Cambridge IAS-Level · Thinka 原創模擬試題

2025 Cambridge IAS-Level Biology (9700) 模擬試題連答案詳解

Thinka Jun 2025 (V4) Cambridge International A Level-Style Mock — Biology (9700)

140 270 分鐘2025
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V4) Cambridge International A Level Biology (9700) paper. Not affiliated with or reproduced from Cambridge.

卷一 (選擇題)

Answer all 40 multiple-choice questions. Each question has four options. Use a soft pencil to mark your choices.
40 題目 · 40
題目 1 · 選擇題
1
An experiment was carried out to investigate the effect of a reversible inhibitor on the rate of reaction of an enzyme. The table shows the maximum rate of reaction (\(V_{max}\)) and the Michaelis-Menten constant (\(K_m\)) of the enzyme in the presence and absence of the inhibitor.

| Condition | \(V_{max}\) / arbitrary units | \(K_m\) / \(\text{mmol dm}^{-3}\) |
| :--- | :---: | :---: |
| No inhibitor | 100 | 2.0 |
| With inhibitor | 50 | 2.0 |

Which statement about this inhibitor is correct?
  1. A.It is a competitive inhibitor that binds to the active site of the enzyme.
  2. B.It is a competitive inhibitor because the value of \(K_m\) remains unchanged.
  3. C.It is a non-competitive inhibitor that binds to an allosteric site of the enzyme.
  4. D.It is a non-competitive inhibitor because the value of \(V_{max}\) has increased.
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解題

A non-competitive inhibitor decreases the maximum rate of reaction (\(V_{max}\)) because it binds to an allosteric site (a site other than the active site) and permanently or temporarily prevents the enzyme from functioning, regardless of substrate concentration. The affinity of the remaining active enzymes for the substrate is unaffected, which is why the Michaelis-Menten constant (\(K_m\)) remains unchanged. Competitive inhibitors, on the other hand, increase \(K_m\) while keeping \(V_{max}\) the same. Thus, the data indicates that this is a non-competitive inhibitor that binds to an allosteric site.

評分準則

Award 1 mark for the correct option C.
題目 2 · 選擇題
1
An enzyme with an optimum temperature of \(37^\circ\text{C}\) was used to catalyze a reaction at \(37^\circ\text{C}\) and at \(50^\circ\text{C}\). All other conditions, including substrate concentration (which was in excess), were kept constant.

Which row correctly describes the initial rate of reaction and the total yield of product after 30 minutes at \(50^\circ\text{C}\) compared to \(37^\circ\text{C}\)?

| Row | Initial rate of reaction at \(50^\circ\text{C}\) | Total yield of product after 30 minutes at \(50^\circ\text{C}\) |
| :--- | :--- | :--- |
| **A** | higher | higher |
| **B** | higher | lower |
| **C** | lower | higher |
| **D** | lower | lower |
  1. A.Row A
  2. B.Row B
  3. C.Row C
  4. D.Row D
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解題

At \(50^\circ\text{C}\), the higher temperature gives the enzyme and substrate molecules more kinetic energy, leading to more frequent successful collisions per unit time. Therefore, the initial rate of reaction is higher at \(50^\circ\text{C}\) than at \(37^\circ\text{C}\). However, over the course of 30 minutes, the high temperature of \(50^\circ\text{C}\) causes the enzyme to denature rapidly as hydrogen bonds holding its tertiary structure break. Once denatured, the enzyme can no longer catalyze the reaction, so the reaction stops early. At \(37^\circ\text{C}\), the enzyme remains active and continues to convert the excess substrate into product. Thus, the total yield after 30 minutes will be lower at \(50^\circ\text{C}\) than at \(37^\circ\text{C}\).

評分準則

Award 1 mark for the correct option B.
題目 3 · 選擇題
1
Which row correctly describes the mechanisms involved in the active loading of sucrose into companion cells and its subsequent movement into sieve tube elements at a source?

| Row | Movement of \(\text{H}^+\) out of companion cell | Movement of sucrose into companion cell | Movement of sucrose from companion cell to sieve tube element |
| :--- | :--- | :--- | :--- |
| **A** | active transport | co-transport against its concentration gradient | diffusion through plasmodesmata |
| **B** | active transport | co-transport down its concentration gradient | active transport through plasmodesmata |
| **C** | facilitated diffusion | co-transport against its concentration gradient | facilitated diffusion through carrier proteins |
| **D** | facilitated diffusion | co-transport down its concentration gradient | active transport through carrier proteins |
  1. A.Row A
  2. B.Row B
  3. C.Row C
  4. D.Row D
查看答案詳解

解題

During sucrose loading:
1. Proton pumps actively transport hydrogen ions (\(\text{H}^+\)) out of the companion cell cytoplasm into the cell wall, establishing an electrochemical gradient of \(\text{H}^+\).
2. \(\text{H}^+\) ions then diffuse back into the companion cell down their concentration gradient through co-transporter proteins. This movement is coupled with the transport of sucrose against its concentration gradient into the companion cell.
3. The high concentration of sucrose in the companion cell allows sucrose to diffuse down its concentration gradient into the adjacent sieve tube element through plasmodesmata.

評分準則

Award 1 mark for the correct option A.
題目 4 · 選擇題
1
A student used a potometer to measure the rate of transpiration of a leafy shoot. Under initial laboratory conditions, the air bubble in the capillary tube moved a distance of \(60\text{ mm}\) in \(15\text{ minutes}\). The student then changed one environmental factor, keeping all other factors constant, and measured the movement of the bubble again. Under these new conditions, the bubble moved a distance of \(40\text{ mm}\) in \(5\text{ minutes}\).

Which change in environmental factor could explain these results?
  1. A.Moving a fan closer to the leafy shoot
  2. B.Placing a clear plastic bag over the leafy shoot
  3. C.Moving the apparatus into a darker room
  4. D.Spraying the leaves with a thin layer of wax
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解題

First, calculate the rate of transpiration under both conditions:
- Initial rate: \(60\text{ mm} / 15\text{ minutes} = 4.0\text{ mm min}^{-1}\).
- New rate: \(40\text{ mm} / 5\text{ minutes} = 8.0\text{ mm min}^{-1}\).
The rate of transpiration increased from \(4.0\text{ mm min}^{-1}\) to \(8.0\text{ mm min}^{-1}\).
We must identify the change that increases transpiration:
- Moving a fan closer (option A) increases air movement, removing water vapor from around the leaves, which increases the water potential gradient and thus increases transpiration.
- Placing a plastic bag over the shoot (option B) increases humidity, decreasing transpiration.
- Moving to a darker room (option C) causes stomata to close, decreasing transpiration.
- Coating leaves with wax (option D) blocks stomata and reduces water loss, decreasing transpiration.

評分準則

Award 1 mark for the correct option A.
題目 5 · 選擇題
1
Which row correctly identifies the type of pathogen, the primary method of transmission, and a key prevention or control method for tuberculosis (TB)?

| Row | Pathogen type | Primary transmission method | Key prevention or control method |
| :--- | :--- | :--- | :--- |
| **A** | bacterium | vector-borne | vaccination with BCG |
| **B** | bacterium | airborne droplets | contact tracing and antibiotic treatment |
| **C** | virus | airborne droplets | vaccination with BCG |
| **D** | protoctist | water-borne | water treatment and improved sanitation |
  1. A.Row A
  2. B.Row B
  3. C.Row C
  4. D.Row D
查看答案詳解

解題

Tuberculosis (TB) is caused by the bacterium *Mycobacterium tuberculosis* or *Mycobacterium bovis*. It is primarily transmitted through the air in droplets produced by coughing or sneezing of infected individuals. A key control method is contact tracing (to identify individuals who have been exposed) and treatment with a combination of antibiotics over several months.

評分準則

Award 1 mark for the correct option B.
題目 6 · 選擇題
1
The tertiary structure of a protein is maintained by several types of chemical bonds.

Which of these bonds are disrupted when a protein is denatured by a decrease in pH to 2.0?

1. Disulfide bonds
2. Hydrogen bonds
3. Ionic bonds
4. Peptide bonds
  1. A.1, 2 and 3
  2. B.2 and 3 only
  3. C.1 and 4 only
  4. D.3 only
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解題

A decrease in pH (highly acidic conditions) increases the concentration of hydrogen ions (\(\text{H}^+\)). This alters the charge of acidic and basic R-groups (for example, \(\text{-COO}^-\) becomes protonated to \(\text{-COOH}\)), disrupting existing ionic bonds. The excess hydrogen ions also compete for hydrogen-bonding sites, disrupting hydrogen bonds. Disulfide bonds are strong covalent bonds and are generally stable across changes in pH (requiring reducing agents to break). Peptide bonds maintain the primary structure and are only broken by enzymatic or acid-catalyzed hydrolysis, not by conditions that cause denaturation. Therefore, only hydrogen bonds (2) and ionic bonds (3) are disrupted.

評分準則

Award 1 mark for the correct option B.
題目 7 · 選擇題
1
Carbon dioxide produced by actively respiring tissues diffuses into red blood cells, where it undergoes a series of reactions.

Which statement correctly describes a reaction or event occurring in the red blood cells in respiring tissues?
  1. A.Carbonic anhydrase catalyzes the conversion of carbonic acid into carbon dioxide and water.
  2. B.Hydrogen ions bind to hemoglobin, reducing its affinity for oxygen and promoting oxygen release.
  3. C.Hydrogencarbonate ions diffuse into the red blood cells from the plasma in exchange for chloride ions.
  4. D.Carbon dioxide binds directly to the heme groups of hemoglobin to form carboxyhemoglobin.
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解題

In actively respiring tissues:
1. Carbon dioxide diffuses into red blood cells, where carbonic anhydrase catalyzes its conversion with water to form carbonic acid (\(\text{H}_2\text{CO}_3\)).
2. Carbonic acid dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)).
3. The \(\text{H}^+\) ions bind to oxyhemoglobin, acting as a buffer. This binding reduces hemoglobin's affinity for oxygen, causing it to release oxygen to the tissues (the Bohr effect). This makes option B correct.
4. Hydrogencarbonate ions diffuse out of the red blood cell into the plasma, while chloride ions move in (the chloride shift) to maintain electrical neutrality. This makes option C incorrect.
5. Carbon dioxide binds to the amine groups of the globin polypeptide chains to form carbaminohemoglobin (not carboxyhemoglobin, which is formed by carbon monoxide binding). This makes option D incorrect.

評分準則

Award 1 mark for the correct option B.
題目 8 · 選擇題
1
A student calibrated an eyepiece graticule using a stage micrometer. At a magnification of \(\times100\), 100 eyepiece graticule units (epu) aligned precisely with a length of \(1.0\text{ mm}\) on the stage micrometer.

The student then changed the objective lens so that the total magnification was increased to \(\times400\).

What is the actual length represented by 1 eyepiece graticule unit (epu) at this new magnification?
  1. A.\(0.25\ \mu\text{m}\)
  2. B.\(2.5\ \mu\text{m}\)
  3. C.\(25.0\ \mu\text{m}\)
  4. D.\(40.0\ \mu\text{m}\)
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解題

At a magnification of \(\times100\):
\(100\text{ epu} = 1.0\text{ mm} = 1000\ \mu\text{m}\).
Therefore, the actual length represented by \(1\text{ epu} = 1000\ \mu\text{m} / 100 = 10\ \mu\text{m}\).

When the magnification is increased to \(\times400\), the magnification is 4 times higher (\(400 / 100 = 4\)).
As magnification increases, the eyepiece graticule remains the same physical size in the eyepiece, but the specimen is magnified. Thus, each eyepiece division covers a smaller actual distance on the specimen. Specifically, the actual distance represented by 1 epu decreases by a factor of 4.
Therefore, at \(\times400\) magnification:
\(1\text{ epu} = 10\ \mu\text{m} / 4 = 2.5\ \mu\text{m}\).

評分準則

Award 1 mark for the correct option B.
題目 9 · 選擇題
1
An enzyme-catalysed reaction is investigated. The rate of reaction is measured at different substrate concentrations in the presence and absence of a chemical substance, \(X\).

