An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V4) Cambridge International A Level Biology (9700) paper. Not affiliated with or reproduced from Cambridge.
卷一 選擇題
Answer all forty questions. For each question, choose from the four possible options A, B, C or D.
40 題目 · 40 分
題目 1 · multiple_choice
1 分
A student calibrates an eyepiece graticule using a stage micrometer. The stage micrometer has scale divisions of \(0.1\text{ mm}\). Under low power (\(\times 10\) objective lens), 40 divisions of the eyepiece graticule align perfectly with 10 divisions of the stage micrometer.
The student then switches to high power (\(\times 40\) objective lens) and measures the length of a plant cell. The cell is found to span 6 divisions of the eyepiece graticule.
What is the actual length of the plant cell?
A.\(9.38\ \mu\text{m}\)
B.\(15.0\ \mu\text{m}\)
C.\(37.5\ \mu\text{m}\)
D.\(150\ \mu\text{m}\)
查看答案詳解收起答案詳解
解題
Under low power: 10 divisions of the stage micrometer = \(10 \times 0.1\text{ mm} = 1.0\text{ mm} = 1000\ \mu\text{m}\). Since 40 eyepiece graticule divisions align with these 10 stage micrometer divisions: \(1\text{ eyepiece division} = 1000\ \mu\text{m} / 40 = 25\ \mu\text{m}\).
When switching from the \(\times 10\) objective to the \(\times 40\) objective, the magnification increases by a factor of 4. Therefore, each eyepiece division now represents a distance 4 times smaller: \(25\ \mu\text{m} / 4 = 6.25\ \mu\text{m}\).
Since the plant cell spans 6 eyepiece graticule divisions under high power, its actual length is: \(6 \times 6.25\ \mu\text{m} = 37.5\ \mu\text{m}\).
評分準則
Award 1 mark for the correct answer (C). - Method: Calculate size of 1 eyepiece division at low power (\(25\ \mu\text{m}\)), scale it down for high power (\(6.25\ \mu\text{m}\)), and multiply by the cell's graticule width (6). - Incorrect options represent arithmetic errors in scaling or misinterpreting the stage micrometer divisions.
題目 2 · multiple_choice
1 分
Four liquid samples were analyzed using the Benedict's test before and after acid hydrolysis. The table below shows the color changes observed:
| Sample | Color with Benedict's test on original sample | Color with Benedict's test after acid hydrolysis and neutralization | | :---: | :---: | :---: | | 1 | Green | Brick-red | | 2 | Blue | Blue | | 3 | Blue | Orange | | 4 | Yellow | Yellow |
Which of these samples contained non-reducing sugars?
A.1 and 3 only
B.2 and 4 only
C.3 only
D.1, 3 and 4
查看答案詳解收起答案詳解
解題
- **Sample 1:** The initial green color indicates a low concentration of reducing sugars. After acid hydrolysis, the color changes to brick-red, representing a much higher concentration. This increase indicates that non-reducing sugars were hydrolyzed to yield additional reducing sugars. - **Sample 2:** Remained blue throughout, meaning neither reducing nor non-reducing sugars were present. - **Sample 3:** Changed from blue (no reducing sugars) to orange (high concentration of reducing sugars), proving that non-reducing sugars were hydrolyzed. - **Sample 4:** Remained yellow throughout, indicating no change in the concentration of reducing sugars, and thus no non-reducing sugars were present.
Therefore, samples 1 and 3 contained non-reducing sugars.
評分準則
Award 1 mark for the correct option (A). - Reject options containing Sample 2 or 4, as they do not show an increase in reducing sugar concentration after acid hydrolysis.
題目 3 · multiple_choice
1 分
The list describes features of two different pathways, 1 and 2, involved in the loading of sucrose from a leaf mesophyll cell into a phloem companion cell:
* **Pathway 1:** Sucrose travels through plasmodesmata directly from the cytoplasm of one cell to the next. * **Pathway 2:** Sucrose is transported from the cell wall space (apoplast) across the cell surface membrane of the companion cell using co-transport proteins with protons (\(\text{H}^+\)).
Which statements about these pathways are correct?
1. Pathway 1 represents the symplastic pathway and does not directly require the hydrolysis of ATP. 2. Pathway 2 requires ATP to drive proton pumps that establish a proton gradient across the companion cell membrane. 3. The concentration of sucrose inside the companion cell must be kept lower than that of the mesophyll cell for Pathway 2 to function.
A.1, 2 and 3
B.1 and 2 only
C.2 and 3 only
D.1 only
查看答案詳解收起答案詳解
解題
- **Statement 1 is correct:** Movement through plasmodesmata represents the symplastic pathway, which occurs passively via diffusion and does not directly require ATP hydrolysis. - **Statement 2 is correct:** Active loading (Pathway 2) depends on proton pumps actively transporting protons out of the companion cell using ATP, creating a proton gradient that drives the cotransport of sucrose. - **Statement 3 is incorrect:** Active transport mechanisms accumulate sucrose in the companion cells against its concentration gradient, meaning the concentration of sucrose inside companion cells is significantly higher than in mesophyll cells.
評分準則
Award 1 mark for the correct option (B). - Reject options A and C as statement 3 is incorrect because sucrose concentration is higher in the companion cell. - Reject option D as statement 2 is correct.
題目 4 · multiple_choice
1 分
A population of bacteria is transferred from an incubator at \(37\,^{\circ}\text{C}\) to a cold room at \(10\,^{\circ}\text{C}\). To survive and maintain optimal cell surface membrane function, the bacteria must adapt to prevent their membranes from becoming too rigid.
Which modification to the bacterial cell surface membrane would help maintain its optimum fluidity at this lower temperature?
A.Increase the proportion of phospholipids containing saturated fatty acid tails.
B.Increase the proportion of phospholipids containing unsaturated fatty acid tails.
C.Increase the average length of the fatty acid tails.
D.Decrease the proportion of glycoproteins and increase peripheral protein concentration.
查看答案詳解收起答案詳解
解題
At low temperatures, membrane fluidity decreases because phospholipid molecules pack closer together. Unsaturated fatty acid tails contain carbon-to-carbon double bonds (\(\text{C}=\text{C}\)) which create 'kinks' in the tails. These kinks prevent the phospholipids from packing closely together, thereby maintaining membrane fluidity at lower temperatures. Saturated fatty acids pack very tightly and would make the membrane too rigid.
評分準則
Award 1 mark for identifying the increase of unsaturated fatty acid tails as the mechanism to prevent rigid packing (B). - Reject A and C because saturated and longer fatty acids increase packing density, reducing fluidity.
題目 5 · multiple_choice
1 分
Which option correctly compares the features of a light microscope with a transmission electron microscope (TEM)?
A.Resolution limit: 200 nm (light), 0.5 nm (TEM); Specimen state: living or dead (light), dead only (TEM)
B.Resolution limit: 2 μm (light), 200 nm (TEM); Specimen state: dead only (light), living or dead (TEM)
D.Radiation source: ultraviolet light (light), electron beam (TEM); Image color: black and white only (light), colored (TEM)
查看答案詳解收起答案詳解
解題
- **A is correct:** The resolution limit of a light microscope is about \(200\text{ nm}\), whereas a TEM can achieve a resolution of up to \(0.5\text{ nm}\). Light microscopes can view specimens in both living and dead states, whereas TEM requires a vacuum and therefore can only view dead (fixed) specimens. - **B is incorrect:** The resolution limits are wrong and the specimen states are swapped. - **C is incorrect:** The focusing mechanisms are swapped; light microscopes use glass lenses, while TEMs use electromagnets. - **D is incorrect:** TEMs use electron beams and generate black and white images, while light microscopes can use white light and generate colored images.
評分準則
Award 1 mark for the correct comparison table row (A). - Reject B because the resolution limits are inaccurate and specimen states are reversed. - Reject C because focusing lenses are swapped. - Reject D because light microscopes produce color images, while TEM produces black-and-white images.
題目 6 · multiple_choice
1 分
A student tested four different liquid food samples, W, X, Y, and Z, with different reagents to identify their biological molecule content. The results are shown below:
* **Sample W:** Blue with Benedict's reagent; orange with Benedict's after acid hydrolysis; purple with Biuret reagent; cloudy emulsion with ethanol and water; yellow-brown with iodine solution. * **Sample X:** Red with Benedict's reagent; red with Benedict's after acid hydrolysis; blue with Biuret reagent; cloudy emulsion with ethanol and water; blue-black with iodine solution. * **Sample Y:** Blue with Benedict's reagent; blue with Benedict's after acid hydrolysis; purple with Biuret reagent; clear with ethanol and water; yellow-brown with iodine solution. * **Sample Z:** Orange with Benedict's reagent; orange with Benedict's after acid hydrolysis; blue with Biuret reagent; clear with ethanol and water; blue-black with iodine solution.
Which sample contains non-reducing sugar, protein, and lipid, but no starch?
A.Sample W
B.Sample X
C.Sample Y
D.Sample Z
查看答案詳解收起答案詳解
解題
Let's analyze the tests: 1. **Non-reducing sugar present:** The original Benedict's test must be negative (blue), but positive (orange/red/green) after acid hydrolysis. This is true for **Sample W** and **Sample Y**. 2. **Protein present:** The Biuret test must turn purple. This is true for **Sample W** and **Sample Y**. 3. **Lipid present:** The ethanol emulsion test must form a cloudy emulsion. This is true for **Sample W** and **Sample X**. 4. **No starch:** The iodine test must remain yellow-brown. This is true for **Sample W** and **Sample Y**.
Only **Sample W** meets all four criteria.
評分準則
Award 1 mark for the correct option (A). - Sample X is incorrect as it contains reducing sugars, starch, lipids, but no protein or non-reducing sugar. - Sample Y is incorrect as it lacks lipids. - Sample Z is incorrect as it lacks non-reducing sugars and proteins.
題目 7 · multiple_choice
1 分
A plant cell has an initial water potential (\(\psi\)) of \(-600\text{ kPa}\). It is placed in a hypertonic solution with a water potential of \(-900\text{ kPa}\).
At the precise point of incipient plasmolysis, what is the cell's hydrostatic pressure potential (\(\psi_p\)) and what is the direction of net water movement up to this point?
A.\(\psi_p = 0\text{ kPa}\); net movement of water is out of the cell.
