Cambridge IGCSE · Thinka 原創模擬試題

2024 Cambridge IGCSE Biology (0610) 模擬試題連答案詳解

Thinka Jun 2024 (V2) Cambridge International A Level-Style Mock — Biology (0610)

160 180 分鐘2024
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V2) Cambridge International A Level Biology (0610) paper. Not affiliated with or reproduced from Cambridge.

卷二 (選擇題)

Answer all forty multiple choice questions. Choose the single best option (A, B, C, or D) and record on the answer sheet.
40 題目 · 40
題目 1 · 選擇題
1
Which changes will increase the rate of diffusion of oxygen into an active muscle cell? 1. An increase in the temperature of the cell. 2. An increase in the surface area of the cell membrane. 3. A decrease in the concentration of oxygen inside the cell.
  1. A.1 and 2 only
  2. B.1 and 3 only
  3. C.2 and 3 only
  4. D.1, 2 and 3
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解題

All three factors will increase the rate of diffusion: 1. Higher temperature increases the kinetic energy of the particles, leading to faster movement. 2. A larger surface area of the cell membrane provides more space for molecules to cross. 3. A decrease in the concentration of oxygen inside the cell increases the concentration gradient between the outside and the inside of the cell, accelerating net movement.

評分準則

Award 1 mark for identifying that all three factors (temperature, surface area, and concentration gradient) positively affect the rate of diffusion.
題目 2 · 選擇題
1
What is the primary function of the amniotic sac and amniotic fluid during human pregnancy?
  1. A.To exchange nutrients and urea between the maternal and fetal blood systems.
  2. B.To protect the developing fetus against physical shocks and mechanical damage.
  3. C.To secrete progesterone to maintain the lining of the uterus during gestation.
  4. D.To prevent the mixing of maternal and fetal blood to avoid immune rejection.
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解題

The amniotic sac secretes amniotic fluid, which cushions and protects the developing fetus against mechanical shocks and physical damage, and helps to maintain a constant temperature. Option A and D refer to the functions of the placenta. Option C is a function of the corpus luteum and later the placenta.

評分準則

Award 1 mark for identifying the protective cushioning role of the amniotic sac and fluid.
題目 3 · 選擇題
1
A person walks out of a dark room into bright sunlight. Which row correctly describes the response of the iris muscles and the change in pupil size?
  1. A.circular muscles: contract | radial muscles: relax | pupil size: constricts
  2. B.circular muscles: contract | radial muscles: relax | pupil size: dilates
  3. C.circular muscles: relax | radial muscles: contract | pupil size: constricts
  4. D.circular muscles: relax | radial muscles: contract | pupil size: dilates
查看答案詳解

解題

In bright light, the pupil must constrict to limit the amount of light entering the eye and prevent damage to the retina. This pupil reflex is achieved by the contraction of the circular muscles of the iris and the relaxation of the radial muscles.

評分準則

Award 1 mark for correctly identifying that circular muscles contract, radial muscles relax, and the pupil constricts in bright light.
題目 4 · 選擇題
1
The rate of photosynthesis in a glasshouse crop is limited by carbon dioxide concentration. Which modification inside the glasshouse would most effectively increase the rate of photosynthesis?
  1. A.Lowering the temperature to reduce respiration rate.
  2. B.Burning paraffin heaters to release carbon dioxide and heat.
  3. C.Installing shading screens to reduce the light intensity.
  4. D.Increasing the humidity to decrease the transpiration rate.
查看答案詳解

解題

Burning paraffin releases carbon dioxide gas as a product of combustion and also releases thermal energy. This increases both the concentration of CO2 (the limiting factor) and the temperature, allowing the rate of photosynthesis to increase.

評分準則

Award 1 mark for selecting the option that directly addresses the limiting factor (carbon dioxide) by burning paraffin.
題目 5 · 選擇題
1
A plant shoot is illuminated from one side. Which statement correctly describes the distribution of auxin and its effect on the cells in this shoot?
  1. A.Auxin accumulates on the shaded side of the shoot, causing cells on that side to elongate.
  2. B.Auxin accumulates on the shaded side of the shoot, inhibiting cell elongation on that side.
  3. C.Auxin accumulates on the illuminated side of the shoot, causing cells on that side to elongate.
  4. D.Auxin accumulates on the illuminated side of the shoot, inhibiting cell elongation on that side.
查看答案詳解

解題

Auxin is a plant hormone made in the shoot tip that stimulates cell elongation. Under unilateral light, auxin moves away from the light and accumulates on the shaded side of the shoot, causing the cells on the shaded side to elongate more than those on the lit side, bending the shoot toward the light.

評分準則

Award 1 mark for identifying that auxin accumulates on the shaded side of a shoot and causes cell elongation.
題目 6 · 選擇題
1
Which row correctly identifies the products of anaerobic respiration in yeast cells and in human muscle cells?
  1. A.Yeast cells: carbon dioxide and water | Human muscle cells: lactic acid
  2. B.Yeast cells: ethanol and carbon dioxide | Human muscle cells: lactic acid
  3. C.Yeast cells: ethanol and carbon dioxide | Human muscle cells: lactic acid and carbon dioxide
  4. D.Yeast cells: lactic acid | Human muscle cells: ethanol and carbon dioxide
查看答案詳解

解題

In yeast, anaerobic respiration (fermentation) produces ethanol and carbon dioxide. In human muscle cells during vigorous exercise, anaerobic respiration produces only lactic acid (lactate) without releasing carbon dioxide.

評分準則

Award 1 mark for correctly matching the products of anaerobic respiration in both yeast and humans.
題目 7 · 選擇題
1
Which structural feature of an intestinal villus is correctly matched with its adaptation for absorption?
  1. A.Lacteal - carries amino acids and glucose directly to the liver
  2. B.Microvilli - secrete enzymes to finalize protein digestion
  3. C.Thin epithelium - reduces the distance over which nutrients must diffuse
  4. D.Blood capillaries - absorb fatty acids and glycerol to transport around the body
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解題

The epithelium of the villus is only one cell thick, which minimizes the diffusion distance for absorbed nutrients like glucose and amino acids. Lacteals absorb fatty acids and glycerol; blood capillaries absorb glucose and amino acids; microvilli provide a massive surface area for absorption but do not secrete enzymes (membrane-bound enzymes on their surface are for digestion, but their structural role is surface area).

評分準則

Award 1 mark for identifying that a thin epithelium minimizes the diffusion distance for nutrient absorption.
題目 8 · 選擇題
1
By which process does carbon dioxide leave an active respiring muscle cell, and how does its concentration in the cell compare to that in the surrounding capillary blood?
  1. A.process: Active transport | concentration in cell: Higher than in capillary
  2. B.process: Active transport | concentration in cell: Lower than in capillary
  3. C.process: Diffusion | concentration in cell: Higher than in capillary
  4. D.process: Diffusion | concentration in cell: Lower than in capillary
查看答案詳解

解題

Carbon dioxide is a small, uncharged gas molecule that moves passively across cell membranes by diffusion. Because the cell is actively respiring, it continuously produces carbon dioxide, maintaining a higher concentration of CO2 inside the cell compared to the capillary blood, allowing it to diffuse down its concentration gradient.

評分準則

Award 1 mark for identifying diffusion as the process and that concentration inside the respiring cell is higher than in the capillary.
題目 9 · multiple_choice
1
The table shows the concentration of a solute inside four different animal cells and in the surrounding extracellular fluid.

$$\begin{array}{|c|c|c|} \hline \text{Cell} & \text{Concentration in cytoplasm / mmol dm}^{-3} & \text{Concentration in extracellular fluid / mmol dm}^{-3} \\\\ \hline \text{W} & 2 & 12 \\\\ \hline \text{X} & 12 & 2 \\\\ \hline \text{Y} & 5 & 8 \\\\ \hline \text{Z} & 8 & 5 \\\\ \hline \end{array}$$

Through which cell membrane will there be the fastest net movement of this solute **into** the cell by diffusion?
  1. A.cell W
  2. B.cell X
  3. C.cell Y
  4. D.cell Z
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解題

Diffusion is the net movement of particles from a region of their higher concentration to a region of their lower concentration down a concentration gradient. For the solute to move into the cell, the concentration of the solute in the extracellular fluid must be higher than in the cytoplasm. This is true for cell W (difference of \(12 - 2 = 10\text{ mmol dm}^{-3}\)) and cell Y (difference of \(8 - 5 = 3\text{ mmol dm}^{-3}\)). The rate of diffusion is proportional to the steepness of the concentration gradient. Therefore, cell W will have the fastest net movement into the cell because it has the largest concentration gradient.

