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Secondary 4 Pure Biology Plant Biology Quiz
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Questions
Secondary 4 Pure Biology Quiz - Plant Biology
Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50
Duration: 45 minutes Total Marks: 50
Instructions:
- Answer ALL questions in the spaces provided.
- Marks are indicated in brackets [ ].
- Where calculations are required, show all working clearly.
- Diagrams should be drawn with a sharp pencil and clearly labelled.
Section A: Short Answer Questions (10 marks)
Answer all questions in this section.
1. Name the cells in the root epidermis that are primarily responsible for the absorption of water and mineral ions from the soil. [1 mark]
2. State the function of xylem tissue in flowering plants. [1 mark]
3. Write the word equation for photosynthesis. [2 marks]
4. Define the term transpiration. [2 marks]
5. Name the two types of vascular tissue found in flowering plants and state the main substance transported by each. [4 marks]
(i) Tissue: _________________ Substance transported: _________________
(ii) Tissue: _________________ Substance transported: _________________
Section B: Structured Questions (24 marks)
Answer all questions in this section.
6. The diagram below shows a cross-section of a dicotyledonous leaf as seen under a light microscope.
[Diagram showing leaf cross-section with labels A, B, C, D:
A - upper epidermis
B - palisade mesophyll
C - spongy mesophyll
D - lower epidermis with stoma]
(a) Identify the tissue layer labelled B and explain how its cells are adapted for photosynthesis. [3 marks]
(b) Structure D contains specialised cells that control gas exchange. Name these cells and describe how they regulate the opening and closing of stomata. [4 marks]
7. A student investigated the effect of light intensity on the rate of transpiration in a leafy shoot using a potometer. The results are shown in the table below.
| Light intensity (arbitrary units) | Rate of water uptake (cm³ per hour) |
|---|---|
| 0 (darkness) | 0.5 |
| 2 | 1.2 |
| 4 | 2.1 |
| 6 | 3.0 |
| 8 | 3.8 |
| 10 | 4.5 |
(a) Using the data in the table, describe the relationship between light intensity and the rate of water uptake. [2 marks]
(b) Explain why increasing light intensity leads to an increase in the rate of transpiration. [3 marks]
(c) The student also measured the rate of water uptake at a light intensity of 12 arbitrary units and obtained a value of 4.6 cm³ per hour. Suggest why the rate of increase in water uptake slows down at higher light intensities. [2 marks]
8. A plant was placed in a solution with a lower water potential than its cell sap.
(a) Describe and explain what would happen to the plant cells under these conditions. [3 marks]
(b) State the term used to describe the condition of a plant cell when the cell membrane pulls away from the cell wall. [1 mark]
(c) Explain why this condition is usually reversible in plant cells but not in animal cells. [2 marks]
9. The graph below shows the rate of photosynthesis in a plant at different carbon dioxide concentrations, measured at two different temperatures (20°C and 30°C).
[Graph description:
X-axis: Carbon dioxide concentration (%)
Y-axis: Rate of photosynthesis (arbitrary units)
Two curves shown:
- 20°C: rate increases with CO₂ concentration, then plateaus at 0.08% CO₂
- 30°C: rate increases more steeply, plateaus at a higher rate at 0.10% CO₂]
(a) Identify the limiting factor for photosynthesis at point X on the 20°C curve, where the rate has stopped increasing despite rising CO₂ concentration. Explain your answer. [2 marks]
(b) Explain why the rate of photosynthesis is higher at 30°C than at 20°C at any given CO₂ concentration below the plateau. [2 marks]
10. A farmer grows tomatoes in a greenhouse. The farmer can control the temperature, light intensity, and carbon dioxide concentration inside the greenhouse. The table below shows the effect of different combinations of these factors on the yield of tomatoes over a growing season.
| Condition | Temperature (°C) | Light intensity (lux) | CO₂ concentration (ppm) | Yield (kg per plant) |
|---|---|---|---|---|
| A | 20 | 10,000 | 400 | 2.5 |
| B | 20 | 20,000 | 400 | 3.8 |
| C | 25 | 20,000 | 400 | 4.6 |
| D | 25 | 20,000 | 1,000 | 6.2 |
| E | 30 | 20,000 | 1,000 | 5.8 |
(a) Compare the yields obtained under conditions B and C. Explain the difference in yield with reference to the factors that affect photosynthesis. [3 marks]
(b) Explain why the yield under condition D is higher than under condition C. [2 marks]
(c) Suggest why the yield under condition E is lower than under condition D, despite having the same light intensity and CO₂ concentration. [2 marks]
(d) The farmer wants to maximise profit. Explain why the farmer might choose condition D rather than condition E, even though both produce high yields. [2 marks]
Section C: Data Analysis and Extended Response (16 marks)
Answer all questions in this section.