B.\(\psi_p = 0\text{ kPa}\); net movement of water is into the cell.
C.\(\psi_p = -300\text{ kPa}\); net movement of water is out of the cell.
D.\(\psi_p = +300\text{ kPa}\); there is no net movement of water.
查看答案詳解收起答案詳解
解題
At the point of incipient plasmolysis, the plant protoplast has shrunk slightly so that it no longer exerts any pressure against the cell wall. Therefore, the hydrostatic pressure potential (\(\psi_p\)) becomes exactly \(0\text{ kPa}\).
Since the water potential of the surrounding solution is lower (more negative, \(-900\text{ kPa}\)) than the initial water potential of the cell (\(-600\text{ kPa}\)), water moves down a water potential gradient out of the cell by osmosis.
評分準則
Award 1 mark for identifying both the correct pressure potential of 0 kPa and net water movement out of the cell (A). - Reject B because water moves out of the cell, not into it. - Reject C and D because pressure potential cannot be negative in this condition and must equal 0 kPa at incipient plasmolysis.
題目 8 · multiple_choice
1 分
The rates of entry of three substances, X, Y, and Z, into a mammalian cell were measured over a range of external concentrations. The following observations were made:
* **Substance X:** The rate of entry increases linearly with increasing external concentration with no limit. * **Substance Y:** The rate of entry increases with concentration but plateaus at higher concentrations. The rate is significantly reduced when a respiratory inhibitor is added. * **Substance Z:** The rate of entry increases with concentration and plateaus at higher concentrations. The rate is unaffected by the addition of a respiratory inhibitor.
Which mechanisms are used for the transport of X, Y, and Z?
A.X = facilitated diffusion; Y = active transport; Z = simple diffusion
B.X = simple diffusion; Y = active transport; Z = facilitated diffusion
C.X = simple diffusion; Y = facilitated diffusion; Z = active transport
D.X = active transport; Y = facilitated diffusion; Z = simple diffusion
查看答案詳解收起答案詳解
解題
- **Substance X** moves by simple diffusion. The rate is directly proportional to the concentration gradient, as there are no transport proteins involved to limit the rate via saturation. - **Substance Y** moves by active transport. The rate plateaus because carrier proteins become saturated at high concentrations, and it is inhibited by respiratory inhibitors because active transport requires ATP produced by aerobic respiration. - **Substance Z** moves by facilitated diffusion. The rate plateaus because transport proteins (channels or carriers) become saturated, but it is passive and does not require ATP, so respiratory inhibitors have no effect.
評分準則
Award 1 mark for matching all three substances to their correct transport mechanisms (B). - Reject other options where transport mechanisms are incorrectly matched to the substances' behaviors.
題目 9 · 選擇題
1 分
A student calibrated an eyepiece graticule using a stage micrometer. At a magnification of \times 100, 100 eyepiece graticule units corresponded to 10 divisions on the stage micrometer. Each division on the stage micrometer was 0.1 mm apart. The student then changed the magnification to \times 400 to view plant cells. A chloroplast was measured to be 3 eyepiece graticule units in length. What is the actual length of this chloroplast?
A.1.875 \mu m
B.7.5 \mu m
C.30.0 \mu m
D.120.0 \mu m
查看答案詳解收起答案詳解
解題
At \times 100 magnification: 10 divisions of the stage micrometer = 10 \times 0.1 mm = 1.0 mm = 1000 \mu m. Therefore, 100 eyepiece units = 1000 \mu m, so 1 eyepiece unit (epu) = 10 \mu m. When magnification is increased to \times 400 (which is 4 times higher), the actual distance represented by each eyepiece unit decreases by a factor of 4. Thus, at \times 400, 1 epu = 10 \mu m / 4 = 2.5 \mu m. The chloroplast was 3 epu long, so its actual length is 3 \times 2.5 \mu m = 7.5 \mu m.
評分準則
1 mark for the correct choice B.
題目 10 · 選擇題
1 分
Which of the following processes are involved in the movement of water up a plant stem? 1. Adhesion between water molecules and the hydrophilic components of xylem vessel walls. 2. Cohesion between water molecules due to hydrogen bonding. 3. The active transport of water molecules into the root xylem. 4. Evaporation of water from the wet cell walls of the spongy mesophyll.
A.1, 2 and 3 only
B.1, 2 and 4 only
C.1 and 4 only
D.2, 3 and 4 only
查看答案詳解收起答案詳解
解題
Adhesion (1) is the attraction between water molecules and the hydrophilic parts of the xylem walls (cellulose and lignin), which helps support the water column. Cohesion (2) is the attraction between water molecules due to hydrogen bonding, keeping the water column intact under tension. Evaporation (4) from spongy mesophyll cells into the air spaces lowers the water potential, drawing water out of the leaf xylem and driving the transpiration pull. Water molecules are never actively transported (3); water movement across cell membranes is entirely passive (osmosis).
評分準則
1 mark for the correct choice B.
題目 11 · 選擇題
1 分
A student is provided with four liquid samples, W, X, Y and Z. The student performs biochemical tests on each sample. The results are shown in the table below. Sample W: Biuret test = violet, Iodine test = blue-black, Benedict's test = blue, Benedict's test after acid hydrolysis = brick-red. Sample X: Biuret test = blue, Iodine test = blue-black, Benedict's test = blue, Benedict's test after acid hydrolysis = brick-red. Sample Y: Biuret test = blue, Iodine test = yellow-brown, Benedict's test = brick-red, Benedict's test after acid hydrolysis = brick-red. Sample Z: Biuret test = violet, Iodine test = yellow-brown, Benedict's test = blue, Benedict's test after acid hydrolysis = blue. Which sample contains both sucrose and starch, but no other biological molecules?
A.W
B.X
C.Y
D.Z
查看答案詳解收起答案詳解
解題
A sample containing only sucrose and starch will test negative for protein (Biuret test remains blue), positive for starch (Iodine test turns blue-black), negative for reducing sugars (initial Benedict's test remains blue), and positive for non-reducing sugars (Benedict's test after acid hydrolysis turns brick-red because sucrose is hydrolysed into glucose and fructose). Sample X matches all these criteria: blue Biuret (no protein), blue-black Iodine (starch present), blue initial Benedict's (no reducing sugars), and brick-red after hydrolysis (sucrose present).
評分準則
1 mark for the correct choice B.
題目 12 · 選擇題
1 分
A plant cell with a solute potential (\psi_s) of -900 kPa and a pressure potential (\psi_p) of +300 kPa is placed in a solution with a water potential (\psi) of -800 kPa. Which row correctly describes the net movement of water and the change in the cell's pressure potential?
A.Net movement of water: out of the cell; Change in pressure potential: decreases
B.Net movement of water: out of the cell; Change in pressure potential: increases
C.Net movement of water: into the cell; Change in pressure potential: decreases
D.Net movement of water: into the cell; Change in pressure potential: increases
查看答案詳解收起答案詳解
解題
The initial water potential of the plant cell is \psi = \psi_s + \psi_p = -900\text{ kPa} + 300\text{ kPa} = -600\text{ kPa}. Since the cell's water potential (-600 kPa) is higher (less negative) than the water potential of the surrounding solution (-800 kPa), there will be a net movement of water out of the cell by osmosis. As water leaves the cell, the volume of the protoplast decreases, which reduces the pressure exerted by the protoplast against the rigid cell wall, causing the pressure potential (\psi_p) to decrease.
評分準則
1 mark for the correct choice A.
題目 13 · 選擇題
1 分
In a double-stranded DNA molecule, 18% of the nucleotides contain the base cytosine. A gene within this DNA molecule consists of 1200 base pairs. How many adenine bases are present in this gene?
A.216
B.384
C.432
D.768
查看答案詳解收起答案詳解
解題
Since the DNA is double-stranded, the percentage of cytosine (C) equals the percentage of guanine (G). Therefore, C = 18% and G = 18%, making up 36% of the total bases. The remaining 64% (100% - 36%) must be shared equally between adenine (A) and thymine (T), so adenine comprises 32% of the total bases. The gene consists of 1200 base pairs, which means it contains a total of 2400 bases. The number of adenine bases is 32% of 2400, which is 0.32 \times 2400 = 768.
評分準則
1 mark for the correct choice D.
題目 14 · 選擇題
1 分
A diploid animal cell has a chromosome number of 2n = 8. Which row correctly identifies the number of chromosomes, sister chromatids and centromeres present in this cell during metaphase of mitosis?
During metaphase of mitosis, the diploid cell contains 8 chromosomes (since 2n = 8). Each chromosome consists of two sister chromatids joined at a single centromere. Therefore, there are 8 chromosomes, 16 sister chromatids (8 \times 2), and 8 centromeres.
評分準則
1 mark for the correct choice B.
題目 15 · 選擇題
1 分
Which of the following statements about the induced-fit hypothesis of enzyme action are correct? 1. The active site of the enzyme is fully complementary to the shape of the substrate before the substrate binds. 2. The substrate induces a conformational change in the shape of the enzyme's active site upon binding. 3. The change in shape of the active site puts strain on the bonds of the substrate, lowering the activation energy. 4. The active site returns to its original shape after the products are released.
A.1 and 3 only
B.2 and 4 only
C.2, 3 and 4 only
D.1, 2, 3 and 4
查看答案詳解收起答案詳解
解題
Statement 1 is incorrect because in the induced-fit model, the active site is not fully complementary to the substrate before binding; rather, it changes shape to become complementary. Statement 2 is correct as the binding of the substrate induces this conformational change. Statement 3 is correct because this alteration of the enzyme's structure strains the substrate bonds, making them easier to break and thus lowering the activation energy. Statement 4 is correct because the enzyme remains chemically unchanged and returns to its original conformation after the reaction.
評分準則
1 mark for the correct choice C.
題目 16 · 選擇題
1 分
Which row correctly shows the distribution of tissues in the structures of the human gas exchange system?
The trachea and bronchi both contain cartilage (to keep airways open), smooth muscle (to regulate airway diameter), and elastic fibres (to allow recoil). Bronchioles do not have cartilage, but they do have smooth muscle and elastic fibres. Alveoli have neither cartilage nor smooth muscle, but contain abundant elastic fibres to facilitate gas exchange and recoil. Therefore, row A is correct.
評分準則
1 mark for the correct choice A.
