Cambridge IGCSE · thinka 原创模拟试题

2023 Cambridge IGCSE Biology (0610) 模拟试题及答案详解

Thinka Jun 2023 (V1) Cambridge IGCSE-Style Mock — Biology (0610)

160 195 分钟2023
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V1) Cambridge IGCSE Biology (0610) paper. Not affiliated with or reproduced from Cambridge.

Paper 21: 選擇題 (Extended)

Answer all 40 multiple-choice questions. Choose the one you consider correct.
40 题目 · 40
题目 1 · 選擇題
1
Plant tissue is placed into a concentrated sucrose solution that has a lower water potential than the cell cytoplasm.

Which row correctly identifies the net direction of water movement and the resulting state of the plant cells?
  1. A.Net movement of water: out of the cell | State of cells: plasmolysed
  2. B.Net movement of water: into the cell | State of cells: turgid
  3. C.Net movement of water: out of the cell | State of cells: turgid
  4. D.Net movement of water: into the cell | State of cells: plasmolysed
查看答案详解

解题

Water moves by osmosis down a water potential gradient from a region of higher water potential (inside the cell cytoplasm) to a region of lower water potential (the concentrated sucrose solution outside). As water leaves the plant cell, the cytoplasm and cell membrane shrink away from the cell wall, causing the cell to become plasmolysed.

评分标准

A is correct [1]; B is incorrect as water moves out down a water potential gradient; C is incorrect as loss of water leads to plasmolysis, not turgidity; D is incorrect because water leaves the cell rather than entering it.
题目 2 · 選擇題
1
Which row correctly compares the products of anaerobic respiration in yeast cells with the products of anaerobic respiration in human muscle cells?
  1. A.Yeast: ethanol and carbon dioxide | Human muscle cells: lactic acid only
  2. B.Yeast: lactic acid and carbon dioxide | Human muscle cells: ethanol only
  3. C.Yeast: lactic acid only | Human muscle cells: ethanol and carbon dioxide
  4. D.Yeast: ethanol only | Human muscle cells: lactic acid and carbon dioxide
查看答案详解

解题

In yeast (and plant cells), anaerobic respiration produces ethanol and carbon dioxide (alcohol fermentation). In human muscle cells during vigorous exercise, anaerobic respiration breaks down glucose into lactic acid without releasing carbon dioxide.

评分标准

A is correct [1]; B, C, and D confuse the products of anaerobic respiration in yeast with those in mammalian muscle cells.
题目 3 · 選擇題
1
What changes occur in the diaphragm and external intercostal muscles during inspiration in humans?
  1. A.Diaphragm: contracts and flattens | External intercostal muscles: contract
  2. B.Diaphragm: relaxes and becomes dome-shaped | External intercostal muscles: contract
  3. C.Diaphragm: contracts and flattens | External intercostal muscles: relax
  4. D.Diaphragm: relaxes and becomes dome-shaped | External intercostal muscles: relax
查看答案详解

解题

During inspiration (breathing in), the diaphragm contracts and moves downwards (flattens), and the external intercostal muscles contract to pull the ribs upwards and outwards. This increases the volume of the thorax and decreases the pressure below atmospheric pressure, causing air to rush into the lungs.

评分标准

A is correct [1]; B is incorrect because the diaphragm contracts and flattens during inspiration; C and D are incorrect because the external intercostal muscles contract during inspiration.
题目 4 · 選擇題
1
A mixture of starch solution and salivary amylase is incubated at \(37\text{ }^\circ\text{C}\) at pH 7.0.

Samples are tested at the beginning of the experiment and after 30 minutes using iodine solution and Benedict's solution.

Which row shows the expected observations?
  1. A.At start: Iodine = blue-black, Benedict's = blue | After 30 min: Iodine = yellow-brown, Benedict's = brick-red
  2. B.At start: Iodine = yellow-brown, Benedict's = brick-red | After 30 min: Iodine = blue-black, Benedict's = blue
  3. C.At start: Iodine = blue-black, Benedict's = brick-red | After 30 min: Iodine = blue-black, Benedict's = blue
  4. D.At start: Iodine = yellow-brown, Benedict's = blue | After 30 min: Iodine = yellow-brown, Benedict's = brick-red
查看答案详解

解题

At the start, starch is present and reducing sugars (maltose/glucose) are absent, giving a blue-black colour with iodine solution and a blue colour (negative) with Benedict's solution. After 30 minutes, amylase hydrolyses the starch into reducing sugars; thus, the iodine test is negative (remains yellow-brown) and the Benedict's test is positive (turns brick-red upon heating).

评分标准

A is correct [1]; B represents the reverse of the true biochemical progression; C and D show incorrect colorimetric combinations for initial or final digestion products.
题目 5 · 選擇題
1
Which statement correctly describes a step in protein synthesis?
  1. A.The sequence of bases in mRNA determines the sequence of amino acids in the protein.
  2. B.DNA leaves the nucleus to attach to ribosomes in the cytoplasm.
  3. C.Ribosomes synthesise mRNA molecules inside the nucleus.
  4. D.The sequence of amino acids determines the sequence of bases in the DNA molecule.
查看答案详解

解题

During protein synthesis, the sequence of bases in mRNA (which is complementary to the template strand of DNA) determines the specific order in which amino acids are joined together at the ribosome to form a polypeptide chain.

评分标准

A is correct [1]; B is incorrect as DNA remains in the nucleus; C is incorrect as mRNA is transcribed in the nucleus by RNA polymerase, not ribosomes; D is incorrect because the base sequence determines the amino acid sequence, not vice versa.
题目 6 · 選擇題
1
Which row correctly compares aerobic respiration with anaerobic respiration in human muscle cells?
  1. A.Aerobic respiration produces lactic acid; anaerobic respiration produces carbon dioxide and water; more energy is released per glucose molecule in anaerobic respiration.
  2. B.Aerobic respiration requires oxygen; anaerobic respiration does not require oxygen; more energy is released per glucose molecule in aerobic respiration.
  3. C.Aerobic respiration does not require oxygen; anaerobic respiration requires oxygen; more energy is released per glucose molecule in aerobic respiration.
  4. D.Aerobic respiration produces carbon dioxide and water; anaerobic respiration produces alcohol; more energy is released per glucose molecule in anaerobic respiration.
查看答案详解

解题

Aerobic respiration requires oxygen and breaks down glucose completely to release a relatively large amount of energy per glucose molecule (producing carbon dioxide and water). Anaerobic respiration in human muscle cells occurs without oxygen, breaks down glucose partially to lactic acid, and releases much less energy per glucose molecule.

评分标准

B [1]
A — incorrect products and energy comparison.
C — oxygen requirements are inverted.
D — alcohol is produced in yeast, not human muscle; energy comparison is incorrect.
题目 7 · 選擇題
1
Four identical cylinders of potato tissue, each with an initial mass of \(5.0\text{ g}\), were placed into test-tubes containing sucrose solutions of different concentrations: \(0.0\text{ mol/dm}^3\), \(0.2\text{ mol/dm}^3\), \(0.4\text{ mol/dm}^3\), and \(0.8\text{ mol/dm}^3\). The water potential of the potato tissue is equivalent to a \(0.3\text{ mol/dm}^3\) sucrose solution. Which row shows the expected final masses of the cylinders after \(2\text{ hours}\)?
  1. A.\(0.0\text{ mol/dm}^3\): \(5.6\text{ g}\) ; \(0.2\text{ mol/dm}^3\): \(5.2\text{ g}\) ; \(0.4\text{ mol/dm}^3\): \(4.7\text{ g}\) ; \(0.8\text{ mol/dm}^3\): \(4.1\text{ g}\)
  2. B.\(0.0\text{ mol/dm}^3\): \(4.1\text{ g}\) ; \(0.2\text{ mol/dm}^3\): \(4.7\text{ g}\) ; \(0.4\text{ mol/dm}^3\): \(5.2\text{ g}\) ; \(0.8\text{ mol/dm}^3\): \(5.6\text{ g}\)
  3. C.\(0.0\text{ mol/dm}^3\): \(5.6\text{ g}\) ; \(0.2\text{ mol/dm}^3\): \(4.7\text{ g}\) ; \(0.4\text{ mol/dm}^3\): \(5.2\text{ g}\) ; \(0.8\text{ mol/dm}^3\): \(4.1\text{ g}\)
  4. D.\(0.0\text{ mol/dm}^3\): \(5.0\text{ g}\) ; \(0.2\text{ mol/dm}^3\): \(5.0\text{ g}\) ; \(0.4\text{ mol/dm}^3\): \(5.0\text{ g}\) ; \(0.8\text{ mol/dm}^3\): \(5.0\text{ g}\)
查看答案详解

解题

The potato tissue has a water potential equivalent to \(0.3\text{ mol/dm}^3\). In solutions with lower solute concentrations (\(0.0\) and \(0.2\text{ mol/dm}^3\)), the solution has a higher water potential than the tissue, so water enters the cells by osmosis, increasing the mass (> \(5.0\text{ g}\)). In solutions with higher solute concentrations (\(0.4\) and \(0.8\text{ mol/dm}^3\)), the solution has a lower water potential than the tissue, so water leaves the cells by osmosis, decreasing the mass (< \(5.0\text{ g}\)). Thus, row A is correct.

评分标准

A [1]
B — inverted mass changes.
C — inconsistent direction of osmosis at 0.2 mol/dm³.
D — represents no net movement of water in all solutions.
题目 8 · 選擇題
1
Which changes occur to the diaphragm and the volume of the thorax during inspiration in humans?
  1. A.The diaphragm contracts and flattens; the volume of the thorax increases.
  2. B.The diaphragm contracts and arches upwards; the volume of the thorax decreases.
  3. C.The diaphragm relaxes and flattens; the volume of the thorax increases.
  4. D.The diaphragm relaxes and arches upwards; the volume of the thorax decreases.
查看答案详解

解题

During inspiration (breathing in), the diaphragm muscles contract, causing the diaphragm to flatten and move downwards. At the same time, the external intercostal muscles contract to lift the ribcage up and out. These movements increase the volume of the thorax, reducing the internal pressure and drawing air into the lungs.