11. Explain why most living organisms, including humans, depend on photosynthesis for their survival. In your answer, you should refer to the products of photosynthesis and their roles in sustaining life. [7 marks]
12. A student placed a potted plant on a balance and measured the mass over several hours under different environmental conditions. The results are shown in the table below.
| Condition | Mass loss (g) after 2 hours |
|---|---|
| Still air, room light | 1.2 |
| Moving air (fan), room light | 2.8 |
| Still air, bright light | 2.1 |
| Moving air (fan), bright light | 4.5 |
(a) Explain why the mass loss is greater in moving air compared to still air under the same light condition. [2 marks]
(b) Explain why the mass loss is greater in bright light compared to room light under the same air condition. [2 marks]
(c) The student concluded that the mass loss was mainly due to transpiration. Suggest how the student could modify the experiment to verify this conclusion. [2 marks]
13. A cross-section of a root was examined under a microscope. The diagram below shows the arrangement of tissues.
[Diagram showing root cross-section with labels:
E - root hair
F - cortex
G - endodermis
H - xylem
I - phloem]
(a) State the function of the root hair labelled E. [1 mark]
(b) Explain how the structure of the root hair cell is adapted for its function. [2 marks]
(c) Describe the pathway taken by water from the root hair to the xylem vessel labelled H. [3 marks]
14. A student investigated the effect of temperature on the rate of photosynthesis in an aquatic plant by counting the number of oxygen bubbles produced per minute. The results are shown in the graph below.
[Graph description:
X-axis: Temperature (°C) from 0 to 50
Y-axis: Rate of photosynthesis (bubbles per minute)
Curve shows: rate increases from 0°C to 25°C, peaks at 25°C, then declines sharply from 30°C to 45°C, reaching zero at 50°C]
(a) Describe the trend shown in the graph. [2 marks]
(b) Explain why the rate of photosynthesis increases as temperature rises from 0°C to 25°C. [2 marks]
(c) Explain why the rate of photosynthesis decreases rapidly at temperatures above 30°C. [2 marks]
15. A variegated leaf (a leaf with green and white patches) was destarched and then exposed to sunlight for several hours. The leaf was then tested for starch using the iodine test.
(a) State the colour change observed when iodine solution is added to a part of the leaf that contains starch. [1 mark]
(b) Predict and explain the results of the iodine test on the green and white patches of the leaf. [3 marks]
(c) Explain why the leaf was destarched before the experiment. [2 marks]
Section D: Extended Response (10 marks)
Answer all questions in this section.
16. Describe the process of water uptake and transport from the soil to the leaves of a flowering plant. In your answer, explain the roles of root pressure, capillary action, and transpiration pull. [6 marks]
17. A student set up an experiment using four test tubes, each containing the same volume of hydrogen carbonate indicator solution and a leaf of the same size. The indicator is red at equilibrium, yellow in high carbon dioxide concentrations, and purple in low carbon dioxide concentrations. The test tubes were placed under different conditions for one hour.
| Test tube | Condition | Colour of indicator after 1 hour |
|---|---|---|
| 1 | Leaf in bright light | Purple |
| 2 | Leaf in darkness | Yellow |
| 3 | No leaf, bright light | Red |
| 4 | Leaf in dim light | Orange-red |
(a) Explain the colour change observed in test tube 1. [2 marks]
(b) Explain the colour change observed in test tube 2. [2 marks]
(c) State the purpose of test tube 3 in this experiment. [1 mark]
(d) Suggest why the indicator in test tube 4 turned orange-red instead of purple or yellow. [1 mark]
18. Compare the structure and function of xylem and phloem tissues in flowering plants. [4 marks]
19. A farmer noticed that plants in a waterlogged field were wilting even though the soil was saturated with water. Explain why the plants were wilting. [2 marks]
20. Explain the importance of nitrate ions and magnesium ions for healthy plant growth. [2 marks]
END OF QUIZ
Check your answers carefully before submitting.
Answers
Secondary 4 Pure Biology Quiz - Plant Biology - ANSWER KEY
Total Marks: 50
Section A: Short Answer Questions (10 marks)
1. Name the cells in the root epidermis that are primarily responsible for the absorption of water and mineral ions from the soil. [1 mark]
Answer: Root hair cells. [1 mark]
Marking note: Accept "root hair cell" (singular). Do not accept "root cap cells" or "epidermal cells" alone.
