubule. B and D are incorrect because urea and mineral ions are excreted in urine. C is incorrect because large proteins cannot pass into the glomerular filtrate.
题目 14 · multiple_choice
1 分
A person walks from a dark cinema into bright sunlight. Which row correctly describes the response of the iris and the change in pupil size?
In bright light, the pupil constricts to protect the retina from damage and reduce light intake. This constriction is caused by the contraction of the circular muscles and the relaxation of the radial muscles in the iris.
评分标准
A is correct [1]. Circular muscles contract, radial muscles relax, and the pupil constricts (narrows). B, C, and D describe incorrect combinations of muscle action or incorrect pupil diameter change.
题目 15 · multiple_choice
1 分
Why do food chains rarely have more than four or five trophic levels?
A.Top predators reproduce at a much faster rate than producers.
B.Energy is lost at each transfer, leaving too little energy to support higher levels.
C.Decomposers consume all available chemical energy before it reaches secondary consumers.
D.Tertiary consumers lose less heat energy during respiration than primary consumers.
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解题
Energy is lost at every trophic level due to metabolic processes such as respiration (heat loss), movement, excretion of waste products, and uneaten or undigested body parts. Because only about 10% of energy is transferred from one level to the next, there is insufficient energy available after 4 or 5 levels to support an additional viable trophic level.
评分标准
B is correct [1]. Energy losses at each trophic level limit the total energy available to sustain additional higher trophic levels. A, C, and D are scientifically incorrect explanations.
题目 16 · 選擇題
1 分
A piece of plant tissue is placed in a concentrated sucrose solution. Which row correctly describes the net movement of water and the final state of the plant cells?
A.water moves into the cells; cells become turgid
B.water moves into the cells; cells become plasmolysed
C.water moves out of the cells; cells become plasmolysed
D.water moves out of the cells; cells become turgid
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解题
In a concentrated sucrose solution, the water potential outside the plant cells is lower than inside the cell cytoplasm and vacuole. Water moves out of the cells down the water potential gradient by osmosis. This loss of water causes the cytoplasm to pull away from the cell wall, resulting in plasmolysis.
评分标准
C is correct [1]
题目 17 · 選擇題
1 分
A sample of food is tested with three different chemical reagents. The results are recorded: biuret test remains blue; Benedict's test turns brick-red; iodine solution remains yellow-brown. Which nutrients are present in this food sample?
A.protein and reducing sugar
B.reducing sugar only
C.starch and protein
D.starch only
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解题
A blue biuret test indicates that no protein is present. A brick-red Benedict's test confirms the presence of reducing sugar. A yellow-brown iodine test indicates the absence of starch. Therefore, only reducing sugar is present in the sample.
评分标准
B is correct [1]
题目 18 · 選擇題
1 分
Which row correctly identifies the site of urea production and the principal organ responsible for the excretion of urea in humans?
A.site of urea production: kidney; principal organ of excretion: bladder
B.site of urea production: kidney; principal organ of excretion: liver
C.site of urea production: liver; principal organ of excretion: kidney
D.site of urea production: liver; principal organ of excretion: lungs
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解题
Urea is formed in the liver by the deamination of excess amino acids. It travels in the blood plasma to the kidneys, where it is filtered from the blood and excreted in the urine.
评分标准
C is correct [1]
题目 19 · 選擇題
1 分
What changes occur in the human eye when focusing on a near object?
A.ciliary muscles contract, suspensory ligaments become slack, lens becomes more convex
B.ciliary muscles contract, suspensory ligaments tighten, lens becomes less convex
C.ciliary muscles relax, suspensory ligaments become slack, lens becomes less convex
D.ciliary muscles relax, suspensory ligaments tighten, lens becomes more convex
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解题
Accommodation for near vision involves the contraction of ciliary muscles. This reduces tension on the suspensory ligaments, allowing them to become slack. Consequently, the elastic lens rounds up, becoming more convex to refract light rays more strongly.
评分标准
A is correct [1]
题目 20 · 選擇題
1 分
Which statement correctly describes the flow of energy through an ecosystem?
A.Energy flows in a closed cycle between organisms and the environment.
B.Energy enters ecosystems mainly as light from the Sun and is transferred into chemical energy.
C.Energy transfer between trophic levels is one hundred percent efficient.
D.Heat energy lost from consumers is trapped again by producers during photosynthesis.
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解题
Light from the Sun provides the initial source of energy for ecosystems, where producers convert it into chemical energy via photosynthesis. Energy flow is non-cyclic because energy is continuously lost as heat to the environment and cannot be recaptured by organisms.
评分标准
B is correct [1]
题目 21 · 選擇題
1 分
A plant cell is placed into a solution that has a higher water potential than the cell sap.
Which row correctly describes the direction of net water movement and the resulting condition of the cell?
A.Direction of net water movement: into the cell; Condition of cell: turgid
B.Direction of net water movement: into the cell; Condition of cell: plasmolysed
C.Direction of net water movement: out of the cell; Condition of cell: flaccid
D.Direction of net water movement: out of the cell; Condition of cell: turgid
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解题
Water enters the cell by osmosis down a water potential gradient from high water potential in the external solution to lower water potential inside the cell sap. As water enters, the vacuole expands and pushes against the rigid cellulose cell wall, making the cell firm and turgid.
评分标准
A is correct [1]. B is incorrect because plasmolysis occurs when water leaves the cell. C is incorrect because water enters the cell when placed in a higher water potential solution. D is incorrect because net movement is into the cell, not out of the cell.
题目 22 · 選擇題
1 分
A sample of liquid food was tested with four different reagents. The observations are recorded below:
A positive Biuret test turns purple, indicating the presence of protein. A positive ethanol emulsion test forms a milky-white emulsion, confirming the presence of lipids (fats). Iodine remains yellow-brown when starch is absent, and Benedict's solution remains blue when reducing sugars are absent.
评分标准
A is correct [1]. B is incorrect because Benedict's and Iodine tests were both negative. C is incorrect because Benedict's test was negative (blue). D is incorrect because Iodine test was negative (yellow-brown).
题目 23 · 選擇題
1 分
In the human nephron, ultrafiltration occurs under high pressure.
Where does ultrafiltration take place, and which substance is NOT filtered into the renal capsule?
A.Site of ultrafiltration: Glomerulus; Substance not filtered into capsule: Plasma proteins
B.Site of ultrafiltration: Glomerulus; Substance not filtered into capsule: Glucose
C.Site of ultrafiltration: Collecting duct; Substance not filtered into capsule: Urea
D.Site of ultrafiltration: Collecting duct; Substance not filtered into capsule: Plasma proteins
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解题
Ultrafiltration occurs in the glomerulus (the knot of capillaries inside the Bowman's capsule). Large molecules such as plasma proteins (e.g. albumen, fibrinogen) and blood cells are too large to pass through the basement membrane and capillary pores, so they remain in the blood.
评分标准
A is correct [1]. B is incorrect because glucose is small and passes easily into the filtrate during ultrafiltration. C is incorrect because ultrafiltration does not take place in the collecting duct, and urea is filtered. D is incorrect because ultrafiltration does not take place in the collecting duct.