At high substrate concentrations, the rate of reaction in the presence of \(X\) reaches the same maximum velocity (\(V_{\max}\)) as without \(X\), but the Michaelis-Menten constant (\(K_m\)) is increased.

What can be concluded about substance \(X\) and the enzyme?
  1. A.X is a competitive inhibitor and its presence decreases the apparent affinity of the enzyme for its substrate.
  2. B.X is a competitive inhibitor and its presence increases the apparent affinity of the enzyme for its substrate.
  3. C.X is a non-competitive inhibitor and its presence decreases the apparent affinity of the enzyme for its substrate.
  4. D.X is a non-competitive inhibitor and its presence increases the apparent affinity of the enzyme for its substrate.
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解題

The Michaelis-Menten constant (\(K_m\)) is the substrate concentration at which the rate of reaction is half of \(V_{\max}\). An increase in \(K_m\) without a change in \(V_{\max}\) is characteristic of a competitive inhibitor. This indicates that the inhibitor competes with the substrate for the active site, which decreases the apparent affinity of the enzyme for its substrate because a higher substrate concentration is required to achieve the same rate of reaction.

評分準則

Award 1 mark for identifying the correct option (A).
- Reject options B, C, D as non-competitive inhibition would decrease \(V_{\max}\), and competitive inhibition decreases, rather than increases, the apparent affinity of the enzyme for its substrate (indicated by an increased \(K_m\)).
題目 10 · 選擇題
1
What is the correct sequence of events that occurs during the active loading of sucrose into phloem companion cells and its subsequent movement into sieve tube elements?

1. Sucrose is co-transported with hydrogen ions into the companion cell.
2. Hydrogen ions are actively pumped out of the companion cell into the cell wall using ATP.
3. Sucrose diffuses through plasmodesmata from the companion cell into the sieve tube element.
4. A high concentration of hydrogen ions is established in the cell wall of the companion cell.
  1. A.2 \(\rightarrow\) 4 \(\rightarrow\) 1 \(\rightarrow\) 3
  2. B.4 \(\rightarrow\) 2 \(\rightarrow\) 1 \(\rightarrow\) 3
  3. C.2 \(\rightarrow\) 1 \(\rightarrow\) 4 \(\rightarrow\) 3
  4. D.4 \(\rightarrow\) 1 \(\rightarrow\) 2 \(\rightarrow\) 3
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解題

Active loading begins when hydrogen ions (protons) are actively pumped out of the companion cell into its cell wall using ATP (step 2). This creates a high concentration gradient of hydrogen ions in the cell wall (step 4). Hydrogen ions then diffuse back into the companion cell down their concentration gradient through co-transporter proteins, bringing sucrose molecules along with them against the sucrose concentration gradient (step 1). Finally, sucrose builds up to a high concentration in the companion cell and diffuses through plasmodesmata into the sieve tube elements (step 3).

評分準則

Award 1 mark for the correct sequence of events (2 -> 4 -> 1 -> 3), corresponding to option A.
- Reject incorrect sequences (B, C, D).
題目 11 · 選擇題
1
Which row correctly describes the pathogen type, method of transmission, and a major control method for cholera?
  1. A.Pathogen: bacterium; Transmission: ingestion of contaminated food or water; Control: safe sewage disposal and clean water supply
  2. B.Pathogen: bacterium; Transmission: inhalation of airborne droplets; Control: systematic vaccination of all children
  3. C.Pathogen: virus; Transmission: ingestion of contaminated food or water; Control: safe sewage disposal and clean water supply
  4. D.Pathogen: virus; Transmission: insect vector (mosquitoes); Control: use of insecticide-treated bed nets
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解題

Cholera is caused by the bacterium *Vibrio cholerae*. It is transmitted via the faecal-oral route through the ingestion of food or water contaminated with the pathogen. Because of this, the most effective global control measures involve modern sanitation, including safe sewage treatment and the provision of clean, treated drinking water.

評分準則

Award 1 mark for selecting option A, which correctly matches all three characteristics of cholera.
- Reject B (transmission is not via airborne droplets and there is no universal/systematic childhood vaccination programme for cholera).
- Reject C and D (the pathogen is a bacterium, not a virus, and it is not transmitted by mosquitoes).
題目 12 · 選擇題
1
Haemoglobin is a globular protein with a quaternary structure. Which statement correctly describes the bonds or interactions stabilizing the structure of a haemoglobin molecule?
  1. A.Hydrophobic interactions between non-polar R-groups are located mainly on the outside of the molecule to increase solubility.
  2. B.Hydrophilic R-groups on the outer surface of the polypeptide chains form hydrogen bonds with water molecules, maintaining solubility.
  3. C.Disulfide bonds between cysteine residues in different polypeptide subunits stabilize the quaternary structure.
  4. D.Ionic bonds between R-groups are the only bonds that stabilize both the tertiary and quaternary structures.
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解題

Haemoglobin is a soluble globular protein. To maintain solubility, its hydrophilic (polar/charged) R-groups are located on the outer surface where they can form hydrogen bonds with water molecules. In contrast, hydrophobic non-polar R-groups point towards the center of the molecule. Haemoglobin does not contain covalent disulfide bonds between its subunits; the quaternary structure is stabilized by weaker, non-covalent interactions (such as hydrophobic interactions, hydrogen bonds, and ionic bonds).

評分準則

Award 1 mark for the correct answer (B).
- Reject A because hydrophobic R-groups are located on the inside, not outside.
- Reject C because haemoglobin does not contain disulfide bonds.
- Reject D because ionic bonds are not the only bonds stabilizing both levels of structure; hydrophobic interactions and hydrogen bonds also play major roles.
題目 13 · 選擇題
1
Which row correctly describes the net movement of fluid and the main reason for this movement at the arteriole end of a capillary bed?
  1. A.Net movement of fluid: Out of the capillary into the tissue space; Main reason: Hydrostatic pressure of the blood is greater than the osmotic pressure of the blood
  2. B.Net movement of fluid: Out of the capillary into the tissue space; Main reason: Osmotic pressure of the blood is greater than the hydrostatic pressure of the blood
  3. C.Net movement of fluid: Into the capillary from the tissue space; Main reason: Hydrostatic pressure of the tissue fluid is greater than the hydrostatic pressure of the blood
  4. D.Net movement of fluid: Into the capillary from the tissue space; Main reason: Osmotic pressure of the tissue fluid is greater than the osmotic pressure of the blood
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解題

At the arteriole end of a capillary, the hydrostatic pressure of the blood is relatively high (e.g., about \(4.6\text{ kPa}\)), which tends to force fluid out of the capillary. Although the blood has a lower water potential than tissue fluid (creating an osmotic pressure / oncotic pressure of about \(3.3\text{ kPa}\) that draws water in), the outward hydrostatic pressure is greater than the inward osmotic pressure. Therefore, there is a net outward movement of fluid into the tissue space.

評分準則

Award 1 mark for option A.
- Reject B, C, and D as they incorrectly describe either the direction of net fluid movement or the relative magnitudes of the pressures involved at the arteriole end.
題目 14 · 選擇題
1
A student uses a light microscope with an eyepiece graticule and a stage micrometer to measure the diameter of a plant cell.

- The stage micrometer has divisions spaced \(0.01\text{ mm}\) apart.
- When the stage micrometer is focused, 100 divisions of the eyepiece graticule align with 25 divisions of the stage micrometer.
- The stage micrometer is replaced with a slide of plant cells, and a cell is measured to be 24 eyepiece graticule units in diameter.

What is the actual diameter of the plant cell in micrometres (\(\mu\text{m}\))?
  1. A.60 \(\mu\text{m\)}
  2. B.6.0 \(\mu\text{m\)}
  3. C.150 \(\mu\text{m\)}
  4. D.240 \(\mu\text{m\)}
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解題

First, convert the stage micrometer divisions into micrometres: \(0.01\text{ mm} = 10\ \mu\text{m}\).
25 divisions of the stage micrometer represent: \(25 \times 10\ \mu\text{m} = 250\ \mu\text{m}\).
These 25 stage micrometer divisions align exactly with 100 eyepiece graticule units (egu).
Therefore, 1 egu represents: \(\frac{250\ \mu\text{m}}{100} = 2.5\ \mu\text{m}\).
The measured cell is 24 egu in diameter. Its actual diameter is: \(24 \times 2.5\ \mu\text{m} = 60\ \mu\text{m}\).

評分準則

Award 1 mark for the correct calculation yielding 60 \(\mu\text{m}\) (Option A).
- Reject B, C, D due to arithmetic errors or incorrect unit conversions.
題目 15 · 選擇題
1
An experiment was carried out to investigate the effect of pH on the activity of salivary amylase, which has an optimum pH of 7.0. Two tubes were prepared:

- Tube P contains salivary amylase and starch buffered at pH 7.0.
- Tube Q contains salivary amylase and starch buffered at pH 2.0.

Both tubes are incubated at \(37^\circ\text{C}\) for 10 minutes.

Which statement correctly explains why the rate of reaction in Tube Q is much lower than in Tube P?
  1. A.The low pH in Tube Q causes the enzyme to denature by disrupting hydrogen and ionic bonds, changing the shape of the active site.
  2. B.The low pH in Tube Q increases the activation energy of the reaction by changing the chemical properties of the starch substrate.
  3. C.Hydrogen ions at pH 2.0 act as competitive inhibitors by binding reversibly to the active site.
  4. D.The low pH in Tube Q reduces the kinetic energy of the enzyme and substrate molecules, leading to fewer successful collisions.
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解題

An extreme pH far from the optimum (such as pH 2.0 compared to the optimum pH 7.0 for salivary amylase) alters the charges on the R-groups of the amino acids making up the enzyme. This disrupts ionic and hydrogen bonds holding the tertiary structure together, causing the enzyme to denature. This permanent change in the 3D shape of the active site prevents the substrate from binding, lowering the rate of reaction.

評分準則

Award 1 mark for identifying option A as the correct explanation.
- Reject B because enzymes lower, not increase, activation energy.
- Reject C because hydrogen ions do not act as competitive inhibitors.
- Reject D because pH changes do not affect the kinetic energy of molecules (which is determined by temperature).
題目 16 · 選擇題
1
An enzyme-catalysed reaction is carried out at \(25^\circ\text{C}\) and at \(37^\circ\text{C}\) (the optimum temperature for this enzyme), with all other conditions kept constant. Both reactions start with the same concentration of substrate.

Which statement correctly describes the curves for the concentration of product against time for these two reactions?
  1. A.The curve for \(37^\circ\text{C}\) has a steeper initial gradient and reaches a plateau at a higher concentration of product than the curve for \(25^\circ\text{C}\).
  2. B.The curve for \(37^\circ\text{C}\) has a steeper initial gradient and reaches a plateau at the same final concentration of product as the curve for \(25^\circ\text{C}\).
  3. C.The curve for \(37^\circ\text{C}\) has a shallower initial gradient and reaches a plateau sooner than the curve for \(25^\circ\text{C}\).
  4. D.The curve for \(37^\circ\text{C}\) has a steeper initial gradient and reaches a plateau later than the curve for \(25^\circ\text{C}\).
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解題

At \(37^\circ\text{C}\) (the optimum temperature), the kinetic energy of the enzyme and substrate molecules is higher than at \(25^\circ\text{C}\). This results in more frequent successful collisions per unit time and a faster initial rate of reaction, which corresponds to a steeper initial gradient on the graph of product concentration against time. Since the initial concentration of substrate is identical in both reactions, the total amount of product formed will eventually be the same once all substrate is converted. Therefore, both curves will eventually reach a plateau at the same final concentration of product, but the reaction at \(37^\circ\text{C}\) will reach it sooner.

評分準則

Award 1 mark for option B.
- Reject A because the total yield of product is determined by the initial substrate concentration, which is identical, so the plateau heights must be equal.
- Reject C because the initial rate is higher (steeper) at the higher temperature.
- Reject D because the reaction at the higher temperature is faster and will reach its plateau sooner (earlier), not later.
題目 17 · 選擇題
1
An investigation was carried out into the effect of two different inhibitors, X and Y, on an enzyme-catalyzed reaction. The rate of reaction was measured at different substrate concentrations.