題目 17 · multiple_choice
1 分
A student calibrates an eyepiece graticule using a stage micrometer. The stage micrometer has scale divisions that are \(0.1\text{ mm}\) apart. At a magnification of \(\times 100\), 40 divisions of the eyepiece graticule coincide exactly with 8 divisions of the stage micrometer. The student then views a plant cell using the same microscope at a magnification of \(\times 400\). Under \(\times 400\) magnification, the cell spans 15 divisions of the eyepiece graticule. What is the actual length of the plant cell?
A.\(7.5\ \mu\text{m}\)
B.\(75.0\ \mu\text{m}\)
C.\(300.0\ \mu\text{m}\)
D.\(750.0\ \mu\text{m}\)
查看答案詳解收起答案詳解
解題
First, determine the value of one eyepiece division (epd) at \(\times 100\) magnification: 1 stage micrometer division (smd) = \(0.1\text{ mm} = 100\ \mu\text{m}\). 8 smd = \(800\ \mu\text{m}\). Since 40 epd = 8 smd, 40 epd = \(800\ \mu\text{m}\). Therefore, 1 epd = \(800\ /\ 40 = 20\ \mu\text{m}\).
Next, adjust for the change in magnification to \(\times 400\): Since the magnification increases by a factor of 4 (from \(\times 100\) to \(\times 400\)), the actual size represented by each epd decreases by a factor of 4. 1 epd at \(\times 400\) = \(20\ \mu\text{m}\ /\ 4 = 5\ \mu\text{m}\).
Finally, calculate the actual length of the cell: The cell spans 15 epd. Actual length = \(15\ \times 5\ \mu\text{m} = 75\ \mu\text{m}\).
評分準則
Award 1 mark for the correct answer (B). - Reject other options because they represent calculation errors (e.g. failing to adjust for the \(\times 400\) magnification change).
題目 18 · multiple_choice
1 分
Which statement correctly describes the limits of resolution and magnification of a modern light microscope compared to a transmission electron microscope (TEM)?
A.The light microscope has a maximum resolution of \(200\text{ nm}\) and a maximum useful magnification of \(\times 1500\), while the TEM has a maximum resolution of \(0.5\text{ nm}\) and a maximum useful magnification of \(\times 500\,000\).
B.The light microscope has a maximum resolution of \(2.0\ \mu\text{m}\) and a maximum useful magnification of \(\times 1500\), while the TEM has a maximum resolution of \(50\text{ nm}\) and a maximum useful magnification of \(\times 250\,000\).
C.The light microscope has a maximum resolution of \(200\text{ nm}\) and a maximum useful magnification of \(\times 15\,000\), while the TEM has a maximum resolution of \(5.0\text{ nm}\) and a maximum useful magnification of \(\times 500\,000\).
D.The light microscope has a maximum resolution of \(20\text{ nm}\) and a maximum useful magnification of \(\times 1500\), while the TEM has a maximum resolution of \(0.5\text{ nm}\) and a maximum useful magnification of \(\times 5\,000\,000\).
查看答案詳解收起答案詳解
解題
A standard modern light microscope has a maximum resolution of approximately \(200\text{ nm}\) (or \(0.2\ \mu\text{m}\)) and a maximum useful magnification of \(\times 1500\). A transmission electron microscope (TEM) uses a beam of electrons with a much shorter wavelength, achieving a much smaller maximum resolution limit of about \(0.5\text{ nm}\) (or even down to \(0.2\text{ nm}\)) and a maximum useful magnification of up to \(\times 500\,000\).
評分準則
Award 1 mark for the correct statement (A). - B is incorrect because light microscope resolution is much finer than \(2.0\ \mu\text{m}\). - C is incorrect because the maximum useful magnification of a light microscope is around \(\times 1500\), not \(\times 15\,000\). - D is incorrect because light microscope resolution is limited to \(200\text{ nm}\), not \(20\text{ nm}\).
題目 19 · multiple_choice
1 分
Plant cells with a water potential of \(-600\text{ kPa}\) were placed in four different sucrose solutions (W, X, Y, and Z). The table shows the concentration of sucrose in each solution and the resulting state of the plant cells after 30 minutes.
* Solution W (\(0.1\text{ mol dm}^{-3}\)): Fully turgid * Solution X (\(0.2\text{ mol dm}^{-3}\)): Turgid * Solution Y (\(0.4\text{ mol dm}^{-3}\)): Incipient plasmolysis * Solution Z (\(0.6\text{ mol dm}^{-3}\)): Fully plasmolysed
Which statement about the water potential of the sucrose solutions is correct?
A.Solution W has a water potential of \(-600\text{ kPa}\).
B.Solution X has a water potential lower (more negative) than \(-600\text{ kPa}\).
C.Solution Y has a water potential approximately equal to \(-600\text{ kPa}\).
D.Solution Z has a water potential higher (less negative) than \(-600\text{ kPa}\).
查看答案詳解收起答案詳解
解題
Incipient plasmolysis (seen in Solution Y) is defined as the stage where the protoplast just starts to pull away from the cell wall. At this point, there is no net movement of water because the water potential of the solution is equal to the water potential of the cell. Since the cell's volume changes very little just up to the point of incipient plasmolysis, the cell's water potential remains approximately equal to its initial value of \(-600\text{ kPa}\). Therefore, Solution Y must have a water potential approximately equal to \(-600\text{ kPa}\).
評分準則
Award 1 mark for selecting C. - A is incorrect because Solution W allows water to enter, meaning its water potential is higher (less negative) than \(-600\text{ kPa}\). - B is incorrect because Solution X allows water to enter, meaning its water potential is higher (less negative) than \(-600\text{ kPa}\). - D is incorrect because Solution Z causes water loss, so its water potential must be lower (more negative) than \(-600\text{ kPa}\).
題目 20 · multiple_choice
1 分
A student tested four food samples for biological molecules. The results are shown in the table:
* Sample 1: Benedict's test (original) is blue, Benedict's test (after acid hydrolysis) is red, Biuret test is purple, Emulsion test is clear, Iodine test is yellow-brown. * Sample 2: Benedict's test (original) is red, Benedict's test (after acid hydrolysis) is red, Biuret test is pale blue, Emulsion test is cloudy, Iodine test is blue-black. * Sample 3: Benedict's test (original) is blue, Benedict's test (after acid hydrolysis) is blue, Biuret test is purple, Emulsion test is cloudy, Iodine test is yellow-brown. * Sample 4: Benedict's test (original) is red, Benedict's test (after acid hydrolysis) is red, Biuret test is purple, Emulsion test is clear, Iodine test is blue-black.
Which sample contains non-reducing sugar and protein, but no starch, lipid, or reducing sugar?
A.Sample 1
B.Sample 2
C.Sample 3
D.Sample 4
查看答案詳解收起答案詳解
解題
Let's analyze Sample 1: 1. Benedict's test (original) is blue, indicating no reducing sugars are present. 2. Benedict's test (after acid hydrolysis) is red, indicating a non-reducing sugar (such as sucrose) was hydrolyzed into reducing sugars. 3. Biuret test is purple, indicating protein is present. 4. Emulsion test is clear, indicating no lipid is present. 5. Iodine test is yellow-brown, indicating no starch is present.
This perfectly matches the required profile.
評分準則
Award 1 mark for A. - Sample 2 contains reducing sugar, lipid, and starch, but no protein. - Sample 3 contains protein and lipid, but no non-reducing sugar. - Sample 4 contains reducing sugar, protein, and starch.
題目 21 · multiple_choice
1 分
Which processes are involved in the movement of water from the soil, across the root cortex, and up to the leaves of a plant?
1. Active transport of mineral ions into xylem vessels to lower water potential. 2. Adhesion of water molecules to the cellulose walls of xylem vessels. 3. Symplastic pathway involving movement through plasmodesmata. 4. Mass flow driven by hydrostatic pressure differences between source and sink.
A.1, 2, and 3
B.1, 2, and 4
C.1, 3, and 4
D.2, 3, and 4
查看答案詳解收起答案詳解
解題
Process 1 is correct: Active transport of mineral ions into the xylem lowering the water potential is responsible for drawing water into the xylem by osmosis, creating root pressure. Process 2 is correct: Adhesion of water molecules to the polar cellulose walls of xylem vessels supports the column of water against gravity. Process 3 is correct: The symplastic pathway involves water moving from cell to cell via cytoplasm connected by plasmodesmata in the root cortex. Process 4 is incorrect: Mass flow between source and sink describes translocation of organic solutes in the phloem, not water transport up the xylem.
評分準則
Award 1 mark for option A (1, 2, and 3 only). - Reject choices containing statement 4 (B, C, and D) because source-to-sink flow is a phloem mechanism.
題目 22 · multiple_choice
1 分
Which statements describe the roles of cholesterol and glycoproteins in cell surface membranes?
1. Cholesterol regulates membrane fluidity by preventing phospholipids from packing too closely together at low temperatures. 2. Cholesterol acts as a barrier to polar molecules, preventing them from diffusing through the membrane. 3. Glycoproteins act as receptor molecules for hormones and neurotransmitters. 4. Glycoproteins help cells adhere to one another to form tissues.
A.1, 2, and 3
B.1, 3, and 4
C.2 and 3 only
D.3 and 4 only
查看答案詳解收起答案詳解
解題
Statement 1 is correct: Cholesterol inserts between fatty acid tails to maintain fluidity and prevent crystallization at low temperatures. Statement 2 is incorrect: The hydrophobic core of the phospholipid bilayer acts as the barrier to polar molecules, not cholesterol. Statement 3 is correct: Glycoproteins have carbohydrate chains that extend outward to serve as specific receptors for signaling molecules. Statement 4 is correct: Glycoproteins are key in cell-to-cell adhesion to stabilize tissue structures.
評分準則
Award 1 mark for option B (1, 3, and 4 only). - Reject options containing statement 2 (A and C) and those omitting statement 1 (D).
題目 23 · multiple_choice
1 分
During the biochemical test for a non-reducing sugar, sucrose is first heated with dilute hydrochloric acid. Sodium hydrogencarbonate is then added to the mixture before heating with Benedict's solution. What is the purpose of adding the sodium hydrogencarbonate?
A.To catalyze the hydrolysis of the glycosidic bond in sucrose.
B.To neutralize the acid, because Benedict's reagent requires alkaline conditions to work.