2. State the function of xylem tissue in flowering plants. [1 mark]
Answer: Xylem transports water and dissolved mineral ions from the roots to the stems and leaves. [1 mark]
Marking note: Must mention both water AND mineral ions for full credit. Accept "transports water and minerals" or equivalent phrasing.
3. Write the word equation for photosynthesis. [2 marks]
Answer: Carbon dioxide + Water → Glucose + Oxygen [1 mark for correct reactants, 1 mark for correct products]
Marking note: The equation must include "light energy" and "chlorophyll" written above/below the arrow for full marks, OR these can be stated separately. Accept:
- Carbon dioxide + Water --(light, chlorophyll)--> Glucose + Oxygen
- Award 1 mark if reactants are correct but products incomplete/incorrect.
- Award 1 mark if products are correct but reactants incomplete/incorrect.
4. Define the term transpiration. [2 marks]
Answer: Transpiration is the loss of water vapour [1 mark] from the aerial parts of a plant, mainly through the stomata of the leaves [1 mark].
Marking note: Must specify "water vapour" (not just "water") and mention stomata or leaves. Accept "evaporation of water from the leaves/stomata."
5. Name the two types of vascular tissue found in flowering plants and state the main substance transported by each. [4 marks]
Answer: (i) Tissue: Xylem [1 mark] Substance transported: Water and dissolved mineral ions [1 mark]
(ii) Tissue: Phloem [1 mark] Substance transported: Sucrose / amino acids / food substances / products of photosynthesis [1 mark]
Marking note: For phloem, accept "sucrose," "food," "organic substances," "products of photosynthesis," or "assimilates." Do not accept "water" for phloem. Award 1 mark for each correct tissue name and 1 mark for each correct substance.
Section B: Structured Questions (24 marks)
6. (a) Identify the tissue layer labelled B and explain how its cells are adapted for photosynthesis. [3 marks]
Answer:
- Tissue B is the palisade mesophyll [1 mark].
- Adaptations:
- Cells are elongated/columnar and closely packed, containing many chloroplasts to maximise light absorption [1 mark].
- Chloroplasts can move within cells to position themselves for optimal light capture / cells are arranged vertically to allow light to penetrate deeper [1 mark].
Marking note: Award 1 mark for correct identification. Award up to 2 marks for adaptations (1 mark per valid adaptation with explanation). Accept: "large number of chloroplasts," "cells arranged perpendicular to leaf surface," "chloroplasts can rotate/move."
(b) Structure D contains specialised cells that control gas exchange. Name these cells and describe how they regulate the opening and closing of stomata. [4 marks]
Answer:
- The specialised cells are guard cells [1 mark].
- Mechanism of stomatal opening and closing:
- During the day/in light, guard cells carry out photosynthesis, producing glucose/solutes [1 mark]. This lowers the water potential inside the guard cells.
- Water enters the guard cells by osmosis from surrounding epidermal cells, causing the guard cells to become turgid and swell [1 mark]. Due to the uneven thickening of the guard cell walls (thinner outer wall, thicker inner wall), the swelling causes the guard cells to curve/bow apart, opening the stoma.
- At night/in darkness, photosynthesis stops, solutes are used up or transported out, water potential increases, water leaves guard cells by osmosis, guard cells become flaccid, and the stoma closes [1 mark].
Marking note: Award 1 mark for naming guard cells. Award up to 3 marks for the explanation. Key points: solute accumulation (potassium ions also acceptable), osmotic water entry, turgidity causing opening due to uneven wall thickness, and reversal at night. Accept reference to potassium ion (K⁺) pump mechanism.
7. (a) Using the data in the table, describe the relationship between light intensity and the rate of water uptake. [2 marks]
Answer: As light intensity increases, the rate of water uptake increases [1 mark]. The rate of increase is greater at lower light intensities and becomes smaller at higher light intensities (the relationship is not linear; the rate of increase slows down) [1 mark].
Marking note: Must describe the trend (positive correlation) AND note that the rate of increase diminishes. Award 1 mark for stating the positive relationship and 1 mark for noting the decreasing gradient/slowing rate of increase.
(b) Explain why increasing light intensity leads to an increase in the rate of transpiration. [3 marks]
Answer:
- Higher light intensity causes the stomata to open wider / more stomata to open [1 mark].
- This increases the diffusion gradient for water vapour between the inside of the leaf (air spaces) and the external atmosphere [1 mark].
- As a result, more water vapour diffuses out of the leaf through the open stomata, increasing the rate of transpiration [1 mark].