評分準則

Award 1 mark for selecting the correct cell (W) which has the greatest concentration gradient favouring movement into the cell (A).
題目 10 · multiple_choice
1
Which pathway correctly shows the route taken by a sperm cell from its site of production until it can fertilise an egg cell?
  1. A.testis \(\rightarrow\) sperm duct \(\rightarrow\) urethra \(\rightarrow\) vagina \(\rightarrow\) cervix \(\rightarrow\) uterus \(\rightarrow\) oviduct
  2. B.testis \(\rightarrow\) ureter \(\rightarrow\) urethra \(\rightarrow\) vagina \(\rightarrow\) uterus \(\rightarrow\) cervix \(\rightarrow\) oviduct
  3. C.ovary \(\rightarrow\) oviduct \(\rightarrow\) uterus \(\rightarrow\) cervix \(\rightarrow\) vagina
  4. D.testis \(\rightarrow\) sperm duct \(\rightarrow\) urethra \(\rightarrow\) vagina \(\rightarrow\) cervix \(\rightarrow\) oviduct \(\rightarrow\) uterus
查看答案詳解

解題

Sperm cells are produced in the testes. During ejaculation, they travel through the sperm duct and out through the urethra. Upon entering the female reproductive tract, they pass through the vagina, past the cervix, through the uterus, and finally into the oviduct, which is where fertilisation normally occurs.

評分準則

Award 1 mark for selecting the correct pathway (A).
題目 11 · multiple_choice
1
A person is looking at a star in the night sky. They then look down to read a message on their mobile phone screen.

Which changes must occur in the ciliary muscles and the suspensory ligaments to focus on the mobile phone screen?
  1. A.The ciliary muscles contract and the suspensory ligaments slacken.
  2. B.The ciliary muscles contract and the suspensory ligaments tighten.
  3. C.The ciliary muscles relax and the suspensory ligaments slacken.
  4. D.The ciliary muscles relax and the suspensory ligaments tighten.
查看答案詳解

解題

To change focus from a distant object (the star) to a near object (the phone screen), the eye undergoes accommodation. The ciliary muscles contract, which decreases their diameter. This relieves tension on the suspensory ligaments, causing them to slacken (become loose). As a result, the elastic lens becomes thicker and more rounded, increasing its refractive power to focus light from the near object onto the retina.

評分準則

Award 1 mark for identifying that ciliary muscles contract and suspensory ligaments slacken (A).
題目 12 · multiple_choice
1
At a very low light intensity, a plant was supplied with an increased concentration of carbon dioxide. However, the rate of photosynthesis did not change.

What was the limiting factor for photosynthesis under these conditions?
  1. A.carbon dioxide concentration
  2. B.chlorophyll concentration
  3. C.light intensity
  4. D.temperature
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解題

A limiting factor is something present in the environment in such short supply that it restricts life processes. Since increasing the concentration of carbon dioxide did not increase the rate of photosynthesis, carbon dioxide is not the limiting factor. Because the light intensity is described as 'very low', it is the light intensity itself that is limiting the rate of photosynthesis.

評分準則

Award 1 mark for identifying light intensity as the limiting factor (C).
題目 13 · multiple_choice
1
A plant shoot is growing with light shining from one side only.

Which statement correctly describes the movement of auxin and the resulting growth of the shoot?
  1. A.Auxin is destroyed on the shaded side of the shoot, causing the lit side to grow faster.
  2. B.Auxin accumulates on the shaded side of the shoot, causing the shaded side to elongate faster.
  3. C.Auxin is destroyed on the lit side of the shoot, causing the lit side to elongate faster.
  4. D.Auxin accumulates on the lit side of the shoot, causing the shaded side to grow slower.
查看答案詳解

解題

In a phototropic response, auxin produced in the shoot tip diffuses downwards and concentrates on the shaded side of the shoot. In shoots, a higher concentration of auxin stimulates cell elongation. Consequently, cells on the shaded side elongate faster than cells on the illuminated side, causing the shoot to bend towards the light source.

評分準則

Award 1 mark for the correct explanation that auxin accumulates on the shaded side, stimulating elongation on that side (B).
題目 14 · multiple_choice
1
Which row correctly identifies the products of anaerobic respiration in human muscle cells and in yeast cells?

$$\begin{array}{|c|c|c|} \hline & \text{Products in human muscle cells} & \text{Products in yeast cells} \\\\ \hline \text{A} & \text{lactic acid only} & \text{alcohol (ethanol) and carbon dioxide} \\\\ \hline \text{B} & \text{lactic acid and carbon dioxide} & \text{alcohol (ethanol) only} \\\\ \hline \text{C} & \text{alcohol (ethanol) and carbon dioxide} & \text{lactic acid only} \\\\ \hline \text{D} & \text{lactic acid only} & \text{lactic acid and carbon dioxide} \\\\ \hline \end{array}$$
  1. A.A
  2. B.B
  3. C.C
  4. D.D
查看答案詳解

解題

In human muscle cells, anaerobic respiration during vigorous exercise breaks down glucose to produce lactic acid only (no carbon dioxide is produced). In yeast cells, anaerobic respiration (also known as fermentation) breaks down glucose to produce alcohol (ethanol) and carbon dioxide.

評分準則

Award 1 mark for selecting the row (A) showing lactic acid only for human muscle and alcohol and carbon dioxide for yeast.
題目 15 · multiple_choice
1
Which row correctly matches a structure inside a villus of the small intestine with the primary nutrient it is adapted to absorb and transport?

$$\begin{array}{|c|c|c|} \hline & \text{Structure of villus} & \text{Nutrient absorbed and transported} \\\\ \hline \text{A} & \text{lacteal} & \text{fatty acids and glycerol} \\\\ \hline \text{B} & \text{lacteal} & \text{glucose and amino acids} \\\\ \hline \text{C} & \text{blood capillary} & \text{fatty acids and glycerol} \\\\ \hline \text{D} & \text{epithelium} & \text{glycogen and proteins} \\\\ \hline \end{array}$$
  1. A.A
  2. B.B
  3. C.C
  4. D.D
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解題

The villi are finger-like projections in the small intestine that increase the surface area for absorption. Inside each villus, there is a lacteal (lymphatic capillary) and a network of blood capillaries. The lacteal is responsible for the absorption and transport of fatty acids and glycerol (products of fat digestion). The blood capillaries absorb and transport water-soluble molecules such as glucose and amino acids.

評分準則

Award 1 mark for selecting the row (A) where the lacteal matches with fatty acids and glycerol.
題目 16 · multiple_choice
1
Which set of alveolar conditions would result in the fastest rate of diffusion of oxygen from the alveoli into the blood capillaries?

$$\begin{array}{|c|c|c|c|} \hline & \text{Surface area of alveoli} & \text{Distance for diffusion} & \text{Concentration gradient of oxygen} \\\\ \hline \text{A} & \text{large} & \text{small} & \text{large} \\\\ \hline \text{B} & \text{large} & \text{large} & \text{small} \\\\ \hline \text{C} & \text{small} & \text{small} & \text{large} \\\\ \hline \text{D} & \text{small} & \text{large} & \text{small} \\\\ \hline \text{/em}\n\end{array}$$
  1. A.A
  2. B.B
  3. C.C
  4. D.D
查看答案詳解

解題

The rate of diffusion is increased by: a larger surface area (more area for gas exchange to occur), a smaller distance for diffusion (thinner barrier for particles to travel through), and a larger (steeper) concentration gradient (greater difference in concentration between the two sides). Therefore, the combination of a large surface area, a small distance, and a large concentration gradient will result in the fastest rate of diffusion.