C.To act as a reducing agent that converts copper(II) ions to copper(I) ions.
D.To prevent the re-formation of sucrose from glucose and fructose.
查看答案詳解收起答案詳解
解題
Dilute hydrochloric acid is added to hydrolyze the glycosidic bonds of non-reducing sugars like sucrose, releasing reducing monomers (glucose and fructose). However, Benedict's reagent is highly alkaline and its chemical reaction (reduction of copper(II) to copper(I) oxide) can only occur under alkaline conditions. Therefore, sodium hydrogencarbonate (an alkali) must be added to neutralize the acid beforehand.
評分準則
Award 1 mark for option B. - A is incorrect because hydrochloric acid, not sodium hydrogencarbonate, catalyzes the hydrolysis. - C is incorrect because reducing sugars act as the reducing agents, not sodium hydrogencarbonate. - D is incorrect because neutralizing the solution does not exist to prevent re-formation of sucrose.
題目 24 · multiple_choice
1 分
A transmission electron micrograph of a plant cell shows a chloroplast with a measured length of \(4.5\text{ cm}\). The magnification of the micrograph is \(\times 15\,000\). What is the actual length of the chloroplast?
A.\(0.3\ \mu\text{m}\)
B.\(3.0\ \mu\text{m}\)
C.\(30.0\ \mu\text{m}\)
D.\(300.0\ \mu\text{m}\)
查看答案詳解收起答案詳解
解題
Using the formula: \(\text{Actual Size (A)} = \frac{\text{Image Size (I)}}{\text{Magnification (M)}}\)
1. Convert the measured image size to micrometers: \(I = 4.5\text{ cm} = 45\text{ mm} = 45\,000\ \mu\text{m}\).
2. Divide by the magnification: \(A = \frac{45\,000\ \mu\text{m}}{15\,000} = 3\ \mu\text{m}\).
評分準則
Award 1 mark for B. - A is incorrect because of an extra division by 10. - C and D are incorrect due to errors in converting cm to \(\mu\text{m}\).
題目 25 · multiple_choice
1 分
An eyepiece graticule has 100 small divisions. It is calibrated using a stage micrometer. The stage micrometer has a total length of \( 2.0\text{ mm} \) with 100 divisions. At a magnification of \( \times 100 \), 40 divisions of the eyepiece graticule coincide with 2 divisions of the stage micrometer. A student measures a mitochondrion using the same eyepiece graticule at a magnification of \( \times 400 \). The mitochondrion is 12 eyepiece graticule divisions in length. What is the actual length of the mitochondrion in micrometres (\( \mu\text{m} \))?
A.0.3
B.3.0
C.12.0
D.48.0
查看答案詳解收起答案詳解
解題
1. Find the distance of one stage micrometer division (smd): \( 2.0\text{ mm} = 2000\text{ }\mu\text{m} \). With 100 divisions, \( 1\text{ smd} = 2000 / 100 = 20\text{ }\mu\text{m} \). 2. At \( \times 100 \) magnification: 40 eyepiece divisions (epd) = 2 smd = \( 2 \times 20\text{ }\mu\text{m} = 40\text{ }\mu\text{m} \). Therefore, \( 1\text{ epd} = 40 / 40 = 1\text{ }\mu\text{m} \). 3. When magnification increases to \( \times 400 \) (which is a 4-fold increase from \( \times 100 \)), the actual distance represented by each eyepiece division decreases by a factor of 4. Therefore, at \( \times 400 \), \( 1\text{ epd} = 1\text{ }\mu\text{m} / 4 = 0.25\text{ }\mu\text{m} \). 4. The length of the mitochondrion is 12 epd. Thus, actual length = \( 12 \times 0.25\text{ }\mu\text{m} = 3.0\text{ }\mu\text{m} \).
評分準則
Award 1 mark for the correct answer (B). 1 mark is awarded for the correct sequence of calculations: calibrating the stage micrometer, finding the value of one eyepiece division at x100, adjusting the value of the eyepiece division for the x400 magnification, and calculating the final actual length.
題目 26 · multiple_choice
1 分
Which statements correctly describe the differences between a transmission electron microscope (TEM) and a scanning electron microscope (SEM)? 1. A TEM produces a three-dimensional image of the surface of a specimen, whereas an SEM produces a two-dimensional image of a thin section. 2. A TEM has a higher resolution than an SEM because the electron beam passes through the specimen. 3. Both microscopes require the specimen to be in a vacuum, meaning living cells cannot be observed. 4. An SEM relies on electrons reflected from the surface of a heavy-metal coated specimen.
A.1, 2 and 3
B.1 and 4 only
C.2, 3 and 4 only
D.2 and 3 only
查看答案詳解收起答案詳解
解題
Statement 1 is incorrect: SEM produces 3D surface images, whereas TEM produces 2D images of thin sections (the statements are reversed). Statement 2 is correct: TEM has a significantly higher resolution than SEM because the transmitted electron beam has a very short wavelength and passes through thin specimens. Statement 3 is correct: Both instruments require high vacuum conditions to operate, which dehydrates and kills living cells. Statement 4 is correct: SEM works by scanning a beam of electrons across the surface of a specimen coated with heavy metal (e.g. gold) and capturing the reflected secondary electrons to map the surface.
評分準則
Award 1 mark for identifying that statements 2, 3, and 4 are correct (C).
題目 27 · multiple_choice
1 分
A sample of an unknown carbohydrate solution was tested to identify its components. The following tests and results were recorded: - Test 1: Benedict's test was performed by heating the sample with Benedict's reagent. The mixture remained blue. - Test 2: A second sample of the solution was heated with dilute hydrochloric acid, neutralised with sodium hydrogencarbonate, and then heated with Benedict's reagent. The mixture turned brick-red. - Test 3: Iodine in potassium iodide solution was added to a third sample. The mixture remained yellow-brown. Which of the following could be the carbohydrate present in the original solution?
A.Fructose
B.Maltose
C.Starch
D.Sucrose
查看答案詳解收起答案詳解
解題
Test 1 resulted in a negative Benedict's test (remained blue), which shows that no reducing sugars are present in the original sample. This rules out fructose and maltose, which are reducing sugars. Test 2 is a positive test for non-reducing sugars (hydrolysis followed by neutralisation and Benedict's test turned brick-red), indicating that a non-reducing disaccharide like sucrose is present. Test 3 resulted in a negative iodine test (remained yellow-brown), showing that starch is not present. This rules out starch. Therefore, the carbohydrate is sucrose.
評分準則
Award 1 mark for identifying sucrose (D) as the correct carbohydrate based on the outcomes of all three biochemical tests.
題目 28 · multiple_choice
1 分
A student wanted to estimate the concentration of glucose in a solution using a semi-quantitative Benedict's test. They prepared five glucose standards of known concentrations and carried out the Benedict's test under identical conditions. To make this investigation semi-quantitative and reliable, which variables must be controlled? 1. The volume of glucose solution used in each test 2. The volume and concentration of Benedict's reagent added 3. The temperature of the water bath and the duration of heating 4. The color of the precipitate formed at the end of the heating period
A.1, 2 and 3 only
B.1, 2 and 4 only
C.1, 3 and 4 only
D.2 and 3 only
查看答案詳解收起答案詳解
解題
To ensure a fair and reliable comparison, all variables that can affect the rate or extent of the chemical reaction must be controlled (kept constant). These include the volume of the sample (1), the volume and concentration of Benedict's reagent (2), and the temperature of the water bath and duration of heating (3). Statement 4 is incorrect because the color of the precipitate is the dependent variable (the result of the test being measured/observed), not a controlled variable.
評分準則
Award 1 mark for choosing option A, identifying that variables 1, 2, and 3 must be controlled, while variable 4 is the dependent variable.
題目 29 · multiple_choice
1 分
Three pathways of membrane transport are described: - Pathway 1: Active transport of sodium ions out of a cell. - Pathway 2: Facilitated diffusion of glucose into a cell. - Pathway 3: Simple diffusion of oxygen into a cell. Which row correctly identifies the characteristics of these pathways?
A.Row A: Pathway 1: requires proteins & dependent on ATP; Pathway 2: requires proteins & not dependent on ATP; Pathway 3: does not require proteins & not dependent on ATP
B.Row B: Pathway 1: requires proteins & dependent on ATP; Pathway 2: does not require proteins & not dependent on ATP; Pathway 3: requires proteins & not dependent on ATP
C.Row C: Pathway 1: does not require proteins & not dependent on ATP; Pathway 2: requires proteins & dependent on ATP; Pathway 3: requires proteins & not dependent on ATP
D.Row D: Pathway 1: requires proteins & not dependent on ATP; Pathway 2: requires proteins & dependent on ATP; Pathway 3: does not require proteins & not dependent on ATP
查看答案詳解收起答案詳解
解題
Pathway 1 (active transport of sodium ions) requires membrane proteins (specifically, a carrier protein or pump like the sodium-potassium pump) and is directly dependent on ATP to move ions against their concentration gradient. Pathway 2 (facilitated diffusion of glucose) requires membrane proteins (carrier or channel proteins) to allow polar glucose molecules to cross, but does NOT depend on ATP because it is a passive process down a concentration gradient. Pathway 3 (simple diffusion of oxygen) does not require membrane proteins (since non-polar oxygen molecules diffuse directly through the hydrophobic core of the phospholipid bilayer) and does NOT depend on ATP.
評分準則
Award 1 mark for the correct option (A), which accurately categorizes the protein and ATP requirements for all three pathways.
題目 30 · multiple_choice
1 分
Which row correctly matches the membrane component with its primary function in the fluid mosaic model?
A.Cholesterol — regulates membrane fluidity and increases stability
B.Glycoprotein — forms a barrier to polar molecules and ions
C.Glycolipid — acts as a channel for facilitated diffusion of water
D.Phospholipid — acts as an active transport pump for sodium ions
查看答案詳解收起答案詳解
解題
Option A is correct: Cholesterol molecules fit between the fatty acid tails of phospholipids to regulate membrane fluidity and stability (preventing the membrane from becoming too fluid at high temperatures and preventing it from freezing at low temperatures). Option B is incorrect: Phospholipids, not glycoproteins, form the barrier to polar/water-soluble substances. Option C is incorrect: Glycolipids function in cell-to-cell recognition and adhesion, not as transport channels. Option D is incorrect: Phospholipids do not act as active transport pumps; this is the function of carrier proteins.