Marking note: Must link light → stomatal opening → increased water vapour loss. Accept reference to increased temperature from light increasing evaporation rate, but stomatal opening is the primary mechanism. Award marks for logical sequence.
(c) The student also measured the rate of water uptake at a light intensity of 12 arbitrary units and obtained a value of 4.6 cm³ per hour. Suggest why the rate of increase in water uptake slows down at higher light intensities. [2 marks]
Answer: At higher light intensities, the stomata are already fully/mostly open [1 mark], so further increases in light intensity cannot cause significantly more stomatal opening. The rate of transpiration becomes limited by other factors such as humidity, wind speed, or temperature [1 mark].
Marking note: Accept any valid limiting factor explanation. Key concept: stomata reach maximum aperture, so transpiration rate plateaus. Accept "stomata are already fully open" or "other factors become limiting."
8. (a) Describe and explain what would happen to the plant cells under these conditions. [3 marks]
Answer:
- The solution has a lower water potential than the cell sap, so water moves out of the plant cells by osmosis [1 mark] down the water potential gradient.
- The vacuole shrinks and the cytoplasm shrinks away from the cell wall [1 mark].
- The cells become plasmolysed/flaccid, and the plant wilts [1 mark].
Marking note: Must mention osmosis, direction of water movement (out of cells), and the outcome (plasmolysis/wilting). Award 1 mark for each key point.
(b) State the term used to describe the condition of a plant cell when the cell membrane pulls away from the cell wall. [1 mark]
Answer: Plasmolysis [1 mark].
Marking note: Accept "plasmolysed." Do not accept "crenation" (this applies to animal cells).
(c) Explain why this condition is usually reversible in plant cells but not in animal cells. [2 marks]
Answer:
- Plant cells have a rigid cell wall that prevents the cell from bursting or collapsing completely [1 mark]. When placed back in a solution with higher water potential, water re-enters by osmosis, and the cell returns to its turgid state.
- Animal cells lack a cell wall. In a solution with lower water potential, animal cells lose water, shrink, and become crenated. If water loss is severe, the cell membrane may be damaged irreversibly, leading to cell death [1 mark].
Marking note: Must contrast the presence/absence of a cell wall and the consequences. Award 1 mark for plant cell explanation (cell wall allows reversibility) and 1 mark for animal cell explanation (no cell wall, irreversible damage).
9. (a) Identify the limiting factor for photosynthesis at point X on the 20°C curve, where the rate has stopped increasing despite rising CO₂ concentration. Explain your answer. [2 marks]
Answer: The limiting factor is temperature [1 mark]. At point X, increasing CO₂ concentration no longer increases the rate of photosynthesis, indicating that CO₂ is no longer limiting. The low temperature (20°C) limits the rate of enzyme-catalysed reactions in the Calvin cycle/light-independent stage, so temperature becomes the limiting factor [1 mark].
Marking note: Must identify temperature and explain that CO₂ is no longer limiting, so another factor (temperature) must be limiting. Accept "light intensity" only if justified with reference to the graph, but temperature is the more direct answer given the two curves.
(b) Explain why the rate of photosynthesis is higher at 30°C than at 20°C at any given CO₂ concentration below the plateau. [2 marks]
Answer: At 30°C, the enzymes involved in photosynthesis (e.g., Rubisco in the Calvin cycle) have higher kinetic energy [1 mark], leading to more frequent and successful enzyme-substrate collisions, thus increasing the rate of reaction [1 mark]. The temperature is closer to the optimum for photosynthetic enzymes.
Marking note: Must refer to enzyme activity and kinetic energy. Award 1 mark for mentioning enzymes/kinetic energy and 1 mark for linking to increased reaction rate.
10. (a) Compare the yields obtained under conditions B and C. Explain the difference in yield with reference to the factors that affect photosynthesis. [3 marks]
Answer: Condition C (25°C) has a higher yield (4.6 kg) than condition B (20°C, 3.8 kg) [1 mark]. Both have the same light intensity and CO₂ concentration, so the difference is due to temperature. At 25°C, the enzymes involved in photosynthesis are closer to their optimum temperature [1 mark], so the rate of photosynthesis is higher, leading to more glucose production and greater yield [1 mark].
Marking note: Must compare the yields and identify temperature as the differing factor. Award 1 mark for comparison, 1 mark for linking to enzyme activity/optimum temperature, and 1 mark for linking to increased photosynthesis/yield.