The results obtained showed:
- Substrate concentration at which half-maximum velocity (\(K_m\)) is achieved:
- Control (no inhibitor): \(2.0 \text{ mmol dm}^{-3}\)
- With inhibitor X: \(2.0 \text{ mmol dm}^{-3}\)
- With inhibitor Y: \(5.5 \text{ mmol dm}^{-3}\)
- The maximum rate of reaction (\(V_{max}\)) achieved:
- Control (no inhibitor): \(100 \text{ arbitrary units (a.u.)}\)
- With inhibitor X: \(60 \text{ a.u.}\)
- With inhibitor Y: \(100 \text{ a.u.}\)

Which statement correctly identifies the types of inhibition shown by X and Y?
  1. A.X is a competitive inhibitor; Y is a non-competitive inhibitor.
  2. B.X is a non-competitive inhibitor; Y is a competitive inhibitor.
  3. C.Both X and Y are competitive inhibitors.
  4. D.Both X and Y are non-competitive inhibitors.
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解題

For a non-competitive inhibitor, the maximum velocity (\(V_{max}\)) decreases because the inhibitor binds to an allosteric site, altering the shape of the active site and rendering some enzyme molecules inactive regardless of substrate concentration. However, the affinity of the remaining active enzymes for the substrate remains unchanged, so the substrate concentration required to reach half-maximum velocity (\(K_m\)) remains the same. This matches the behavior of inhibitor X.

For a competitive inhibitor, the inhibitor competes with the substrate for the active site. This increases the apparent \(K_m\) (requiring a higher substrate concentration to reach half of the maximum velocity), but \(V_{max}\) can still be achieved at very high substrate concentrations where substrate molecules outcompete the inhibitor. This matches the behavior of inhibitor Y.

評分準則

Award 1 mark for the correct option (B).
- Reject A, C, and D as they incorrectly classify the competitive and non-competitive behaviors of inhibitors X and Y based on changes to \(K_m\) and \(V_{max}\).
題目 18 · 選擇題
1
Which process occurs during the active loading of sucrose into companion cells in the phloem of a photosynthesizing leaf?
  1. A.Hydrogen ions are actively pumped out of the companion cells into the cell wall, establishing an electrochemical gradient.
  2. B.Hydrogen ions diffuse into the sieve tube elements through carrier proteins down a concentration gradient.
  3. C.Sucrose is actively transported out of the mesophyll cells into the companion cells via symport proteins.
  4. D.Sucrose and hydrogen ions are cotransported out of the companion cells into the apoplast.
查看答案詳解

解題

During active phloem loading via the apoplastic pathway, hydrogen ions (protons) are actively pumped out of the companion cells into the cell wall (apoplast) using ATP. This active transport establishes a steep electrochemical gradient of hydrogen ions. The hydrogen ions then diffuse back down their gradient into the companion cells through cotransporter (symport) proteins, bringing sucrose molecules with them against their concentration gradient.

評分準則

Award 1 mark for identifying that hydrogen ions are pumped out of the companion cells into the cell wall (A).
- Reject B because protons diffuse back into companion cells, not sieve tubes directly, and cotransporter proteins are involved rather than simple carrier proteins.
- Reject C and D as they outline incorrect directions and mechanisms of sucrose transport.
題目 19 · 選擇題
1
Which row correctly matches the pathogen, the transmission method, and the primary site of action of the pathogen for cholera?
  1. A.Pathogen: *Vibrio cholerae*; Transmission: Water-borne; Primary site of action: Epithelial cells of the small intestine
  2. B.Pathogen: *Vibrio cholerae*; Transmission: Vector-borne; Primary site of action: Endothelial cells of blood vessels
  3. C.Pathogen: *Vibrio cholerae*; Transmission: Water-borne; Primary site of action: Hepatocytes of the liver
  4. D.Pathogen: *Vibrio cholerae*; Transmission: Droplet-borne; Primary site of action: Epithelial cells of the large intestine
查看答案詳解

解題

Cholera is caused by the Gram-negative bacterium *Vibrio cholerae*. It is transmitted primarily via contaminated drinking water or food (water-borne transmission). Once ingested, the bacteria colonize the small intestine, releasing a toxin that stimulates the secretion of chloride ions into the lumen, causing severe watery diarrhea.

評分準則

Award 1 mark for the correct row (A).
- Reject B because cholera is not vector-borne.
- Reject C because *Mycobacterium tuberculosis* causes tuberculosis, not cholera.
- Reject D because the primary site of action is the small intestine, not the large intestine, and the pathogen is *Vibrio cholerae*.
題目 20 · 選擇題
1
Which of the following features describe the quaternary structure of a molecule of haemoglobin?

1. The presence of four polypeptide chains.
2. The presence of ionic bonds, hydrogen bonds, and hydrophobic interactions holding the subunits together.
3. The presence of four haem groups, each containing an iron ion (\(\text{Fe}^{2+}\)).
4. The alpha-helical regions within each individual polypeptide chain.
  1. A.1, 2, 3 and 4
  2. B.1, 2 and 3 only
  3. C.1 and 2 only
  4. D.3 and 4 only
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解題

Quaternary structure is defined by the arrangement, interaction, and binding of multiple polypeptide subunits (and any associated non-protein prosthetic groups) in a multi-subunit protein complex. Haemoglobin is composed of four polypeptide chains (1) which are held together by hydrophobic interactions, hydrogen bonds, and ionic bonds (2). Each subunit contains a prosthetic haem group containing an iron ion (3). The alpha-helical regions (4) describe the folding of individual polypeptide chains, which is a feature of the secondary structure, not the quaternary structure.

評分準則

Award 1 mark for selecting B (1, 2, and 3 only).
- Reject options A, C, and D because statement 4 is a secondary structure characteristic and must be excluded, whereas 1, 2, and 3 are correct components of the quaternary level of organization.
題目 21 · 選擇題
1
The table shows the hydrostatic and oncotic pressures of the blood inside a capillary and the tissue fluid outside the capillary at both the arteriole and venule ends of a capillary bed.

| Pressure type | At the arteriole end / kPa | At the venule end / kPa |
| :--- | :--- | :--- |
| Hydrostatic pressure of blood | +4.6 | +2.3 |
| Hydrostatic pressure of tissue fluid | +1.2 | +1.2 |
| Oncotic pressure of blood | -3.3 | -3.3 |
| Oncotic pressure of tissue fluid | -0.2 | -0.2 |

*Note: Positive (+)* values represent pressures forcing fluid out of the capillary; *negative (-)* values represent pressures drawing fluid into the capillary.

What is the net filtration pressure at the arteriole end and the net reabsorption pressure at the venule end?
  1. A.Net filtration pressure at arteriole end = +0.3 kPa; Net reabsorption pressure at venule end = 2.0 kPa
  2. B.Net filtration pressure at arteriole end = +1.3 kPa; Net reabsorption pressure at venule end = 1.0 kPa
  3. C.Net filtration pressure at arteriole end = +3.4 kPa; Net reabsorption pressure at venule end = 1.1 kPa
  4. D.Net filtration pressure at arteriole end = +3.1 kPa; Net reabsorption pressure at venule end = 3.1 kPa
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解題

To calculate the net pressure at each end, we sum the forces acting to move fluid out of the capillary (blood hydrostatic pressure and tissue fluid oncotic pressure) and subtract the forces acting to move fluid back in (tissue fluid hydrostatic pressure and blood oncotic pressure).

- **At the arteriole end:**
- Outward forces = \(4.6 \text{ (blood hydrostatic)} + 0.2 \text{ (tissue oncotic)} = 4.8 \text{ kPa}\)
- Inward forces = \(1.2 \text{ (tissue hydrostatic)} + 3.3 \text{ (blood oncotic)} = 4.5 \text{ kPa}\)
- Net pressure = \(4.8 - 4.5 = +0.3 \text{ kPa}\) (filtration)

- **At the venule end:**
- Outward forces = \(2.3 \text{ (blood hydrostatic)} + 0.2 \text{ (tissue oncotic)} = 2.5 \text{ kPa}\)
- Inward forces = \(1.2 \text{ (tissue hydrostatic)} + 3.3 \text{ (blood oncotic)} = 4.5 \text{ kPa}\)
- Net pressure = \(2.5 - 4.5 = -2.0 \text{ kPa}\) (reabsorption with magnitude of \(2.0 \text{ kPa}\))

評分準則

Award 1 mark for the correct calculations of net filtration (\(+0.3 \text{ kPa}\)) and net reabsorption (\(2.0 \text{ kPa}\)) pressures (A).
- Reject B, C, and D as they represent incorrect arithmetic or misunderstandings of which pressures act inwards versus outwards.
題目 22 · 選擇題
1
A student calibrated an eyepiece graticule using a stage micrometer.

The stage micrometer had scale divisions of \(0.1\text{ mm}\).

At a magnification of \(\times 100\), 40 divisions of the eyepiece graticule aligned perfectly with 8 divisions of the stage micrometer.

The student then replaced the stage micrometer with a slide of plant tissue and observed a cell using a magnification of \(\times 400\).

The cell was measured to be 15 eyepiece graticule divisions in width.

What is the actual width of the plant cell in micrometres (\(\mu\text{m}\))?
  1. A.18.75 \(\mu\text{m}\)
  2. B.75.0 \(\mu\text{m}\)
  3. C.300.0 \(\mu\text{m}\)
  4. D.1200.0 \(\mu\text{m}\)
查看答案詳解

解題

1. First, determine the value of 1 eyepiece graticule division (epd) at \(\times 100\):
- Each stage micrometer division (smd) \(= 0.1\text{ mm} = 100\ \mu\text{m}\).
- 8 smd \(= 8 \times 100\ \mu\text{m} = 800\ \mu\text{m}\).
- 40 epd \(= 800\ \mu\text{m}\).
- Therefore, 1 epd at \(\times 100 = \frac{800\ \mu\text{m}}{40} = 20\ \mu\text{m}\).

2. Determine the value of 1 epd at \(\times 400\):
- When increasing the magnification from \(\times 100\) to \(\times 400\) (a \(4\)-fold increase), the field of view shrinks, meaning each epd represents a distance that is 4 times smaller.
- 1 epd at \(\times 400 = \frac{20\ \mu\text{m}}{4} = 5\ \mu\text{m}\).

3. Calculate the actual size of the plant cell:
- Actual width \(= 15\text{ epd} \times 5\ \mu\text{m/epd} = 75\ \mu\text{m}\).

評分準則

Award 1 mark for the correct option (B).
- Reject A: resulting from incorrectly scaling the division value (dividing by 16 instead of 4).
- Reject C: resulting from forgetting to scale the calibration factor for the new magnification of \(\times 400\) (using \(20\ \mu\text{m}\) instead of \(5\ \mu\text{m}\)).
- Reject D: resulting from multiplying the calibration factor by 4 instead of dividing.
題目 23 · 選擇題
1
A sample of double-stranded DNA was analyzed and found to contain 34% cytosine.

During transcription of a section of this DNA, an mRNA molecule was synthesized. The template strand of the DNA used for transcription contained 18% adenine.

What is the percentage of uracil in the resulting mRNA molecule?
  1. A.14%
  2. B.16%
  3. C.18%
  4. D.34%
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解題

During transcription, complementary RNA nucleotides pair with the template strand of DNA. Specifically, Adenine (A) bases on the DNA template strand form hydrogen bonds with Uracil (U) on the growing mRNA chain. Therefore, the proportion of uracil in the synthesized mRNA is directly complementary and equal to the proportion of adenine in the template DNA strand. Since the template strand contains 18% adenine, the mRNA will contain 18% uracil. The overall composition of cytosine (34%) in the double-stranded DNA is distractor information and is not required for this calculation.