评分标准

A [1]
B — the diaphragm flattens, increasing thorax volume.
C — the diaphragm contracts during inspiration, not relaxes.
D — describes changes during expiration.
题目 9 · 選擇題
1
An investigation was carried out to measure the rate of photosynthesis of an aquatic plant at various light intensities. The temperature was maintained at \(20\text{ }^\circ\text{C}\) and the carbon dioxide concentration was \(0.04\%\). Above \(50\text{ arbitrary units}\) of light intensity, the rate of photosynthesis reached a plateau. When the carbon dioxide concentration was increased to \(0.10\%\) at \(60\text{ arbitrary units}\), the rate of photosynthesis increased significantly. What was the limiting factor for photosynthesis at \(60\text{ arbitrary units}\) of light intensity and \(0.04\%\) carbon dioxide?
  1. A.Carbon dioxide concentration
  2. B.Light intensity
  3. C.Oxygen concentration
  4. D.Water availability
查看答案详解

解题

A limiting factor is something present in the environment in such short supply that it restricts life processes. Because increasing the carbon dioxide concentration caused an increase in the rate of photosynthesis at \(60\text{ arbitrary units}\), carbon dioxide concentration was the factor limiting the rate.

评分标准

A [1]
B — light intensity was already in excess (in plateau region).
C — oxygen is a product, not a limiting factor for photosynthesis.
D — the plant is aquatic, so water is abundant.
题目 10 · 選擇題
1
Which row correctly identifies a digestive enzyme, the organ that secretes it into the alimentary canal, and the main products of its chemical reaction?
  1. A.Amylase | Pancreas | Fatty acids and glycerol
  2. B.Lipase | Liver | Fatty acids and glycerol
  3. C.Lipase | Pancreas | Fatty acids and glycerol
  4. D.Protease | Salivary glands | Amino acids
查看答案详解

解题

Lipase is synthesized and secreted by the pancreas into the duodenum (small intestine), where it hydrolyses lipids (fats and oils) into fatty acids and glycerol. The liver produces bile (not enzymes), salivary glands produce amylase (which digests starch to maltose), and proteases in the stomach produce peptides/amino acids.

评分标准

C [1]
A — amylase breaks down starch to maltose/glucose, not lipids to fatty acids and glycerol.
B — the liver secretes bile (no digestive enzymes); lipase is secreted by the pancreas.
D — salivary glands secrete amylase, not protease.
题目 11 · 選擇題
1
The table compares features of aerobic respiration with anaerobic respiration in human muscle cells. Which row correctly identifies features of aerobic respiration? [Row A: Oxygen used: Yes | Lactic acid produced: No | Relative energy released per glucose molecule: High] [Row B: Oxygen used: Yes | Lactic acid produced: Yes | Relative energy released per glucose molecule: Low] [Row C: Oxygen used: No | Lactic acid produced: No | Relative energy released per glucose molecule: High] [Row D: Oxygen used: No | Lactic acid produced: Yes | Relative energy released per glucose molecule: Low]
  1. A.Oxygen used: Yes; Lactic acid produced: No; Relative energy released: High
  2. B.Oxygen used: Yes; Lactic acid produced: Yes; Relative energy released: Low
  3. C.Oxygen used: No; Lactic acid produced: No; Relative energy released: High
  4. D.Oxygen used: No; Lactic acid produced: Yes; Relative energy released: Low
查看答案详解

解题

Aerobic respiration uses oxygen to completely break down glucose, producing carbon dioxide and water rather than lactic acid, and releasing a relatively large amount of energy per molecule of glucose.

评分标准

A is correct [1]; B is incorrect because lactic acid is not produced in aerobic respiration; C and D are incorrect because aerobic respiration requires oxygen.
题目 12 · 選擇題
1
Cylinders of potato tissue of equal initial mass were placed in sucrose solutions of different concentrations for 2 hours. In a 0.6 mol dm\(^{-3}\) sucrose solution, the mass of the potato cylinder decreased by 5.5%. Which statement explains this change in mass?
  1. A.Water moved into the potato cells by active transport.
  2. B.Water moved out of the potato cells by osmosis because the solution had a lower water potential than the cells.
  3. C.Sucrose moved into the potato cells by diffusion down a concentration gradient.
  4. D.Sucrose moved out of the potato cells by osmosis because the cells had a lower water potential than the solution.
查看答案详解

解题

The potato tissue lost mass because water moved out of the cells by osmosis down a water potential gradient (from a higher water potential inside the cells to a lower water potential in the concentrated sucrose solution).

评分标准

B is correct [1]; A is incorrect because water moves by osmosis, not active transport; C and D are incorrect because the movement of water (not sucrose) accounts for the rapid mass change across selectively permeable cell membranes.
题目 13 · 選擇題
1
Which set of actions occurs during inspiration (breathing in) in a human?
  1. A.External intercostal muscles contract, diaphragm contracts and flattens, volume of the thorax increases.
  2. B.External intercostal muscles relax, diaphragm relaxes and domes upwards, volume of the thorax increases.
  3. C.Internal intercostal muscles contract, diaphragm contracts and flattens, pressure in the thorax increases.
  4. D.Internal intercostal muscles relax, diaphragm relaxes and domes upwards, pressure in the thorax decreases.
查看答案详解

解题

During inspiration, the external intercostal muscles contract (pulling the ribs up and out) and the diaphragm contracts and flattens. This increases the volume of the thorax and decreases the internal pressure below atmospheric pressure, causing air to enter the lungs.

评分标准

A is correct [1]; B is incorrect because the diaphragm flattens when it contracts, it does not dome during inspiration; C and D are incorrect because the internal intercostal muscles relax during normal inspiration and thoracic pressure decreases.
题目 14 · 選擇題
1
An experiment investigated the rate of photosynthesis in an aquatic plant. Carbon dioxide was supplied in excess. At a low light intensity of 400 lux, increasing the temperature from 15 °C to 25 °C produced no change in the rate of photosynthesis. What was the limiting factor for photosynthesis at 400 lux?
  1. A.Carbon dioxide concentration
  2. B.Chlorophyll concentration
  3. C.Light intensity
  4. D.Temperature
查看答案详解

解题

A limiting factor is something present in the environment in such short supply that it restricts life processes. Since carbon dioxide was in excess and changing the temperature did not increase the rate, light intensity is the factor restricting the rate.

评分标准

C is correct [1]; A is incorrect because carbon dioxide was provided in excess; B is incorrect because chlorophyll is not a variable tested here; D is incorrect because increasing temperature did not change the rate.
题目 15 · 選擇題
1
Which row correctly matches a digestive enzyme, its site of secretion, its substrate, and its main product?
  1. A.Enzyme: Amylase | Site of secretion: Salivary glands | Substrate: Starch | Product: Maltose
  2. B.Enzyme: Lipase | Site of secretion: Stomach | Substrate: Fatty acids | Product: Glycerol
  3. C.Enzyme: Pepsin | Site of secretion: Pancreas | Substrate: Protein | Product: Amino acids
  4. D.Enzyme: Maltase | Site of secretion: Liver | Substrate: Maltose | Product: Glucose
查看答案详解

解题

Amylase is secreted by salivary glands (and pancreas), acts on starch, and breaks it down into maltose. Lipase is secreted by the pancreas (not stomach) and acts on fats to produce fatty acids and glycerol. Pepsin is secreted by the stomach, not the pancreas.

评分标准

A is correct [1]; B is incorrect because lipase is secreted by the pancreas and its substrate is fats/lipids; C is incorrect because pepsin is secreted by the gastric glands of the stomach; D is incorrect because maltase is located on the epithelial lining of the small intestine, not secreted by the liver.
题目 16 · 選擇題
1
During vigorous exercise, human muscle cells respire anaerobically.

Which row shows the product formed in the muscle cells and explains why an oxygen debt occurs?
  1. A.Product: ethanol and carbon dioxide | Explanation: oxygen is needed to oxidise lactic acid in the muscles
  2. B.Product: lactic acid | Explanation: oxygen is needed to aerobically break down lactic acid in the liver
  3. C.Product: lactic acid | Explanation: oxygen is needed to convert glucose into glycogen in the muscles
  4. D.Product: carbon dioxide and water | Explanation: oxygen is needed to replace expired air in the lungs
查看答案详解

解题

During anaerobic respiration in human muscle cells, glucose is converted to lactic acid without oxygen. The accumulation of lactic acid creates an oxygen debt because extra oxygen is subsequently required to transport lactic acid to the liver and break it down aerobically (or convert it back to glucose/glycogen).

评分标准

B is correct [1]

A is incorrect as carbon dioxide and ethanol are produced in yeast, not human muscles.
C is incorrect as oxygen debt is required to metabolise lactic acid in the liver, not convert glucose to glycogen.
D is incorrect as carbon dioxide and water are products of aerobic respiration.
题目 17 · 選擇題
1
Four identical potato cylinders were placed into test-tubes containing sucrose solutions of different concentrations. After two hours, the change in mass of each cylinder was recorded.