题目 24 · 選擇題
1 分
A person is looking at a distant bird in the sky and then shifts their gaze to read text in a book held close to their eyes.
Which row correctly describes the changes that occur in the eye to focus on the book?
A.Ciliary muscles: contract; Suspensory ligaments: slacken; Lens: becomes more convex
D.Ciliary muscles: relax; Suspensory ligaments: tighten; Lens: becomes more convex
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解题
To focus on a near object (accommodation for near vision), the ciliary muscles contract. This releases tension on the suspensory ligaments (they slacken), allowing the elastic lens to recoil and become more convex (thicker/fatter), increasing its refractive power.
评分标准
A is correct [1]. B is incorrect because contracting ciliary muscles cause suspensory ligaments to slacken, not tighten, and the lens becomes thicker. C is incorrect because ciliary muscles contract during near accommodation. D is incorrect because ciliary muscle relaxation and ligament tightening are for distant vision.
题目 25 · 選擇題
1 分
A young shoot of a seedling is placed horizontally in the dark.
Which row correctly identifies where auxin accumulates and its effect on the cells of the shoot?
A.Side with higher auxin concentration: Lower side; Effect on shoot cells: Stimulates cell elongation
B.Side with higher auxin concentration: Upper side; Effect on shoot cells: Stimulates cell elongation
C.Side with higher auxin concentration: Lower side; Effect on shoot cells: Inhibits cell elongation
D.Side with higher auxin concentration: Upper side; Effect on shoot cells: Inhibits cell elongation
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解题
In a horizontally oriented shoot, gravity causes auxin to accumulate on the lower side. In shoots, high concentrations of auxin stimulate cell elongation. The cells on the lower side elongate faster than those on the upper side, causing the shoot to bend upwards (negative gravitropism).
评分标准
A is correct [1]. B is incorrect because auxin accumulates on the lower side due to gravity. C is incorrect because auxin stimulates cell elongation in shoots (it only inhibits elongation at high concentrations in roots). D is incorrect on both accumulation location and effect in shoots.
题目 26 · multiple_choice
1 分
Plant tissue cylinders of equal mass were placed into four different test-tubes containing sucrose solutions of different concentrations. In which solution will the plant cells become plasmolysed?
A.a solution with a higher water potential than the cell sap
B.a solution with the same water potential as the cell sap
C.a solution with a lower water potential than the cell sap
D.pure distilled water
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解题
Plasmolysis occurs when plant cells are placed in a hypertonic solution that has a lower water potential than the cell sap inside the vacuole. Water moves out of the cells by osmosis down the water potential gradient, causing the cytoplasm and cell membrane to pull away from the cell wall.
评分标准
C is correct [1 mark]. A and D are incorrect because water enters cells when placed in a solution with a higher water potential, making them turgid. B is incorrect because there is no net movement of water in isotonic solutions.
题目 27 · multiple_choice
1 分
A student carried out food tests on a liquid sample. The results are shown in the table.
- Biuret test: solution remains blue - Iodine solution test: colour turns blue-black - Ethanol emulsion test: cloudy white layer forms - Benedict's test (heated): solution remains blue
Which biological molecules are present in the sample?
A.reducing sugar and starch
B.protein and lipid
C.starch and lipid
D.reducing sugar and protein
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解题
A positive iodine test turns blue-black, indicating the presence of starch. A positive ethanol emulsion test produces a white cloudy emulsion, indicating the presence of lipids (fats). Negative Biuret (blue) indicates absence of protein, and negative Benedict's (blue) indicates absence of reducing sugars.
评分标准
C is correct [1 mark]. A is incorrect because Benedict's test is negative (no reducing sugars). B is incorrect because Biuret test is negative (no proteins). D is incorrect because both Biuret and Benedict's tests are negative.
题目 28 · multiple_choice
1 分
Which row correctly identifies the site of urea production and the main organ from which urea is excreted from the human body?
A.site of production: kidney; organ of excretion: liver
B.site of production: liver; organ of excretion: kidney
C.site of production: liver; organ of excretion: bladder
D.site of production: bladder; organ of excretion: kidney
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解题
Urea is produced in the liver through the deamination of excess amino acids. It is transported in blood plasma to the kidneys, where it is filtered out of the blood and excreted in urine.
评分标准
B is correct [1 mark]. A is incorrect as the kidneys excrete urea, they do not produce it. C and D are incorrect because the bladder only stores urine temporarily before urination and does not excrete or filter it from the bloodstream.
题目 29 · multiple_choice
1 分
A person looks at a distant tree and then focuses on a book held close to their eyes. Which changes take place in the eye to focus light from the book onto the retina?
A.ciliary muscles contract, suspensory ligaments slacken, lens becomes more convex
D.ciliary muscles relax, suspensory ligaments tighten, lens becomes more convex
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解题
To focus on a near object (accommodation for near vision), the ciliary muscles contract. This reduces tension on the suspensory ligaments (they slacken), allowing the elastic lens to become more convex (fatter/thicker), which refracts light rays more strongly onto the retina.
评分标准
A is correct [1 mark]. B is incorrect because contracting ciliary muscles slackens the suspensory ligaments, not tightens them. C is incorrect because ciliary muscles contract during near accommodation. D is incorrect because relaxing ciliary muscles and tight ligaments cause the lens to become thinner, which is used for distant vision.
题目 30 · multiple_choice
1 分
Which statement explains why only approximately 10% of energy is transferred from one trophic level to the next in an ecosystem?
A.Energy is lost as metabolic heat during respiration and in undigested waste.
B.Herbivores photosynthesise to replace energy lost by producers.
C.Decomposers recycle all lost energy directly back to the primary producers.
D.Total energy increases towards the top of the food chain due to bioaccumulation.
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解题
Energy transfer between trophic levels is inefficient because most energy is lost as heat released during cellular respiration, through metabolic waste (excretion and egestion), and via uneaten organism parts.
评分标准
A is correct [1 mark]. B is incorrect because herbivores cannot photosynthesise. C is incorrect because decomposers consume dead matter and lose energy as heat; they do not recycle energy to producers. D is incorrect because energy does not accumulate at higher trophic levels (it decreases).
题目 31 · 選擇題
1 分
A plant cell is placed into a concentrated sucrose solution and left for 30 minutes until it becomes plasmolysed. Which substance fills the space between the cell wall and the shrunken cell membrane?
A.air
B.cell sap
C.concentrated sucrose solution
D.pure water
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解题
Plant cell walls are fully permeable to small solute molecules, allowing the external concentrated sucrose solution to pass straight through. However, the cell membrane is partially permeable and loses water by osmosis, causing the cytoplasm and vacuole to shrink away from the wall. Therefore, the space between the wall and the membrane is filled with the concentrated sucrose solution.
评分标准
C [1] — concentrated sucrose solution fills the space because the cell wall is freely permeable.