評分準則

Award 1 mark for the correct percentage (C).
- Reject A: calculated as \(32\% - 18\%\) under the false assumption that template strand bases must sum to a particular subgroup.
- Reject B: calculated as the overall percentage of Adenine in the double-stranded DNA (which is \(16\%\)), rather than using the template strand value.
- Reject D: copy of the cytosine percentage.
題目 24 · 選擇題
1
Which row correctly identifies an example of passive artificial immunity and the reason for its duration of protection?
  1. A.Example: Injection of anti-venom antibodies; Duration: Short-term; Reason: No memory cells are produced and the injected antibodies are eventually broken down.
  2. B.Example: Transfer of maternal antibodies across the placenta; Duration: Short-term; Reason: The infant's body does not produce the antibodies itself.
  3. C.Example: Injection of a vaccine containing weakened pathogens; Duration: Long-term; Reason: The vaccine contains antigens that trigger the production of memory cells.
  4. D.Example: Injection of monoclonal antibodies; Duration: Long-term; Reason: Injected antibodies replicate in the body to maintain constant protection.
查看答案詳解

解題

Passive immunity involves the transfer of pre-formed antibodies into an individual. Because these antibodies are injected via medical intervention, it is classified as artificial passive immunity. Since no host immune response occurs, no memory cells are produced, and the injected antibodies are gradually cleared and broken down, resulting in short-term protection. This is exemplified by anti-venom injections (A). Maternal antibodies via the placenta (B) represent natural passive immunity. Vaccines (C) represent active artificial immunity. Monoclonal antibodies (D) provide short-term protection, not long-term protection, as antibodies do not replicate.

評分準則

Award 1 mark for identifying the correct example, duration, and reason for passive artificial immunity (A).
- Reject B because placental transfer of antibodies is natural, not artificial.
- Reject C because vaccines stimulate active, not passive, immunity.
- Reject D because antibodies do not replicate in the body, and passive therapies do not yield long-term protection.
題目 25 · 選擇題
1
An investigation was carried out to determine the effect of two different inhibitors, X and Y, on the activity of an enzyme. The table shows the maximum rate of reaction (\(V_{max}\)) and the Michaelis-Menten constant (\(K_m\)) values obtained.

| Condition | \(V_{max}\) / arbitrary units | \(K_m\) / \(\text{mmol dm}^{-3}\) |
|---|---|---|
| Enzyme alone | 100 | 2.5 |
| Enzyme + Inhibitor X | 100 | 5.0 |
| Enzyme + Inhibitor Y | 50 | 2.5 |

Which statement correctly identifies the types of inhibition shown by X and Y?
  1. A.X is a competitive inhibitor; Y is a non-competitive inhibitor.
  2. B.X is a non-competitive inhibitor; Y is a competitive inhibitor.
  3. C.Both X and Y are competitive inhibitors.
  4. D.Both X and Y are non-competitive inhibitors.
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解題

Competitive inhibitors compete with the substrate for the active site of the enzyme, which increases the apparent \(K_m\) value (a higher substrate concentration is needed to reach half the maximum velocity) but does not change the \(V_{max}\) because high substrate concentrations can overcome the inhibition. Inhibitor X shows these features (\(V_{max}\) remains 100, \(K_m\) increases from 2.5 to 5.0). Non-competitive inhibitors bind to an allosteric site on the enzyme, reducing the overall rate of reaction and decreasing \(V_{max}\), but they do not affect the binding affinity of the substrate to the active sites of remaining functional enzymes, meaning the \(K_m\) value remains unchanged. Inhibitor Y shows these features (\(V_{max}\) decreases from 100 to 50, \(K_m\) remains 2.5).

評分準則

Award 1 mark for the correct option (A). Reject other options based on the biochemical definitions of competitive and non-competitive inhibitors in relation to \(K_m\) and \(V_{max}\).
題目 26 · 選擇題
1
Which row correctly describes the changes in the companion cell and the sieve tube element shortly after the application of a metabolic poison that inhibits cellular respiration?
  1. A.pH of companion cell wall increases; concentration of sucrose in sieve tube element decreases
  2. B.pH of companion cell wall decreases; concentration of sucrose in sieve tube element increases
  3. C.pH of companion cell cytoplasm increases; concentration of sucrose in sieve tube element increases
  4. D.pH of companion cell cytoplasm decreases; concentration of sucrose in sieve tube element remains constant
查看答案詳解

解題

Active transport of hydrogen ions (protons) out of the companion cell cytoplasm into the cell wall requires ATP produced by respiration. If a metabolic poison inhibits respiration, ATP production ceases, proton pumps stop working, and fewer protons are moved into the cell wall, causing the pH of the companion cell wall to increase (become less acidic). Consequently, the co-transport of sucrose along with protons back into the companion cell and subsequently into the sieve tube element stops, leading to a decrease in the concentration of sucrose within the sieve tube element.

評分準則

Award 1 mark for the correct option (A). Correctly linking respiration inhibition with decreased proton pumping (increased pH of cell wall) and reduced sucrose loading into the sieve tube element.
題目 27 · 選擇題
1
Which of the following options correctly pairs an infectious disease with its causative pathogen, the cellular nature of that pathogen, and its principal method of transmission?
  1. A.Cholera | Vibrio cholerae | Prokaryotic cell, possesses flagella | Water-borne, contaminated food or water
  2. B.Malaria | Plasmodium falciparum | Eukaryotic cell, possesses cell wall | Vector-borne, female Anopheles mosquito
  3. C.Tuberculosis | Mycobacterium tuberculosis | Eukaryotic cell, contains flagella | Airborne droplets
  4. D.HIV/AIDS | Human Immunodeficiency Virus | Non-cellular, contains DNA genome | Sexual contact and needle sharing
查看答案詳解

解題

Cholera is caused by the bacterium Vibrio cholerae, which is a prokaryote containing flagella, and it is transmitted via contaminated food and water. Option B is incorrect because Plasmodium falciparum (a protoctist) does not possess a cell wall. Option C is incorrect because Mycobacterium tuberculosis is a bacterium (prokaryotic) and does not contain flagella. Option D is incorrect because the Human Immunodeficiency Virus (HIV) contains an RNA genome, not DNA.

評分準則

Award 1 mark for identifying the row containing the correct match of cholera, Vibrio cholerae, prokaryotic flagellated structure, and water-borne transmission (A).
題目 28 · 選擇題
1
Which of the statements about the structural properties of haemoglobin and collagen are correct?

1. Haemoglobin is a globular protein whereas collagen is a fibrous protein.
2. Both proteins contain a non-protein prosthetic group essential for their function.
3. Every third amino acid in the primary structure of a collagen polypeptide is glycine, whereas haemoglobin does not have this repeating pattern.
4. High temperatures break the covalent peptide bonds in haemoglobin but not in collagen.
  1. A.1 and 3 only
  2. B.1, 2 and 3
  3. C.2 and 4 only
  4. D.1, 3 and 4
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解題

Statement 1 is correct; haemoglobin is a soluble globular protein while collagen is an insoluble fibrous structural protein. Statement 2 is incorrect; haemoglobin contains a non-protein haem prosthetic group, but collagen does not have a prosthetic group. Statement 3 is correct; glycine is the smallest amino acid and occurs at every third position in the collagen polypeptide chain to allow tight helical packing. Statement 4 is incorrect; moderate high temperatures denature proteins by disrupting weaker non-covalent interactions (like hydrogen bonds, ionic bonds, hydrophobic interactions) but do not break covalent peptide bonds. Therefore, only statements 1 and 3 are correct.

評分準則

Award 1 mark for selecting option A (1 and 3 only).
題目 29 · 選擇題
1
Which of the following describes the correct sequence of events occurring in red blood cells as blood flows through actively respiring muscle tissue?
  1. A.Carbon dioxide diffuses in \(\to\) carbonic anhydrase catalyzes the formation of carbonic acid \(\to\) carbonic acid dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\) \(\to\) \(\text{HCO}_3^-\) diffuses out and \(\text{Cl}^-\) diffuses in.
  2. B.Carbon dioxide diffuses in \(\to\) \(\text{H}^+\) and \(\text{HCO}_3^-\) combine to form carbonic acid \(\to\) carbonic anhydrase breaks down carbonic acid \(\to\) \(\text{Cl}^-\) diffuses out and \(\text{HCO}_3^-\) diffuses in.
  3. C.Oxygen diffuses out \(\to\) haemoglobin binds to carbon dioxide to form carboxyhaemoglobin \(\to\) carbonic anhydrase converts \(\text{CO}_2\) to \(\text{HCO}_3^-\).
  4. D.Carbon dioxide diffuses in \(\to\) carbonic anhydrase catalyzes the direct conversion of \(\text{CO}_2\) into \(\text{HCO}_3^-\) \(\to\) \(\text{H}^+\) binds to haemoglobin to form haemoglobinic acid \(\to\) \(\text{Cl}^-\) diffuses out.
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解題

In active tissues, carbon dioxide diffuses into red blood cells where carbonic anhydrase catalyzes its reaction with water to form carbonic acid. This acid dissociates into hydrogen ions (which bind to haemoglobin, causing it to release oxygen) and hydrogencarbonate ions. Hydrogencarbonate ions diffuse out of the red blood cell, and to maintain electrical neutrality, chloride ions diffuse into the cell (the chloride shift).

評分準則

Award 1 mark for the correct chronological physiological pathway (A).
題目 30 · 選擇題
1
A photomicrograph of a plant cell nucleus shows a nucleolus with a measured diameter of \(6\text{ mm}\). If the actual diameter of the nucleolus is \(1.5\ \mu\text{m}\), what is the magnification of this image, and which type of microscope would be required to resolve two distinct structures inside this nucleolus that are only \(50\text{ nm}\) apart?

| | Magnification | Type of microscope |
|---|---|---|
| **A** | \(\times 4000\) | Electron microscope |
| **B** | \(\times 4000\) | Light microscope |
| **C** | \(\times 400\) | Electron microscope |
| **D** | \(\times 400\) | Light microscope |
  1. A.Magnification: \(\times 4000\) | Type of microscope: Electron microscope
  2. B.Magnification: \(\times 4000\) | Type of microscope: Light microscope
  3. C.Magnification: \(\times 400\) | Type of microscope: Electron microscope
  4. D.Magnification: \(\times 400\) | Type of microscope: Light microscope
查看答案詳解

解題

First, calculate magnification: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\). Convert the image size to micrometer units: \(6\text{ mm} = 6000\ \mu\text{m}\). \(\text{Magnification} = \frac{6000\ \mu\text{m}}{1.5\ \mu\text{m}} = \times 4000\). Second, determine the type of microscope: The limit of resolution of a light microscope is approximately \(200\text{ nm}\). Because the two points are only \(50\text{ nm}\) apart, they cannot be resolved by a light microscope. An electron microscope (with a resolution limit of up to \(0.5\text{ nm}\)) is required.

評分準則

Award 1 mark for showing both correct magnification calculation (\(\times 4000\)) and selecting the correct microscope type (Electron microscope) based on resolution limits (A).
題目 31 · 選擇題
1
A double-stranded DNA molecule is found to contain 28% thymine bases. This DNA is allowed to replicate semi-conservatively. What will be the percentage of guanine bases in the newly synthesized complementary strands?
  1. A.22%
  2. B.28%
  3. C.44%
  4. D.56%
查看答案詳解

解題

According to Chargaff's rules of base pairing, in double-stranded DNA: \(\text{Adenine (A)} = \text{Thymine (T)}\) and \(\text{Guanine (G)} = \text{Cytosine (C)}\). Given \(\text{T} = 28\%\), \(\text{A}\) must also be \(28\%\). Combined, \(\text{A} + \text{T} = 56\%\). The remaining percentage of bases must be \(100\% - 56\% = 44\%\), which is shared equally between guanine and cytosine. Therefore, \(\text{G} = 22\%\) and \(\text{C} = 22\%\). During semi-conservative replication, the newly synthesized complementary strands will have the exact complementary base composition to their templates, preserving the overall percentage of guanine at 22%.