評分準則

Award 1 mark for selecting the correct row (A) indicating large surface area, small distance, and large concentration gradient.
題目 17 · multiple_choice
1
Four identical agar cubes containing universal indicator are placed into beakers of hydrochloric acid under different conditions. Which combination of conditions will result in the indicator changing color throughout the cube in the shortest time?
  1. A.Acid concentration: \(0.1\text{ mol/dm}^3\), Temperature: \(20^\circ\text{C}\)
  2. B.Acid concentration: \(1.0\text{ mol/dm}^3\), Temperature: \(20^\circ\text{C}\)
  3. C.Acid concentration: \(0.1\text{ mol/dm}^3\), Temperature: \(40^\circ\text{C}\)
  4. D.Acid concentration: \(1.0\text{ mol/dm}^3\), Temperature: \(40^\circ\text{C}\)
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解題

The rate of diffusion is increased by both a higher concentration gradient and a higher temperature. A higher concentration of hydrochloric acid (\(1.0\text{ mol/dm}^3\) compared to \(0.1\text{ mol/dm}^3\)) provides a steeper concentration gradient, increasing the net movement of acid particles into the agar. A higher temperature (\(40^\circ\text{C}\) compared to \(20^\circ\text{C}\)) gives the acid particles more kinetic energy, causing them to move and diffuse faster. Therefore, the combination of high concentration and high temperature (Option D) results in the fastest diffusion.

評分準則

Award 1 mark for the correct option (D).
- Reject A, B, and C as they do not combine both the highest concentration and highest temperature.
題目 18 · multiple_choice
1
A cell is surrounded by a solution containing dissolved oxygen. The concentration of oxygen outside the cell is greater than the concentration inside the cell. Which change would decrease the rate of diffusion of oxygen into the cell?
  1. A.Decreasing the temperature of the solution
  2. B.Increasing the surface area of the cell membrane
  3. C.Decreasing the oxygen concentration inside the cell
  4. D.Increasing the oxygen concentration outside the cell
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解題

Diffusion is the net movement of particles from a region of their higher concentration to a region of their lower concentration down a concentration gradient, as a result of their random movement. Decreasing the temperature of the solution (Option A) reduces the kinetic energy of the oxygen molecules, which decreases their speed of random movement and thus slows down the rate of diffusion. Increasing the surface area (Option B), decreasing the internal concentration (Option C, which steepens the gradient), or increasing the external concentration (Option D, which also steepens the gradient) would all increase the rate of diffusion.

評分準則

Award 1 mark for the correct option (A).
- Reject B, C, and D because these changes would increase, rather than decrease, the rate of diffusion.
題目 19 · multiple_choice
1
Which pathway correctly describes the route taken by sperm from their site of production until they exit the male body?
  1. A.testis \(\rightarrow\) urethra \(\rightarrow\) sperm duct
  2. B.testis \(\rightarrow\) sperm duct \(\rightarrow\) urethra
  3. C.sperm duct \(\rightarrow\) testis \(\rightarrow\) urethra
  4. D.urethra \(\rightarrow\) sperm duct \(\rightarrow\) testis
查看答案詳解

解題

Sperm are produced in the testes (testis). They then travel through the sperm duct (vas deferens) and enter the urethra, which conducts them out of the body through the penis during ejaculation. Therefore, the correct pathway is testis \(\rightarrow\) sperm duct \(\rightarrow\) urethra.

評分準則

Award 1 mark for the correct option (B).
- Reject A because sperm must pass through the sperm duct before reaching the urethra.
- Reject C and D because sperm production begins in the testis, not the sperm duct or urethra.
題目 20 · multiple_choice
1
A person walks from a dark room into bright sunlight. Which row correctly describes the response of the iris muscles and the pupil?
  1. A.Circular muscles: contract; Radial muscles: relax; Pupil: constricts
  2. B.Circular muscles: relax; Radial muscles: contract; Pupil: constricts
  3. C.Circular muscles: contract; Radial muscles: relax; Pupil: dilates
  4. D.Circular muscles: relax; Radial muscles: contract; Pupil: dilates
查看答案詳解

解題

In bright light, the pupil constricts to protect the retina from damage by reducing the amount of light entering the eye. This pupil reflex is controlled by antagonistic muscles in the iris: the circular muscles contract, and the radial muscles relax. Therefore, option A is correct.

評分準則

Award 1 mark for the correct option (A).
- Reject B because circular muscles contract and radial muscles relax to constrict the pupil.
- Reject C and D because the pupil constricts, not dilates, in bright light.
題目 21 · multiple_choice
1
An experiment was set up to investigate the rate of photosynthesis in Elodea (water plant) at different light intensities. Carbon dioxide concentration and temperature were kept constant. At high light intensities, the rate of photosynthesis remained constant even when light intensity was increased further. What could explain this observation?
  1. A.Light intensity has become a limiting factor.
  2. B.Temperature or carbon dioxide concentration is now a limiting factor.
  3. C.The plant has run out of oxygen.
  4. D.The rate of respiration has decreased to zero.
查看答案詳解

解題

When light intensity is increased but the rate of photosynthesis no longer increases, light intensity is no longer the limiting factor. At this point, some other factor that is kept constant—such as temperature or carbon dioxide concentration—limits the rate of photosynthesis. Therefore, Option B is correct.

評分準則

Award 1 mark for the correct option (B).
- Reject A because light intensity is no longer limiting when its increase does not affect the rate.
- Reject C and D as they are biologically incorrect descriptions of limiting factors under these conditions.
題目 22 · multiple_choice
1
A seedling is placed horizontally in a dark box. After several days, the shoot grows upwards and the root grows downwards. Which statement correctly explains the role of auxin in the downward growth of the root?
  1. A.Auxin accumulates on the lower side of the root and stimulates cell elongation there.
  2. B.Auxin accumulates on the lower side of the root and inhibits cell elongation there.
  3. C.Auxin accumulates on the upper side of the root and stimulates cell elongation there.
  4. D.Auxin accumulates on the upper side of the root and inhibits cell elongation there.
查看答案詳解

解題

Due to gravity, auxin accumulates on the lower side of both the shoot and the root. In roots, a high concentration of auxin inhibits cell elongation. Consequently, cells on the upper side of the root (where there is less auxin) elongate more rapidly than those on the lower side, causing the root to bend downwards (positive gravitropism). Therefore, Option B is correct.

評分準則

Award 1 mark for the correct option (B).
- Reject A because auxin inhibits, rather than stimulates, cell elongation in roots.
- Reject C and D because gravity causes auxin to accumulate on the lower side of the root, not the upper side.
題目 23 · multiple_choice
1
Which row correctly shows the products of anaerobic respiration in yeast and in human muscle cells?
  1. A.Yeast: alcohol and carbon dioxide; Human muscle: lactic acid only
  2. B.Yeast: lactic acid only; Human muscle: alcohol and carbon dioxide
  3. C.Yeast: alcohol and carbon dioxide; Human muscle: lactic acid and carbon dioxide
  4. D.Yeast: lactic acid and carbon dioxide; Human muscle: lactic acid only
查看答案詳解

解題

Anaerobic respiration in yeast (fermentation) produces alcohol (ethanol) and carbon dioxide. Anaerobic respiration in human muscle cells during vigorous exercise produces lactic acid only (no carbon dioxide is produced during this pathway in humans). Thus, Option A is correct.

評分準則

Award 1 mark for the correct option (A).
- Reject B because the products are reversed.
- Reject C because human muscles do not produce carbon dioxide during anaerobic respiration.
- Reject D because yeast does not produce lactic acid.
題目 24 · multiple_choice
1
Which feature of a villus is correctly matched with its role in the absorption of digested food?
  1. A.Lacteal — transports glucose and amino acids
  2. B.Microvilli — decrease the surface area for diffusion
  3. C.One-cell thick epithelium — provides a short diffusion pathway
  4. D.Capillary network — transports fatty acids and glycerol
查看答案詳解

解題

The epithelium of a villus is only one-cell thick, which minimizes the distance over which nutrients must diffuse, providing a short diffusion pathway (Option C). A lacteal absorbs and transports fatty acids and glycerol, not glucose and amino acids (Option A is incorrect). Microvilli increase, rather than decrease, the surface area for diffusion (Option B is incorrect). The blood capillary network absorbs and transports water-soluble molecules like glucose and amino acids, whereas fatty acids and glycerol are mainly absorbed into the lacteal (Option D is incorrect).