题目 32 · 選擇題
1 分
A student carried out standard food tests on four liquid samples, P, Q, R, and S. The results are shown below: Sample P gave a yellow-brown colour with iodine, a purple colour with Biuret reagent, and a brick-red precipitate with heated Benedict's solution. Sample Q gave a blue-black colour with iodine, a purple colour with Biuret reagent, and remained blue with heated Benedict's solution. Sample R gave a yellow-brown colour with iodine, remained blue with Biuret reagent, and gave a brick-red precipitate with heated Benedict's solution. Sample S gave a blue-black colour with iodine, remained blue with Biuret reagent, and remained blue with heated Benedict's solution. Which sample contains protein and reducing sugar, but no starch?
A.sample P
B.sample Q
C.sample R
D.sample S
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解题
A positive test for protein is a purple colour with Biuret reagent. A positive test for reducing sugar is a brick-red precipitate with Benedict's solution after heating. A negative test for starch is a yellow-brown colour with iodine solution (remaining unchanged). Sample P is yellow-brown with iodine (no starch), purple with Biuret (protein present), and brick-red with Benedict's (reducing sugar present).
评分标准
A [1] — Sample P indicates absence of starch (yellow-brown), presence of protein (purple), and presence of reducing sugar (brick-red).
题目 33 · 選擇題
1 分
In the kidney of a healthy human, glucose is filtered from the blood into the Bowman's capsule. What happens to this glucose as the glomerular filtrate passes through the proximal convoluted tubule?
A.It is completely reabsorbed back into the blood by active transport.
B.It is converted into glycogen and stored in the walls of the collecting duct.
C.It is broken down into urea and carbon dioxide by tubule cells.
D.It remains in the tubule fluid and is excreted in the urine.
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解题
In a healthy kidney, all filtered glucose is selectively reabsorbed back into the capillary blood from the proximal convoluted tubule by active transport against its concentration gradient. Consequently, no glucose is found in the urine of a healthy person.
评分标准
A [1] — all glucose is reabsorbed back into the blood by active transport in the proximal convoluted tubule.
题目 34 · 選擇題
1 分
A person reading a book looks up to focus on an aeroplane flying in the distance. Which changes occur in the eye to bring the distant object into sharp focus on the retina?
A.ciliary muscles contract, suspensory ligaments become slack, lens becomes more convex
B.ciliary muscles contract, suspensory ligaments become taut, lens becomes less convex
C.ciliary muscles relax, suspensory ligaments become slack, lens becomes more convex
D.ciliary muscles relax, suspensory ligaments become taut, lens becomes less convex
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解题
When focusing on a distant object, the ciliary muscles relax. This pulls the suspensory ligaments tight (taut), which pulls on the lens, causing the lens to become thinner and less convex (flatter), reducing its refractive power.
评分标准
D [1] — ciliary muscles relax, suspensory ligaments become taut, lens becomes less convex.
题目 35 · 選擇題
1 分
A growing shoot tip is exposed to unilateral light coming from the right-hand side. Which statement correctly describes the movement of auxin and the resulting growth response?
A.Auxin accumulates on the shaded side; cells on the shaded side elongate faster; shoot bends to the right.
B.Auxin accumulates on the illuminated side; cells on the illuminated side elongate faster; shoot bends to the left.
C.Auxin accumulates on the shaded side; cells on the shaded side elongate slower; shoot bends to the left.
D.Auxin accumulates on the illuminated side; cells on the illuminated side elongate slower; shoot bends to the right.
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解题
Auxin is produced in the shoot tip and diffuses down into the elongation zone, moving away from light towards the shaded (left-hand) side. In shoots, higher auxin concentration stimulates greater cell elongation. Because cells on the shaded side elongate faster than cells on the illuminated side, the shoot bends towards the light source (to the right).
评分标准
A [1] — auxin accumulates on the shaded side, stimulating faster elongation of cells on the shaded side, causing the shoot to bend towards the light.
题目 36 · multiple_choice
1 分
Four identical potato cylinders were placed into test-tubes containing sucrose solutions of different concentrations: \(0.0\text{ mol dm}^{-3}\), \(0.2\text{ mol dm}^{-3}\), \(0.6\text{ mol dm}^{-3}\) and \(1.0\text{ mol dm}^{-3}\).
After two hours, which cylinder will show the greatest percentage increase in mass?
A.The cylinder in \(0.0\text{ mol dm}^{-3}\) sucrose solution
B.The cylinder in \(0.2\text{ mol dm}^{-3}\) sucrose solution
C.The cylinder in \(0.6\text{ mol dm}^{-3}\) sucrose solution
D.The cylinder in \(1.0\text{ mol dm}^{-3}\) sucrose solution
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解题
Pure water (\(0.0\text{ mol dm}^{-3}\) sucrose solution) has the highest water potential among all the solutions. This creates the steepest water potential gradient into the potato cells, resulting in the highest net rate of water entry by osmosis, leading to the greatest percentage increase in mass.
评分标准
A is correct [1]. B is incorrect because the water potential gradient is less steep than in pure water. C and D are incorrect because the sucrose solutions have lower water potentials than the potato tissue, causing water to leave the cells by osmosis and resulting in a decrease in mass.
题目 37 · multiple_choice
1 分
A sample of food was tested using four standard food test reagents. The results were recorded as follows:
- Benedict's test: remained blue after heating - Biuret test: turned purple - Ethanol emulsion test: remained colourless - Iodine solution: turned blue-black
Which biological molecules are present in the food sample?
A.protein and reducing sugar only
B.protein and starch only
C.reducing sugar and lipid only
D.starch and lipid only
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解题
A purple result with the Biuret test confirms the presence of protein. A blue-black colour with iodine solution confirms the presence of starch. A blue result in the Benedict's test means reducing sugars are absent, and a colourless result in the ethanol emulsion test indicates lipids are absent.
评分标准
B is correct [1]. A is incorrect because reducing sugar is absent (Benedict's test remained blue). C is incorrect because reducing sugar and lipids are both absent. D is incorrect because lipids are absent (ethanol emulsion test showed no emulsion/remained colourless).
题目 38 · multiple_choice
1 分
In which organ is urea produced in the human body, and what is the process of its formation?
A.kidney, by the breakdown of fatty acids
B.kidney, by the breakdown of excess glucose
C.liver, by the deamination of excess amino acids
D.liver, by the synthesis of mineral ions
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解题
Urea is formed in the liver through the process of deamination, which is the removal of the nitrogen-containing amino group from excess amino acids. It is then transported in blood plasma to the kidneys for excretion.
评分标准
C is correct [1]. A and B are incorrect because urea is produced in the liver, not the kidney (the kidney filters and excretes it). D is incorrect because urea is synthesized from excess amino acids, not mineral ions.
题目 39 · multiple_choice
1 分
Which row correctly describes the state of the ciliary muscles, suspensory ligaments, and lens shape when the human eye focuses on a near object?
A.ciliary muscles contract; suspensory ligaments become slack; lens becomes more convex
B.ciliary muscles relax; suspensory ligaments become taut; lens becomes less convex
C.ciliary muscles contract; suspensory ligaments become taut; lens becomes less convex
D.ciliary muscles relax; suspensory ligaments become slack; lens becomes more convex
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解题
When focusing on a near object (accommodation for near vision), the ciliary muscles contract. This reduces tension on the suspensory ligaments, allowing them to become slack. Consequently, the elastic lens recoils into a more convex (fatter/thicker) shape to increase light refraction.