Which cylinder showed the greatest net movement of water out of its cells by osmosis?
  1. A.a cylinder in \(0.2\text{ mol/dm}^{3}\) sucrose solution with a change in mass of \(+0.4\text{ g}\)
  2. B.a cylinder in \(0.4\text{ mol/dm}^{3}\) sucrose solution with a change in mass of \(+0.1\text{ g}\)
  3. C.a cylinder in \(0.6\text{ mol/dm}^{3}\) sucrose solution with a change in mass of \(-0.3\text{ g}\)
  4. D.a cylinder in \(0.8\text{ mol/dm}^{3}\) sucrose solution with a change in mass of \(-0.7\text{ g}\)
查看答案详解

解题

When plant tissue is placed in a hypertonic solution (lower water potential than the cells), water moves out of the cells by osmosis down a water potential gradient, causing a decrease in mass. The cylinder with the greatest decrease in mass (\(-0.7\text{ g}\)) experienced the greatest net loss of water.

评分标准

D is correct [1]

A and B represent a net gain in mass (water entered the cells).
C represents water loss, but of a smaller magnitude (\(-0.3\text{ g}\)) than D (\(-0.7\text{ g}\)).
题目 18 · 選擇題
1
Which row correctly describes the actions of the diaphragm and external intercostal muscles, and the resulting air pressure inside the lungs, during inspiration in a human?
  1. A.Diaphragm: relaxes and domes upward | External intercostal muscles: relax | Air pressure inside lungs: increases above atmospheric pressure
  2. B.Diaphragm: relaxes and domes upward | External intercostal muscles: contract | Air pressure inside lungs: decreases below atmospheric pressure
  3. C.Diaphragm: contracts and flattens | External intercostal muscles: contract | Air pressure inside lungs: decreases below atmospheric pressure
  4. D.Diaphragm: contracts and flattens | External intercostal muscles: relax | Air pressure inside lungs: increases above atmospheric pressure
查看答案详解

解题

During inspiration, the external intercostal muscles contract (pulling the ribcage upwards and outwards) and the diaphragm contracts and flattens. This increases the volume of the thorax, causing the air pressure inside the lungs to decrease below atmospheric pressure, drawing air in.

评分标准

C is correct [1]

A and B are incorrect because the diaphragm contracts and flattens during inspiration.
D is incorrect because the external intercostal muscles contract, not relax, and pressure decreases below atmospheric pressure.
题目 19 · 選擇題
1
A student investigated the effect of light intensity on the rate of photosynthesis in an aquatic plant. As light intensity increased above a certain level, the rate of photosynthesis remained constant.

Which statement explains why the rate remained constant?
  1. A.Another factor, such as carbon dioxide concentration or temperature, has become limiting.
  2. B.Chlorophyll molecules in the chloroplasts have denatured.
  3. C.Light intensity is still the only factor limiting the rate of photosynthesis.
  4. D.Water has been completely depleted from the leaf cells of the plant.
查看答案详解

解题

When the rate of photosynthesis reaches a plateau despite increasing light intensity, light is no longer the limiting factor. Another environmental factor, such as carbon dioxide concentration or temperature, has become the limiting factor.

评分标准

A is correct [1]

B is incorrect because light intensity within normal experimental ranges does not denature chlorophyll.
C is incorrect because if light intensity were still limiting, increasing it would increase the rate.
D is incorrect because aquatic plants are submerged in water, so water availability is not depleted.
题目 20 · 選擇題
1
Which row correctly identifies where bile is produced, where it is stored, and its main role in the digestive system?
  1. A.Site of production: gall bladder | Site of storage: liver | Main role: chemically digests fats into fatty acids and glycerol
  2. B.Site of production: liver | Site of storage: gall bladder | Main role: emulsifies fats to increase the surface area for lipase
  3. C.Site of production: pancreas | Site of storage: gall bladder | Main role: neutralises stomach acid and digests starch
  4. D.Site of production: liver | Site of storage: pancreas | Main role: provides an acidic pH for pepsin to function
查看答案详解

解题

Bile is produced by the liver, stored in the gall bladder, and released into the duodenum. Its main role is to emulsify fats (breaking large fat globules into small droplets), which increases the surface area for the action of lipase enzymes, as well as neutralising acidic chyme from the stomach.

评分标准

B is correct [1]

A is incorrect because bile is produced in the liver, not the gall bladder, and emulsifies fats rather than chemically digesting them.
C is incorrect because bile is not produced by the pancreas and does not digest starch.
D is incorrect because bile is stored in the gall bladder and provides an alkaline, not acidic, pH.
题目 21 · 選擇題
1
Which row correctly identifies the products of aerobic respiration in human muscle cells and anaerobic respiration in yeast cells?
  1. A.Aerobic in muscle: lactic acid only | Anaerobic in yeast: ethanol and carbon dioxide
  2. B.Aerobic in muscle: carbon dioxide and water | Anaerobic in yeast: ethanol and carbon dioxide
  3. C.Aerobic in muscle: carbon dioxide and water | Anaerobic in yeast: lactic acid only
  4. D.Aerobic in muscle: ethanol and carbon dioxide | Anaerobic in yeast: lactic acid only
查看答案详解

解题

Aerobic respiration in human cells breaks down glucose completely in the presence of oxygen to produce carbon dioxide and water:
\(\text{glucose} + \text{oxygen} \rightarrow \text{carbon dioxide} + \text{water}\).

Anaerobic respiration in yeast (alcohol fermentation) breaks down glucose in the absence of oxygen to produce ethanol (alcohol) and carbon dioxide:
\(\text{glucose} \rightarrow \text{ethanol} + \text{carbon dioxide}\).

Therefore, row B is correct.

评分标准

B [1]
题目 22 · 選擇題
1
Cylinders of potato tissue of identical initial mass were placed in four sucrose solutions of different concentrations, P, Q, R, and S. After 60 minutes, the percentage change in mass was measured:

• Solution P: \(+8.5\%\)
• Solution Q: \(-4.2\%\)
• Solution R: \(0.0\%\)
• Solution S: \(-12.1\%\)

Which sequence places the solutions in order from the lowest water potential to the highest water potential?
  1. A.P \(\rightarrow\) R \(\rightarrow\) Q \(\rightarrow\) S
  2. B.S \(\rightarrow\) Q \(\rightarrow\) R \(\rightarrow\) P
  3. C.P \(\rightarrow\) Q \(\rightarrow\) R \(\rightarrow\) S
  4. D.S \(\rightarrow\) R \(\rightarrow\) Q \(\rightarrow\) P
查看答案详解

解题

Water moves by osmosis down a water potential gradient from a region of higher water potential to a region of lower water potential.
- A potato cylinder loses the most mass in the solution with the lowest (most negative) water potential because water moves out of the cells by osmosis (Solution S: \(-12.1\%\)).
- Solution Q caused a smaller mass loss (\(-4.2\%\)), meaning it has a higher water potential than S.
- Solution R caused no net change (\(0.0\%\)), meaning its water potential equals that of the potato cells.
- Solution P caused mass gain (\(+8.5\%\)), showing water entered the cells, so Solution P has the highest water potential.

Order from lowest to highest water potential: S \(\rightarrow\) Q \(\rightarrow\) R \(\rightarrow\) P.

评分标准

B [1]
题目 23 · 選擇題
1
What happens to the diaphragm and the external intercostal muscles during expiration at rest?
  1. A.Diaphragm contracts; external intercostal muscles contract
  2. B.Diaphragm contracts; external intercostal muscles relax
  3. C.Diaphragm relaxes; external intercostal muscles contract
  4. D.Diaphragm relaxes; external intercostal muscles relax
查看答案详解

解题

During quiet expiration at rest:
- The diaphragm relaxes and returns to its dome shape, pushing up into the thorax.
- The external intercostal muscles relax, allowing the ribs to move downwards and inwards under gravity and tissue elasticity.

Both muscle groups relax, decreasing thoracic volume and increasing pressure to force air out.

评分标准

D [1]
题目 24 · 選擇題
1
A sample of double-stranded DNA contains \(1200\) base pairs. Analysis shows that \(28\%\) of all the nitrogenous bases are cytosine (C).

How many adenine (A) bases are present in this sample of DNA?
  1. A.264
  2. B.336
  3. C.528
  4. D.672
查看答案详解

解题

1. Double-stranded DNA with \(1200\) base pairs contains \(1200 \times 2 = 2400\) total bases.
2. According to complementary base pairing, \(\%\text{Cytosine (C)} = \%\text{Guanine (G)} = 28\%\).
3. Total percentage of \(\text{C} + \text{G} = 28\% + 28\% = 56\%\).
4. Therefore, total percentage of \(\text{Adenine (A)} + \text{Thymine (T)} = 100\% - 56\% = 44\%\).
5. Since \(\%\text{A} = \%\text{T}\), the percentage of adenine is \(44\% \div 2 = 22\%\).
6. Number of adenine bases \(= 22\% \times 2400 = 0.22 \times 2400 = 528\).

评分标准

C [1]
题目 25 · 選擇題
1
A photomicrograph shows an organelle with an image length of \(18\text{ mm}\). The magnification of the image is \(\times 3000\).

What is the actual length of the organelle in micrometres (\(\mu\text{m}\))?
  1. A.\(0.006\,\mu\text{m}\)
  2. B.\(0.17\,\mu\text{m}\)
  3. C.\(6.0\,\mu\text{m}\)
  4. D.\(60\,\mu\text{m}\)
查看答案详解

解题

Using the magnification formula:
\(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}}\)

Convert image size to micrometres (\(\mu\text{m}\)):
\(18\text{ mm} = 18 \times 1000\,\mu\text{m} = 18000\,\mu\text{m}\)

Calculate actual size:
\(\text{Actual size} = \frac{18000\,\mu\text{m}}{3000} = 6.0\,\mu\text{m}\)

评分标准

C [1]
题目 26 · 選擇題
1
Which statement describes what happens to lactic acid produced in human muscle cells during vigorous exercise?
  1. A.It is filtered out by the kidneys and excreted directly in urine.
  2. B.It is transported in the blood to the liver where it is broken down or converted to glucose.
  3. C.It is transported to the lungs and exhaled as a gas.
  4. D.It remains permanently in the muscle tissue until new muscle fibres develop.
查看答案详解

解题

During vigorous exercise, muscles respire anaerobically producing lactic acid. Lactic acid diffuses into the bloodstream and is transported to the liver, where it is broken down aerobically or converted back into glucose/glycogen, helping to pay back the oxygen debt.