評分準則

Award 1 mark for the correct option (C).
- Reject A because lacteals transport fats/lipids.
- Reject B because microvilli increase surface area.
- Reject D because capillaries primarily transport amino acids and glucose, while lacteals transport fatty acids and glycerol.
題目 25 · 選擇題
1
Four agar cubes of different sizes are prepared using an alkaline solution containing phenolphthalein indicator, which turns pink in alkaline conditions. The cubes are then placed into a beaker of dilute hydrochloric acid. The cube sizes (side lengths) are 1 cm, 2 cm, 3 cm, and 4 cm respectively. Which cube will be the first to turn completely colorless?
  1. A.The cube with side length 1 cm
  2. B.The cube with side length 2 cm
  3. C.The cube with side length 3 cm
  4. D.The cube with side length 4 cm
查看答案詳解

解題

The rate of diffusion of the acid is constant. The time taken to reach the center of the cube depends on the distance the acid has to travel (diffusion distance) and the surface area-to-volume ratio of the cube. The 1 cm cube has the smallest side length, meaning it has the shortest distance to its center and the largest surface area-to-volume ratio. Therefore, it will become completely colorless first.

評分準則

1 mark: Correctly identifies the 1 cm cube as the fastest to fully decolorize due to the shortest diffusion distance and highest surface area to volume ratio.
題目 26 · 選擇題
1
Which combination of factors will result in the fastest rate of diffusion across a cell membrane?
  1. A.A thin membrane, a steep concentration gradient, and a high temperature
  2. B.A thin membrane, a shallow concentration gradient, and a low temperature
  3. C.A thick membrane, a steep concentration gradient, and a high temperature
  4. D.A thick membrane, a shallow concentration gradient, and a low temperature
查看答案詳解

解題

The rate of diffusion is increased by a shorter diffusion distance (thin membrane), a greater concentration difference (steep concentration gradient), and higher kinetic energy of particles (high temperature).

評分準則

1 mark: Correctly identifies that a thin membrane, steep concentration gradient, and high temperature maximize the rate of diffusion.
題目 27 · 選擇題
1
Which statement correctly compares human sperm cells and human egg cells?
  1. A.Sperm cells are flagellated and motile, whereas egg cells are non-motile and are moved along the oviduct by ciliated cells.
  2. B.Sperm cells contain 46 chromosomes, whereas egg cells contain only 23 chromosomes.
  3. C.Egg cells contain smaller food reserves in their cytoplasm compared to sperm cells.
  4. D.Sperm cells and egg cells are produced in equal numbers throughout a person's lifetime.
查看答案詳解

解題

Sperm cells are flagellated and motile, while eggs are passive (non-motile) and transported via cilia in the oviduct. Both gametes are haploid (23 chromosomes). Eggs contain a much larger nutrient reserve in their cytoplasm compared to sperm. Millions of sperm are produced daily, whereas egg cells are released in much smaller quantities (typically one per month).

評分準則

1 mark: Correct comparison of sperm and egg cell motility.
題目 28 · 選擇題
1
When a person walks from a dark room into bright sunlight, how do the muscles in the iris of the eye respond to protect the retina from damage?
  1. A.The circular muscles contract and the radial muscles relax.
  2. B.The circular muscles relax and the radial muscles contract.
  3. C.Both circular and radial muscles contract simultaneously.
  4. D.Both circular and radial muscles relax simultaneously.
查看答案詳解

解題

In bright light, the pupil must constrict (get smaller) to limit light entry. This is achieved by the contraction of the circular muscles and the relaxation of the radial muscles in the iris.

評分準則

1 mark: Correct identification of circular muscle contraction and radial muscle relaxation during pupil constriction in bright light.
題目 29 · 選擇題
1
An experiment is conducted using an aquatic plant, Elodea, to measure the rate of photosynthesis by counting the number of oxygen bubbles released per minute. The distance of the light source from the plant is varied. Why does the rate of bubble production decrease as the light source is moved further away?
  1. A.The water temperature decreases, reducing photosynthetic enzyme activity.
  2. B.The concentration of dissolved carbon dioxide in the water decreases.
  3. C.Light intensity decreases, which makes light the limiting factor.
  4. D.The pH of the water increases, denaturing photosynthetic enzymes.
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解題

Moving the light source further away decreases the light intensity reaching the plant. Because light intensity is directly linked to the rate of photosynthesis, a lower light intensity causes a slower rate of oxygen production, meaning light intensity is the limiting factor.

評分準則

1 mark: Correctly identifies light intensity as the limiting factor affecting the rate of bubble production.
題目 30 · 選擇題
1
A plant shoot is placed horizontally in a dark box. After several days, the shoot is observed growing upwards. Which statement explains this gravitropic response?
  1. A.Auxin accumulates on the lower side of the shoot, inhibiting cell elongation on that side.
  2. B.Auxin accumulates on the lower side of the shoot, stimulating cell elongation on that side.
  3. C.Auxin accumulates on the upper side of the shoot, stimulating cell elongation on that side.
  4. D.Auxin is destroyed on the lower side of the shoot, causing the upper side to grow faster.
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解題

Gravity causes auxin to accumulate on the lower side of the horizontal shoot. In shoots, a higher concentration of auxin stimulates cell elongation. Consequently, cells on the lower side elongate more than those on the upper side, causing the shoot to bend upwards.

評分準則

1 mark: Correctly identifies that auxin accumulates on the lower side of the shoot and stimulates elongation on that side.
題目 31 · 選擇題
1
Which row correctly identifies the products of anaerobic respiration in yeast and in human muscle cells?
  1. A.Yeast: carbon dioxide and ethanol | Human muscle cells: lactic acid
  2. B.Yeast: lactic acid | Human muscle cells: carbon dioxide and ethanol
  3. C.Yeast: carbon dioxide and water | Human muscle cells: lactic acid and water
  4. D.Yeast: ethanol only | Human muscle cells: carbon dioxide and lactic acid
查看答案詳解

解題

Anaerobic respiration in yeast produces carbon dioxide and ethanol (fermentation), whereas anaerobic respiration in human muscle cells produces only lactic acid.

評分準則

1 mark: Correctly identifies products of anaerobic respiration for both yeast (ethanol and carbon dioxide) and human muscles (lactic acid).
題目 32 · 選擇題
1
Which feature of a villus in the small intestine is correctly matched with its adaptation for absorption?
  1. A.Microvilli — decrease the diffusion distance of nutrients
  2. B.Lacteal — absorbs glucose and amino acids into the bloodstream
  3. C.Capillary network — absorbs fatty acids and glycerol into the lymphatic system
  4. D.One-cell-thick epithelium — provides a short distance for rapid diffusion of nutrients
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解題

A one-cell-thick epithelium provides a short distance for rapid diffusion of digested nutrients. Microvilli increase surface area rather than decrease diffusion distance. Lacteals absorb fatty acids and glycerol, while the capillary network absorbs glucose and amino acids.

評分準則

1 mark: Correctly matches the single-cell layer thickness of the epithelium with a short diffusion distance.
題目 33 · 選擇題
1
An investigation is carried out to study the diffusion of a dye into cubes of agar. Which combination of factors will result in the shortest time for the dye to diffuse completely to the center of an agar cube?
  1. A.Large cube size, low dye concentration, low temperature
  2. B.Large cube size, high dye concentration, high temperature
  3. C.Small cube size, high dye concentration, high temperature
  4. D.Small cube size, low dye concentration, low temperature Gold-level speed and efficiency is achieved through minimized diffusion distance, maximised gradient and kinetic energy of particles.
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解題

To achieve the shortest time for diffusion to the center of the cube, we need the fastest rate of diffusion and the shortest distance to travel. A smaller agar cube reduces the diffusion distance. High dye concentration increases the concentration gradient, which increases the rate of diffusion. High temperature increases the kinetic energy of the dye molecules, leading to faster movement. Therefore, a small cube under high concentration and high temperature conditions is correct.

評分準則

Award 1 mark for the correct option (C). Any other choice receives 0 marks.
題目 34 · 選擇題
1
Which statement correctly describes the process of diffusion?
  1. A.The active movement of particles down a concentration gradient using energy from respiration.
  2. B.The net movement of particles from a region of their higher concentration to a region of their lower concentration as a result of their random movement.
  3. C.The net movement of water molecules from a region of higher water potential to a region of lower water potential through a partially permeable membrane.
  4. D.The movement of substances against a concentration gradient using energy from respiration.
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解題

Diffusion is defined as the net movement of particles from a region of their higher concentration to a region of their lower concentration as a result of their random movement. It is a passive process and does not require energy from respiration.

評分準則

Award 1 mark for the correct option (B). Any other choice receives 0 marks.
題目 35 · 選擇題
1
What is the correct pathway taken by sperm cells from their site of production to where they leave the male body?
  1. A.testis → sperm duct → urethra
  2. B.testis → urethra → sperm duct
  3. C.prostate gland → sperm duct → urethra
  4. D.sperm duct → testis → urethra
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解題

Sperm cells are produced in the testes (testis). They then travel through the sperm duct (vas deferens) and enter the urethra, through which they exit the body during ejaculation.