评分标准
A is correct [1]. B describes accommodation for viewing distant objects. C is incorrect because contraction of ciliary muscles causes suspensory ligaments to slacken, not become taut. D is incorrect because relaxation of ciliary muscles tightens suspensory ligaments, making the lens less convex.
题目 40 · multiple_choice
1 分
Why is only approximately 10% of the energy in a trophic level transferred to the next trophic level in a food chain?
A.Carnivores cannot digest any of the material consumed by herbivores.
B.Organisms lose energy as heat from respiration, in excretion, and in unconsumed parts.
C.Carnivores obtain most of their required energy directly from sunlight.
D.All energy stored in biomass is released as light energy into the surrounding environment.
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解题
Most of the energy consumed by organisms at a trophic level is lost to the environment as heat released during cellular respiration, or remains in uneaten parts, faeces, and excretory products. Therefore, only around 10% of the energy is converted into new biomass and available to the next trophic level.
评分标准
B is correct [1]. A is incorrect because energy loss occurs primarily via respiration and excretion, not an inability of carnivores to digest cellulose. C is incorrect because carnivores are heterotrophs and cannot obtain energy directly from sunlight. D is incorrect because metabolic energy is lost primarily as thermal energy (heat), not light.
Answer all questions. Write your answers in the spaces provided on the question paper.
7 题目 · 92.33 分
题目 1 · structured
13 分
1 (a) Define the term osmosis. .......................................................................................................................................................................................................................................................................................................................................... [3]
(b) A student investigated the change in mass of sweet potato cylinders placed in different concentrations of sucrose solution. Cylinders of sweet potato of equal initial length and mass were submerged in six different sucrose solutions for 45 minutes. The percentage change in mass of each cylinder was determined. The results are shown in Table 1.1.
Table 1.1 concentration of sucrose solution / mol per dm³ | initial mass / g | final mass / g | percentage change in mass / % 0.0 | 4.20 | 4.75 | +13.1 0.2 | 4.18 | 4.51 | +7.9 0.4 | 4.22 | 4.31 | +2.1 0.6 | 4.15 | 4.02 | -3.1 0.8 | 4.25 | 3.91 | -8.0 1.0 | 4.19 | 3.73 | .....................
(i) Calculate the percentage change in mass for the sweet potato cylinder in the 1.0 mol per dm³ sucrose solution. Give your answer to one decimal place. Space for working.
percentage change in mass = ..................................................... % [2]
(ii) Using data from Table 1.1, estimate the concentration of sucrose solution that is isotonic to the sweet potato tissue. Explain your reasoning. concentration = ..................................................... mol per dm³ explanation: .......................................................................................................................................................................................................................................................................................................................................... [3]
(iii) Explain, in terms of water potential, why the sweet potato cylinder gained mass in the 0.0 mol per dm³ sucrose solution. .................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. [3]
(iv) State two variables that the student should have kept constant in this investigation. 1. .................................................................................................................................................... 2. .................................................................................................................................................... [2]
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解题
(a) Osmosis is the net movement of water molecules from an area of higher water potential (dilute solution) to an area of lower water potential (concentrated solution) through a partially permeable membrane.
(b)(i) Initial mass = 4.19 g, Final mass = 3.73 g. Change in mass = 3.73 - 4.19 = -0.46 g. Percentage change in mass = (-0.46 / 4.19) * 100 = -10.978... % = -11.0% (to 1 d.p.).
(b)(ii) Between 0.4 and 0.6 mol per dm³, the percentage change crosses zero (from +2.1% to -3.1%). Linear interpolation gives approximately 0.47 mol per dm³ (acceptable range 0.44 to 0.50 mol per dm³). At this point, no net osmosis occurs because the water potential of the tissue is equal to that of the external sucrose solution.
(b)(iii) The pure water (0.0 mol per dm³) has a higher water potential than the cytoplasm/cell sap of the sweet potato cells. Water enters the cells by osmosis down the water potential gradient across the partially permeable cell surface membrane, increasing cell volume/mass and making cells turgid.
(b)(iv) Any two controlled variables: temperature, surface area/diameter/length of potato cylinders, volume of sucrose solution, source/type of sweet potato, duration of immersion.
评分标准
(a) 1. net movement of water (molecules); 2. from a region of higher water potential to a region of lower water potential / down a water potential gradient; 3. through a partially permeable membrane;
(b)(i) 1. (3.73 - 4.19) / 4.19 * 100 OR -0.46 / 4.19 * 100; 2. -11.0 (%); [Allow ecf for correct rounding if calculation method is shown; minus sign required unless stated as 'decrease of 11.0%']
(b)(ii) 1. value between 0.44 and 0.50 (mol per dm³); 2. point where there is no net movement of water / percentage change in mass is 0 (%); 3. water potential of the sucrose solution is equal to the water potential of the sweet potato cells / tissue;
(b)(iii) 1. 0.0 mol per dm³ / pure water has a higher water potential than the potato cells / cell sap / cytoplasm (ora); 2. water moved into the cells by osmosis; 3. (down a water potential gradient) causing cells to expand / become turgid / gain mass;
(b)(iv) Any two from: - temperature; - surface area / dimensions / length / diameter of cylinders; - volume of sucrose solution; - immersion time / 45 minutes; - same batch / variety of sweet potato; [R 'size of potato' unqualified / 'amount of solution']
题目 2 · structured
13 分
2 Biological molecules are essential for the structure and metabolism of living organisms.
(a) Complete Table 2.1 by identifying the smaller basic units (monomers) that make up each large biological molecule.
Table 2.1 large biological molecule | basic sub-units (monomers) protein | ..................................................... glycogen | ..................................................... lipid | ..................................................... [3]
(b) Describe how you would test a sample of liquid food for the presence of reducing sugars. .................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. [4]
(c) Describe the test used to confirm the presence of fats and oils (lipids) in a solid food sample, and state the positive result. test: ............................................................................................................................................................................................................................................................................................................................................................. positive result: ................................................................................................................................................... [3]
(d) DNA is a biological molecule found in the nucleus of eukaryotic cells. (i) State the shape of a DNA molecule. .................................................................................................................................................... [1] (ii) State the four base pairs that occur in DNA and describe the rule of complementary base pairing. .......................................................................................................................................................................................................................................................................................................................................... [2]
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解题
(a) Proteins are built from amino acids; glycogen is built from glucose molecules; lipids are constructed from glycerol and three fatty acid chains. (b) To test for reducing sugars, add an equal volume of Benedict's solution to the liquid food sample in a test-tube. Heat the mixture in a thermostatically controlled hot water bath (above 80 °C) for 3–5 minutes. If reducing sugars are present, the solution changes from blue to green, yellow, orange, or brick-red precipitate depending on concentration. (c) The ethanol emulsion test: mix the solid food sample with ethanol and shake to dissolve lipids, then filter/decant the liquid into a tube containing water. A cloudy white/milky emulsion confirms lipids. (d)(i) Double helix. (d)(ii) The four bases are Adenine (A), Thymine (T), Cytosine (C), and Guanine (G). Complementary base pairing means A always bonds with T, and C always bonds with G.