评分标准

B [1 mark] - Correctly identifies that lactic acid is transported via the blood to the liver for conversion/breakdown.
题目 27 · 選擇題
1
Four fresh potato cylinders, each measuring \(50\text{ mm}\) in length, were placed into four test-tubes containing sucrose solutions of different concentrations: P, Q, R, and S.

After 3 hours, their lengths were measured:
- Solution P: \(54\text{ mm}\)
- Solution Q: \(50\text{ mm}\)
- Solution R: \(47\text{ mm}\)
- Solution S: \(43\text{ mm}\)

Which statement is correct?
  1. A.Solution P has a lower water potential than the potato cells.
  2. B.Solution Q has a higher water potential than Solution P.
  3. C.Solution R has a lower water potential than Solution Q.
  4. D.Solution S has the highest water potential of all four solutions.
查看答案详解

解题

In solution P, water enters the potato cells by osmosis, increasing cylinder length; so P has a higher water potential than the tissue. In Q, no net movement occurs (isotonic). In R and S, water leaves the tissue by osmosis causing shrinkage, with S losing the most water. Therefore, water potential order from highest to lowest is P > Q > R > S. Thus, solution R has a lower water potential than solution Q.

评分标准

C [1 mark] - Deduces the correct relative water potential based on change in cylinder length.
题目 28 · 選擇題
1
Which row correctly shows the actions of the muscles and the change in volume of the thorax during inspiration in a human?
  1. A.External intercostals: contract | Internal intercostals: relax | Diaphragm: contracts and moves downwards | Volume of thorax: increases
  2. B.External intercostals: relax | Internal intercostals: contract | Diaphragm: relaxes and curves upwards | Volume of thorax: increases
  3. C.External intercostals: contract | Internal intercostals: relax | Diaphragm: relaxes and curves upwards | Volume of thorax: decreases
  4. D.External intercostals: relax | Internal intercostals: contract | Diaphragm: contracts and moves downwards | Volume of thorax: decreases
查看答案详解

解题

During inspiration (breathing in), external intercostal muscles contract (pulling ribs up and out) while internal intercostal muscles relax. The diaphragm contracts and flattens (moves downwards), causing the volume of the thorax to increase, which lowers thoracic pressure below atmospheric pressure.

评分标准

A [1 mark] - Correct combination of muscle states and thoracic volume change for inspiration.
题目 29 · 選擇題
1
An investigation was carried out to determine the effect of temperature on the rate of an enzyme-catalysed reaction.

Which statement explains why the rate of reaction decreases rapidly at temperatures significantly above the optimum temperature?
  1. A.The kinetic energy of the enzyme and substrate molecules decreases, reducing collision frequency.
  2. B.The high temperature alters the shape of the active site so the substrate can no longer bind.
  3. C.The substrate molecules are permanently converted into inhibitors by heat.
  4. D.The activation energy of the reaction is lowered too much for the reaction to proceed.
查看答案详解

解题

At temperatures well above the optimum, excess kinetic energy disrupts the bonds maintaining the three-dimensional tertiary structure of the enzyme. This alters the shape of the active site (denaturation), preventing substrate molecules from binding complementary to form enzyme-substrate complexes.

评分标准

B [1 mark] - Recognises that high temperatures cause denaturation by changing the active site shape.
题目 30 · 選擇題
1
A student measured the rate of photosynthesis in an aquatic plant at different light intensities while keeping the temperature constant at \(20^\circ\text{C}\) and carbon dioxide concentration constant at \(0.03\%\).

At high light intensities, further increases in light intensity did not increase the rate of photosynthesis.

Which factor is limiting the rate of photosynthesis at these high light intensities?
  1. A.Light intensity
  2. B.Carbon dioxide concentration
  3. C.Oxygen concentration
  4. D.Glucose concentration
查看答案详解

解题

When an increase in light intensity no longer increases the rate of photosynthesis, light intensity is no longer the limiting factor. The rate is now limited by another environmental condition present in shortest supply, such as the carbon dioxide concentration or temperature.

评分标准

B [1 mark] - Correctly identifies carbon dioxide concentration as the limiting factor when the light response curve plateaus.
题目 31 · multiple_choice
1
Which row correctly compares anaerobic respiration in yeast cells with anaerobic respiration in human muscle cells?
  1. A.Yeast produces alcohol and carbon dioxide; human muscle produces lactic acid; both release much less energy per glucose molecule than aerobic respiration.
  2. B.Yeast produces lactic acid; human muscle produces alcohol and carbon dioxide; both release much less energy per glucose molecule than aerobic respiration.
  3. C.Yeast produces alcohol and carbon dioxide; human muscle produces lactic acid; both release much more energy per glucose molecule than aerobic respiration.
  4. D.Yeast produces lactic acid and water; human muscle produces alcohol; both release the same amount of energy per glucose molecule as aerobic respiration.
查看答案详解

解题

In yeast (a fungus), anaerobic respiration (fermentation) converts glucose into alcohol (ethanol) and carbon dioxide. In human muscle cells during vigorous exercise, glucose is broken down into lactic acid without producing carbon dioxide. In both cases, only a small fraction of the total chemical energy in glucose is released compared to aerobic respiration.

评分标准

A ; [1]
题目 32 · multiple_choice
1
Four identical potato cylinders, each with an initial mass of 2.00 g, were placed into four different concentrations of sucrose solution (P, Q, R, and S) for 60 minutes. The final masses were recorded:
• Solution P: 2.45 g
• Solution Q: 2.00 g
• Solution R: 1.65 g
• Solution S: 1.85 g

Which solution had the highest water potential?
  1. A.Solution S
  2. B.Solution R
  3. C.Solution P
  4. D.Solution Q
查看答案详解

解题

Water moves into plant cells by osmosis down a water potential gradient (from higher water potential to lower water potential). The potato cylinder in Solution P gained the most mass (+0.45 g), which means water entered the cells at the fastest net rate because Solution P had the highest water potential of all four solutions.

评分标准

C ; [1]
题目 33 · multiple_choice
1
Which set of changes occurs in the human thorax during inspiration at rest?
  1. A.Internal intercostal muscles contract; diaphragm becomes dome-shaped; pressure in thorax decreases
  2. B.External intercostal muscles contract; diaphragm flattens; pressure in thorax decreases
  3. C.External intercostal muscles contract; diaphragm becomes dome-shaped; pressure in thorax increases
  4. D.Internal intercostal muscles contract; diaphragm flattens; pressure in thorax increases
查看答案详解

解题

During inspiration, the external intercostal muscles contract (pulling the ribcage upwards and outwards) and the diaphragm contracts and flattens. This increases the volume of the thorax, causing the pressure inside the thorax and lungs to decrease below atmospheric pressure, drawing air in.

评分标准

B ; [1]
题目 34 · multiple_choice
1
An investigation was carried out to measure the rate of photosynthesis in an aquatic plant. Light intensity and dissolved carbon dioxide concentration were maintained at very high levels, but the temperature was maintained at \(10\ ^\circ\text{C}\). Further increases in light intensity did not increase the rate of photosynthesis.

What was the limiting factor under these conditions?
  1. A.Carbon dioxide concentration
  2. B.Light intensity
  3. C.Temperature
  4. D.Water availability
查看答案详解

解题

A limiting factor is something present in the environment in such short supply that it restricts life processes. Since light intensity and carbon dioxide are in excess/very high, the relatively low temperature (\(10\ ^\circ\text{C}\)) limits the kinetic energy of photosynthetic enzymes and substrate molecules, making temperature the limiting factor.

评分标准

C ; [1]
题目 35 · multiple_choice
1
Which row correctly identifies the organ of secretion, the substrate, and the optimum pH for the enzyme pepsin?
  1. A.Pancreas | protein | pH 8
  2. B.Stomach | lipid | pH 2
  3. C.Salivary glands | starch | pH 7
  4. D.Stomach | protein | pH 2
查看答案详解

解题

Pepsin is a protease enzyme secreted by the stomach wall (gastric glands). It hydrolyses proteins into polypeptides in the acidic environment of the stomach, where hydrochloric acid provides an optimum pH of approximately 1.5 to 2.0.

评分标准

D ; [1]
题目 36 · 選擇題
1
During vigorous exercise, anaerobic respiration takes place in human muscle cells. What happens to the lactic acid produced by this process during the recovery period?
  1. A.It is transported in the blood to the liver where it is broken down using oxygen.
  2. B.It is excreted directly by the kidneys into the urine.
  3. C.It is transported in the blood to the lungs to be exhaled.
  4. D.It remains permanently stored inside muscle fibres.
查看答案详解

解题

During vigorous exercise, lactic acid builds up in muscle cells due to anaerobic respiration. In the recovery period, lactic acid is transported in the blood to the liver, where it is broken down aerobically (converted to glucose or oxidised to carbon dioxide and water), which requires extra oxygen to repay the oxygen debt.