評分準則

Award 1 mark for the correct option (A). Any other choice receives 0 marks.
題目 36 · 選擇題
1
A person walks out of a dark cinema into bright afternoon sunlight. Which row correctly describes the responses of the iris muscles and the change in pupil size?
  1. A.Circular muscles: contract | Radial muscles: relax | Pupil size: becomes smaller
  2. B.Circular muscles: relax | Radial muscles: contract | Pupil size: becomes smaller
  3. C.Circular muscles: contract | Radial muscles: relax | Pupil size: becomes larger
  4. D.Circular muscles: relax | Radial muscles: contract | Pupil size: becomes larger
查看答案詳解

解題

In bright light, the pupil must constrict (become smaller) to prevent too much light from entering and damaging the retina. This pupil reflex is controlled by the antagonistic iris muscles: the circular muscles contract and the radial muscles relax.

評分準則

Award 1 mark for the correct option (A). Any other choice receives 0 marks.
題目 37 · 選擇題
1
A plant is kept at a constant favorable temperature of \(20\ ^\circ\text{C}\) with high light intensity, but at a low atmospheric carbon dioxide concentration of \(0.04\%\). The rate of photosynthesis is found to be constant and does not increase when light intensity is increased further. What is the limiting factor for photosynthesis under these conditions?
  1. A.light intensity
  2. B.carbon dioxide concentration
  3. C.temperature
  4. D.water availability
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解題

Since the light intensity is already high and increasing it further has no effect on the rate of photosynthesis, light intensity is not the limiting factor. Given that the temperature is constant and favorable, the rate is restricted by the low availability of carbon dioxide. Therefore, carbon dioxide concentration is the limiting factor.

評分準則

Award 1 mark for the correct option (B). Any other choice receives 0 marks.
題目 38 · 選擇題
1
A plant shoot is placed horizontally in a dark room. Which statement correctly explains the subsequent growth of the shoot?
  1. A.Auxin concentrates on the lower side of the shoot, inhibiting cell elongation, causing the shoot to grow downwards.
  2. B.Auxin concentrates on the upper side of the shoot, stimulating cell elongation, causing the shoot to grow upwards.
  3. C.Auxin concentrates on the lower side of the shoot, stimulating cell elongation, causing the shoot to grow upwards.
  4. D.Auxin remains evenly distributed throughout the shoot, causing it to grow straight ahead horizontally.
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解題

Due to gravity, auxin accumulates on the lower side of the horizontally placed shoot. In plant shoots, a higher concentration of auxin stimulates cell elongation. Consequently, the cells on the lower side elongate more than those on the upper side, causing the shoot to bend and grow upwards (negative gravitropism).

評分準則

Award 1 mark for the correct option (C). Any other choice receives 0 marks.
題目 39 · 選擇題
1
Which statement about anaerobic respiration is correct?
  1. A.In yeast cells, anaerobic respiration produces lactic acid.
  2. B.It releases more energy per glucose molecule than aerobic respiration.
  3. C.In human muscle cells, it produces alcohol and carbon dioxide.
  4. D.It releases less energy per glucose molecule than aerobic respiration because glucose is only partially broken down.
查看答案詳解

解題

Anaerobic respiration does not require oxygen and results in the incomplete breakdown of glucose. Because glucose is only partially broken down, much less energy is released per glucose molecule compared to aerobic respiration.

評分準則

Award 1 mark for the correct option (D). Any other choice receives 0 marks.
題目 40 · 選擇題
1
Which structure in a villus of the small intestine is correctly matched to the substance(s) it absorbs?
  1. A.capillary → fatty acids and glycerol
  2. B.lacteal → glucose and amino acids
  3. C.capillary → glucose and amino acids
  4. D.lacteal → starch and proteins
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解題

The blood capillaries in the villi absorb water-soluble molecules such as glucose and amino acids. The lacteal (lymphatic capillary) absorbs fat-soluble molecules such as fatty acids and glycerol.

評分準則

Award 1 mark for the correct option (C). Any other choice receives 0 marks.

Paper 4 (Written Theory)

Answer all structured questions in the spaces provided. Show calculations and use appropriate units.
6 題目 · 79.98
題目 1 · Structured
13.33
A student investigated the rate of diffusion using agar cubes containing phenolphthalein indicator and sodium hydroxide solution. The indicator turns pink in alkaline conditions and colorless in acidic conditions. The student prepared two cubes: Cube A (1 cm x 1 cm x 1 cm) and Cube B (2 cm x 2 cm x 2 cm). Both cubes were submerged in a beaker of dilute hydrochloric acid. (a) Calculate the surface area to volume ratio (SA:V) for both Cube A and Cube B. Show your working. (b) Predict and explain the difference in the time taken for the pink color to completely disappear in Cube A compared to Cube B. (c) State three other factors, besides surface area to volume ratio, that influence the rate of diffusion of substances into or out of living cells.
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解題

(a) For Cube A: Surface Area = \(6 \times (1\text{ cm} \times 1\text{ cm}) = 6\text{ cm}^2\); Volume = \(1\text{ cm} \times 1\text{ cm} \times 1\text{ cm} = 1\text{ cm}^3\). The surface area to volume ratio is 6:1 (or 6). For Cube B: Surface Area = \(6 \times (2\text{ cm} \times 2\text{ cm}) = 24\text{ cm}^2\); Volume = \(2\text{ cm} \times 2\text{ cm} \times 2\text{ cm} = 8\text{ cm}^3\). The surface area to volume ratio is 24:8 which simplifies to 3:1 (or 3). (b) The pink color in Cube A will completely disappear much faster than in Cube B. Hydrochloric acid neutralizes the sodium hydroxide. Acid moves into the agar cubes by diffusion. Because Cube A has a larger surface area relative to its volume, and a smaller overall distance to its center (0.5 cm compared to 1.0 cm in Cube B), diffusion occurs across a relatively larger surface, neutralizing the volume more quickly. (c) Three other factors that influence diffusion rate: 1. Temperature (higher temperature increases kinetic energy, increasing diffusion rate); 2. Concentration gradient (steeper gradient increases rate); 3. Diffusion distance (thinner membranes/shorter distances increase rate).

評分準則

(a) [3 marks] 1 mark for correct Cube A SA:V ratio (6:1 or 6), 1 mark for correct Cube B SA:V ratio (3:1 or 3), 1 mark for showing correct working of surface area and volume. (b) [6 marks] 1 mark for predicting Cube A is faster / Cube B is slower; 1 mark for identifying that acid neutralizes the alkaline sodium hydroxide; 1 mark for identifying diffusion as the process; 1 mark for stating Cube A has a larger surface area to volume ratio; 1 mark for stating Cube A has a shorter diffusion distance to the center; 1 mark for explaining that a larger SA:V allows faster diffusion per unit volume. (c) [4 marks] 1 mark for each valid factor up to 4 marks: Temperature, Concentration gradient, Diffusion distance / cell membrane thickness, Surface area.
題目 2 · Structured
13.33
The process of human sexual reproduction relies on specialized gametes that undergo fertilization to form a zygote. (a) Describe how the structure of a human sperm cell is adapted to its function of reaching and fertilizing an egg. (b) Explain the process of fertilization, starting from the moment the sperm contacts the jelly coat of the egg cell to the formation of a diploid zygote. (c) State where fertilization normally occurs in the female reproductive system, and describe how the resulting embryo reaches the uterus and implants.
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解題

(a) Human sperm cell adaptations: 1. Acrosome at the head contains digestive enzymes to break down the egg's jelly coat; 2. Flagellum (tail) allows active swimming towards the egg; 3. Spiral mitochondria in the midpiece provide ATP energy for swimming; 4. Haploid nucleus containing 23 chromosomes to restore diploidy. (b) Fertilization process: When the sperm reaches the egg, it contacts the jelly coat and releases enzymes from its acrosome (acrosome reaction). These enzymes digest a path through the jelly coat. The sperm cell membrane then fuses with the egg cell membrane, allowing the sperm nucleus to enter. The egg membrane immediately undergoes changes to block any other sperm (preventing polyspermy). Finally, the haploid sperm nucleus fuses with the haploid egg nucleus to form a diploid zygote. (c) Fertilization occurs in the oviduct (fallopian tube). The zygote begins dividing by mitosis to form an embryo. It is moved along the oviduct toward the uterus by the beating of ciliated cells and peristaltic contractions of the oviduct wall. Once in the uterus, the embryo (as a blastocyst) implants itself into the thickened uterus lining (endometrium).