評分準則
Award 1 mark for selecting option A, which has the correct component matched with its accurate biochemical function.
題目 31 · multiple_choice
1 分
During the movement of water from the soil to the xylem of a root, water can follow different pathways. Which statement about these pathways is correct?
A.In the apoplast pathway, water moves through the cytoplasm and plasmodesmata, which offers high resistance to flow.
B.In the symplast pathway, water moves through cell walls, which is a faster route because there are no membranes to cross.
C.At the endodermis, the apoplast pathway is blocked by the Casparian strip, forcing water into the symplast pathway.
D.The Casparian strip consists of a layer of hydrophilic cellulose that increases the rate of water absorption into the xylem.
查看答案詳解收起答案詳解
解題
Option C is correct: At the endodermis, the cell walls contain a band of suberin called the Casparian strip. Suberin is a waterproof substance that blocks the apoplast pathway, forcing water to pass through the selectively permeable cell surface membrane of the endodermal cells into the symplast pathway. Option A is incorrect because the apoplast pathway is through cell walls and intercellular spaces (not cytoplasm/plasmodesmata). Option B is incorrect because the symplast pathway involves water moving through the cytoplasm and plasmodesmata. Option D is incorrect because the Casparian strip consists of hydrophobic suberin, not hydrophilic cellulose.
評分準則
Award 1 mark for identifying C as the correct statement regarding water movement pathways in the root.
題目 32 · multiple_choice
1 分
Active loading of sucrose into phloem companion cells and sieve tube elements involves several steps. Which sequence of events correctly describes this process?
A.Protons are actively pumped out of companion cells into the cell wall \(\rightarrow\) protons diffuse back into companion cells down their electrochemical gradient via co-transporter proteins \(\rightarrow\) sucrose is co-transported against its concentration gradient with the protons.
B.Sucrose is actively pumped out of companion cells \(\rightarrow\) protons diffuse back in through co-transporter proteins \(\rightarrow\) water potential inside the companion cell decreases, causing water to enter by osmosis.
C.Protons are actively pumped into the companion cells from the cell wall \(\rightarrow\) sucrose diffuses out of the companion cells into the sieve tube elements via plasmodesmata.
D.Sucrose is actively transported into the cell wall by carrier proteins \(\rightarrow\) protons diffuse out down their concentration gradient \(\rightarrow\) sucrose moves into the sieve tube elements by facilitated diffusion.
查看答案詳解收起答案詳解
解題
Active loading begins when protons (\( H^+ \) ions) are actively pumped out of the companion cells into the surrounding cell wall using energy from ATP. This creates a high concentration of protons outside the cell. Protons then diffuse back into the companion cell down their concentration gradient through a co-transporter protein. As they do so, sucrose is co-transported (carried along) against its concentration gradient. Sucrose then diffuses into the sieve tube elements via plasmodesmata.
評分準則
Award 1 mark for the correct sequence of events describing active phloem loading (A).
題目 33 · 選擇題
1 分
A student calibrated an eyepiece graticule using a stage micrometer. The stage micrometer had a scale of \(2.0\text{ mm}\) length divided into \(100\) equal divisions. At \(\times 100\) magnification, \(40\) eyepiece graticule divisions coincided with \(16\) stage micrometer divisions. The student then replaced the stage micrometer with a slide of plant tissue and observed some guard cells at \(\times 400\) magnification. One guard cell was measured to be \(30\) eyepiece graticule divisions in length. What is the actual length of the guard cell in micrometres (\(\mu\text{m}\))?
A.\(15\ \mu\text{m}\)
B.\(60\ \mu\text{m}\)
C.\(240\ \mu\text{m}\)
D.\(300\ \mu\text{m}\)
查看答案詳解收起答案詳解
解題
First, calculate the value of one stage micrometer division (smd): \(2.0\text{ mm} / 100 = 0.02\text{ mm} = 20\ \mu\text{m}\). Next, determine the calibration at \(\times 100\) magnification: \(40\text{ eyepiece divisions (epd)} = 16\text{ smd} = 16 \times 20\ \mu\text{m} = 320\ \mu\text{m}\). Therefore, \(1\text{ epd} = 320 / 40 = 8\ \mu\text{m}\). Since magnification is increased from \(\times 100\) to \(\times 400\) (a four-fold increase), the actual value represented by each eyepiece division decreases by a factor of 4: \(\text{value of 1 epd at } \times 400 = 8\ \mu\text{m} / 4 = 2\ \mu\text{m}\). The guard cell measures \(30\text{ epd}\) at \(\times 400\), so its actual length is \(30 \times 2\ \mu\text{m} = 60\ \mu\text{m}\).
評分準則
Correctly calculates the stage micrometer division to be 20 micrometres, calibrates 1 eyepiece division at x100 to be 8 micrometres, adjusts the calibration for x400 magnification to be 2 micrometres, and multiplies 30 eyepiece divisions by 2 micrometres to obtain 60 micrometres.
題目 34 · 選擇題
1 分
A student is testing a food sample of unknown composition. They perform three separate tests on samples of the solution: 1. Biuret test: remains blue. 2. Benedict's test (direct): remains blue. 3. Acid hydrolysis, neutralization, followed by Benedict's test: turns brick-red. Which biological molecules are present or absent in the food sample based on these results?
A blue Biuret test result indicates that protein is absent. A direct Benedict's test remaining blue indicates that no reducing sugars are present. Acid hydrolysis followed by neutralization breaks glycosidic bonds in non-reducing sugars (such as sucrose) into monomeric reducing sugars (glucose and fructose). The subsequent positive (brick-red) Benedict's test confirms that a non-reducing sugar was originally present in the sample. Thus, non-reducing sugar is present, while reducing sugar and protein are absent.
評分準則
Correctly interprets the negative Biuret test as indicating the absence of protein, the negative direct Benedict's test as indicating the absence of reducing sugar, and the positive acid-hydrolysed Benedict's test as indicating the presence of non-reducing sugar.
題目 35 · 選擇題
1 分
An experiment was carried out to investigate the effect of temperature on the leakage of pigment from beetroot cells. Red pigment is stored within the large central vacuole, which is surrounded by the tonoplast membrane. The cell is surrounded by the cell surface membrane. Which statement correctly explains why there is a large increase in the leakage of pigment at temperatures above \(60\ ^\circ\text{C}\)?
A.The active transport proteins in the cell surface membrane are denatured, allowing the pigment to be transported out rapidly.
B.Phospholipids in both the cell surface membrane and the tonoplast gain kinetic energy and become more fluid, while membrane proteins denature, creating gaps for pigment diffusion.
C.The water potential of the external solution becomes much lower than the vacuole, causing water and pigment to leave by osmosis.
D.Pigment molecules are broken down into smaller molecules that can freely pass through intact phospholipid bilayers.
查看答案詳解收起答案詳解
解題
At temperatures above \(60\ ^\circ\text{C}\), the high thermal energy causes phospholipids to gain kinetic energy and move more rapidly, increasing the fluidity and permeability of both the cell surface membrane and the tonoplast. Additionally, proteins embedded in these membranes denature, losing their tertiary structure and creating large gaps. This allows the red vacuolar pigment to freely diffuse out of the cell.
評分準則
Identifies that increased temperature increases the kinetic energy of phospholipids, increasing membrane fluidity and disruption, and recognizes that membrane proteins denature, disrupting the membrane structure and allowing passive diffusion of pigment.
題目 36 · 選擇題
1 分
In a transpiring plant, water moves from the soil, through the root, up the stem, and out of the leaves. Which of the following describes the correct sequence of water potentials (\(\psi\)) in the pathway of water movement from the soil to the atmosphere?
Water always moves down a water potential gradient, from a region of higher (less negative) water potential to a region of lower (more negative) water potential. Therefore, for continuous water movement, the water potential must decrease progressively along the pathway: \(\psi_{\text{soil}} > \psi_{\text{root cell}} > \psi_{\text{xylem}} > \psi_{\text{leaf cell}} > \psi_{\text{atmosphere}}\).
評分準則
Explains that water moves down a water potential gradient (from less negative to more negative) and arranges the components of the transpiration stream in the correct hierarchical sequence.
題目 37 · 選擇題
1 分
An electron micrograph of a plant cell contains a chloroplast with a measured length of \(24\text{ mm}\). The magnification of the micrograph is stated as \(\times 6000\). What is the actual length of the chloroplast in micrometres (\(\mu\text{m}\))?
A.0.25 \(\mu\text{m}\)
B.4.0 \(\mu\text{m}\)
C.40 \(\mu\text{m}\)
D.144 \(\mu\text{m}\)
查看答案詳解收起答案詳解
解題
Using the formula \(\text{Actual size } (A) = \frac{\text{Image size } (I)}{\text{Magnification } (M)}\). First convert the image size from millimetres to micrometres: \(24\text{ mm} = 24 \times 1000\ \mu\text{m} = 24,000\ \mu\text{m}\). Then, divide by the magnification: \(A = \frac{24,000\ \mu\text{m}}{6000} = 4.0\ \mu\text{m}\).
評分準則
Correctly converts the image measurement from millimetres to micrometres (24,000 micrometres) and applies the formula A = I / M to compute the actual length as 4.0 micrometres.
題目 38 · 選擇題
1 分
Triglycerides and amylose are both energy-storage molecules in living organisms. Which of the statements below correctly describes a difference between a triglyceride molecule and an amylose molecule?
B.Triglycerides are macromolecules, whereas amylose is a small monomer.
C.Triglycerides are soluble in water, whereas amylose is insoluble.
D.Triglycerides contain nitrogen and phosphorus, whereas amylose contains only carbon, hydrogen, and oxygen.
查看答案詳解收起答案詳解
解題
Triglycerides are formed from one glycerol molecule and three fatty acids linked together by ester bonds. Amylose is a linear polymer of glucose units linked by \(\alpha\)-1,4-glycosidic bonds. This is a correct difference. Option B is incorrect because amylose is a large polymer (macromolecule). Option C is incorrect because triglycerides are highly hydrophobic and insoluble. Option D is incorrect because triglycerides consist only of C, H, and O, not N or P.