評分準則

Award 1 mark for the correct option (A). Calculate: \(100 - (28 \times 2) = 44\%\), then divide by 2 to find \(\%\text{G} = 22\%\).
題目 32 · 選擇題
1
A patient is bitten by a venomous snake and is immediately injected with an antivenom containing specific antibodies. Two months later, the patient is bitten by the same species of snake but is not immune and requires another injection. What type of immunity is provided by the antivenom, and why is long-term immunity not achieved?
  1. A.Artificial passive; no memory cells were produced in response to the injected antibodies.
  2. B.Artificial active; the injected antibodies did not trigger clonal expansion of T-lymphocytes.
  3. C.Natural passive; the patient's immune system rapidly degraded the foreign antivenom.
  4. D.Artificial passive; the injected antigens were cleared too quickly to stimulate B-lymphocytes.
查看答案詳解

解題

The injection of pre-formed antibodies (antivenom) provides artificial passive immunity. It is 'artificial' because it is medically acquired, and 'passive' because the patient's own body does not manufacture the antibodies. Because the patient's own immune system did not undergo clonal selection, clonal expansion, and differentiation of B-lymphocytes, no memory cells were produced. Consequently, long-term immunity is not achieved and the protection is temporary as the injected antibodies are cleared from the blood.

評分準則

Award 1 mark for identifying the correct immunity type (artificial passive) and the correct biological reason (no memory cells produced) (A).
題目 33 · 選擇題
1
An investigation was carried out to determine the effect of two different inhibitors, X and Y, on the activity of an enzyme. The rate of reaction was measured at different substrate concentrations.
The table shows the values of \(V_{\text{max}}\) and \(K_{\text{m}}\) obtained:

| Condition | \(V_{\text{max}}\) / arbitrary units | \(K_{\text{m}}\) / mmol dm\({}^{-3}\) |
|---|---|---|
| Enzyme alone | 100 | 2.5 |
| Enzyme + Inhibitor X | 100 | 5.0 |
| Enzyme + Inhibitor Y | 50 | 2.5 |

Which statement correctly identifies the type of inhibition and explains its effect?
  1. A.Inhibitor X is a non-competitive inhibitor because it increases the value of \(K_{\text{m}}\).
  2. B.Inhibitor X is a competitive inhibitor because it increases the value of \(K_{\text{m}}\) and does not change the \(V_{\text{max}}\).
  3. C.Inhibitor Y is a competitive inhibitor because it decreases the value of \(V_{\text{max}}\).
  4. D.Inhibitor Y is a non-competitive inhibitor because it decreases the value of \(V_{\text{max}}\) by binding to the active site of the enzyme.
查看答案詳解

解題

Inhibitor X increases the Michaelis-Menten constant (\(K_{\text{m}}\)) from 2.5 to 5.0 mmol dm\({}^{-3}\) but leaves the maximum velocity (\(V_{\text{max}}\)) unchanged at 100 arbitrary units. This is characteristic of competitive inhibition, where the inhibitor competes with the substrate for the active site. The apparent affinity of the enzyme for its substrate is reduced, requiring a higher substrate concentration to reach half-maximum velocity. Inhibitor Y is a non-competitive inhibitor because it reduces the \(V_{\text{max}}\) without affecting the \(K_{\text{m}}\).

評分準則

Award 1 mark for selecting option B. Reject all other options.
- Option A is incorrect because inhibitor X is a competitive (not non-competitive) inhibitor.
- Option C is incorrect because inhibitor Y is non-competitive and does not increase \(K_{\text{m}}\).
- Option D is incorrect because non-competitive inhibitors do not bind to the active site.
題目 34 · 選擇題
1
During the active loading of sucrose into a phloem companion cell, several transport steps are involved. Which row correctly describes these processes?
  1. A.Mechanism of \(\text{H}^+\) transport out of companion cell: active transport; Protein involved in \(\text{H}^+\) transport into companion cell: co-transporter protein; Direction of sucrose movement relative to its concentration gradient: against its concentration gradient
  2. B.Mechanism of \(\text{H}^+\) transport out of companion cell: active transport; Protein involved in \(\text{H}^+\) transport into companion cell: channel protein; Direction of sucrose movement relative to its concentration gradient: down its concentration gradient
  3. C.Mechanism of \(\text{H}^+\) transport out of companion cell: facilitated diffusion; Protein involved in \(\text{H}^+\) transport into companion cell: co-transporter protein; Direction of sucrose movement relative to its concentration gradient: against its concentration gradient
  4. D.Mechanism of \(\text{H}^+\) transport out of companion cell: facilitated diffusion; Protein involved in \(\text{H}^+\) transport into companion cell: carrier protein; Direction of sucrose movement relative to its concentration gradient: down its concentration gradient
查看答案詳解

解題

Active loading of sucrose into phloem sieve tube elements begins with proton pumps actively transporting protons (\(\text{H}^+\) ions) out of the companion cell into the cell wall space, creating an electrochemical gradient. Protons then flow back down their electrochemical gradient into the companion cell via a co-transporter protein. This movement of protons drives the transport of sucrose against its concentration gradient into the companion cell through the same co-transporter.

評分準則

Award 1 mark for option A. Reject all other options as they misidentify the direction of sucrose transport or the nature of the transport proteins involved.
題目 35 · 選擇題
1
Which row correctly matches the infectious disease with its pathogen type, method of transmission, and primary site of infection in the human body?
  1. A.Disease: Cholera; Pathogen type: Bacterium; Method of transmission: Vector-borne; Primary site of infection: Small intestine
  2. B.Disease: Malaria; Pathogen type: Protoctist; Method of transmission: Vector-borne; Primary site of infection: Red blood cells and liver cells
  3. C.Disease: Tuberculosis; Pathogen type: Virus; Method of transmission: Airborne droplets; Primary site of infection: Alveoli of the lungs
  4. D.Disease: HIV/AIDS; Pathogen type: Virus; Method of transmission: Contaminated water; Primary site of infection: T-helper lymphocytes
查看答案詳解

解題

Malaria is caused by the protoctist *Plasmodium*, which is transmitted by the female *Anopheles* mosquito vector and infects both red blood cells and hepatocytes (liver cells). Cholera is water-borne, Tuberculosis is caused by a bacterium (not a virus), and HIV is transmitted through direct bodily fluids (not contaminated water).

評分準則

Award 1 mark for option B.
- Option A: Cholera is transmitted via contaminated water/food (water-borne), not a vector.
- Option C: Tuberculosis is caused by a bacterium (*Mycobacterium*), not a virus.
- Option D: HIV is transmitted by exchange of bodily fluids, not water.
題目 36 · 選擇題
1
Which of the following describes a structural feature of collagen that distinguishes it from haemoglobin?
  1. A.It contains hydrogen bonds that stabilize the secondary structure of the polypeptide chains.
  2. B.It is composed of three polypeptide chains that form a tight triple helix called tropocollagen.
  3. C.It consists of a high proportion of hydrophilic amino acids on the outer surface of the molecule.
  4. D.Its tertiary structure is spherical and easily soluble in water.
查看答案詳解

解題

Collagen is a fibrous protein composed of three helical polypeptide chains wound tightly around each other to form a triple helix (tropocollagen). In contrast, haemoglobin is a globular protein made of four polypeptide chains (two alpha and two beta chains), each associated with a prosthetic haem group.

評分準則

Award 1 mark for option B. Reject options A, C, and D because they describe either shared features or features of globular proteins like haemoglobin rather than collagen.
題目 37 · 選擇題
1
At which point in the mammalian cardiac cycle does the left atrioventricular (bicuspid) valve close, and at which point does the left semilunar (aortic) valve open?
  1. A.Atrioventricular valve closes: when atrial pressure becomes higher than ventricular pressure; Semilunar valve opens: when ventricular pressure becomes higher than aortic pressure
  2. B.Atrioventricular valve closes: when ventricular pressure becomes higher than atrial pressure; Semilunar valve opens: when ventricular pressure becomes higher than aortic pressure
  3. C.Atrioventricular valve closes: when ventricular pressure becomes higher than atrial pressure; Semilunar valve opens: when aortic pressure becomes higher than ventricular pressure
  4. D.Atrioventricular valve closes: when atrial pressure becomes higher than ventricular pressure; Semilunar valve opens: when aortic pressure becomes higher than ventricular pressure
查看答案詳解

解題

The atrioventricular valve closes when the pressure in the left ventricle rises above the pressure in the left atrium, preventing backflow. The semilunar valve opens when the pressure inside the left ventricle rises above the pressure in the aorta, allowing blood to flow from the ventricle into the aortic arch.

評分準則

Award 1 mark for option B. Reject options A, C, and D as they incorrectly identify the pressure relationships that drive valve opening and closing.
題目 38 · 選擇題
1
An electron micrograph shows a mitochondrion that has an image length of 60 mm.
If the actual length of the mitochondrion is 1.5 \(\mu\text{m}\), what is the magnification of the micrograph?
  1. A.\(\times 40\)
  2. B.\(\times 400\)
  3. C.\(\times 4,000\)
  4. D.\(\times 40,000\)
查看答案詳解

解題

Using the formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\)
Convert image size from mm to \(\mu\text{m}\):
\(60\text{ mm} = 60,000\text{ }\mu\text{m}\)
Now calculate magnification:
\(\text{Magnification} = \frac{60,000\text{ }\mu\text{m}}{1.5\text{ }\mu\text{m}} = \times 40,000\)

評分準則

Award 1 mark for option D. Reject options A, B, and C as they represent incorrect unit conversion factors or mathematical errors.
題目 39 · 選擇題
1
Bacteria were grown in a medium containing heavy nitrogen (\({}^{15}\text{N}\)) until all of their DNA was labeled with \({}^{15}\text{N}\).
They were then transferred to a medium containing light nitrogen (\({}^{14}\text{N}\)) and allowed to replicate their DNA.
What would be the composition of the DNA molecules after two rounds of semi-conservative replication in the \({}^{14}\text{N}\) medium?
  1. A.50% hybrid DNA (\({}^{15}\text{N}\)-\({}^{14}\text{N}\)) and 50% light DNA (\({}^{14}\text{N}\)-\({}^{14}\text{N}\))
  2. B.100% hybrid DNA (\({}^{15}\text{N}\)-\({}^{14}\text{N}\))
  3. C.25% heavy DNA (\({}^{15}\text{N}\)-\({}^{15}\text{N}\)) and 75% light DNA (\({}^{14}\text{N}\)-\({}^{14}\text{N}\))
  4. D.25% hybrid DNA (\({}^{15}\text{N}\)-\({}^{14}\text{N}\)) and 75% light DNA (\({}^{14}\text{N}\)-\({}^{14}\text{N}\))
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解題

In generation 0, all DNA is heavy (\({}^{15}\text{N}\)-\({}^{15}\text{N}\)).
After the first replication (generation 1) in \({}^{14}\text{N}\), all DNA molecules consist of one heavy and one light strand (100% hybrid \({}^{15}\text{N}\)-\({}^{14}\text{N}\)).
After the second replication (generation 2) in \({}^{14}\text{N}\), the hybrid strands separate and serve as templates. The original \({}^{15}\text{N}\) strands produce 2 hybrid molecules (\({}^{15}\text{N}\)-\({}^{14}\text{N}\)), while the \({}^{14}\text{N}\) strands produce 2 light molecules (\({}^{14}\text{N}\)-\({}^{14}\text{N}\)).
Therefore, 50% of the DNA is hybrid and 50% is light.

評分準則

Award 1 mark for option A. Reject other options as they miscalculate the strand distribution according to the semi-conservative model.
題目 40 · 選擇題
1
A child is bitten by a venomous snake. At the hospital, they are immediately injected with antivenom containing pre-formed antibodies against the venom.
Several years later, the child is bitten by the same species of snake but is not immune and requires another injection of antivenom.
Which type of immunity is provided by the antivenom, and why does it not provide long-term protection?
  1. A.Type of immunity: Active artificial; Reason for no long-term protection: The child's own B-lymphocytes do not divide to form plasma cells.
  2. B.Type of immunity: Passive artificial; Reason for no long-term protection: No memory cells are produced because the child's immune system was not stimulated.
  3. C.Type of immunity: Active natural; Reason for no long-term protection: The injected antibodies are recognized as foreign and destroyed by phagocytes.
  4. D.Type of immunity: Passive natural; Reason for no long-term protection: The venom prevents the activation of memory T-helper cells.
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解題

Antivenom provides passive artificial immunity because pre-formed antibodies are introduced artificially. Passive immunity is temporary because the recipient's own immune system is not stimulated to undergo clonal selection or produce memory cells, and the injected antibodies eventually degrade.