(b) 1. add Benedict's reagent / solution; 2. heat / warm (in a water bath) to >70 °C / to boiling point; 3. color change from blue to green / yellow / orange / brick-red / red precipitate; 4. color intensity / specific color indicates concentration of reducing sugar;
(c) test: 1. chop / grind / crush food sample AND add ethanol / alcohol; 2. pour / decant the ethanol solution into water; positive result: 3. white / milky emulsion / cloudy appearance;
(d)(i) double helix;
(d)(ii) 1. adenine (A), thymine (T), cytosine (C), guanine (G) [all four named or symbols given]; 2. A pairs with T AND C pairs with G / adenine pairs with thymine AND cytosine pairs with guanine;
题目 3 · structured
14 分
3 (a) Excretion is an essential characteristic of all living organisms. (i) Define the term excretion. .......................................................................................................................................................................................................................................................................................................................................... [3] (ii) State two toxic excretory substances produced by humans and name the organ that excretes each substance. substance 1: .................................................... organ 1: .................................................... substance 2: .................................................... organ 2: .................................................... [2]
(b) Urea is produced during the breakdown of excess amino acids. (i) Name the organ in which urea is produced. .................................................................................................................................................... [1] (ii) Name the process by which excess amino acids are converted into urea. .................................................................................................................................................... [1]
(c) People with severe kidney failure may require treatment with a kidney dialysis machine. Fig. 3.1 represents a section of a dialysis machine.
[ BLOOD IN ] -------> [ Dialysis tubing (partially permeable) ] -------> [ BLOOD OUT ] || [ Fresh Dialysis Fluid In ] || [ Used Dialysis Fluid Out ]
(i) Explain why the dialysis fluid must contain the same concentration of glucose and mineral ions as normal blood plasma. .......................................................................................................................................................................................................................................................................................................................................... [3]
(ii) Explain why dialysis fluid contains no urea. .......................................................................................................................................................................................................................................................................................................................................... [2]
(iii) Suggest two advantages of a kidney transplant compared to regular kidney dialysis. 1. .................................................................................................................................................... 2. .................................................................................................................................................... [2]
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解题
(a)(i) Excretion is the removal from organisms of toxic materials, the waste products of metabolism (chemical reactions in cells including respiration), and substances in excess of requirements. (a)(ii) Carbon dioxide excreted by the lungs; urea excreted by the kidneys (excess mineral salts/water also accepted). (b)(i) Liver. (b)(ii) Deamination (the removal of the nitrogen-containing part of amino acids to form urea). (c)(i) Dialysis fluid contains the exact physiological concentration of glucose and essential ions so there is no net concentration gradient for these vital substances, preventing them from diffusing out of the blood into the dialysis fluid, thus maintaining correct blood levels and ensuring energy supply. (c)(ii) Having 0% urea creates a steep concentration gradient between blood (high urea) and the dialysis fluid (zero urea), maximising the rate of diffusion of urea out of the blood. (c)(iii) A successful kidney transplant restores normal renal function full-time without requiring prolonged hospital visits multiple times per week, allows a less restrictive diet, and provides a higher long-term quality of life.
评分标准
(a)(i) 1. removal (from organisms) of toxic materials / toxic substances; 2. (removal of) waste products of metabolism / cellular reactions; 3. (removal of) substances in excess of requirements;
(a)(ii) 1. carbon dioxide AND lungs; 2. urea / excess salts / excess water AND kidney(s); [R skin for urea]
(b)(i) liver; (b)(ii) deamination;
(c)(i) 1. no concentration gradient for glucose / ions / same concentration; 2. prevents diffusion / loss of glucose and ions from blood into dialysis fluid; 3. ensures patient maintains blood glucose level / has glucose for respiration;
(c)(ii) 1. creates / maintains a steep concentration gradient (between blood and fluid); 2. (so) urea diffuses rapidly / effectively out of the blood into the dialysis fluid;
(c)(iii) Any two from: - no need for regular / frequent / time-consuming dialysis sessions; - patient can return to normal / less restricted diet / fluid intake; - improved quality of life / feels healthier / more energetic; - long-term cost is lower than continuous dialysis; [A permanent cure / long-term solution]
题目 4 · structured
13 分
4 Fig. 4.1 shows a diagram of a horizontal section through a human eye.
(a) Identify the part of the eye that: (i) contains the highest concentration of cone cells: ..................................................... [1] (ii) changes shape to focus light onto the retina: ..................................................... [1] (iii) lacks all photoreceptor cells (rods and cones): ..................................................... [1]
(b) Describe the functions of rod cells and cone cells in the human retina. rod cells: .......................................................................................................................................................................................................................................................................................................................................... cone cells: .......................................................................................................................................................................................................................................................................................................................................... [4]
(c) A student is reading a book and then looks up at a distant tree. Explain the changes that take place in the eye to focus light from the distant tree onto the retina. .................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. [4]
(d) Describe the pupil reflex when a person moves from a dimly lit room into bright sunlight. .......................................................................................................................................................................................................................................................................................................................................... [2]
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解题
(a)(i) Fovea. (a)(ii) Lens. (a)(iii) Blind spot. (b) Rod cells function in low light intensity (dim light), providing night vision and black-and-white (monochrome) images with low visual acuity. Cone cells operate only under high light intensities (bright light), provide sharp focus/high visual acuity, and allow color vision via three different types of cones (sensitive to red, green, and blue light). (c) To focus on a distant object (accommodation): the ciliary muscles relax, causing the suspensory ligaments to be pulled tight (taut). This tension pulls the elastic lens, causing it to become thinner/less convex/flatter. The thinner lens refracts light rays less, accurately focusing the nearly parallel light rays onto the fovea/retina. (d) In bright light, circular muscles of the iris contract while radial muscles relax. This makes the pupil diameter smaller (pupil constricts), reducing the amount of light entering the eye to prevent damage to photoreceptors.