評分準則
Identifies that ester bonds are present in triglycerides and glycosidic bonds are present in amylose, and distinguishes between the chemical composition and physical properties of lipids and polysaccharides.
題目 39 · 選擇題
1 分
The descriptions show three different pathways, P, Q, and R, by which substances can cross a cell surface membrane: - Pathway P: Substances pass directly through the phospholipid bilayer down a concentration gradient. - Pathway Q: Substances cross through a membrane protein, which changes shape using energy from ATP to move the substance against its concentration gradient. - Pathway R: Substances pass through a water-filled channel protein down a concentration gradient. Which pathways represent active transport, simple diffusion, and facilitated diffusion?
A.Active transport: Q; Simple diffusion: P; Facilitated diffusion: R
B.Active transport: P; Simple diffusion: Q; Facilitated diffusion: R
C.Active transport: Q; Simple diffusion: R; Facilitated diffusion: P
Pathway P involves direct movement through the bilayer down a concentration gradient, which is simple diffusion. Pathway Q uses ATP to pump substances against a concentration gradient via conformational changes in a carrier protein, representing active transport. Pathway R utilizes a channel protein for movement down a concentration gradient, representing facilitated diffusion. This matches Option A.
評分準則
Identifies Pathway P as simple diffusion, Pathway Q as active transport, and Pathway R as facilitated diffusion.
題目 40 · 選擇題
1 分
Which features of companion cells are adaptations that support the active loading of sucrose into phloem sieve tube elements? 1. Many mitochondria to produce ATP for active transport of hydrogen ions. 2. Plasmodesmata to allow passive movement of sucrose and water into the sieve tube. 3. Co-transporter proteins in the cell surface membrane to transport sucrose along with hydrogen ions. 4. Lignified walls to prevent the collapse of the companion cell under high pressure.
A.1, 2 and 3
B.1 and 3 only
C.2 and 4 only
D.1, 2, 3 and 4
查看答案詳解收起答案詳解
解題
Active loading of sucrose relies on proton pumps in the companion cell membrane pumping \(\text{H}^+\) ions into the cell wall using ATP (provided by the many mitochondria, so feature 1 is correct). The \(\text{H}^+\) ions diffuse back into the companion cell through co-transporter proteins, bringing sucrose with them (feature 3 is correct). Sucrose then diffuses into the sieve tube elements via plasmodesmata (feature 2 is correct). Lignified walls (feature 4) are found in xylem vessels and sclerenchyma, not companion cells or phloem tissue.
評分準則
Identifies that many mitochondria (1), plasmodesmata (2), and co-transporter proteins (3) are key structural adaptations of companion cells for active phloem loading, while excluding lignification (4).
卷二 AS Level 結構題
Answer all six structured questions in the spaces provided on the question paper.
6 題目 · 60 分
題目 1 · 結構題
10 分
The magnification and resolution of microscopes are fundamental to studying cell structures.
(a) Explain the difference between magnification and resolution. [3]
(b) A student calibrated an eyepiece graticule using a stage micrometer. The stage micrometer had scale divisions of 0.1 mm. At \(\times 100\) magnification, 40 divisions of the eyepiece graticule coincided with 8 divisions of the stage micrometer. (i) Calculate the value of 1 eyepiece graticule unit (epu) in micrometres (\(\mu\text{m}\)). Show your working. [2] (ii) The student then observed a plant cell under \(\times 400\) magnification using the same microscope and eyepiece graticule. The length of a chloroplast was measured as 3.5 eyepiece graticule units. Calculate the actual length of the chloroplast in \(\mu\text{m}\). Show your working. [3]
(c) State one advantage of using an electron microscope instead of a light microscope to study cell organelles. [2]
查看答案詳解收起答案詳解
解題
(a) Magnification is the number of times larger an image is compared to the actual size of the specimen, whereas resolution is the ability to distinguish between two close objects or points as separate entities. Magnification is determined by the lenses used, while resolution is determined by the wavelength of the radiation source.
(b) (i) 8 divisions of the stage micrometer = \(8 \times 0.1\text{ mm} = 0.8\text{ mm} = 800\ \mu\text{m}\). Since 40 epu = \(800\ \mu\text{m}\), 1 epu = \(800 / 40 = 20\ \mu\text{m}\). (ii) When magnification increases by a factor of 4 (from \(\times 100\) to \(\times 400\)), each epu represents a distance 4 times smaller. Thus, 1 epu at \(\times 400 = 20\ \mu\text{m} / 4 = 5\ \mu\text{m}\). Actual length of the chloroplast = \(3.5 \text{ epu} \times 5\ \mu\text{m/epu} = 17.5\ \mu\text{m}\).
(c) Electron microscopes use an electron beam which has a much shorter wavelength than light. This gives a much higher resolution (up to 0.5 nm), allowing the detailed internal ultrastructure of organelles (such as cristae in mitochondria or grana in chloroplasts) to be clearly seen.
評分準則
(a) [Max 3 marks] - Magnification is how much larger the image is than the actual specimen [1] - Resolution is the ability to distinguish between two close points as separate [1] - Resolution is dependent on the wavelength of radiation used / magnification is not [1]
(b) (i) [2 marks] - \(8 \text{ divisions} = 0.8\text{ mm}\) or \(800\ \mu\text{m}\) [1] - \(800 / 40 = 20\ \mu\text{m}\) [1] (ii) [3 marks] - Calculation of epu at \(\times 400\) as \(5\ \mu\text{m}\) (by dividing by 4) [1] - Multiply \(3.5 \times 5\) to get correct final answer [1] - Correct final answer of \(17.5\ \mu\text{m}\) with unit [1]
(c) [2 marks] - Shorter wavelength of electrons (compared to light) [1] - Higher resolution / detailed ultrastructure of organelles can be seen [1]
題目 2 · 結構題
10 分
Transmission electron microscopes (TEM) allow detailed observation of cellular organelles.
(a) Describe three structural differences between a mitochondrion and a chloroplast that can be observed using a TEM. [3]
(b) An electron micrograph of a plant mesophyll cell shows a chloroplast with a measured length of 48 mm on the print. The magnification of the micrograph is stated as \(\times 8000\). (i) Calculate the actual length of the chloroplast in micrometres (\(\mu\text{m}\)). Show your working. [2] (ii) Outline the steps required to prepare a temporary wet mount of a plant epidermal tissue to view with a light microscope. [3]
(c) Define the term 'resolution' in the context of microscopy and state the limit of resolution of a typical transmission electron microscope. [2]
查看答案詳解收起答案詳解
解題
(a) A chloroplast has internal membranes called thylakoids stacked into grana, whereas a mitochondrion has an inner membrane folded into cristae. Chloroplasts are typically larger and can contain large starch grains, whereas mitochondria do not contain starch grains. Chloroplasts have a smooth inner envelope membrane, whereas mitochondria have a folded inner membrane.
(b) (i) Formula: \(A = I / M\). Measured image size \(I = 48\text{ mm} = 48000\ \mu\text{m}\). Magnification \(M = 8000\). Actual size \(A = 48000 / 8000 = 6\ \mu\text{m}\). (ii) Peel a single layer of epidermal cells (e.g. from onion) using forceps. Place the tissue flat on the center of a clean glass slide. Add a drop of water or iodine stain. Lower a coverslip slowly at an angle using a mounted needle to avoid trapping air bubbles.
(c) Resolution is the ability to distinguish two separate points that are close together as distinct entities. The limit of resolution of a typical TEM is approximately 0.2 nm to 0.5 nm.
評分準則
(a) [Max 3 marks] - Chloroplast has thylakoids / grana / stroma whereas mitochondrion has cristae / matrix [1] - Chloroplast is larger / longer than mitochondrion [1] - Chloroplast may contain starch grains [1] - Chloroplast has smooth inner membrane / envelope whereas mitochondrion has folded inner membrane [1]
(b) (i) [2 marks] - Conversion: \(48\text{ mm} = 48000\ \mu\text{m}\) [1] - Final answer: \(6\ \mu\text{m}\) [1] (ii) [3 marks] - Use forceps to peel a thin, one-cell-thick epidermal layer [1] - Place flat on slide and add water or iodine stain [1] - Lower coverslip at an angle / use mounted needle to prevent air bubbles [1]
(c) [2 marks] - Ability to distinguish two separate points as separate entities [1] - Value in range: \(0.2\text{ to }0.5\text{ nm}\) (accept \(2 \times 10^{-10}\text{ m}\)) [1]
題目 3 · 結構題
10 分
Water movement in plants is vital for transport and support.
(a) Describe the pathways and mechanisms by which water moves across the root cortex from the root hair cells to the xylem. [4]
(b) Explain how the structure of xylem vessel elements is adapted to their function of transporting water. [3]
(c) A potometer is used to estimate transpiration rate by measuring the rate of water uptake. State two reasons why the rate of water uptake measured by a potometer might not exactly equal the rate of transpiration of the plant. [2]
(d) State the role of the Casparian strip in the endodermis of the root. [1]
查看答案詳解收起答案詳解
解題
(a) Water travels across the root cortex via two main pathways: 1. The apoplast pathway, where water moves through the cell walls and intercellular spaces by mass flow. 2. The symplast pathway, where water moves through the cytoplasm and plasmodesmata by osmosis down a water potential gradient. When water reaches the endodermis, the apoplast pathway is blocked by the waterproof Casparian strip, forcing all water into the symplast pathway.
(b) Xylem vessels are hollow tubes made of dead cells with no cytoplasm or organelles, minimizing resistance to water flow. The lack of end walls between cells allows for a continuous column of water. Additionally, the walls are reinforced with lignin, which prevents the vessel from collapsing under negative pressure/tension and provides structural support.
(c) The potometer measures water uptake, which may not equal transpiration because: 1. Some water is used by the plant cells for photosynthesis. 2. Some water is used to maintain cell turgidity (turgor pressure) and cell expansion.
(d) The Casparian strip is made of suberin (a waterproof substance) and acts as a barrier that blocks the apoplast pathway, forcing water and dissolved minerals to pass through the selectively permeable cell membrane of the endodermal cells, allowing selective uptake.