評分準則

(a) [4 marks] 1 mark each for any four correct adaptation-function pairs: Acrosome containing digestive enzymes, Flagellum/tail for swimming, Mitochondria in midpiece providing energy/ATP, Haploid nucleus to restore diploid number. (b) [5 marks] 1 mark for acrosome releasing enzymes, 1 mark for enzymes digesting the jelly coat, 1 mark for fusion of sperm and egg membranes, 1 mark for egg membrane changing/hardening to prevent other sperm entering, 1 mark for fusion of haploid nuclei to form a diploid zygote. (c) [4 marks] 1 mark for stating oviduct, 1 mark for action of cilia/beating of ciliated cells or peristalsis, 1 mark for division of zygote by mitosis (to form an embryo), 1 mark for implantation into the endometrium/lining of the uterus.
題目 3 · Structured
13.33
A person walking out of a dark cinema room into bright, direct sunlight experiences immediate changes in their eyes to adjust to the light levels. They then focus on a distant sailboat on the horizon. (a) Describe the reflex pathway and pupil changes that occur as the person steps into the bright sunlight, explaining the action of the muscles in the iris. (b) Explain how the focus of the eye changes when the person looks from a near object to a distant sailboat. (c) Contrast the functions of rod cells and cone cells in the retina.
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解題

(a) Pupil reflex: The high light intensity is detected by photoreceptors in the retina. Electrical impulses travel along the sensory neurones in the optic nerve to the brain. The brain coordinates a reflex response, sending impulses along motor neurones to the effector muscles in the iris. The circular muscles contract and the radial muscles relax, causing the pupil to constrict (become smaller), which reduces the amount of light entering the eye to protect the retina from damage. (b) Accommodation for distant objects: When looking at a distant sailboat, the ciliary muscles in the ciliary body relax. This increases the tension on the suspensory ligaments, pulling them tight (taut). The suspensory ligaments pull on the lens, causing the lens to become thinner and flatter. A thinner lens refracts (bends) light rays less, which is necessary because light rays from a distant object are almost parallel and require less refraction to be focused precisely onto the retina (fovea). (c) Rods vs Cones: Rod cells are highly sensitive to low light intensities and do not detect color, providing black-and-white vision in dim light. Cone cells are less sensitive to light, requiring high light intensities (bright light) to function, and they detect color (there are three types sensitive to red, green, and blue light) providing high-acuity color vision.

評分準則

(a) [5 marks] 1 mark for detection of light by photoreceptors / retina; 1 mark for pathway: sensory neurone / optic nerve to brain, and motor neurone to iris; 1 mark for circular muscles contracting; 1 mark for radial muscles relaxing; 1 mark for pupil constricting / becoming smaller. (b) [5 marks] 1 mark for ciliary muscles relaxing; 1 mark for suspensory ligaments becoming tight / taut; 1 mark for lens becoming thinner / flatter; 1 mark for refracting / bending light less; 1 mark for focusing light rays onto the retina / fovea. (c) [3 marks] 1 mark for rods functioning in dim light / low light intensity; 1 mark for cones functioning in bright light / high light intensity; 1 mark for rods providing monochromatic/black-and-white vision and cones providing color vision (or identifying three types of cones: red, green, blue).
題目 4 · Structured
13.33
An experiment was set up to investigate the effect of light intensity on the rate of photosynthesis in the aquatic plant Elodea. (a) Write the balanced chemical equation for photosynthesis. (b) Explain why measuring the rate of gas bubble production is only an approximation of the actual rate of photosynthesis. (c) Describe how the student could adapt this experimental setup to investigate the effect of temperature on the rate of photosynthesis, indicating three variables that must be kept constant.
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解題

(a) The balanced chemical equation for photosynthesis is: \(6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2\). (b) Counting gas bubbles is only an approximation because: 1. The gas collected is not pure oxygen, it also contains some nitrogen and other gases dissolved in the water; 2. Some of the oxygen produced by photosynthesis is immediately used by the plant's mitochondria for aerobic respiration; 3. Some oxygen dissolves into the surrounding water instead of forming bubbles; 4. Bubbles can vary in volume and size, so counting bubbles does not give an exact volume of gas produced. (c) To investigate temperature: Place the Elodea in water baths set at different temperatures (e.g., 10°C, 20°C, 30°C, 40°C, and 50°C). Allow the plant to adjust/equilibrate to each temperature for 5 minutes before taking measurements. Count the number of bubbles produced per minute (or measure gas volume using a gas syringe). Variables to keep constant: 1. Light intensity (keep the light source at a fixed distance from the plant); 2. Carbon dioxide concentration (add a fixed mass of sodium hydrogencarbonate to the water); 3. The length, species, and surface area of the Elodea shoot used.

評分準則

(a) [3 marks] 1 mark for correct formulas of reactants (CO2 and H2O), 1 mark for correct formulas of products (C6H12O6 and O2), 1 mark for correct balancing (6, 6, 1, 6). (b) [4 marks] 1 mark for noting the gas is oxygen; 1 mark for explaining some oxygen is consumed by aerobic respiration; 1 mark for explaining some oxygen dissolves in the water; 1 mark for noting bubbles vary in size / volume (accept: gas may contain other gases like nitrogen). (c) [6 marks] 1 mark for using different/controlled temperatures (using water baths); 1 mark for describing how rate is measured (counting bubbles or gas volume per unit time); 1 mark for allowing equilibration time at each temperature; 1 mark each for any three constant variables up to 3 marks: light intensity/distance of lamp, carbon dioxide concentration (use of sodium hydrogencarbonate), same plant specimen/mass/length, same pH of water.
題目 5 · Structured
13.33
Plants use hormones to coordinate their growth responses to external stimuli. (a) Define the term phototropism. (b) State and explain the expected phototropic response of an intact oat coleoptile that is exposed to unilateral light from the right side. (c) Explain the role of auxin in coordinating this phototropic response, describing where it is synthesized, how it moves, and how it affects cell development on different sides of the shoot.
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解題

(a) Phototropism is a plant growth response where the direction of growth is determined by the direction of light. (b) The intact oat coleoptile will bend and grow towards the right (towards the source of unilateral light). This is a positive phototropic response. It functions to position the leaves and shoot in the direction of maximum light to maximize the rate of photosynthesis. (c) Role of Auxin: 1. Auxin is a plant hormone synthesized in the tip of the growing shoot. 2. It diffuses downwards into the zone of elongation. 3. When exposed to unilateral light from the right, auxin is redistributed and moves away from the light, accumulating on the shaded (left) side of the shoot. 4. Auxin stimulates cell elongation by causing the cell walls to loosen and take up more water. 5. Due to the higher concentration of auxin on the shaded (left) side, cells on this side elongate much faster and to a greater extent than cells on the illuminated (right) side. 6. This unequal rate of growth causes the shoot to bend towards the right (towards the light source).