(b) rod cells (max 2): 1. function in dim light / low light intensity; 2. give night vision / black-and-white vision / do not detect color; cone cells (max 2): 3. function in bright light / high light intensity; 4. detect color / three types (red, green, blue); 5. provide high visual acuity / detailed images;
(c) 1. ciliary muscles relax; 2. suspensory ligaments become tight / taut / pulled; 3. lens becomes thinner / less convex / flatter; 4. light rays are refracted less (by the lens); 5. light is focused sharply onto the retina / fovea; [max 4]
5 Plants respond to external stimuli by directional growth responses known as tropisms.
(a) (i) Define the term phototropism. .......................................................................................................................................................................................................................................................................................................................................... [2] (ii) State the name of the plant hormone responsible for controlling phototropism and gravitropism. .................................................................................................................................................... [1]
(b) A young seedling was grown with unidirectional light shining from the right side. (i) Describe and explain how the shoot tip responds to unidirectional light. .................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. [5]
(ii) Explain the selective advantage to a plant shoot of positive phototropism. .......................................................................................................................................................................................................................................................................................................................................... [2]
(c) Describe how auxin controls the gravitropic response in a horizontally placed plant root. .................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. [3]
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解题
(a)(i) Phototropism is a directional growth response where the direction of growth is determined by the direction of the light stimulus. (a)(ii) Auxin. (b)(i) Auxin is produced in the shoot tip and diffuses downwards. When exposed to light from one side (unidirectional light), auxin moves to and accumulates on the shaded side of the shoot. In shoots, high auxin concentration stimulates cell elongation. Consequently, cells on the shaded side elongate more rapidly than cells on the illuminated side, causing the shoot to bend towards the light source (positive phototropism). (b)(ii) Bending towards the light allows leaves to absorb maximum light energy for photosynthesis, leading to increased synthesis of carbohydrates and improved survival/growth. (c) In a horizontally placed root, gravity causes auxin to accumulate on the lower side. In root cells, unlike shoot cells, high concentrations of auxin inhibit cell elongation. The cells on the upper side continue to elongate at a normal/faster rate than those on the lower side, resulting in downward curvature of the root into the soil (positive gravitropism) to anchor the plant and absorb water/minerals.
评分标准
(a)(i) 1. growth response of a plant / plant organ; 2. in response to a directional light stimulus / direction of light;
(a)(ii) auxin; [1]
(b)(i) 1. auxin is made / synthesised in the shoot tip; 2. auxin diffuses (downwards) / moves to the shaded / dark side of the shoot; 3. unequal distribution / higher concentration of auxin on the shaded side; 4. auxin stimulates cell elongation (in shoots); 5. cells on shaded side elongate / grow faster than cells on the illuminated side / shoot bends towards the light; [max 5]
(b)(ii) 1. absorbs more light (energy); 2. for (increased rate of) photosynthesis / greater production of glucose;
(c) 1. auxin accumulates on the lower side of the root (due to gravity); 2. (high concentration of) auxin inhibits cell elongation in roots; 3. cells on the upper side elongate more / faster than cells on lower side (causing root to bend downwards);
题目 6 · structured
13 分
(a) Define the term *excretion*. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3]
(b) Urea is produced in the liver from excess amino acids. Describe the process of deamination and explain why it is necessary. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3]
(c) The functional unit of the human kidney is the nephron. (i) State the part of the nephron where ultrafiltration occurs and name one component of blood that does not pass into the filtrate during this process. part of nephron: ........................................................................................................................ component: ............................................................................................................................ [2]
(ii) In a healthy person, all filtered glucose is reabsorbed from the proximal convoluted tubule back into the blood. Explain how glucose is reabsorbed across the tubule wall. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3]
(d) People with kidney failure can be treated using haemodialysis. Describe how dialysis fluid is formulated and maintained to remove excess urea from blood without removing essential nutrients such as glucose. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2]
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解题
(a) Excretion is the removal of the waste products of metabolism, toxic materials, and substances in excess of requirements from an organism.
(b) Excess amino acids cannot be stored in the body. In the liver, the nitrogen-containing amino group is removed (deamination) and converted into urea. The remaining keto acid group can be used for energy / respiration or converted to carbohydrates / fats. This prevents the buildup of toxic ammonia.
(c)(i) Ultrafiltration occurs across the glomerulus into the Bowman's (renal) capsule. Large proteins and blood cells (red cells, white cells, platelets) are too large to pass through the basement membrane and remain in the blood.
(ii) Glucose moves from the nephron lumen across the epithelial cells into blood capillaries by active transport (and facilitated diffusion). Active transport involves transport proteins / carrier proteins in the cell membrane and uses energy / ATP released from respiration to move glucose against its concentration gradient.
(d) Dialysis fluid contains zero urea, creating a steep concentration gradient so urea diffuses rapidly out of the blood into the fluid. It contains the normal physiological concentration of glucose and salts, preventing net diffusion of glucose out of the blood.
评分标准
(a) any three from: 1. removal of toxic materials / substances; 2. removal of waste products of metabolism / cellular reactions; 3. removal of substances in excess of requirements; 4. from an organism; [max 3]
(b) any three from: 1. removal of the nitrogen-containing part / amino group of amino acids; 2. to form urea; 3. (remaining part converted to) carbohydrate / glycogen / used in respiration; 4. amino acids cannot be stored (in the body); 5. prevents accumulation of toxic ammonia / urea is less toxic than ammonia; [max 3]
(c)(i) part of nephron: glomerulus / Bowman's capsule / renal capsule; [1] component: red blood cells / erythrocytes / white blood cells / leucocytes / platelets / large blood proteins / albumen / fibrinogen / globulins; [1] [I: platelets alone unless specified as cell fragments; R: small molecules like ions, urea, water]
(c)(ii) any three from: 1. active transport; 2. (movement) against a concentration gradient; 3. requires energy / ATP (from cellular respiration); 4. involving protein carriers / carrier molecules / protein pumps (in the cell membrane); 5. diffusion / facilitated diffusion (from cell to capillary); [max 3]
(d) any two from: 1. dialysis fluid contains no urea / lower concentration of urea than blood (so urea diffuses out of blood down concentration gradient); 2. dialysis fluid contains the same / equal concentration of glucose (as healthy blood); 3. so no net loss / movement of glucose from blood (by diffusion); 4. constant replacement / flow of fresh dialysis fluid maintains concentration gradient; [max 2]
题目 7 · structured
13.33 分
(a) Define the term *excretion*. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3]
(b) Urea is produced in the liver from excess amino acids. Describe the process of deamination and explain why it is necessary. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3]
(c) The functional unit of the human kidney is the nephron. (i) State the part of the nephron where ultrafiltration occurs and name one component of blood that does not pass into the filtrate during this process. part of nephron: ........................................................................................................................ component: ............................................................................................................................ [2]
(ii) In a healthy person, all filtered glucose is reabsorbed from the proximal convoluted tubule back into the blood. Explain how glucose is reabsorbed across the tubule wall. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3]
(d) People with kidney failure can be treated using haemodialysis. Describe how dialysis fluid is formulated and maintained to remove excess urea from blood without removing essential nutrients such as glucose. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2]
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解题
(a) Excretion is defined as the removal of toxic materials, the waste products of metabolism, and substances in excess of requirements from an organism. (b) Deamination occurs in the liver: the nitrogen-containing amino group (-NH2) is removed from surplus amino acids and converted into urea. This is essential because the body cannot store amino acids, and free ammonia is highly toxic. (c)(i) Ultrafiltration occurs across the glomerulus into the Bowman's capsule. Large plasma proteins and cellular components (red blood cells, white blood cells, platelets) are too large to pass through the basement membrane. (ii) Glucose is reabsorbed by active transport across the epithelial cell membranes of the proximal convoluted tubule. This requires carrier proteins and energy in the form of ATP produced by aerobic respiration in the tubule cells to transport glucose against its concentration gradient. (d) Dialysis fluid is maintained with zero urea concentration, establishing a concentration gradient that allows urea to diffuse out of the patient's blood. It contains the exact physiological concentration of glucose found in healthy blood, preventing any net loss of glucose by diffusion.