評分準則
(a) [Max 4 marks] - Apoplast pathway: water moves through cell walls / intercellular spaces [1] - Symplast pathway: water moves through cytoplasm / plasmodesmata [1] - Water moves down a water potential gradient / by osmosis in the symplast [1] - Casparian strip blocks the apoplast pathway at the endodermis [1] - Forces water to cross selectively permeable plasma membranes of endodermal cells [1]
(b) [Max 3 marks] - Lignified walls prevent collapse under negative pressure / tension [1] - Dead cells / no cytoplasm / no organelles creates a hollow lumen to allow uninterrupted flow [1] - No end walls between cells creates a continuous tube [1] - Pits allow lateral movement of water [1]
(c) [2 marks] - Water used in photosynthesis [1] - Water used to maintain turgor / cell hydration / growth [1] - Respiration produces water [1] (any two, max 2 marks)
(d) [1 mark] - Blocks the apoplast pathway / forces water into the symplast [1]
題目 4 · 結構題
10 分
Biochemical testing allows for the identification and quantification of biological molecules.
(a) Describe how a student would carry out a biochemical test to show that an unknown solution contains a non-reducing sugar, such as sucrose, but no reducing sugars. [4]
(b) In an investigation, serial dilutions of a 1.0% glucose solution were prepared to create a calibration curve using a colorimeter. (i) Describe how a student could prepare five concentration levels (0.5%, 0.25%, 0.125%, 0.0625%, and 0.03125%) from a starting solution of 1.0% glucose using a serial dilution method. [3] (ii) Explain how a colorimeter can be used to determine the concentration of reducing sugar in an unknown sample after performing a Benedict's test on standard glucose solutions. [3]
查看答案詳解收起答案詳解
解題
(a) 1. Perform a Benedict's test on the original sample by adding Benedict's reagent and heating. The sample must remain blue, confirming the absence of reducing sugars. 2. Take a fresh sample of the solution and add dilute hydrochloric acid. Heat in a boiling water bath for a few minutes to hydrolyse the glycosidic bonds of the non-reducing sugar into reducing sugars. 3. Neutralize the acid by adding sodium hydrogencarbonate until effervescence stops. 4. Add Benedict's reagent and heat again. A color change from blue to green/yellow/orange/brick-red indicates the presence of non-reducing sugar.
(b) (i) 1. Set up five test tubes and add 5 \(\text{cm}^3\) of distilled water to each. 2. Add 5 \(\text{cm}^3\) of the 1.0% glucose solution to the first tube, mix thoroughly to obtain a 0.5% solution. 3. Transfer 5 \(\text{cm}^3\) of this 0.5% solution to the second tube, mix to obtain a 0.25% solution. Repeat this 1:1 transfer sequentially for the remaining tubes to get 0.125%, 0.0625%, and 0.03125% solutions.
(ii) 1. Perform the Benedict's test on each standard solution and the unknown. Centrifuge the mixtures to pellet the red precipitate, leaving a blue supernatant. 2. Calibrate the colorimeter using a blank (distilled water) to 100% transmission / 0 absorbance using a red filter. 3. Measure the absorbance of each standard supernatant, plot a calibration curve of absorbance against concentration, and read the concentration of the unknown from the curve.
評分準則
(a) [Max 4 marks] - Perform Benedict's test on original sample first to show it remains blue [1] - Heat fresh sample with dilute acid (HCl) to hydrolyse glycosidic bonds [1] - Neutralize with alkali (sodium hydrogencarbonate) [1] - Heat with Benedict's reagent and observe change to green/yellow/orange/brick-red [1]
(b) (i) [3 marks] - Prepare tubes with equal volumes (e.g. \(5\text{ cm}^3\)) of distilled water [1] - Transfer equal volume of 1.0% glucose into first tube and mix [1] - Repeat sequential 1:1 dilution transfers for remaining concentrations [1] (ii) [3 marks] - Centrifuge reaction mixtures to obtain clear supernatant [1] - Calibrate colorimeter / set to zero absorbance / use red filter [1] - Plot calibration curve of absorbance vs concentration and read unknown concentration from curve [1]
題目 5 · 結構題
10 分
The cell-surface membrane is a dynamic barrier regulating movement in and out of cells.
(a) Outline the role of the following components in the fluid mosaic membrane: (i) Cholesterol [2] (ii) Glycoproteins [2]
(b) The permeability of beetroot cell membranes can be investigated by measuring the leakage of the red vacuolar pigment betalain. Explain the effect of increasing temperature from \(20^\circ\text{C}\) to \(70^\circ\text{C}\) on the permeability of the beetroot cell membranes. [4]
(c) Distinguish between active transport and facilitated diffusion. [2]
查看答案詳解收起答案詳解
解題
(a) (i) Cholesterol regulates membrane fluidity. At high temperatures, it stabilizes the membrane and prevents it from becoming too fluid by restricting phospholipid movement. At low temperatures, it prevents phospholipids from packing too closely and solidifying. (ii) Glycoproteins act as cell-surface receptors for signaling molecules (like hormones) and cell-recognition sites/antigens. They also help stabilize membrane structure by forming hydrogen bonds with surrounding water molecules.
(b) As temperature increases, the kinetic energy of the phospholipids increases, making the bilayer more fluid and less stable. At high temperatures (above \(50^\circ\text{C}\)), membrane proteins (such as transport proteins) denature, creating large gaps in the membrane. This loss of structural integrity of both the tonoplast and cell-surface membranes allows the betalain pigment to leak out rapidly by diffusion, causing the surrounding solution to turn dark red.
(c) Active transport requires metabolic energy in the form of ATP to move substances against their concentration gradient (from low to high concentration), whereas facilitated diffusion is a passive process that does not require ATP and moves substances down their concentration gradient (from high to low concentration) via channel or carrier proteins.
評分準則
(a) (i) [2 marks] - Regulates membrane fluidity [1] - Prevents phospholipids packing too closely at low temperatures / reduces movement at high temperatures [1] (ii) [2 marks] - Act as cell-surface receptors / cell signaling / cell recognition [1] - Stabilize membrane structure / form H-bonds with water [1]
(b) [Max 4 marks] - Increasing temperature increases kinetic energy of phospholipids [1] - Bilayer becomes more fluid / starts to disrupt [1] - Membrane proteins / transport proteins denature [1] - Loss of membrane/tonoplast integrity creates gaps [1] - Betalain pigment leaks out by diffusion down concentration gradient [1]
(c) [2 marks] - Active transport requires ATP / facilitated diffusion is passive [1] - Active transport is against concentration gradient / facilitated diffusion is down concentration gradient [1]
題目 6 · 結構題
10 分
Proteins are polymer molecules with highly specific three-dimensional structures.
(a) Describe the structure of a collagen molecule and explain how its structure relates to its high tensile strength. [4]
(b) Contrast the structure and function of collagen with that of haemoglobin. [4]
(c) Explain how a peptide bond is formed between two amino acids. [2]
查看答案詳解收起答案詳解
解題
(a) A collagen molecule consists of three polypeptide chains wound around each other to form a tight triple helix (tropocollagen). Glycine is every third amino acid in the chain; its small size (R group is a hydrogen atom) allows tight packing. Hydrogen bonds hold the three chains together. Multiple tropocollagen molecules lie parallel to each other and are linked by covalent cross-links to form fibrils, which aggregate into high-tensile-strength collagen fibres.
(b) Differences in structure and function: 1. Collagen is a fibrous, elongated protein, whereas haemoglobin is a globular, compact, spherical protein. 2. Collagen is insoluble in water, whereas haemoglobin is highly soluble. 3. Collagen consists of three polypeptide chains, whereas haemoglobin consists of four polypeptide chains. 4. Collagen has no prosthetic group, whereas haemoglobin contains four iron-containing haem groups. 5. Collagen serves a structural function, whereas haemoglobin functions to transport oxygen in the blood.
(c) A peptide bond is formed by a condensation reaction between the amine group (\(-\text{NH}_2\)) of one amino acid and the carboxyl group (\(-\text{COOH}\)) of another amino acid. This reaction results in the loss of a water molecule (\(\text{H}_2\text{O}\)) and the creation of a covalent peptide linkage (\(-\text{CONH}-\)).
評分準則
(a) [Max 4 marks] - Triple helix / 3 polypeptide chains wound together [1] - Glycine as every third amino acid allows tight packing [1] - Hydrogen bonds hold the chains together [1] - Covalent cross-links between tropocollagen molecules form strong fibrils [1]
(b) [Max 4 marks] - Collagen is fibrous/long whereas haemoglobin is globular/spherical [1] - Collagen is insoluble whereas haemoglobin is soluble [1] - Collagen has 3 chains whereas haemoglobin has 4 chains [1] - Collagen has no prosthetic group whereas haemoglobin has haem/iron groups [1] - Collagen has structural role whereas haemoglobin has oxygen transport role [1]
(c) [2 marks] - Condensation reaction with release of water [1] - Between amine (\(-\text{NH}_2\)) group of one amino acid and carboxyl (\(-\text{COOH}\)) group of another [1]
Paper 3 Advanced Practical Skills 2
Carry out all laboratory activities, record observations, perform calculations, and answer both diagnostic drawing questions.
2 題目 · 40 分
題目 1 · Practical Investigation
20 分
Practical Investigation 1: Water Potential of Plant Tissue
A student investigated the effect of different concentrations of sucrose solution on potato tuber tissue (Solanum tuberosum). The student prepared five concentrations of sucrose solution: \(0.2, 0.4, 0.6, 0.8,\) and \(1.0\text{ mol dm}^{-3}\) from a stock solution of \(1.0\text{ mol dm}^{-3}\) sucrose solution.