评分标准

A is correct [1]; B is incorrect as lactic acid is processed by the liver, not filtered out in urine [0]; C is incorrect as lactic acid is not transported directly to the lungs for exhalation [0]; D is incorrect as it requires oxygen (aerobic metabolism) in the liver [0].
题目 37 · 選擇題
1
Cylinders of potato tissue with an internal water potential of \(-600\text{ kPa}\) were placed into four different sucrose solutions. In which sucrose solution will the plant cells become plasmolysed and show the greatest decrease in mass?
  1. A.\(-200\text{ kPa}\)
  2. B.\(-400\text{ kPa}\)
  3. C.\(-600\text{ kPa}\)
  4. D.\(-900\text{ kPa}\)
查看答案详解

解题

For cells to become plasmolysed and decrease in mass, water must leave the cells by osmosis down a water potential gradient (from higher water potential inside to lower water potential outside). The external solution with the lowest water potential is \(-900\text{ kPa}\), which provides the steepest water potential gradient for net water loss from the cells.

评分标准

D is correct [1]; A and B have higher water potentials than the tissue, causing water entry and mass increase [0]; C has the same water potential, causing no net water movement [0].
题目 38 · 選擇題
1
Which row correctly identifies the state of the muscles and the volume change in the thorax during normal expiration at rest?
  1. A.External intercostal muscles contract; diaphragm relaxes; volume of thorax increases
  2. B.External intercostal muscles relax; diaphragm contracts; volume of thorax increases
  3. C.External intercostal muscles relax; diaphragm relaxes; volume of thorax decreases
  4. D.External intercostal muscles contract; diaphragm contracts; volume of thorax decreases
查看答案详解

解题

During normal expiration at rest, the external intercostal muscles relax (causing the ribs to move downwards and inwards) and the diaphragm relaxes (becoming dome-shaped). As a result, the volume of the thorax decreases, which increases pressure and forces air out of the lungs.

评分标准

C is correct [1]; A and D incorrectly state that external intercostal muscles contract [0]; B incorrectly states that the diaphragm contracts and the thorax expands [0].
题目 39 · 選擇題
1
An investigation was conducted into the rate of photosynthesis of an aquatic plant at different light intensities and temperatures. Carbon dioxide concentration was kept high and constant throughout.

At \(10^\circ\text{C}\), the rate of photosynthesis reached a maximum at a light intensity of \(40\text{ arbitrary units}\) and did not increase further when light intensity was increased to \(60\text{ arbitrary units}\).

At \(25^\circ\text{C}\), the rate of photosynthesis continued to increase up to \(80\text{ arbitrary units}\).

What was the factor limiting the rate of photosynthesis at a light intensity of \(60\text{ arbitrary units}\) at \(10^\circ\text{C}\)?
  1. A.Carbon dioxide concentration
  2. B.Light intensity
  3. C.Oxygen concentration
  4. D.Temperature
查看答案详解

解题

At a light intensity of \(60\text{ arbitrary units}\) at \(10^\circ\text{C}\), an increase in light intensity does not increase the rate of photosynthesis. However, increasing the temperature to \(25^\circ\text{C}\) increases the rate. Therefore, temperature is the limiting factor at that point.

评分标准

D is correct [1]; A is incorrect because carbon dioxide was provided in high/excess concentration [0]; B is incorrect because increasing light intensity beyond 40 units at 10 °C did not increase the rate [0]; C is incorrect because oxygen is a product, not a limiting factor of photosynthesis [0].
题目 40 · 選擇題
1
Which statement correctly describes the sequence of events during protein synthesis in a cell?
  1. A.An mRNA copy is made from DNA in the nucleus, moves to the cytoplasm, and attaches to a ribosome where amino acids are linked.
  2. B.DNA leaves the nucleus, enters the cytoplasm, and binds directly to amino acids to form a protein.
  3. C.mRNA is produced on ribosomes, moves into the nucleus, and copies the DNA sequence.
  4. D.Ribosomes produce DNA molecules in the cytoplasm which directly combine to make proteins.
查看答案详解

解题

During protein synthesis, a molecule of mRNA is transcribed from DNA in the nucleus. The mRNA then leaves the nucleus and passes into the cytoplasm, where it binds to a ribosome. At the ribosome, amino acids are assembled in the sequence dictated by the base triplets on the mRNA.

评分标准

A is correct [1]; B is incorrect because DNA remains inside the nucleus [0]; C is incorrect because mRNA is made in the nucleus, not on ribosomes [0]; D is incorrect because ribosomes do not make DNA copies [0].

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Paper 41: Theory (Extended)

Answer all questions in the spaces provided. Show all calculations where required.
6 题目 · 79.65
题目 1 · Structured
13.33
1 Respiration is a fundamental metabolic process occurring in all living cells.

(a) Define the term aerobic respiration. [2]

(b) An investigation was carried out to determine the effect of temperature on anaerobic respiration in yeast (Saccharomyces cerevisiae). Suspensions of yeast in 5% glucose solution were incubated in water baths at five different temperatures. The volume of carbon dioxide gas evolved in 10 minutes was recorded.

Table 1.1 shows the results.

Table 1.1

| Temperature / °C | Volume of carbon dioxide collected in 10 min / cm³ |
| :--- | :--- |
| 20 | 12 |
| 30 | 28 |
| 40 | 42 |
| 50 | 22 |
| 60 | 0 |

(i) State the balanced chemical equation for anaerobic respiration in yeast. [2]

(ii) Using the data in Table 1.1, calculate the percentage increase in the volume of carbon dioxide collected when the temperature increased from 20 °C to 40 °C.

Space for working.

........................................................... % [2]

(iii) Explain why the volume of carbon dioxide evolved decreased when the temperature was increased from 40 °C to 60 °C. [3]

(c) During strenuous sprint exercise, human muscle cells respire anaerobically, producing lactic acid.

Describe what happens to lactic acid in the body after exercise has ceased to remove the oxygen debt. [4]

[Total: 13]
查看答案详解

解题

(a) Aerobic respiration is the chemical reactions in cells that use oxygen to break down nutrient molecules to release energy.

(b)(i) Glucose breaks down into ethanol and carbon dioxide: \(\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2\).

(b)(ii) Initial volume at 20 °C = 12 cm³; Final volume at 40 °C = 42 cm³.
Change = 42 - 12 = 30 cm³.
Percentage increase = \(\frac{30}{12} \times 100 = 250\%\).

(b)(iii) The optimum temperature for yeast enzymes is around 40 °C. Above this temperature, high thermal energy causes excessive vibrations that break bonds in the active site of enzymes, leading to denaturation. The active site shape changes so it is no longer complementary to the glucose substrate, preventing enzyme-substrate complexes from forming, resulting in zero gas output at 60 °C.

(c) Lactic acid travels in the blood from muscles to the liver. In the liver, aerobic respiration oxidises part of the lactic acid to \(\text{CO}_2\) and water, while the rest is converted back into glucose/glycogen. Fast breathing persists after exercise to provide the required oxygen to metabolise this lactic acid (paying back oxygen debt).

评分标准

(a) chemical reactions in cells that use oxygen ; to break down nutrient / glucose molecules to release energy ;
[2]

(b)(i) \(\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2\)
correct formulas for glucose, ethanol, and \(\text{CO}_2\) ;
correct balancing ;
[2]

(b)(ii) 250 ;
(Allow 1 mark for correct working: \(\frac{42 - 12}{12} \times 100\) or \(\frac{30}{12} \times 100\))
[2]

(b)(iii) any three from:
- temperature above optimum (40 °C) ;
- enzymes / active sites denatured ;
- change in shape of active site / active site no longer complementary to substrate ;
- substrate can no longer fit / bind / no enzyme-substrate complexes formed ;
- at 60 °C all enzymes completely denatured / reaction stops ;
[max 3]

(c) any four from:
- lactic acid diffuses / moves into blood / plasma ;
- transported to the liver ;
- oxidised / broken down aerobically to carbon dioxide and water ;
- converted to glucose / glycogen ;
- continued high breathing rate / heavy breathing supplies extra oxygen required ;
- oxygen debt repaid ;
[max 4]
题目 2 · Structured
13.33
2 Plant tissues interact dynamically with their surrounding aqueous environment via osmosis.

(a) Define the term osmosis. [3]

(b) Cylinders of fresh potato tissue were cut to exactly 50 mm in length and weighed. Each cylinder had an initial mass of 5.00 g. Cylinders were immersed in sucrose solutions of concentrations 0.0, 0.2, 0.4, 0.6, and 0.8 mol/dm³ for 60 minutes.

(i) Explain why the potato cylinder placed in the 0.0 mol/dm³ (pure water) solution gained mass. [3]

(ii) The cylinder placed in 0.8 mol/dm³ sucrose solution had a final mass of 4.15 g.

Calculate the percentage change in mass of this potato cylinder.

Space for working.

........................................................... % [2]

(iii) State the condition of the cells in the potato cylinder after 60 minutes in 0.8 mol/dm³ sucrose solution, and describe their microscopic appearance. [2]

(c) Explain how water moves continuously from the root hair cells through the root cortex into the xylem vessels. [3]

[Total: 13]
查看答案详解

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) The water potential of pure water (0.0 mol/dm³) is 0, which is higher than the water potential inside the potato cell vacuoles. Water enters the cells by osmosis across the partially permeable cell membrane down the water potential gradient, causing an increase in volume and mass.

(b)(ii) Initial mass = 5.00 g, Final mass = 4.15 g.
Change in mass = 4.15 - 5.00 = -0.85 g.
Percentage change = \(\frac{-0.85}{5.00} \times 100 = -17.0\%\).

(b)(iii) The cells become plasmolysed (or flaccid). The cell membrane and cytoplasm shrink and pull away from the rigid cellulose cell wall because water has left the vacuole.

(c) Water enters root hair cells by osmosis because cell sap has lower water potential than soil water. This raises the water potential of root hair cells relative to neighboring cortex cells. Water moves by osmosis down the water potential gradient from cell to cell across the cortex into xylem vessels.