评分标准
(a) any three from: 1. removal of toxic materials / substances; 2. removal of waste products of metabolism / cellular reactions; 3. removal of substances in excess of requirements; 4. from an organism; [max 3]
(b) any three from: 1. removal of the nitrogen-containing part / amino group of amino acids; 2. to form urea; 3. (remaining part converted to) carbohydrate / glycogen / used in respiration; 4. amino acids cannot be stored (in the body); 5. prevents accumulation of toxic ammonia / urea is less toxic than ammonia; [max 3]
(c)(i) part of nephron: glomerulus / Bowman's capsule / renal capsule; [1] component: red blood cells / erythrocytes / white blood cells / leucocytes / platelets / large blood proteins / albumen / fibrinogen / globulins; [1]
(c)(ii) any three from: 1. active transport; 2. (movement) against a concentration gradient; 3. requires energy / ATP (from cellular respiration); 4. involving protein carriers / carrier molecules / protein pumps (in the cell membrane); 5. diffusion / facilitated diffusion (from cell to capillary); [max 3]
(d) any two from: 1. dialysis fluid contains no urea / lower concentration of urea than blood (so urea diffuses out of blood down concentration gradient); 2. dialysis fluid contains the same / equal concentration of glucose (as healthy blood); 3. so no net loss / movement of glucose from blood (by diffusion); 4. constant replacement / flow of fresh dialysis fluid maintains concentration gradient; [max 2]
Paper 61 (Alternative to Practical)
Answer all questions. Write your answers in the spaces provided.
2 题目 · 40 分
题目 1 · alternative to practical
20 分
1 A student investigated the effect of sucrose solutions of different concentrations on the mass of cylinders cut from a sweet potato (\textit{Ipomoea batatas}).
The student used the following method: • Six cylinders of sweet potato tissue were cut using a cork borer. • Each cylinder was trimmed to a length of \(40\text{ mm}\). • The initial mass of each cylinder was recorded. • Each cylinder was placed into a test-tube containing \(20\text{ cm}^3\) of one of six sucrose solutions: \(0.0\text{ mol/dm}^3\), \(0.2\text{ mol/dm}^3\), \(0.4\text{ mol/dm}^3\), \(0.6\text{ mol/dm}^3\), \(0.8\text{ mol/dm}^3\), and \(1.0\text{ mol/dm}^3\). • The test-tubes were left at room temperature for \(60\text{ minutes}\). • After \(60\text{ minutes}\), the cylinders were removed, blotted dry with paper towels, and their final masses recorded.
(a) (i) Calculate the change in mass and the percentage change in mass for the sweet potato cylinders placed in \(0.4\text{ mol/dm}^3\) and \(0.8\text{ mol/dm}^3\) sucrose solutions. Write your answers in the spaces in Table 1.1. Give your percentage change values to one decimal place. [3]
(ii) Plot a line graph on the grid of the percentage change in mass against the concentration of sucrose solution. Draw a line of best fit. [4]
(iii) Use your graph to estimate the concentration of sucrose solution that has the same water potential as the sweet potato cell sap. Show clearly on your graph how you obtained your answer.
(b) (i) State why the student calculated the percentage change in mass rather than comparing the change in mass alone. [1]
(ii) State one variable, other than temperature and immersion time, that was kept constant in this investigation. Explain why this variable was kept constant. [2]
(c) State two sources of experimental error in this method and suggest an improvement for each error. [4]
(d) Sweet potato contains reducing sugars. Describe how you would carry out a test on sweet potato tissue to show the presence of reducing sugars, and state the observation for a positive result. [4]
(ii) Graph plotting: - \(x\)-axis: concentration of sucrose solution / \(\text{mol/dm}^3\) - \(y\)-axis: percentage change in mass / \(\%\) - Scales: even, linear, covering more than half the grid area in both directions (with \(0\%\) centrally located on \(y\)-axis to accommodate positive and negative values). - Points: correctly plotted within \(\pm 0.5\) small square. - Line: single smooth line of best fit or point-to-point joined with neat straight lines passing through points.
(iii) The concentration at which the line crosses the \(x\)-axis (\(0\%\) change in mass) is determined by reading the intercept: approximately \(0.46\text{ mol/dm}^3\) (acceptable range: \(0.44\text{ to }0.48\text{ mol/dm}^3\)). Clear construction lines shown on the graph.
(b) (i) The cylinders had different starting/initial masses, so percentage change allows a valid comparison.
(ii) Constant variable: Volume of sucrose solution \((20\text{ cm}^3)\) / length or diameter of cylinders \((40\text{ mm})\). Explanation: Differing surface area would alter the rate of osmosis; differing volume might cause changes in external concentration.
(c) Error 1: Uneven blotting of sweet potato cylinders before weighing. Improvement: Standardise blotting (e.g. roll each cylinder twice on a paper towel with uniform pressure). Error 2: Evaporation of water from open test-tubes altering concentration. Improvement: Cover tubes with bungs/parafilm/stoppers.
(d) Method: 1. Crush/grate sweet potato tissue with a small volume of distilled water to release cell contents (or cut thin slices). 2. Add Benedict's reagent/solution to the sample in a test-tube. 3. Heat the mixture in a hot water bath (at least \(80\text{ }^\circ\text{C}\)) for 3–5 minutes. 4. Positive result: colour changes from blue to green / yellow / orange / brick-red.
评分标准
(a)(i) [3] • \(+0.08\) and \(-0.43\) (both required for change in mass); [1] • \(+1.9\) (or \(+1.88\)) for \(0.4\text{ mol/dm}^3\); [1] • \(-10.0\) for \(0.8\text{ mol/dm}^3\); [1] (Guidance: + and - signs required; A \(+1.88\) and \(-10\))
(a)(ii) [4] • \(\mathbf{A}\) (axes): concentration of sucrose solution / \(\text{mol/dm}^3\) on \(x\)-axis AND percentage change in mass / \(\%\) on \(y\)-axis; [1] • \(\mathbf{S}\) (scale): suitable linear scales with origin labelled, occupying \(\ge 50\%\) of grid in both dimensions; [1] • \(\mathbf{P}\) (plotting): all 6 points plotted correctly to within \(\pm\) half a small square; [1] • \(\mathbf{L}\) (line): single, continuous, clean line of best fit or neat point-to-point lines; [1] (R thick / double lines / extrapolating beyond \(1.0\text{ mol/dm}^3\))
(a)(iii) [2] • Value read correctly from candidate's graph where percentage change is \(0\%\) (range \(0.44\text{--}0.48\text{ mol/dm}^3\)); [1] • Construction lines / clear indication on the graph showing intercept at \(y = 0\); [1]
(b)(i) [1] • Cylinders had different initial masses / to allow a fair / valid comparison (between cylinders); [1]
(b)(ii) [2] • Surface area / diameter / length / volume of solution / source of sweet potato; [1] • To ensure rate of water movement is not affected / ensure concentration gradient is unaffected / avoid a confounding variable; [1]
(c) [4] (Any two error + matching improvement pairs for 2 marks each) • Error: blotting variation / water remaining on surface / pressing too hard; [1] Improvement: standardise blotting technique (e.g. roll each cylinder three times on paper towel); [1] • Error: evaporation from open test-tubes; [1] Improvement: place bungs / stoppers / paraffin film over the test-tubes; [1] • Error: only one cylinder tested at each concentration / no repeats; [1] Improvement: repeat test 3 times at each concentration and calculate mean; [1]
(d) [4] • Grind / crush sweet potato tissue with distilled water / make liquid extract; [1] • Add Benedict's solution / reagent; [1] • Heat in a water bath / temperature \(>75\text{ }^\circ\text{C}\) / boiling water bath for \(2\text{--}5\text{ minutes}\); [1] • Positive observation: (colour changes from blue to) green / yellow / orange / red / brick-red (precipitate); [1] (R if heating step is omitted or if acid hydrolysis is mentioned without neutralisation)
题目 2 · alternative to practical
20 分
2 Fig. 2.1 is a photograph showing the cross-section of a mature fruit of a garden pea (\textit{Pisum sativum}) showing an attached developing seed.