(a) (i) Describe how the student would prepare \(50\text{ cm}^3\) of each of the concentrations \(0.2, 0.4, 0.6,\) and \(0.8\text{ mol dm}^{-3}\) from the stock solution using a simple dilution method. Show the volumes of stock sucrose solution and distilled water required for each. [3]
The student cut potato cylinders to an initial length of \(50.0\text{ mm}\). Three potato cylinders were placed into each concentration. After 60 minutes, the final lengths of the cylinders were measured. The raw results are shown in the table below:
Sucrose concentration / \(\text{mol dm}^{-3}\)Cylinder 1 final length / \(\text{mm}\)Cylinder 2 final length / \(\text{mm}\)Cylinder 3 final length / \(\text{mm}\)0.251.552.051.50.450.550.051.00.648.549.048.00.847.047.547.01.045.546.045.5
(a) (ii) Prepare a single complete results table showing the raw final lengths, the calculated mean final length, and the calculated mean percentage change in length for each concentration. Show your working for the percentage change calculation of the \(0.2\text{ mol dm}^{-3}\) solution. [6]
(a) (iii) Describe how you would use these mean percentage changes to construct a graph to estimate the isotonic concentration of the potato tissue. State the expected shape of the graph and how the isotonic concentration would be determined from it. [4]
(a) (iv) Identify two systematic or random sources of error in this practical procedure and describe how you would modify the experiment to minimize each of these errors. [4]
(b) After the experiment, the student wanted to test the liquid remaining in the test tubes containing potato cylinders to confirm if any non-reducing sugars had leaked from the cells. Describe a biochemical test procedure that would confirm the presence of non-reducing sugars in these samples. [3]
(a) (ii) Table of results showing mean final length and mean percentage change in length:
Sucrose concentration / \(\text{mol dm}^{-3}\)Cylinder 1 final length / \(\text{mm}\)Cylinder 2 final length / \(\text{mm}\)Cylinder 3 final length / \(\text{mm}\)Mean final length / \(\text{mm}\)Mean percentage change in length / %0.251.552.051.551.7+3.40.450.550.051.050.5+1.00.648.549.048.048.5-3.00.847.047.547.047.2-5.61.045.546.045.545.7-8.6 Working for \(0.2\text{ mol dm}^{-3}\): \(\text{Mean length} = (51.5 + 52.0 + 51.5) / 3 = 51.7\text{ mm}\). \(\text{Percentage change} = [(\text{Mean Final Length} - \text{Initial Length}) / \text{Initial Length}] \times 100\) \(\text{Percentage change} = [(51.7 - 50.0) / 50.0] \times 100 = (1.7 / 50.0) \times 100 = +3.4\%\).
(a) (iii) Plot a graph with sucrose concentration on the x-axis and mean percentage change in length on the y-axis. Draw a line of best fit that passes through both positive and negative values of percentage change. The curve/line will cross the x-axis (0% change in length). This intercept corresponds to the isotonic concentration of the potato tissue (approx. \(0.45\text{ mol dm}^{-3}\)), where there is no net movement of water by osmosis.
(a) (iv) Errors and Improvements: 1. Error: Evaporation of water from the open tubes, increasing the concentration of sucrose solutions during the experiment. Improvement: Cover the test tubes with rubber bungs or aluminium foil during the 60-minute incubation. 2. Error: Excess surface liquid on the potato cylinders when measuring final length, adding external length/mass or causing slipping errors. Improvement: Gently blot the cylinders dry with a paper towel using a standardized procedure before measuring.
(b) Non-reducing sugar test: 1. Heat the sample with dilute hydrochloric acid in a boiling water bath for 2 minutes to hydrolyse the glycosidic bonds. 2. Cool and neutralize the acid by adding sodium hydrogencarbonate until effervescence ceases. 3. Add Benedict's reagent and heat in a boiling water bath for 5 minutes. A colour change from blue to green/yellow/orange/red indicates the presence of non-reducing sugars.
評分準則
(a) (i) [3 marks total] - 1 mark for calculating correct volumes of stock sucrose solution (40, 30, 20, 10 cm^3). - 1 mark for calculating correct volumes of distilled water (10, 20, 30, 40 cm^3). - 1 mark for expressing the dilution ratio or using the dilution formula properly.
(a) (ii) [6 marks total] - 1 mark for a single, complete table with appropriate column headers and units: 'Sucrose concentration / mol dm-3', 'Length / mm', 'Mean final length / mm', and 'Mean percentage change / %'. (Accept forward slash but reject brackets containing units like (mm) if slashes are not used consistently). - 1 mark for calculating all mean final lengths correctly (51.7, 50.5, 48.5, 47.2, 45.7). - 1 mark for calculating all mean percentage changes correctly (+3.4%, +1.0%, -3.0%, -5.6%, -8.6%) with '+' and '-' signs to indicate gain or loss. - 1 mark for showing correct working for 0.2 mol dm-3 calculation. - 1 mark for writing all raw data and calculated values consistently to appropriate decimal places (raw lengths to 1 d.p., means to 1 d.p., percentage changes to 1 d.p.). - 1 mark for clean presentation of raw and processed values without units inside individual table cells.
(a) (iii) [4 marks total] - 1 mark for stating that sucrose concentration is plotted on the x-axis and percentage change in length on the y-axis. - 1 mark for describing the shape: a curve or straight line sloping downwards from positive to negative values. - 1 mark for stating that the isotonic concentration is the x-intercept / the point where percentage change is zero. - 1 mark for explaining that at this point, water potential inside and outside the cells is equal, resulting in no net water movement.
(a) (iv) [4 marks total] - 1 mark for identifying a valid error (e.g., evaporation of solution, excess surface fluid, variations in cylinder diameter). - 1 mark for matching sensible improvement (e.g., cover tubes with foil, standardized blotting technique, using a single cork borer and measuring with callipers). - 1 mark for identifying a second valid error. - 1 mark for matching second sensible improvement.
(b) [3 marks total] - 1 mark for heating the solution with dilute hydrochloric acid (HCl) to hydrolyse non-reducing sugars into reducing sugars. - 1 mark for neutralizing the acid using sodium hydrogencarbonate (NaHCO3) before adding Benedict's reagent. - 1 mark for heating the mixture with Benedict's reagent and observing a color change from blue to green/yellow/orange/red.
題目 2 · Practical Investigation
20 分
Practical Investigation 2: Microscope Studies and Tissue Structures
Figure 2.1 shows a transverse section through a typical dicotyledonous stem under a light microscope.
(a) (i) Describe the essential rules you must follow when drawing a low-power plan diagram of this transverse section to represent the tissue distribution of a single vascular bundle and its surrounding tissues. Your response should outline what features must be included and what must be avoided. [5]
(a) (ii) Describe how you would prepare a high-power drawing of a small group of three adjacent xylem vessel elements from the center of a vascular bundle. What key features of the cell walls and lumens must be clearly illustrated? [4]
(b) (i) A student calibrated an eyepiece graticule using a stage micrometer. The stage micrometer scale has divisions spaced exactly \(0.1\text{ mm}\) apart. Using a \(\times 40\) objective lens, the student observed that \(40\) eyepiece graticule units aligned perfectly with \(1\) division of the stage micrometer scale. Calculate the actual value of one eyepiece graticule unit under this magnification in micrometres (\(\mu\text{m}\)). Show your working. [3]
(b) (ii) Using the same magnification, the student measured the internal diameter of a large xylem vessel element. The diameter was recorded as \(16.5\) eyepiece graticule units. Calculate the actual internal diameter of the xylem vessel element in micrometres (\(\mu\text{m}\)). Show your working. [2]
(c) State three structural differences between xylem vessel elements and phloem sieve tube elements that can be observed using a light microscope. Present your answer in a comparative table. [6]
查看答案詳解收起答案詳解
解題
(a) (i) Rules for a low-power plan diagram: 1. Draw only tissue boundaries; individual cells must NOT be drawn. 2. Use a sharp HB pencil to draw clean, continuous lines without sketching, gaps, or overlapping lines. 3. Ensure correct proportions of the tissues (e.g., relative thickness of epidermis, cortex, and vascular bundle). 4. Do not shade or color any part of the drawing. 5. Draw straight, uncrossed label lines using a ruler, pointing directly to the tissues (e.g., epidermis, cortex, xylem, phloem).
(a) (ii) Rules for a high-power cellular drawing: 1. Draw only the specified number of cells (exactly three adjacent xylem elements). 2. Draw double lines for cell walls to show their relative thickness. 3. Keep lumens completely empty (no cytoplasm, nuclei, or organelles, as xylem vessels are dead, hollow tubes). 4. Show correct polygonal/angular shapes and the relative sizes of the cells where they touch.
FeatureXylem vessel elementsPhloem sieve tube elementsCell wall thickness & compositionThick, lignified cell wallsThin, non-lignified cellulose wallsEnd wallsNo end walls (open, continuous tubes)Sieve plates with sieve pores presentCell lumen contentsCompletely hollow (no cytoplasm or organelles)Peripheral cytoplasm with no nucleus or ribosomes
評分準則
(a) (i) [5 marks total] - 1 mark for stating that individual cells must not be drawn (only tissue outlines). - 1 mark for specifying the use of single, sharp, continuous pencil lines with no gaps or overlapping. - 1 mark for stating that no shading or coloring is allowed. - 1 mark for requiring correct proportions and relative sizes of tissue layers. - 1 mark for describing clear, non-crossing label lines drawn with a ruler, pointing directly to the tissue layer.<
(a) (ii) [4 marks total] - 1 mark for drawing only the requested number of cells (exactly three adjacent cells). - 1 mark for drawing double lines for cell walls to represent thickness. - 1 mark for leaving the lumen of the cells completely empty. - 1 mark for representing the characteristic polygonal/angular shapes of xylem vessels accurately where they contact each other.
(b) (i) [3 marks total] - 1 mark for converting stage micrometer division to micrometres: \(0.1\text{ mm} = 100\ \mu\text{m}\). - 1 mark for dividing stage micrometer size by the number of graticule units (\(100 / 40\)). - 1 mark for the correct final answer of \(2.5\ \mu\text{m}\) with appropriate units.
(b) (ii) [2 marks total] - 1 mark for multiplying the measured units by the calibration factor: \(16.5 \times 2.5\). - 1 mark for the correct final answer of \(41.25\ \mu\text{m}\) (accept \(41.3\ \mu\text{m}\)) with units.
(c) [6 marks total] - 1 mark for constructing a structured table with clear comparison headings. - 1 mark for comparing cell wall thickness/lignification (thick/lignified in xylem vs. thin/cellulose in phloem). - 1 mark for comparing end walls (no end walls in xylem vs. sieve plates in phloem). - 1 mark for comparing cell lumen contents (completely empty/dead in xylem vs. peripheral cytoplasm/living in phloem). - 1 mark for comparing companion cells (absent in xylem vs. present next to phloem sieve tubes). - 1 mark for comparative statements being aligned directly opposite each other in the table rows.
想知道自己有幾分把握？
Thinka 是 DSE 學生用的 AI 練習應用程式，有無限量練習題、即時自動批改和詳細解題步驟。逾 100,000 名學生用它確認自己真的識，而不只是「以為識」。