评分标准

(a) net movement of water molecules ;
from a region of higher water potential to a region of lower water potential / down a water potential gradient ;
through a partially permeable membrane ;
[3]

(b)(i) pure water has higher water potential than inside potato cell / vacuole / cytoplasm ;
water enters by osmosis ;
down water potential gradient / across partially permeable cell membrane ;
[3]

(b)(ii) -17 / -17.0 (%) ;;
(Allow 1 mark for correct working: \(\frac{4.15 - 5.00}{5.00} \times 100\) or \(\frac{0.85}{5.00} \times 100\))
(Note: award 1 mark if magnitude 17% is given without minus sign or without indicating loss/decrease)
[2]

(b)(iii) plasmolysed / flaccid ;
cytoplasm / cell membrane pulled away from cell wall / vacuole shrunk ;
[2]

(c) any three from:
- water enters root hair cell by osmosis ;
- root hair cell has higher water potential than adjacent cortex cell ;
- water moves down water potential gradient across cortex cells ;
- by osmosis ;
- enters xylem down water potential / pressure gradient / pulled by transpiration stream ;
[max 3]
题目 3 · Structured
13.33
3 The human gas exchange system is adapted to allow efficient diffusion of oxygen and carbon dioxide.

(a) Describe the mechanism of inspiration (breathing in) in a healthy human. [4]

(b) State three structural features of alveoli and explain how each feature adapts them for efficient gas exchange. [3]

(c) Cigarette smoke contains many harmful components, including carbon monoxide and tar.

(i) Describe the physiological effect of inhaling carbon monoxide on oxygen transport in the human body. [2]

(ii) Explain how tar damages the lining of the respiratory tract and leads to chronic bronchitis. [4]

[Total: 13]
查看答案详解

解题

(a) During inspiration:
1. External intercostal muscles contract while internal intercostal muscles relax.
2. Ribs move upwards and outwards.
3. The diaphragm muscle contracts and flattens downwards.
4. These movements increase the volume of the thorax / thoracic cavity.
5. The pressure inside the thorax/lungs drops below atmospheric pressure, causing air to be drawn in.

(b) Alveolar adaptations:
- Millions of microscopic alveoli provide a very large surface area for diffusion.
- Alveolar wall is one cell thick (thin epithelium) giving a short diffusion distance.
- Dense capillary network with continuous blood flow maintains a steep concentration gradient.

(c)(i) Carbon monoxide binds with haemoglobin with higher affinity than oxygen, forming carboxyhaemoglobin. This permanently prevents oxygen from binding, significantly reducing blood oxygen transport.

(c)(ii) Tar irritates the airway lining, stimulating goblet cells to produce excessive sticky mucus. Concurrently, chemicals in tar paralyse and destroy the cilia on epithelial cells. Consequently, mucus cannot be moved up to the throat and pools in the bronchi, providing a breeding ground for bacteria and causing chronic inflammation and narrowing of the airways (chronic bronchitis).

评分标准

(a) any four from:
- external intercostal muscles contract ;
- internal intercostal muscles relax ;
- ribs move up and out(wards) ;
- diaphragm contracts / flattens / moves down ;
- volume of thorax / thoracic cavity / chest increases ;
- pressure inside thorax / lungs decreases / drops below atmospheric pressure ;
- air is drawn into lungs ;
[max 4]

(b) any three pairs (feature + explanation) from:
- large surface area + allows more gas molecules to diffuse at once / increases diffusion rate ;
- thin walls / one cell thick + provides short diffusion pathway / distance ;
- surrounded by dense capillary network / good blood supply + maintains steep concentration gradient ;
- moist lining + allows gases to dissolve before diffusing ;
- well ventilated + maintains steep concentration gradient ;
[max 3]

(c)(i) carbon monoxide binds irreversibly / with high affinity to haemoglobin ;
forms carboxyhaemoglobin / reduces oxygen-carrying capacity of blood / red blood cells ;
[2]

(c)(ii) any four from:
- tar stimulates goblet cells to produce excess / more mucus ;
- damages / destroys / paralyses cilia / ciliated cells ;
- mucus is not swept / cleared away from airways / accumulates ;
- bacteria / pathogens become trapped in mucus and multiply ;
- causes frequent / persistent infection ;
- airways / bronchi become inflamed / narrowed (causing chronic bronchitis / cough) ;
[max 4]
题目 4 · Structured
13.33
4 Photosynthesis is the process by which plants manufacture carbohydrates from raw materials using light energy.

(a) State the balanced chemical equation for photosynthesis. [2]

(b) A student investigated the effect of light intensity on the rate of photosynthesis in the water plant Cabomba caroliniana.

The plant was placed in a beaker containing dilute sodium hydrogencarbonate solution. An LED lamp was placed at various distances from the beaker, and the number of oxygen bubbles released per minute was counted.

(i) State the independent variable and the dependent variable in this investigation. [2]

(ii) State two variables that must be controlled in this experiment and describe how each is kept constant. [4]

(c) In another experiment, the rate of photosynthesis of leaf discs was measured under varying light intensities at two different carbon dioxide concentrations (0.04% and 0.12%) at a constant temperature of 25 °C.

Describe and explain the effect of increasing light intensity on the rate of photosynthesis at both carbon dioxide concentrations. [5]

[Total: 13]
查看答案详解

解题

(a) Balanced equation: \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\).

(b)(i) Independent variable: Light intensity (manipulated by changing the distance from the lamp). Dependent variable: Rate of bubble production per unit time (number of bubbles per minute).

(b)(ii) Controlled variables:
1. Temperature: Maintain using a thermostatically controlled water bath or heat screen between light and beaker.
2. Concentration of dissolved \(\text{CO}_2\): Maintain by using a standardized concentration of sodium hydrogencarbonate solution (e.g. 0.2%).

(c) At low light intensities, light is the limiting factor: increasing light energy increases ATP and NADPH production in chlorophyll, accelerating the rate linearly. Above a certain light intensity, light is no longer the limiting factor. At 0.04% \(\text{CO}_2\), the rate reaches a lower plateau because \(\text{CO}_2\) availability limits the dark reactions. At 0.12% \(\text{CO}_2\), more substrate is present for carbon fixation, enabling the rate to reach a significantly higher maximum plateau until another factor (such as temperature or enzyme availability) becomes limiting.

评分标准

(a) \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)
correct formulas for reactants and products ;
correct balancing ;
[2]

(b)(i) independent: light intensity / distance of lamp (from plant) ;
dependent: rate of photosynthesis / number of (oxygen) bubbles per minute / volume of gas per unit time ;
[2]

(b)(ii) any two pairs (variable + method) from:
- temperature + (controlled) water bath / heat shield / LED lamp ;
- carbon dioxide concentration + use same concentration of sodium hydrogencarbonate solution ;
- species / mass / length / age of plant + use the same piece of plant throughout ;
- wavelength / colour of light + use the same light source / filter ;
[max 4]

(c) any five from:
- at low light intensity, rate increases as light intensity increases (for both curves) ;
- light intensity is the limiting factor at low light intensities ;
- at high light intensity, rate reaches a plateau / levels off / becomes constant ;
- light intensity is no longer the limiting factor ;
- rate is higher at 0.12% \(\text{CO}_2\) than at 0.04% \(\text{CO}_2\) at higher light intensities ;
- at 0.04% \(\text{CO}_2\), carbon dioxide concentration is the limiting factor ;
- more \(\text{CO}_2\) provides more substrate / carbon for glucose production ;
[max 5]
题目 5 · Structured
13.33
5 The human alimentary canal processes ingested food through mechanical and chemical digestion prior to absorption.

(a) State the precise site of production and the site of action for each of the following substances:

(i) Salivary amylase [2]

(ii) Hydrochloric acid [2]

(b) Explain the functions of bile in the digestion and processing of dietary lipids in the duodenum. [3]

(c) The ileum is the main site of nutrient absorption.

(i) State the names of two internal structures found inside an individual villus that transport absorbed nutrients away from the small intestine. [2]

(ii) Describe four adaptations of the small intestine that ensure rapid and efficient absorption of digested food molecules. [4]

[Total: 13]
查看答案详解

解题

(a)(i) Salivary amylase is produced by the salivary glands and acts in the mouth (oral/buccal cavity).

(a)(ii) Hydrochloric acid is produced by parietal/gastric cells in the stomach wall and acts in the lumen of the stomach.

(b) Bile has two key roles:
1. It is alkaline, containing hydrogencarbonate ions that neutralise acidic gastric juices entering the duodenum, establishing the optimal alkaline pH (around pH 7-8) for pancreatic lipase.
2. Bile salts emulsify large fat globules into tiny fat droplets (mechanical digestion), greatly increasing the surface area for lipase to hydrolyse triglycerides into fatty acids and glycerol.

(c)(i) The internal transport vessels in a villus are the blood capillary (for monosaccharides, amino acids, minerals, and water) and the lacteal (lymph vessel, for absorbed fatty acids and glycerol).

(c)(ii) Adaptations of small intestine:
- Inner wall is folded and lined with millions of finger-like villi, whose epithelial cells have microvilli, creating a massive surface area.
- Epithelial wall is single-layered (one cell thick), minimizing the diffusion distance.
- Rich capillary blood supply and lymphatic lacteals rapidly remove absorbed products, maintaining steep concentration gradients.
- Presence of transport proteins / mitochondria in epithelial cells provides energy for active transport of glucose and amino acids.