Fig. 2.1 [A high-detail macro photograph shows the open green pod containing a round seed. A straight line AB is drawn across the maximum diameter of the seed. Line AB measures \(42\text{ mm}\). The magnification of the image is \(\times 3.5\).]
(a) (i) Make a large, labelled diagram of the seed and its stalk (funicle) shown in Fig. 2.1. [4]
(ii) The line \(\text{AB}\) on Fig. 2.1 represents the maximum diameter of the seed. Measure the length of line \(\text{AB}\) on Fig. 2.1 in millimetres.
length of line \(\text{AB}\) = .................................................... \(\text{mm}\)
Calculate the actual diameter of the seed. Give your answer to one decimal place.
Space for working.
actual diameter = .................................................... \(\text{mm}\) [3]
(b) A student tested the crushed pea seed for the presence of protein and lipid.
(i) State the reagent used to test for protein and state the colour observed if protein is present.
(ii) Describe the method for the emulsion test used to confirm the presence of lipids in the seed extract, and state the positive result. [3]
(c) Fresh pea pods contain vitamin C (ascorbic acid). When vegetables are cooked in boiling water, some vitamin C is destroyed by heat.
Plan an investigation to determine the effect of heating temperature on the concentration of vitamin C remaining in pea extract.
You are provided with fresh pea extract and DCPIP solution.
In your plan, include: • how the independent variable will be changed • the dependent variable and how it will be measured • the variables that must be kept constant and how they will be controlled • how you will ensure the results are reliable • relevant safety precautions. [6]
(d) State the colour change observed when DCPIP reacts with vitamin C. [1]
(e) Describe a control experiment that could be carried out for the vitamin C test with DCPIP and explain its purpose. [1]
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解题
(a) (i) Biological drawing: - Clear, continuous single outlines without shading, hatching, or sketchy lines. - Size occupies at least half of the designated response space. - Accurate representation of features (round seed with outer layer and distinct funicle/stalk attached). - One correct label with a straight line touching the structure (e.g. testa / seed coat / stalk / funicle).
(ii) Measurement of line AB: \(42\text{ mm}\) (allow \(41\text{--}43\text{ mm}\)). Formula: \(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}}\) $$\text{Actual diameter} = \frac{42\text{ mm}}{3.5} = 12.0\text{ mm}$$
(ii) Emulsion test: 1. Add ethanol to the crushed seed sample and shake/mix thoroughly to dissolve lipids. 2. Decant/pour the ethanol solution into a tube containing cold distilled water (or add water to ethanol extract). 3. Positive observation: A milky-white / cloudy emulsion forms.
(c) Investigation plan: - Independent variable: Heat samples of pea extract at a minimum of 5 different temperatures across the range \(20\text{ }^\circ\text{C}\text{ to }100\text{ }^\circ\text{C}\) (e.g. \(20, 40, 60, 80, 100\text{ }^\circ\text{C}\)) using thermostatically controlled water baths for a standardised duration (e.g. \(10\text{ min}\)). - Dependent variable: Measure the volume (or number of drops) of heated pea extract required to completely decolourise a fixed volume (e.g. \(1.0\text{ cm}^3\)) of DCPIP solution from blue to colourless, using a graduated syringe, burette, or dropping pipette. - Standardised variables: Volume and concentration of DCPIP solution, initial volume/batch of pea extract, heating time before titration, mixing/swirling after each drop. - Reliability: Perform 3 replicates at each temperature and calculate the mean volume of extract required. - Safety: Wear safety goggles to protect eyes from chemicals/hot liquid, use tongs/heatproof gloves when handling hot tubes.
(d) Colour change: Blue to colourless (or blue to pink).
(e) Control: Add distilled water dropwise to \(1.0\text{ cm}^3\) of DCPIP solution instead of pea extract. Purpose: To prove that DCPIP does not decolourise on its own or in the presence of water without vitamin C.
评分标准
(a)(i) [4] • \(\mathbf{O}\) (outline): single, sharp, unbroken lines without sketchy overlaps or shading; [1] • \(\mathbf{S}\) (size): drawing occupies \(\ge 50\%\) of the provided space; [1] • \(\mathbf{D}\) (detail): realistic representation of the seed attached to the stalk (funicle), with testa clearly distinguishable; [1] • \(\mathbf{L}\) (label): at least one correct label with straight guideline touching the structure (testa / seed coat / funicle / stalk / cotyledon); [1]
(a)(ii) [3] • Measurement of line \(\text{AB}\): \(42\text{ mm}\) (allow \(41\text{--}43\text{ mm}\)); [1] • Correct substitution into formula: \(\frac{\text{measured length}}{3.5}\); [1] • Correct calculation to 1 d.p. (e.g. \(12.0\text{ mm}\)); [1] (ecf applies if measurement is within \(41\text{--}43\text{ mm}\))
(b)(ii) [3] • Add ethanol to sample and shake / mix; [1] • Add water / pour ethanol solution into water; [1] • Positive result: white emulsion / cloudy / milky; [1]
(c) [6] (Maximum of 6 marks from the following points) • \(\mathbf{IV}\): at least 5 different temperatures named / specified range between \(20\text{ }^\circ\text{C}\) and \(100\text{ }^\circ\text{C}\) using water baths; [1] • \(\mathbf{DV}\): titration method: record volume / count number of drops of extract added to decolourise DCPIP; [1] • \(\mathbf{End\text{-}point}\): blue to colourless (or pink); [1] • \(\mathbf{CV}\) (any two): fixed volume of DCPIP, fixed concentration of DCPIP, same heating time (e.g. \(10\text{ min}\)), same source/batch of pea extract; [max 2] • \(\mathbf{Replication}\): repeat at least 3 times at each temperature AND calculate mean; [1] • \(\mathbf{Safety}\): eye protection / heat-resistant gloves / use test-tube holders with hot liquids; [1]
(d) [1] • Blue to colourless / pink; [1]
(e) [1] • Use water instead of pea extract to show that water does not cause decolourisation / changes are due to vitamin C only; [1]
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