評分準則

(a) [2 marks] 1 mark for growth response/movement to a stimulus; 1 mark for stating light as the stimulus. (b) [3 marks] 1 mark for predicting bending/growing towards the right/light source; 1 mark for identifying this as positive phototropism; 1 mark for explaining it increases light absorption for photosynthesis. (c) [8 marks] 1 mark for stating auxin is made in the shoot tip; 1 mark for stating auxin diffuses downwards; 1 mark for stating unilateral light causes auxin to migrate/redistribute to the shaded side; 1 mark for identifying a higher concentration of auxin on the shaded side; 1 mark for stating auxin stimulates cell elongation; 1 mark for explaining cells on the shaded side elongate more than cells on the light side; 1 mark for stating this unequal elongation causes the shoot to bend; 1 mark for logical flow and scientific terminology (e.g., cell wall elasticity, turgor pressure).
題目 6 · Structured
13.33
Respiration can occur anaerobically when oxygen levels are limiting. (a) Write the word equation for anaerobic respiration in yeast and the word equation for anaerobic respiration in human muscle cells. (b) During intense exercise, lactic acid builds up in human muscle tissues. Explain how the body processes and removes this lactic acid after exercise has ceased, referring to the concept of oxygen debt. (c) Describe how the anaerobic respiration of yeast is utilized in the bread-making process, explaining the fate of each product.
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解題

(a) Anaerobic respiration word equations: Yeast: \(\text{glucose} \rightarrow \text{ethanol} + \text{carbon dioxide}\). Human muscle cells: \(\text{glucose} \rightarrow \text{lactic acid}\). (b) During strenuous exercise, anaerobic respiration produces lactic acid, which causes muscle fatigue. After exercise stops, lactic acid is transported via the blood stream from the muscles to the liver. The body must pay back an oxygen debt, which is the volume of extra oxygen required to completely react with and break down the accumulated lactic acid. In the liver, lactic acid is oxidized aerobically to carbon dioxide and water, or converted into glucose and stored as glycogen. To provide this extra oxygen, breathing rate and depth remain high, and the heart rate stays elevated for some time after exercise has finished. (c) Yeast is added to dough to perform anaerobic respiration (fermentation). 1. Yeast utilizes the sugars present in the flour/dough. 2. Carbon dioxide gas is produced, which becomes trapped in the gluten network of the dough, forming bubbles that expand and cause the dough to rise. 3. Ethanol is also produced during fermentation, but it is volatile and completely evaporates during the baking process in the hot oven, while the high heat also kills the yeast cells.

評分準則

(a) [4 marks] 1 mark for glucose as reactant in both; 1 mark for ethanol and carbon dioxide in yeast; 1 mark for lactic acid in humans; 1 mark for correct chemical word equation format (reactants -> products with no energy/ATP as a chemical product). (b) [5 marks] 1 mark for stating lactic acid is carried by blood to liver; 1 mark for defining oxygen debt as the extra oxygen needed after exercise; 1 mark for stating lactic acid is oxidized to carbon dioxide and water (or converted to glucose/glycogen); 1 mark for explaining why breathing remains deep/fast; 1 mark for explaining why heart rate remains high. (c) [4 marks] 1 mark for stating yeast ferments sugars in the flour; 1 mark for carbon dioxide gas bubbles expanding to make dough rise; 1 mark for stating ethanol is produced; 1 mark for explaining that ethanol evaporates and yeast is killed during the baking process.

Paper 6 (Alternative to Practical)

Answer all questions. Plan experiments, execute calculations, plot data, and outline biological drawings.
2 題目 · 40
題目 1 · Practical Skills Tasks
20
### Part 1: Diffusion in Agar Cubes

A student investigated the effect of temperature on the rate of diffusion using agar cubes containing universal indicator. The agar cubes are initially red (acidic). When placed in an alkaline sodium hydroxide solution, the sodium hydroxide diffuses into the cubes, changing the color of the universal indicator from red to purple.

The student used five agar cubes, each measuring \(10\text{ mm} \times 10\text{ mm} \times 10\text{ mm}\).
Each cube was placed in a beaker containing \(50\text{ cm}^3\) of \(0.1\text{ mol/dm}^3\) sodium hydroxide solution at a different temperature: 20°C, 30°C, 40°C, 50°C, and 60°C.
After exactly 10 minutes, the cubes were removed and carefully sliced in half.
The distance that the purple color had penetrated from the outer surface of each cube was measured using a ruler.

Figure 1.1 shows a diagram representing the cross-section of the agar cube that was incubated at 30°C. The central shaded region represents the remaining red (acidic) agar, and the outer border represents the purple region where sodium hydroxide has diffused.

*(Note: Assume the outer square of Figure 1.1 in your printed exam sheet measures exactly \(40\text{ mm}\) on each side, and the inner square measures \(20\text{ mm}\).)*

**(a)** (i) Draw a large, clean, outline diagram of the cross-section of the agar cube shown in Figure 1.1. On your drawing, use a label line and the letter **D** to indicate the region where the sodium hydroxide has diffused. [3]

(ii) The actual size of the agar cube's side was \(10\text{ mm}\). Using the provided print dimension of \(40\text{ mm}\) for the outer square side, calculate the magnification of Figure 1.1. Show your working. [2]

**(b)** Table 1.1 shows the results of the student's investigation.

**Table 1.1**
| Temperature (°C) | Distance diffused (mm) | Rate of diffusion (mm/min) |
|---|---|---|
| 20 | 1.5 | 0.15 |
| 30 | 2.5 | 0.25 |
| 40 | 3.5 | **[Calculate]** |
| 50 | 4.5 | 0.45 |
| 60 | 5.5 | 0.55 |

(i) Calculate the rate of diffusion at 40°C. Write down the value and show your working. [1]

(ii) Plot a line graph on grid paper of the rate of diffusion (y-axis) against temperature (x-axis). [5]

(iii) State the relationship shown by your graph between temperature and the rate of diffusion. [1]

**(c)** (i) Identify two variables that were kept constant in this investigation. [2]

(ii) State one safety precaution that should be taken when handling sodium hydroxide solution, and give a reason for this precaution. [1]

**(d)** A student wants to find out how the surface area to volume ratio affects the rate of diffusion. Plan an investigation to determine the effect of the surface area to volume ratio of agar cubes on the rate of diffusion of sodium hydroxide. [5]
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解題

**(a) (i)**
- Clear outline drawing with a single line (no shading, sketching, or broken lines).
- Drawing occupies more than 50% of the available space and shows a smaller square centered inside a larger square.
- Label line and letter **D** pointing clearly to the outer border area between the two squares.

**(a) (ii)**
- \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} = \frac{40\text{ mm}}{10\text{ mm}} = \times 4\) (or \(\times 4.0\)).

**(b) (i)**
- \(\text{Rate} = \frac{\text{Distance}}{\text{Time}} = \frac{3.5\text{ mm}}{10\text{ min}} = 0.35\text{ mm/min}\).

**(b) (ii)**
- **Axes**: x-axis labeled 'Temperature / °C' and y-axis labeled 'Rate of diffusion / mm/min' (or 'mm/min').
- **Scale**: Linear, even scale filling at least 50% of the grid (e.g., x-axis from 0 to 60 or 70, y-axis from 0 to 0.60).
- **Plotting**: All 5 points plotted accurately within half a small grid square.
- **Line**: Points connected with clean, thin, ruled straight lines or a line of best fit. No double lines.

**(b) (iii)**
- As temperature increases, the rate of diffusion increases.

**(c) (i)**
- Concentration of sodium hydroxide solution (\(0.1\text{ mol/dm}^3\)).
- Volume of sodium hydroxide solution (\(50\text{ cm}^3\)).
- Dimensions of the agar cubes (\(10\text{ mm} \times 10\text{ mm} \times 10\text{ mm}\)).
- Time of immersion (10 minutes).
*(Any two of the above)*

**(c) (ii)**
- Wear safety goggles / gloves because sodium hydroxide is corrosive / alkaline and can cause damage to eyes / skin.

**(d)**
- **Independent Variable**: Prepare agar cubes of at least three different sizes (e.g., \(5 \times 5 \times 5\text{ mm}\), \(10 \times 10 \times 10\text{ mm}\), and \(20 \times 20 \times 20\text{ mm}\)) to vary the surface area to volume ratio.
- **Dependent Variable**: Cut the cubes in half after a fixed time and measure the distance diffused with a ruler, or calculate the percentage of the volume that changed color.
- **Controlled Variables**: Keep the temperature constant (e.g., in a water bath), use the same concentration of sodium hydroxide, and leave all cubes in the solution for the same length of time (e.g., 10 minutes).
- **Method Detail**: Calculate the surface area and volume of each cube size prior to immersion. Submerge the cubes, remove them simultaneously, blot dry, slice in half, and measure the distance.
- **Repetition**: Repeat the experiment at least three times for each cube size to calculate mean values and identify anomalies.