評分準則

Award 1 mark for option B.
- Option A and C are incorrect because passive (not active) immunity is provided.
- Option D is incorrect because antivenom is artificial passive, not natural passive (which refers to antibodies crossing the placenta or in breast milk).

卷二 (AS Structured)

Answer all questions. Show your working in calculations and write your responses in the spaces provided.
6 題目 · 60
題目 1 · structured-short-answer
10
(a) Describe how competitive and non-competitive inhibitors affect the Michaelis-Menten constant (\(K_m\)) and the maximum rate of reaction (\(V_{max}\)). [4]

(b) A student investigated the rate of reaction of amylase with different concentrations of starch, with and without the addition of a chemical called phaseolamin. The results are shown in Table 1.1.

Table 1.1:
- Starch concentration / %: 0.1, 0.2, 0.4, 0.6, 0.8, 1.0
- Rate of reaction without phaseolamin / arbitrary units (a.u.): 12, 22, 38, 45, 48, 48
- Rate of reaction with phaseolamin / arbitrary units (a.u.): 4, 8, 15, 22, 28, 32

(i) Explain whether phaseolamin acts as a competitive or a non-competitive inhibitor, using the data in Table 1.1 to support your answer. [3]

(ii) Outline how the student could determine the \(K_m\) value for amylase in the absence of the inhibitor. [3]
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解題

(a) Competitive inhibitors increase the Michaelis-Menten constant (\(K_m\)) because they compete with the substrate for the active sites, lowering the affinity. They do not alter the maximum rate of reaction (\(V_{max}\)) because the inhibition can be overcome by high substrate concentrations. Non-competitive inhibitors do not affect the affinity of the enzyme for the substrate, so \(K_m\) remains unchanged. However, they decrease \(V_{max}\) because they bind elsewhere on the enzyme, permanently reducing the concentration of active enzyme.

(b)(i) Phaseolamin is a competitive inhibitor. In the presence of phaseolamin, the rate of reaction continues to rise with increasing starch concentration and has not reached a plateau even at 1.0% starch (32 a.u.). This indicates that the inhibition is being overcome by increasing the substrate concentration, showing that the reaction could eventually reach the same maximum velocity (\(V_{max}\) of 48 a.u.) as the uninhibited reaction.

(b)(ii) To determine the \(K_m\) value without the inhibitor:
1. Plot a graph of the rate of reaction (y-axis) against the starch concentration (x-axis).
2. Determine the maximum rate of reaction (\(V_{max}\)) from the plateau of the curve, which is 48 a.u.
3. Calculate half of the maximum rate: \(1/2 V_{max} = 24\) a.u.
4. Find the starch concentration on the x-axis that corresponds to a rate of 24 a.u. on the y-axis. This starch concentration is the \(K_m\) value.

評分準則

(a) [4 marks max]
1. Competitive inhibitor increases \(K_m\) / reduces substrate affinity [1]
2. Competitive inhibitor does not change \(V_{max}\) [1]
3. Non-competitive inhibitor does not change \(K_m\) [1]
4. Non-competitive inhibitor decreases \(V_{max}\) [1]

(b)(i) [3 marks max]
1. Phaseolamin is a competitive inhibitor [1]
2. With the inhibitor, the rate of reaction is still increasing as starch concentration increases / has not reached a plateau [1]
3. Increasing substrate concentration overcomes the inhibition / would eventually reach the original \(V_{max}\) of 48 a.u. [1]

(b)(ii) [3 marks max]
1. Plot a graph of rate of reaction against starch concentration [1]
2. Find \(V_{max}\) (48 a.u.) and calculate half of \(V_{max}\) (24 a.u.) [1]
3. Read the substrate (starch) concentration at half \(V_{max}\) from the x-axis [1]
題目 2 · structured-short-answer
10
(a) Describe three structural features of a sieve tube element that adapt it for the mass flow of organic solutes. [3]

(b) Explain how sucrose is loaded into the companion cells and sieve tube elements at a source. [5]

(c) State two differences between the transport of water in xylem vessels and the transport of organic solutes in phloem sieve tubes. [2]
查看答案詳解

解題

(a) Sieve tube elements are adapted for mass flow by:
1. Having minimal cytoplasm, no nucleus, and no vacuole, which minimizes resistance to the bulk flow of phloem sap.
2. Having sieve plates with open sieve pores at their end walls, which allow continuous flow of organic solutes between adjacent cells.
3. Having cellulose cell walls that provide mechanical strength to withstand the high hydrostatic pressure within the sieve tube.

(b) Sucrose is actively loaded at a source as follows:
1. Proton pumps in the companion cell membrane actively transport hydrogen ions (protons, \(H^+\)) out of the companion cell cytoplasm into the cell wall of surrounding tissues, using energy from ATP hydrolysis.
2. This creates a high concentration/electrochemical gradient of \(H^+\) ions outside the companion cell.
3. \(H^+\) ions diffuse back into the companion cell down their concentration gradient through a specific co-transporter protein.
4. This co-transporter protein simultaneously carries sucrose molecules into the companion cell against their concentration gradient.
5. Once inside the companion cell, sucrose diffuses into the sieve tube element through interconnecting plasmodesmata.

(c) Differences between xylem and phloem transport:
1. Xylem transport is entirely passive (driven by transpiration pull), whereas phloem transport involves active loading of solutes requiring ATP.
2. Xylem transport is strictly unidirectional (upwards from roots to leaves), whereas phloem transport is bidirectional (from source to sink depending on metabolic needs).
3. Xylem vessels consist of dead, lignified cells with no end walls, whereas phloem sieve tubes consist of living cells with sieve plates.

評分準則

(a) [3 marks max]
1. Sieve plates with pores to allow unobstructed flow of phloem sap [1]
2. Minimal cytoplasm / absence of nucleus / absence of vacuole / absence of ribosomes to reduce resistance to flow [1]
3. Cellulose cell wall to withstand high hydrostatic pressure [1]
4. Elongated cells joined end-to-end to form a continuous tube [1]

(b) [5 marks max]
1. \(H^+\) / protons actively pumped out of companion cell [1]
2. Uses ATP as an energy source [1]
3. Establishes a proton concentration gradient (higher concentration in cell wall/outside) [1]
4. \(H^+\) diffuses back into companion cell through a co-transporter protein [1]
5. Sucrose enters companion cell with \(H^+\) against its concentration gradient [1]
6. Sucrose moves into sieve tube element via plasmodesmata [1]

(c) [2 marks max]
Accept any two from:
1. Xylem: passive transport / phloem: active (loading) [1]
2. Xylem: unidirectional / phloem: bidirectional [1]
3. Xylem: dead cells / phloem: living cells [1]
4. Xylem: transports water and minerals / phloem: transports organic solutes/sucrose [1]
題目 3 · structured-short-answer
10
(a) Describe how the bacterium Vibrio cholerae is transmitted from an infected person to an uninfected person. [2]

(b) Explain the sequence of events that occurs once V. cholerae colonises the small intestine, leading to severe watery diarrhoea. [5]

(c) Oral rehydration therapy (ORT) is used to treat cholera. Explain how ORT rehydrates the body, referencing the components of the rehydration solution. [3]
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解題

(a) Vibrio cholerae is transmitted via the fecal-oral route. This occurs when food or water contaminated with the feces of an infected person is ingested by an uninfected person, often due to poor sanitation and sewage treatment.

(b) Once V. cholerae colonises the small intestine:
1. The bacteria secrete an enterotoxin called cholera toxin (choleragen).
2. This toxin binds to specific receptors on the cell surface membrane of the intestinal epithelial cells.
3. The binding activates the enzyme adenylate cyclase inside the epithelial cells, which converts ATP to cyclic AMP (cAMP).
4. High levels of intracellular cAMP cause channel proteins (CFTR channels) in the cell membrane to open.
5. Chloride ions (\(Cl^-\)) diffuse out of the epithelial cells into the lumen of the small intestine.
6. This lowers the water potential of the intestinal lumen below that of the epithelial cells and surrounding blood capillaries.
7. Water moves out of the cells and blood capillaries into the lumen of the intestine by osmosis down a water potential gradient, resulting in severe watery diarrhoea.

(c) Oral rehydration therapy (ORT) contains a precise mixture of water, glucose, and sodium ions (\(Na^+\)):
1. Sodium ions and glucose are co-transported into the intestinal epithelial cells via specific co-transporter proteins.
2. The absorption of these solutes lowers the water potential inside the epithelial cells.
3. This creates a water potential gradient, drawing water from the gut lumen back into the epithelial cells and subsequently into the blood by osmosis, rehydrating the patient.

評分準則

(a) [2 marks max]
1. Fecal-oral route [1]
2. Ingestion of water or food contaminated with feces of infected individual [1]

(b) [5 marks max]
1. Vibrio cholerae secretes cholera toxin / enterotoxin [1]
2. Toxin binds to receptors on intestinal epithelial cells [1]
3. Activates adenylate cyclase / increases cyclic AMP (cAMP) levels [1]
4. CFTR / chloride (\(Cl^-\)) channels open [1]
5. Chloride ions flow out of epithelial cells into lumen [1]
6. Water potential of the lumen decreases / becomes more negative [1]
7. Water moves out of cells/blood into lumen by osmosis down water potential gradient [1]

(c) [3 marks max]
1. ORT solution contains sodium ions (\(Na^+\)) and glucose [1]
2. Sodium and glucose are co-transported into intestinal epithelial cells [1]
3. This lowers the water potential inside the cells [1]
4. Water moves from lumen into cells/blood by osmosis down a water potential gradient [1]
題目 4 · structured-short-answer
10
(a) State three differences between the quaternary structure of collagen and the quaternary structure of haemoglobin. [3]

(b) Explain how the primary structure of collagen determines its secondary and tertiary structures, and how this relates to its function. [5]

(c) Suggest why the replacement of a hydrophilic amino acid (such as glutamic acid) with a hydrophobic amino acid (such as valine) on the surface of a haemoglobin molecule (as in sickle cell anaemia) affects its solubility. [2]
查看答案詳解

解題

(a) Quaternary structure differences:
1. Collagen consists of three polypeptide chains wound into a triple helix, whereas haemoglobin consists of four polypeptide chains (two alpha and two beta chains).
2. Collagen is a fibrous protein with an elongated structure, whereas haemoglobin is a compact, spherical globular protein.
3. Collagen does not contain any prosthetic groups, whereas haemoglobin contains four prosthetic haem groups, each containing an iron ion (\(Fe^{2+}\)).
4. Collagen molecules are covalently cross-linked to adjacent collagen molecules, whereas the subunits of haemoglobin are held together by weaker, non-covalent interactions (hydrophobic, ionic, hydrogen bonds).

(b) Connection of collagen structure to function:
1. The primary structure of collagen consists of a highly repetitive sequence where every third amino acid is glycine.
2. Glycine is the smallest amino acid, with only a single hydrogen atom as its R-group.
3. This small size allows the three polypeptide chains to pack very tightly together to form a tight triple helix (secondary structure).
4. The tertiary structure is stabilized by numerous hydrogen bonds forming between the peptide groups of adjacent chains, giving the molecule high tensile strength.
5. The tight triple helix molecules (tropocollagen) lie parallel to each other and are covalently cross-linked to form fibrils, which assemble into strong fibres, enabling collagen to withstand huge stretching forces in tendons, skin, and bones.

(c) Hydrophilic amino acids (such as glutamic acid) on the surface of normal haemoglobin form hydrogen bonds with surrounding water molecules, keeping the molecule soluble. When replaced by a hydrophobic amino acid (such as valine), the non-polar R-group cannot interact with water. Instead, the hydrophobic R-groups on adjacent haemoglobin molecules associate with each other (hydrophobic interactions) to exclude water, causing the haemoglobin molecules to aggregate and precipitate out of solution, especially under low oxygen conditions.