评分标准

(a)(i) production: salivary glands ;
action: mouth / oral cavity / buccal cavity ;
[2]

(a)(ii) production: stomach wall / gastric glands / stomach lining / stomach mucosa ;
action: stomach (lumen) ;
[2]

(b) any three from:
- neutralises acidic chyme / gastric juice / acid from stomach ;
- provides optimum / alkaline pH for (pancreatic) enzymes / lipase ;
- emulsifies fats / lipids / breaks large fat globules into smaller droplets ;
- increases surface area of lipids for lipase action / digestion ;
[max 3]

(c)(i) capillary / blood vessel / blood capillary ;
lacteal / lymphatic vessel ;
[2]

(c)(ii) any four from:
- presence of villi / microvilli gives very large surface area ;
- thin epithelium / wall is one cell thick providing short diffusion distance ;
- dense capillary network maintains steep concentration gradient / removes nutrients quickly ;
- presence of lacteals for absorbing fatty acids and glycerol ;
- long length of small intestine allows time for complete absorption ;
- epithelial cells have many mitochondria to supply ATP for active transport ;
- presence of membrane carrier proteins / channel proteins for active transport / facilitated diffusion ;
[max 4]
题目 6 · Structured
13
1 (a) State two features of the gas exchange surface in humans. [2]

(b) A student investigated the ventilation of an athlete at rest and during vigorous exercise.

Table 1.1 shows the data collected.

Table 1.1

| Parameter | At rest | During vigorous exercise |
| :--- | :--- | :--- |
| breathing rate / breaths per minute | 12 | 32 |
| tidal volume / dm\(^3\) | 0.50 | 2.25 |
| minute ventilation / dm\(^3\) per minute | 6.00 | X |

(i) Calculate the percentage increase in tidal volume during vigorous exercise compared with at rest.

Space for working.

\(\text{percentage increase} =\) ........................................................... % [2]

(ii) Calculate the minute ventilation, X, during vigorous exercise.

Space for working.

\(\text{minute ventilation} =\) ........................................................... dm\(^3\) per minute [1]

(c) Describe the mechanism that causes inspiration (inhalation) in humans. [4]

(d) Explain why an athlete's breathing rate and breathing depth remain higher than resting values for several minutes after vigorous exercise has stopped. [4]

[Total: 13]
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解题

(a) Features of gas exchange surfaces:
State any two recognized features of the alveoli/lungs:
- Large surface area
- Thin wall / one cell thick / short diffusion pathway
- Good blood supply / surrounded by extensive capillary network
- Well ventilated / maintenance of steep concentration gradient

(b)(i) Percentage increase calculation:
\(\text{Increase} = 2.25\text{ dm}^3 - 0.50\text{ dm}^3 = 1.75\text{ dm}^3\)
\(\text{Percentage increase} = \frac{1.75}{0.50} \times 100 = 350\%\)

(b)(ii) Minute ventilation calculation:
\(\text{Minute ventilation} = \text{breathing rate} \times \text{tidal volume}\)
\(\mathbf{X} = 32\text{ breaths/min} \times 2.25\text{ dm}^3 = 72.0\text{ dm}^3\text{ per minute}\)

(c) Mechanism of inspiration:
- External intercostal muscles contract.
- Ribs move upwards and outwards.
- Diaphragm contracts and flattens / moves downwards.
- Volume of the thorax / thoracic cavity increases.
- Pressure inside the lungs / thorax decreases to below atmospheric pressure.
- Air enters the lungs down a pressure gradient.

(d) Explanation of post-exercise ventilation (Oxygen debt):
- During vigorous exercise, muscles respire anaerobically as well as aerobically.
- Lactic acid is produced and accumulates in muscles and blood.
- An oxygen debt is created.
- Lactic acid is transported in the blood to the liver.
- Extra oxygen is required to break down / oxidise lactic acid to carbon dioxide and water (or convert it into glucose/glycogen).
- Higher breathing rate and depth maintain a supply of oxygen to clear lactic acid and expel higher amounts of carbon dioxide.

评分标准

(a) any two from:
- large surface area ;
- thin / one cell thick / short diffusion distance ;
- good blood supply / dense capillary network ;
- good ventilation / maintains concentration gradient ;
[max 2]

(b)(i)
- \((2.25 - 0.50) / 0.50\) OR \(1.75 / 0.50\) OR \((2.25 / 0.50) \times 100 - 100\) ;
- \(350\) (\%)[correct answer alone gains 2 marks] ;
[2]

(b)(ii)
- \(72\) / \(72.0\) ;
[1]

(c) any four from:
- diaphragm contracts ;
- (diaphragm) flattens / moves down / descends ;
- external intercostal muscles contract ;
- ribs move upwards / outwards / up and out ;
- volume of thorax / lungs / chest increases ;
- pressure in thorax / lungs decreases / becomes lower than atmospheric pressure ;
- air is drawn in / moves into lungs down a pressure gradient ;
[max 4]

(d) any four from:
- anaerobic respiration occurred (during exercise) ;
- lactic acid / lactate produced / built up (in muscles / blood) ;
- idea of oxygen debt (needs to be repaid) ;
- lactic acid transported to the liver (in blood) ;
- oxygen required to break down / oxidise lactic acid (to carbon dioxide and water) / convert lactic acid to glucose / glycogen ;
- extra oxygen supplied by higher breathing rate / deeper breaths ;
- to remove excess carbon dioxide (produced) ;
[max 4]

Paper 51: Practical Test

Complete all practical tasks, record all experimental observations in the spaces provided, and complete the design task.
2 题目 · 40
题目 1 · Practical
20
1 A student investigated the effect of glucose concentration on the rate of anaerobic respiration in yeast (Saccharomyces cerevisiae).

The student used the following method:
• Step 1: Prepared five test-tubes, each containing \(10\text{ cm}^3\) of yeast suspension.
• Step 2: Added \(10\text{ cm}^3\) of glucose solution of known concentration (\(0.0\), \(2.0\), \(4.0\), \(6.0\), and \(8.0\%\)) to each test-tube.
• Step 3: Placed a layer of liquid paraffin oil on top of each mixture.
• Step 4: Fitted a bung with a delivery tube connected to an inverted measuring cylinder filled with water.
• Step 5: Kept all test-tubes in a water-bath at \(35^\circ\text{C}\) for \(15\text{ minutes}\).
• Step 6: Recorded the volume of gas collected in the measuring cylinder after \(15\text{ minutes}\).

The results are shown in Table 1.1.

Table 1.1


glucose concentration / %
volume of gas collected in \(15\text{ minutes}\) / \(\text{cm}^3\)
mean volume of gas collected / \(\text{cm}^3\)
rate of gas production / \(\text{cm}^3\text{ per minute}\)


trial 1
trial 2
trial 3


0.0
0.3
0.2
0.4
0.3
0.02


2.0
3.1
3.4
3.1
3.2
0.21


4.0
5.9
6.2
6.5
6.2
0.41


6.0
8.4
8.7
8.4
8.5
[A]


8.0
9.2
8.9
9.2
9.1
0.61



(a) (i) Calculate the rate of gas production for the \(6.0\%\) glucose concentration at [A] in Table 1.1. Give your answer to two decimal places. [1]

(ii) Plot a line graph on the grid of glucose concentration on the x-axis against the rate of gas production on the y-axis. [4]

(iii) Describe the relationship between glucose concentration and the rate of gas production shown in Table 1.1. [2]

(iv) State the purpose of the layer of liquid paraffin oil added in Step 3. [1]

(v) State two variables that were kept constant in this investigation. [2]

(vi) Identify one source of experimental error in Step 6 and suggest an improvement to minimise this error. [2]

(vii) State the chemical reagent used to test for the gas produced by yeast respiration and describe the positive observation. [2]

(b) Plan an investigation to determine the effect of pH on the rate of respiration in yeast. [6]
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解题

(a) (i)
\(\text{Rate of gas production} = \frac{\text{Mean volume of gas}}{\text{Time}} = \frac{8.5\text{ cm}^3}{15\text{ min}} = 0.5667\text{ cm}^3/\text{min}\)
Rounding to 2 decimal places gives \(0.57\text{ cm}^3\text{ per minute}\).

(ii)
A (Axes): x-axis labelled 'glucose concentration / %' and y-axis labelled 'rate of gas production / \(\text{cm}^3\text{ per minute}\)'
S (Scale): Linear, sensible scales where data occupies more than half of the grid in both dimensions (e.g. x-axis: 0 to 10 with 2 units per major division; y-axis: 0 to 0.7 with 0.1 per major division).
P (Plotting): All 5 points plotted accurately to within half a small square: (0.0, 0.02), (2.0, 0.21), (4.0, 0.41), (6.0, 0.57), (8.0, 0.61).
L (Line): Sharp, clear points connected point-to-point with straight lines using a ruler or drawn as a smooth curve of best fit.

(iii)
• As glucose concentration increases from \(0.0\%\) to \(8.0\%\), the rate of gas production increases.
• The increase is steepest / linear between \(0.0\%\) and \(6.0\%\), and begins to level off between \(6.0\%\) and \(8.0\%\).

(iv)
To exclude atmospheric oxygen / prevent oxygen from diffusing into the suspension to ensure anaerobic conditions.

(v)
Any two from:
• Temperature (\(35^\circ\text{C}\))
• Volume of yeast suspension (\(10\text{ cm}^3\))
• Concentration / source of yeast suspension
• Volume of glucose solution added (\(10\text{ cm}^3\))
• Time allowed for reaction (\(15\text{ minutes}\))

(vi)
Error: Gas escapes when inserting the rubber bung into the test-tube / measuring cylinder scale resolution is low / gas dissolves slightly in water.
Improvement: Use a sealed flask with a side-arm and a syringe / use a gas syringe to measure gas volume directly.

(vii)
• Reagent: Limewater (calcium hydroxide solution)
• Observation: Turns from colourless to cloudy / milky / white precipitate forms.