評分準則

**(a) (i)** [Max 3 marks]
- **Drawing Quality**: Single sharp lines, no sketching or shading, square-in-square proportions sensible. [1]
- **Size**: Drawing occupies at least 50% of the space. [1]
- **Labeling**: Correct label line and letter **D** pointing specifically to the outer border region. [1]

**(a) (ii)** [Max 2 marks]
- Correct calculation: \(40 / 10\). [1]
- Correct magnification with \(\times\) symbol (e.g., \(\times 4\) or \(\times 4.0\)). [1]

**(b) (i)** [Max 1 mark]
- Correct value of \(0.35\text{ mm/min}\) with working shown. [1]

**(b) (ii)** [Max 5 marks]
- **A (Axes)**: Correctly labeled with units on both axes. [1]
- **S (Scale)**: Linear scale, occupying at least half the grid. [1]
- **P (Plotting)**: All 5 points plotted accurately to within \(\pm 0.5\) small square. [1]
- **L (Line)**: Clean, thin lines connecting points or single line of best fit. [1]
- **K (Key/General Neatness)**: Correct symbols used for points (crosses or circled dots), neat layout. [1]

**(b) (iii)** [Max 1 mark]
- State that as temperature increases, the rate of diffusion increases. [1]

**(c) (i)** [Max 2 marks]
- Any two controlled variables correctly identified (e.g., concentration of NaOH, volume of NaOH, time, cube size). [1 mark each]

**(c) (ii)** [Max 1 mark]
- Goggles / gloves AND correct reason (corrosive / irritant to eyes or skin). [1]

**(d)** [Max 5 marks]
- Method of preparing at least three different sizes of agar cubes to change surface area to volume ratio. [1]
- Identification of dependent variable: measuring distance diffused or percentage volume color change. [1]
- At least two controlled variables stated (e.g., temperature, concentration of NaOH, immersion time). [1]
- Replicates: repeat each size at least three times and calculate mean. [1]
- Safety precaution: wearing safety goggles due to corrosive nature of NaOH. [1]
題目 2 · Practical Skills Tasks
20
### Part 2: Photosynthesis in Pondweed

A student investigated the effect of light intensity on the rate of photosynthesis in *Elodea* (pondweed).

The apparatus was set up as follows:
- A piece of pondweed was placed upside down in a boiling tube containing 0.2% sodium hydrogencarbonate solution.
- A light source (lamp) was positioned at different distances from the boiling tube: 10 cm, 20 cm, 30 cm, 40 cm, and 50 cm.
- At each distance, the pondweed was allowed to adjust for 5 minutes, and then the number of bubbles of gas released from the cut stem was counted for 3 minutes.

**(a)** State the purpose of using sodium hydrogencarbonate solution in this investigation rather than distilled water. [1]

**(b)** Table 2.1 shows the results of the investigation.

**Table 2.1**
| Distance from light source (cm) | Number of bubbles in 3 minutes (Replicates) | Mean number of bubbles per minute |
|---|---|---|
| 10 | 138, 141, 135 | 46.0 |
| 20 | 93, 14 (anomaly), 87 | **[Calculate]** |
| 30 | 54, 56, 52 | 18.0 |
| 40 | 30, 29, 31 | 10.0 |
| 50 | 15, 14, 16 | 5.0 |

(i) Explain why the student should exclude the reading of 14 bubbles at a distance of 20 cm when calculating the mean. [1]

(ii) Calculate the mean number of bubbles per minute for a distance of 20 cm, excluding the anomalous value. Show your working. [2]

(iii) Calculate the relative light intensity (\(I\)) at a distance of 20 cm (which is \(0.2\text{ m}\)) using the formula:

\[I = \frac{1}{d^2}\]

where \(d\) is the distance in meters. Show your working and state the unit. [2]

**(c)** (i) Explain why counting bubbles is not an entirely accurate method of measuring the volume of gas produced during photosynthesis. [1]

(ii) Describe how the student could modify the apparatus to measure the volume of gas released more accurately. [2]

(iii) State the chemical test used to confirm that the gas produced is oxygen. [2]

**(d)** Plan an investigation to determine the effect of temperature on the rate of photosynthesis in pondweed. [6]

**(e)** Figure 2.1 shows a photomicrograph of a palisade mesophyll cell from the leaf of *Elodea*.
The actual length of this palisade cell is \(0.05\text{ mm}\).
The student made a drawing of this cell, which measured \(45\text{ mm}\) in length.

Calculate the magnification of the student's drawing. Show your working. [3]
查看答案詳解

解題

**(a)**
- Sodium hydrogencarbonate provides a source of carbon dioxide for the pondweed (which is a raw material needed for photosynthesis).

**(b) (i)**
- It is an anomaly because it is significantly lower than the other replicates (93 and 87) at that distance, and does not fit the trend.

**(b) (ii)**
- Sum of non-anomalous replicates = \(93 + 87 = 180\) bubbles in 3 minutes.
- Mean number of bubbles in 3 minutes = \(\frac{180}{2} = 90\) bubbles.
- Mean number of bubbles per minute = \(\frac{90}{3\text{ minutes}} = 30\) bubbles per minute (or \(30.0\)).

**(b) (iii)**
- \(d = 0.2\text{ m}\)
- \(I = \frac{1}{(0.2)^2} = \frac{1}{0.04} = 25\)
- Unit: arbitrary units (or au).

**(c) (i)**
- Bubbles can be different sizes / human error in counting bubbles that are released very fast / some bubbles may go undetected.

**(c) (ii)**
- Collect the gas using a gas syringe or a graduated capillary tube with a syringe attached to measure the actual volume of the gas in \(\text{cm}^3\) or \(\text{mm}^3\).

**(c) (iii)**
- Collect the gas in a test tube and insert a glowing splint. The splint will relight if oxygen is present.

**(d)**
- **Independent Variable**: Use at least 5 different temperatures (e.g., 10°C, 20°C, 30°C, 40°C, 50°C) achieved using thermostatically controlled water baths (or ice and hot water mixes).
- **Dependent Variable**: Measure the volume of gas produced per minute (or count bubbles per minute) using a gas syringe/capillary tube.
- **Controlled Variables**: Keep the light intensity constant by placing the lamp at a fixed distance (e.g., 20 cm) from the pondweed. Use a heat shield (e.g., a glass beaker of water between the lamp and pondweed) to prevent lamp heat from altering water temperature. Use the same length/type of pondweed and the same concentration of sodium hydrogencarbonate solution.
- **Method**: Allow the pondweed to equilibrate to the target temperature for 5 minutes before counting bubbles or collecting gas.
- **Repetition**: Perform at least three replicates at each temperature and calculate a mean.

**(e)**
- Formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\)
- Conversion: \(\text{Image size} = 45\text{ mm}\); \(\text{Actual size} = 0.05\text{ mm}\) (both in same units).
- Calculation: \(\frac{45}{0.05} = \times 900\).
- State the magnification as \(\times 900\) (or 900 times).

評分準則

**(a)** [Max 1 mark]
- To provide carbon dioxide (for photosynthesis). [1]

**(b) (i)** [Max 1 mark]
- Mention that 14 is anomalous / outlier / does not fit the pattern of the other replicates. [1]

**(b) (ii)** [Max 2 marks]
- Sum of non-anomalous values divided by 2 (90 bubbles in 3 minutes) OR rates of non-anomalous replicates shown: \(93 / 3 = 31\) and \(87 / 3 = 29\). [1]
- Correct calculation to get \(30\) bubbles/min (accept 30). [1]

**(b) (iii)** [Max 2 marks]
- Correct calculation showing value as \(25\). [1]
- Unit: arbitrary units / au. [1]

**(c) (i)** [Max 1 mark]
- State that bubbles are of unequal volume/size, or that bubbles can emerge too quickly to count accurately. [1]

**(c) (ii)** [Max 2 marks]
- Use of a gas syringe / graduated capillary tube. [1]
- Collecting gas over a specified time to measure volume directly. [1]

**(c) (iii)** [Max 2 marks]
- Glowing splint. [1]
- Relights / bursts into flame. [1]

**(d)** [Max 6 marks]
- **Independent Variable**: At least 5 different temperatures using water baths. [1]
- **Dependent Variable**: Volume of gas collected per unit time (or bubble rate). [1]
- **Controlled Variable 1**: Distance of light source constant (light intensity). [1]
- **Controlled Variable 2**: CO2 concentration (sodium hydrogencarbonate solution) constant. [1]
- **Equilibration**: Allow pondweed 5 minutes to adapt to temperature before recording. [1]
- **Safety / Precision**: Use a heat shield/glass beaker of water to block heat from lamp, or repeat each temperature 3 times to calculate a mean. [1]

**(e)** [Max 3 marks]
- Correct formula or substitution: \(\text{Magnification} = 45 / 0.05\). [1]
- Correct unit alignment (both in mm, or both converted to \(\mu\text{m}\): \(45000 / 50\)). [1]
- Correct final calculation: \(\times 900\) (accept \(\times 900\) or 900, reject if units like mm are appended). [1]

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