評分準則

(a) [3 marks max]
1. Collagen has three polypeptide chains vs. haemoglobin has four polypeptide chains [1]
2. Collagen has a fibrous/elongated shape vs. haemoglobin is globular/spherical [1]
3. Collagen has no prosthetic group vs. haemoglobin has four haem groups [1]
4. Collagen contains covalent bonds between molecules vs. haemoglobin chains are held by non-covalent bonds (hydrophobic/ionic/hydrogen) [1]

(b) [5 marks max]
1. Primary structure has glycine at every third position [1]
2. Glycine is the smallest amino acid / has only H as R-group [1]
3. Allows the three polypeptide chains to lie close together / pack tightly [1]
4. Forms a tight triple helix / secondary structure [1]
5. Stabilized by hydrogen bonds between adjacent chains [1]
6. High tensile strength / forms fibrils with covalent cross-links for structural support [1]

(c) [2 marks max]
1. Glutamic acid is polar/hydrophilic and interacts with water to keep the molecule soluble [1]
2. Valine is non-polar/hydrophobic, causing hydrophobic R-groups to stick together/exclude water, leading to aggregation/precipitation [1]
題目 5 · structured-short-answer
10
(a) Define the terms systole and diastole. [2]

(b) In an investigation, the pressure changes in the left side of the heart of a mammal were monitored over a single cardiac cycle lasting 0.8 seconds. The following events were observed:
- At 0.15 seconds, the pressure in the left ventricle rises above the pressure in the left atrium.
- At 0.35 seconds, the pressure in the left ventricle rises above the pressure in the aorta.
- At 0.55 seconds, the pressure in the left ventricle falls below the pressure in the aorta.
- At 0.70 seconds, the pressure in the left ventricle falls below the pressure in the left atrium.

(i) State which valve opens or closes at:
- 0.15 seconds: [1]
- 0.35 seconds: [1]
- 0.55 seconds: [1]

(ii) Calculate the heart rate of this mammal in beats per minute (bpm). Show your working. [2]

(c) Explain why the maximum pressure reached in the left ventricle is much higher than the maximum pressure reached in the right ventricle. [3]
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解題

(a) Systole refers to the stage of the cardiac cycle when the muscle of the heart chambers contracts to pump blood. Diastole refers to the stage of the cardiac cycle when the muscle of the heart chambers relaxes and fills with blood.

(b)(i) Valvular events:
- 0.15 seconds: The bicuspid / mitral (atrioventricular) valve closes. (As pressure in the ventricle becomes higher than in the atrium, it forces the AV valve shut to prevent backflow).
- 0.35 seconds: The semi-lunar (aortic) valve opens. (As ventricular pressure exceeds aortic pressure, blood is forced into the aorta).
- 0.55 seconds: The semi-lunar (aortic) valve closes. (As ventricular pressure drops below aortic pressure, blood starts to flow back, catching and closing the pockets of the semi-lunar valve).

(b)(ii) Calculation:
- One complete cardiac cycle = 0.8 seconds.
- Number of beats in 60 seconds (1 minute) = \(60 / 0.8 = 75\) bpm.
- Working: \(60 \div 0.8\) [1 mark]; Answer = 75 [1 mark].

(c) The maximum pressure in the left ventricle is much higher because: 1. The left ventricle has a much thicker muscular wall (myocardium) than the right ventricle. 2. The left ventricle must pump blood throughout the entire body (systemic circulation), which is a much longer distance and offers higher resistance. 3. The right ventricle only pumps blood to the lungs (pulmonary circulation), which is a shorter distance with lower resistance, requiring less pressure to prevent damage to the delicate lung capillaries.

評分準則

(a) [2 marks max]
1. Systole: contraction of heart muscle [1]
2. Diastole: relaxation of heart muscle [1]

(b)(i) [3 marks max]
1. 0.15 s: Atrioventricular / bicuspid / mitral valve closes [1]
2. 0.35 s: Semi-lunar / aortic valve opens [1]
3. 0.55 s: Semi-lunar / aortic valve closes [1]

(b)(ii) [2 marks max]
1. Show correct working: \(60 / 0.8\) [1]
2. Correct answer: 75 (bpm) [1]

(c) [3 marks max]
1. Left ventricle has a thicker muscular wall / thicker myocardium [1]
2. Left ventricle pumps blood to systemic circulation / rest of the body, which is a longer distance / higher resistance [1]
3. Right ventricle pumps blood to pulmonary circulation / lungs, which is a shorter distance / lower resistance [1]
4. Lower pressure in right ventricle prevents damage to capillaries in the lungs [1]
題目 6 · structured-short-answer
10
(a) An electron micrograph of a liver cell shows a mitochondrion. The actual length of this mitochondrion is 2.5 \(\mu\text{m}\). In the micrograph, the mitochondrion measures 45 mm in length.

Calculate the magnification of this micrograph. Show your working and give your answer to the nearest whole number. [2]

(b) Explain why an electron microscope can resolve details of cell structure that are not visible using a light microscope. [3]

(c) Describe the roles of the rough endoplasmic reticulum (RER) and the Golgi apparatus in the production and secretion of extracellular enzymes. [5]
查看答案詳解

解題

(a) Magnification calculation:
1. Use the formula: \(\text{Magnification} = \text{Image size} / \text{Actual size}\)
2. Convert the image size from millimetres (mm) to micrometres (\(\mu\text{m}\)): \(45\text{ mm} \times 1000 = 45000\ \mu\text{m}\).
3. Calculate magnification: \(45000\ \mu\text{m} / 2.5\ \mu\text{m} = 18000\).
4. Magnification = \(18000\times\) (or \(\times 18000\)).

(b) Resolution differences:
1. Resolution is the ability to distinguish between two points that are close together. 2. The wavelength of a beam of electrons is much shorter than the wavelength of light. 3. Shorter wavelengths allow a much higher resolution to be achieved. 4. Consequently, the electron microscope has a resolution of about 0.5 nm (compared to 200 nm for a light microscope), allowing tiny internal structures (like ribosomes, nuclear pores, or membranes) to be clearly distinguished.

(c) Roles of RER and Golgi:
1. Ribosomes on the surface of the rough endoplasmic reticulum (RER) are the site of protein synthesis (translation). 2. The synthesized polypeptide chains are transported into the lumen of the RER, where they fold into their specific tertiary structures. 3. The proteins are packaged into transport vesicles that bud off the RER. 4. These vesicles travel to and fuse with the cis-face of the Golgi apparatus. 5. In the Golgi apparatus, proteins are modified (e.g., carbohydrates are added to form glycoproteins / glycosylation). 6. The modified enzymes are sorted and packaged into secretory vesicles at the trans-face of the Golgi. 7. Secretory vesicles move to and fuse with the cell surface membrane, releasing the enzymes outside the cell via exocytosis.

評分準則

(a) [2 marks max]
1. Converts 45 mm to 45000 \(\mu\text{m}\) OR sets up equation: \(45\text{ mm} / 0.0025\text{ mm}\) [1]
2. Correct final magnification: 18000 (accept \(18000\times\) or \(\times 18000\)) [1]

(b) [3 marks max]
1. Resolution is the ability to distinguish between two close points [1]
2. Electron beam has a much shorter wavelength than light [1]
3. Shorter wavelength gives higher resolution [1]
4. Light microscope limit is 200 nm, electron microscope limit is 0.5 nm / detail of organelles can be seen [1]

(c) [5 marks max]
1. Ribosomes on RER synthesize polypeptides/proteins [1]
2. Polypeptides enter RER lumen and fold into tertiary structure [1]
3. Transport vesicles containing proteins bud off the RER [1]
4. Vesicles fuse with the Golgi apparatus [1]
5. Golgi modifies proteins (e.g. adds carbohydrate chains / glycosylation) [1]
6. Golgi packages modified proteins into secretory vesicles [1]
7. Secretory vesicles fuse with cell membrane to release proteins by exocytosis [1]

Paper 3 (Advanced Practical Skills)

Complete two major practical exercises. Section 1 involves active laboratory manipulation, dilution creation, and graph plotting. Section 2 involves microscopy and biological drawings.
2 題目 · 40
題目 1 · practical-manipulation-analysis
22
### Question 1: Investigation of Enzyme Inhibition

Catalase is an enzyme found in many living organisms, including yeast. It catalyzes the breakdown of hydrogen peroxide (\(\text{H}_2\text{O}_2\)) into oxygen and water:

\[2\text{H}_2\text{O}_2 \xrightarrow{\text{Catalase}} 2\text{H}_2\text{O} + \text{O}_2\text{ (g)}\]

Copper sulfate (\(\text{CuSO}_4\)) can act as an inhibitor of catalase. In this investigation, you will investigate the effect of different concentrations of copper sulfate solution on the rate of catalase activity.

You are provided with:
- **Y**: A 2.0% suspension of active yeast cells (source of catalase)
- **H**: A 3.0% hydrogen peroxide solution
- **I**: A 1.0% copper sulfate solution (the inhibitor)
- **W**: Distilled water

---

### Part (a) Preparing dilutions

(i) You need to prepare 10.0 cm³ of each of the following concentrations of copper sulfate solution: 0.8%, 0.6%, 0.4%, and 0.2%, using the 1.0% inhibitor solution **I** and distilled water **W**.

Complete **Table 1.1** below to show the volumes of **I** and **W** you will use to prepare each concentration. Ensure all volumes are recorded to one decimal place. [3 marks]

**Table 1.1**
| Final concentration of inhibitor / % | Volume of 1.0% inhibitor solution **I** / cm³ | Volume of distilled water **W** / cm³ | Total volume / cm³ |
| :---: | :---: | :---: | :---: |
| 1.0 | 10.0 | 0.0 | 10.0 |
| 0.8 | | | 10.0 |
| 0.6 | | | 10.0 |
| 0.4 | | | 10.0 |
| 0.2 | | | 10.0 |

(ii) State the volume of distilled water **W** and 1.0% inhibitor solution **I** needed to prepare 10.0 cm³ of the 0.0% control solution. [1 mark]

---

### Part (b) Investigation

The rate of catalase activity can be measured using a paper disc flotation method. Paper discs are soaked in the yeast suspension containing the inhibitor, then dropped into hydrogen peroxide. As oxygen gas is produced, bubbles become trapped in the fibers of the paper disc, causing it to float to the surface. The time taken (\(t\)) for the disc to rise is inversely proportional to the rate of reaction.

**Procedure:**
1. Label six test-tubes with the concentrations of inhibitor: 1.0%, 0.8%, 0.6%, 0.4%, 0.2%, and 0.0%.
2. Add 2.0 cm³ of yeast suspension **Y** and 2.0 cm³ of the corresponding inhibitor solution to each test-tube. Swirl gently and leave to incubate for 3 minutes.
3. Label another six test-tubes with the same concentrations and add 10.0 cm³ of hydrogen peroxide solution **H** to each.
4. Using forceps, submerge a clean filter paper disc in the 1.0% yeast-inhibitor mixture for exactly 10 seconds.
5. Transfer the disc to the bottom of the corresponding tube of hydrogen peroxide **H**. Start a stopwatch immediately.
6. Measure the time taken (\(t\)), in seconds, for the disc to rise to the surface of the liquid. If the disc does not rise within 180 seconds, stop and record '>180'.
7. Repeat steps 4–6 for each concentration.

(i) State one control variable (other than temperature, concentration/volume of yeast, or concentration/volume of hydrogen peroxide) that must be kept constant to ensure valid results, and describe how you would control it. [2 marks]

(ii) Prepare a table in the space below and record all your results from your practical work. [5 marks]

---

### Part (c) Analysis and explanation

(i) Explain the biological effect of increasing copper sulfate concentration on the time taken for the filter paper disc to rise. [3 marks]

---

### Part (d) Graph plotting and interpretation

In a separate investigation, the effect of pH on the rate of reaction of yeast catalase was studied. The results are shown in **Table 1.2**.