(b) Plan:
Independent variable: Use at least five different pH values (e.g., pH 4, 5, 6, 7, 8) produced using buffer solutions.
Dependent variable: Measure the volume of gas (\(\text{CO}_2\)) collected in a gas syringe (or measuring cylinder) over a fixed time period (e.g., \(10\text{ minutes}\)) OR record time to produce a set volume of gas.
Controlled variables:
1. Temperature kept constant using a thermostatically controlled water-bath (e.g. \(35^\circ\text{C}\)).
2. Glucose concentration and volume kept constant (e.g. \(10\text{ cm}^3\) of \(5\%\) glucose).
3. Yeast suspension volume and concentration kept constant (e.g. \(10\text{ cm}^3\)).
Repetition: Repeat the experiment at least 3 times at each pH value and calculate the mean rate to identify anomalies and improve reliability.
Safety: Wear safety goggles / lab coat / gloves when handling acid/alkali buffer solutions.

评分标准

(a)(i) [1 mark]
\(0.57\) ;

(a)(ii) [4 marks]
A: axes labelled with quantity and unit: glucose concentration / % and rate of gas production / \(\text{cm}^3\text{ per minute}\) ;
S: suitable linear scale where plotted data occupies \(>50\%\) of the grid in both directions ;
P: all 5 points correctly plotted within \(\pm\) half a small square ;
L: points joined neatly with straight lines between points using a ruler OR smooth best-fit curve with no feathering ;

(a)(iii) [2 marks]
as glucose concentration increases, the rate of gas production increases ;
rate of increase slows down / levels off above \(6.0\%\) / greatest increase between \(0.0\) and \(6.0\%\) ;

(a)(iv) [1 mark]
prevents oxygen entering / maintains anaerobic conditions ;

(a)(v) [2 marks]
any two from:
• temperature (of water-bath) ;
• volume of yeast suspension ;
• concentration / type of yeast ;
• volume of glucose solution ;
• incubation time / duration ;

(a)(vi) [2 marks]
error: gas lost before bung is sealed / measuring cylinder difficult to read accurately / gas dissolving in water ;
improvement: use a gas syringe / use a delivery tube connected via a two-hole stopper with tap funnel / repeat readings ;

(a)(vii) [2 marks]
limewater / calcium hydroxide solution ;
turns milky / cloudy / forms white precipitate ;

(b) [6 marks max]
MP1 (IV): at least 5 different pH values specified OR using buffer solutions ;
MP2 (DV): measure volume of gas collected in a specified time (e.g. 5–15 min) OR measure time to collect a fixed volume of gas ;
MP3 (CV1): temperature controlled using a water-bath ;
MP4 (CV2): constant volume/concentration of glucose solution AND constant volume/concentration of yeast suspension ;
MP5 (Reliability): repeat at each pH at least 3 times and calculate mean ;
MP6 (Safety): wear eye protection / gloves to avoid skin/eye contact with acidic or alkaline buffers ;
题目 2 · Practical
20
2 Plant tissues can be investigated to determine cell structure, nutrient content, and water relations.

(a) Fig. 2.1 is a photomicrograph of a cross-section through a plant stem showing vascular bundles and cortical tissue.


[ PHOTOMICROGRAPH: Fig. 2.1 ]
Outer epidermis ---
Cortex ------------
Vascular bundle ---
Pith (central) ----
Line AB = 48 mm


(i) Make a large, clear biological line drawing of the sector of the stem showing the arrangement of tissues from the outer epidermis to the central pith. Do not draw individual cells. [4]

(ii) The line AB on Fig. 2.1 represents the actual diameter across one vascular bundle.
The measured length of line AB is \(48\text{ mm}\).
The actual width of the vascular bundle is \(0.60\text{ mm}\).

Calculate the magnification of Fig. 2.1.

Use the formula:
\[\text{magnification} = \frac{\text{length of line } \mathbf{AB}}{\text{actual width}}\]

Give your answer to two significant figures. [2]

(b) A student tested a sample of the plant stem tissue for two biological molecules: reducing sugars and starch.

Complete Table 2.1 to state the test reagent used, the procedure, and the positive observation for each biological molecule.

Table 2.1


biological molecule
reagent / test solution
method / procedure
positive result


reducing sugar
................................................
................................................
................................................


starch
................................................
add drops to sample at room temperature
................................................


[4]

(c) Another student investigated osmosis in plant tissue by placing cylinders of potato tissue of equal initial length (\(50\text{ mm}\)) into different concentrations of sucrose solution for \(60\text{ minutes}\).

The results are shown in Table 2.2.

Table 2.2


concentration of sucrose solution / \(\text{mol/dm}^3\)
initial mass / \(\text{g}\)
final mass / \(\text{g}\)
change in mass / \(\text{g}\)
percentage change in mass / \(\%\)


0.0
3.20
3.68
+0.48
+15.0


0.2
3.15
3.40
+0.25
+7.9


0.4
3.25
3.25
0.00
[X]


0.6
3.30
3.04
-0.26
-7.9


0.8
3.10
2.67
-0.43
[Y]



(i) Calculate the percentage change in mass for:
• \(0.4\text{ mol/dm}^3\) sucrose solution at [X]
• \(0.8\text{ mol/dm}^3\) sucrose solution at [Y]. Give your answer to one decimal place.

Space for working.

[X] .................................... %
[Y] .................................... %
[2]

(ii) Using the data in Table 2.2, state the estimated concentration of sucrose inside the potato cells. Explain your answer. [2]

(iii) Explain why the student calculated the percentage change in mass rather than using only the change in mass. [1]

(iv) State two variables that were kept constant in this osmosis investigation. [2]

(v) Suggest why the student blotted the potato cylinders with a paper towel before weighing them, and explain how failing to do so would affect the calculated percentage change in mass. [3]
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解题

(a) (i)
Biological drawing criteria:
Size: Drawing occupies more than \(50\%\) of the provided box/space.
Outline quality: Clear, unbroken, continuous single lines; no sketching, cross-hatching, or shading.
Proportions and features: Correct sector drawn displaying layers in proper relative proportions: epidermis on outer edge, cortex layer, distinct vascular bundle (xylem and phloem regions indicated), and pith.
No individual cells drawn: Tissue plan drawing only.

(ii)
\[\text{Magnification} = \frac{\text{measured size}}{\text{actual size}} = \frac{48\text{ mm}}{0.60\text{ mm}} = \times 80\]
Value is given to two significant figures: \(\times 80\) (or \(80\)).

(b)
Reducing sugar:
- Reagent: Benedict's solution (or Benedict's reagent)
- Method: Add Benedict's solution and heat in a hot water-bath (\(>80^\circ\text{C}\)) for 3–5 minutes.
- Positive result: Colour change from blue to green / yellow / orange / brick-red.
Starch:
- Reagent: Iodine in potassium iodide solution (or iodine solution)
- Positive result: Colour change from brown / yellow-brown to blue-black.

(c) (i)
For [X] (\(0.4\text{ mol/dm}^3\)):
\[\% \text{ change} = \frac{0.00}{3.25} \times 100 = 0.0\%\]
For [Y] (\(0.8\text{ mol/dm}^3\)):
\[\% \text{ change} = \frac{-0.43}{3.10} \times 100 = -13.8709...\% = -13.9\%\]

(ii)
• Estimated concentration: \(0.4\text{ mol/dm}^3\)
• Explanation: At \(0.4\text{ mol/dm}^3\), there is no net movement of water into or out of the potato cells (change in mass is \(0.0\%\)), meaning the water potential of the sucrose solution is equal to the water potential of the potato cell sap (isotonic).

(iii)
The initial masses of the potato cylinders were different (ranging from \(3.10\text{ g}\) to \(3.30\text{ g}\)). Calculating percentage change allows a valid / fair comparison between cylinders.

(iv)
Any two from:
• Temperature of solutions
• Volume of sucrose solution in each test tube
• Surface area / diameter / length of potato cylinders
• Time potato cylinders spent in solution (\(60\text{ minutes}\))
• Variety / source / age of potato

(v)
Why blotted: To remove excess liquid / surface solution adhering to the outside of the cylinder that is not part of the cell mass.
Effect if not blotted: The final mass measured would be higher than the true mass, resulting in a calculated percentage change that is too high (or less negative in hypertonic solutions).

评分标准

(a)(i) [4 marks]
• clear, single, unbroken outlines with no shading or sketching ;
• size: occupies at least \(50\%\) of space provided ;
• correct sector structure showing epidermis, cortex, vascular bundle, and central pith ;
• correct tissue plan (no individual cells drawn) ;

(a)(ii) [2 marks]
• \(48 / 0.60\) (correct working) ;
• \(\times 80\) / \(80\) ;

(b) [4 marks]
• Benedict's (solution / reagent) ;
• heat / warm in a water-bath (\(>80^\circ\text{C}\)) ;
• (colour change to) green / yellow / orange / brick-red / red precipitate ;
• iodine (solution) AND blue-black / black ;

(c)(i) [2 marks]
[X]: \(0.0\) (%) ;
[Y]: \(-13.9\) (%) (allow \(13.9\%\) decrease) ;

(c)(ii) [2 marks]
• \(0.4\) (\(\text{mol/dm}^3\)) ;
• no change in mass / net movement of water is zero / water potential is equal (isotonic) ;

(c)(iii) [1 mark]
potato cylinders had different starting / initial masses / to allow fair comparison ;

(c)(iv) [2 marks]
any two from:
• temperature ;
• volume of sucrose solution ;
• surface area / dimensions / diameter / length of potato cylinder ;
• duration / time in solution (60 min) ;
• source / type of potato ;

(c)(v) [3 marks]
• to remove excess / surface liquid (that has not been absorbed) ;
• unblotted liquid increases measured final mass ;
• percentage change in mass will be artificially higher / less negative / inaccurate ;

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