**Table 1.2**
| pH of buffer solution | Rate of reaction / arbitrary units (a.u.) |
| :---: | :---: |
| 4.0 | 6.5 |
| 5.0 | 15.0 |
| 6.0 | 32.5 |
| 7.0 | 48.0 |
| 8.0 | 22.0 |
| 9.0 | 4.5 |

(i) Plot a graph of the data in Table 1.2 on a grid, using suitable scales. [4 marks]

(ii) Based on your graph, estimate the optimum pH for this catalase and describe how the investigator could modify the experimental method to determine the optimum pH more precisely. [2 marks]

---

### Part (e) Evaluation

Identify two potential sources of error in the method used in Part (b) that could affect the accuracy or reliability of your results, and suggest an improvement for each. [2 marks]
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解題

### Detailed Worked Solution

#### Part (a)
(i) To prepare 10.0 cm³ of different concentrations from a stock of 1.0% solution, use the formula:
\(C_1 V_1 = C_2 V_2\)
For 0.8%:
\(1.0\% \times V_1 = 0.8\% \times 10.0\text{ cm}^3 \implies V_1 = 8.0\text{ cm}^3\) of **I**.
Volume of **W** = \(10.0 - 8.0 = 2.0\text{ cm}^3\).

Complete calculations:
- **0.8%**: 8.0 cm³ of **I**, 2.0 cm³ of **W**
- **0.6%**: 6.0 cm³ of **I**, 4.0 cm³ of **W**
- **0.4%**: 4.0 cm³ of **I**, 6.0 cm³ of **W**
- **0.2%**: 2.0 cm³ of **I**, 8.0 cm³ of **W**

(ii) For 0.0% control solution: 0.0 cm³ of **I** and 10.0 cm³ of **W**.

#### Part (b)
(i) **Control variable**: Size/thickness/source of the paper discs.
**How to control**: Cut all discs using the same hole punch from the same sheet of standard filter paper.
*(Alternatively: The time the disc is submerged in the yeast-inhibitor mixture. Controlled by timing exactly 10.0 seconds using a stopwatch.)*

(ii) **Results Table Criteria**:
- Table with fully drawn borders separating all cells.
- Heading row containing: "Concentration of copper sulfate / %" and "Time taken for disc to rise, \(t\) / s" (correct slash notation for units).
- All results recorded to whole seconds (or consistently to 1 decimal place).
- Raw results showing the expected trend: as the concentration of inhibitor increases, the time taken for the disc to rise increases (e.g., 0.0% takes ~10s, 1.0% takes >100s or does not rise).
- Replicate trials shown with calculated mean column.

#### Part (c)
(i) **Biological explanation**:
- Copper sulfate acts as an inhibitor of catalase. Copper ions (\(\text{Cu}^{2+}\)) are heavy metals that can bind to the enzyme, altering its tertiary structure and changing the shape of the active site (denaturation or non-competitive inhibition).
- This prevents the substrate (hydrogen peroxide) from binding, reducing the number of active sites available and decreasing the rate of enzyme-substrate complex (ESC) formation.
- This reduces the rate of oxygen gas production, meaning it takes a longer time for enough gas to be trapped in the disc to make it buoyant.

#### Part (d)
(i) **Graph features**:
- **Axes (A)**: Independent variable (pH) on the x-axis, dependent variable (Rate of reaction / a.u.) on the y-axis. Both must be fully labeled with units.
- **Scale (S)**: Linear scale, where the data points fill more than half the grid along both axes (e.g., x-axis from 3.0 to 10.0, y-axis from 0 to 50).
- **Plotting (P)**: All six points plotted precisely within 0.5 small square, represented by a small cross or circled dot.
- **Line (L)**: Connected point-to-point with straight, sharp, ruled lines, or a smooth symmetric curve that passes through all points with no double lines.

(ii) **Optimum and precision modification**:
- **Optimum pH**: Estimated at pH 7.0 (or according to the peak of the drawn line).
- **Modification**: Test more intermediate pH values at smaller intervals (e.g., pH 6.2, 6.4, 6.6, 6.8, 7.2, 7.4, 7.6, 7.8) to pinpoint the actual maximum rate of reaction.

#### Part (e)
- **Source of Error 1**: Discs sticking to the side of the test-tube walls, which adds friction and slows down the rise.
- **Improvement 1**: Use a wider test-tube or boiling tube to minimize contact with the walls.
- **Source of Error 2**: Subjective determination of when the disc starts or finishes rising.
- **Improvement 2**: Record a video of the trials next to a digital timer and play it back in slow-motion to verify exact times.

評分準則

### Marking Scheme

#### Part (a) [Total: 4 marks]
- **(i)** [3 marks]
- Award 1 mark if all volumes of **I** and **W** are calculated correctly.
- Award 1 mark if all values are written to 1 decimal place (e.g., 8.0, 2.0).
- Award 1 mark for correct column headers with units: `Volume of 1.0% inhibitor solution I / cm3` and `Volume of distilled water W / cm3` (accept `/ cm³`).
- **(ii)** [1 mark]
- Award 1 mark for: 0.0 cm³ of **I** and 10.0 cm³ of **W**.

#### Part (b) [Total: 7 marks]
- **(i)** [2 marks]
- Award 1 mark for identifying a valid control variable (e.g., paper disc diameter/source, or soaking time).
- Award 1 mark for describing a precise control method (e.g., use a single hole puncher, use a digital stopwatch to submerge for exactly 10s).
- **(ii)** [5 marks]
- **M1 (Table structure)**: Grid with fully enclosed borders, no open-ended tables.
- **M2 (Headings & Units)**: Columns clearly labeled: `Concentration of copper sulfate / %` and `Time / s` (must use `/` or `( )` for units, no units in body cells).
- **M3 (Data collection)**: Results recorded for all 6 concentrations (0.0%, 0.2%, 0.4%, 0.6%, 0.8%, 1.0%).
- **M4 (Trend)**: Higher inhibitor concentration yields a higher time taken (or no rise/longer than 180s).
- **M5 (Precision)**: All times recorded as whole seconds (no fractions of a second, reflecting human reaction time with stopwatches).

#### Part (c) [Total: 3 marks]
- **(i)** [3 marks]
- Award 1 mark for stating that copper ions (\(\text{Cu}^{2+}\)) act as inhibitors / bind to catalase / denature the active site.
- Award 1 mark for explaining that this changes the shape of the active site, reducing the formation of enzyme-substrate complexes (ESCs).
- Award 1 mark for linking this to a slower rate of oxygen production, meaning it takes longer for the paper disc to gain buoyancy.

#### Part (d) [Total: 6 marks]
- **(i)** [4 marks]
- **A**: Axes correctly oriented, x-axis as `pH of buffer solution` (no units), y-axis as `Rate of reaction / arbitrary units` or `a.u.`.
- **S**: Scale is linear and appropriate (plotted points must occupy more than half the grid on both x and y axes).
- **P**: All 6 points plotted correctly to within half a small square, using small, sharp crosses or circled dots.
- **L**: Clean, thin, ruled lines connecting point-to-point (or a smooth, single-line curve with no double lines or sketching).
- **(ii)** [2 marks]
- Award 1 mark for identifying pH 7.0 as the optimum (or peak value from the graph).
- Award 1 mark for suggesting narrower pH intervals (e.g., steps of 0.2 or 0.5 pH units) between pH 6.0 and 8.0.

#### Part (e) [Total: 2 marks]
- Award 1 mark for each valid Pair of (Error + Improvement), up to a maximum of 2 marks:
- **Error**: Discs sticking to test-tube walls.
**Improvement**: Use a wider vessel (e.g., boiling tube).
- **Error**: Reaction temperature fluctuates during the experiment.
**Improvement**: Use a thermostatically-controlled water bath to maintain constant temperature.
- **Error**: High variation / subjectivity in determining when the disc starts or completes rising.
**Improvement**: Use video analysis / high-definition video recording with a digital timer.
題目 2 · microscopy-drawing
18
Section 2: Microscopy and drawings. M1 is a slide of a transverse section through a herbaceous dicotyledonous stem. (a) Draw a large plan diagram of a sector of the stem on M1 to show the distribution of tissues. Your drawing should represent a wedge-shaped sector containing at least two entire vascular bundles. Do not draw any individual cells. [5] (b) Using the high-power lens of your microscope, select a group of three adjacent xylem vessel elements from one of the vascular bundles in M1. Make a large drawing of this group of three cells. Label one xylem vessel element on your drawing. [4] (c) An eyepiece graticule can be calibrated using a stage micrometer. At high-power magnification, a student determined that 1 eyepiece graticule unit (epu) is equal to 2.4 micrometers. (i) The student used the same high-power magnification to measure the thickness of the sclerenchyma cap (supporting tissue) of a vascular bundle on M1. The thickness was found to be 32 epu. Calculate the actual thickness of the sclerenchyma cap in micrometers. Show your working. [2] (ii) The student made a high-power drawing of the vascular bundle. In this drawing, the sclerenchyma cap was measured to be 38 mm thick. Calculate the magnification of the student's drawing. Show your working and present your final answer to the nearest whole number. [3] (d) Figure 1.1 is a photomicrograph of a transverse section through the stem of a different plant species, Zea mays (a monocotyledon). In Zea mays, the vascular bundles are scattered throughout the ground tissue, there is no cambium, and there is no distinct division into cortex and pith. In contrast, the stem on M1 has vascular bundles arranged in a ring, contains cambium, and has a distinct cortex and central pith. Prepare a table to compare the observable structures of the stem in M1 with those in Figure 1.1. [4]
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解題

Part (a): Draw a wedge-shaped plan diagram. Ensure lines are sharp and unshaded. Draw at least two vascular bundles. Show the layers: outer epidermis, cortex, vascular bundles (with sclerenchyma cap, phloem, cambium, xylem), and central pith. Do not draw cells. Part (b): Draw three adjacent xylem cells with polygonal shapes and double cell walls. Label one 'xylem vessel element'. Part (c)(i): Actual thickness = 32 epu * 2.4 micrometers/epu = 76.8 micrometers. Part (c)(ii): Convert 38 mm to micrometers: 38 mm * 1000 = 38000 micrometers. Magnification = drawing size / actual size = 38000 / 76.8 = 494.79. To the nearest whole number, this is x495 (accept x494 or x495). Part (d): Create a table comparing M1 and Figure 1.1. Feature 1: Bundle arrangement (M1: arranged in a ring; Fig 1.1: scattered). Feature 2: Cambium (M1: present; Fig 1.1: absent). Feature 3: Cortex and pith (M1: distinct cortex and central pith; Fig 1.1: no distinct cortex and pith/ground tissue only).

評分準則

Part (a) [Max 5 marks]: MP1: Quality of drawing: clear, single lines, no shading AND occupies at least 50% of available space. [1] MP2: Correct plan diagram style: no individual cells drawn anywhere. [1] MP3: Draws at least two vascular bundles showing realistic proportions (xylem region larger than phloem region, sclerenchyma cap on the outer edge). [1] MP4: Shows correct tissue layers in correct order: epidermis on outside, cortex, vascular bundle (with regions), central pith. [1] MP5: Draws cambium as a distinct layer/line between the xylem and phloem regions. [1] Part (b) [Max 4 marks]: MP1: Sharp, unbroken lines for cell walls AND no shading. [1] MP2: Draws exactly three adjacent cells with double cell walls (showing thickness of the walls). [1] MP3: Cells have an angular/polygonal shape with lumen shown. [1] MP4: Correct label 'xylem vessel' or 'xylem vessel element' with a straight leader line pointing to the lumen or wall of one cell (no arrowheads). [1] Part (c) [Max 5 marks]: (i) [2 marks] MP1: Shows multiplication of 32 by 2.4. [1] MP2: Correct final answer: 76.8 micrometers (must have units). [1] (ii) [3 marks] MP1: Conversion of drawing size from mm to micrometers (38000 micrometers) OR actual size to mm (0.0768 mm). [1] MP2: Correct formula or division: 38000 / 76.8. [1] MP3: Correct final magnification to the nearest whole number: x495 (accept x494 to x495). [1] Part (d) [Max 4 marks]: MP1: Table format with headings: 'Feature', 'Stem on M1' (or 'dicotyledon'), 'Stem in Figure 1.1' (or 'monocotyledon'). [1] MP2: Correctly identifies difference in arrangement of vascular bundles (ring vs scattered). [1] MP3: Correctly identifies presence vs absence of cambium / distinct cortex and pith. [1] MP4: Identifies another valid structural difference, e.g., sclerenchyma cap on one side of bundles in M1 vs bundle sheath surrounding bundles in Figure 1.1. [1]

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