(b) Explain why the yield under condition D is higher than under condition C. [2 marks]
Answer: Condition D has a higher CO₂ concentration (1,000 ppm) compared to condition C (400 ppm) [1 mark]. CO₂ is a substrate for photosynthesis; increasing its concentration increases the rate of carbon fixation in the Calvin cycle, leading to a higher rate of photosynthesis and greater yield [1 mark].
Marking note: Must identify CO₂ as the differing factor and explain its role as a substrate/reactant in photosynthesis.
(c) Suggest why the yield under condition E is lower than under condition D, despite having the same light intensity and CO₂ concentration. [2 marks]
Answer: Condition E has a higher temperature (30°C) compared to condition D (25°C) [1 mark]. At 30°C, the temperature may be above the optimum for some photosynthetic enzymes, causing them to denature partially or function less efficiently, reducing the rate of photosynthesis and yield [1 mark]. Alternatively, higher temperature increases photorespiration, reducing net photosynthesis.
Marking note: Must identify temperature as the differing factor and explain that it exceeds the optimum, leading to enzyme denaturation/reduced activity. Accept reference to photorespiration.
(d) The farmer wants to maximise profit. Explain why the farmer might choose condition D rather than condition E, even though both produce high yields. [2 marks]
Answer: Condition D (25°C) requires less energy for heating compared to condition E (30°C) [1 mark], reducing energy costs. The yield under condition D (6.2 kg) is higher than under condition E (5.8 kg), so condition D provides both higher yield and lower operating costs, maximising profit [1 mark].
Marking note: Must discuss cost (energy/heating) and/or yield difference. Award 1 mark for identifying lower heating costs and 1 mark for linking to higher profit/higher yield.
Section C: Data Analysis and Extended Response (16 marks)
11. Explain why most living organisms, including humans, depend on photosynthesis for their survival. In your answer, you should refer to the products of photosynthesis and their roles in sustaining life. [7 marks]
Answer: Photosynthesis produces two essential products: glucose and oxygen.
- Glucose is the primary source of chemical energy for plants and, indirectly, for all other organisms in the food chain. Plants use glucose for respiration to release energy for growth and other life processes. Excess glucose is converted into starch for storage, cellulose for cell walls, and other organic compounds like proteins and lipids. Herbivores obtain glucose by eating plants, and carnivores obtain it by eating herbivores. Thus, photosynthesis is the ultimate source of energy for nearly all life.
- Oxygen is released as a by-product of photosynthesis. It is essential for aerobic respiration in most organisms, including plants, animals, and decomposers. Aerobic respiration releases energy from glucose efficiently. Without photosynthesis, atmospheric oxygen would be depleted, and aerobic life could not be sustained.
- Additionally, photosynthesis removes carbon dioxide from the atmosphere, helping to regulate the Earth's climate and maintaining the carbon cycle. The organic matter produced by photosynthesis also forms the basis of fossil fuels, which humans depend on for energy.
Marking note: Award marks for: (1) identifying glucose and oxygen as products; (2) explaining glucose as an energy source/food for plants and other organisms; (3) explaining oxygen's role in aerobic respiration; (4) linking to food chains/energy flow; (5) mentioning carbon dioxide removal/climate regulation; (6) clear, logical structure; (7) use of appropriate biological terminology. Maximum 7 marks.
12. (a) Explain why the mass loss is greater in moving air compared to still air under the same light condition. [2 marks]
Answer: Moving air (wind) removes water vapour from the area surrounding the leaf/stomata [1 mark], maintaining a steep diffusion gradient for water vapour between the leaf's air spaces and the external atmosphere. This increases the rate of transpiration, leading to greater mass loss [1 mark].
Marking note: Must mention diffusion gradient/concentration gradient. Award 1 mark for removing water vapour and 1 mark for maintaining a steep gradient/increasing transpiration rate.
(b) Explain why the mass loss is greater in bright light compared to room light under the same air condition. [2 marks]
Answer: Bright light causes the stomata to open wider [1 mark], reducing the resistance to water vapour diffusion out of the leaf. This increases the rate of transpiration, leading to greater mass loss [1 mark].
Marking note: Must link light to stomatal opening and increased transpiration. Accept reference to increased temperature from bright light increasing evaporation.
(c) The student concluded that the mass loss was mainly due to transpiration. Suggest how the student could modify the experiment to verify this conclusion. [2 marks]
Answer: The student could cover the leaves with a waterproof substance (e.g., petroleum jelly) to block the stomata [1 mark] and repeat the experiment. If the mass loss is significantly reduced, it confirms that transpiration through the stomata was the main cause of mass loss [1 mark]. Alternatively, use a control setup without a plant to measure evaporation from the soil.
Marking note: Must suggest a method to prevent transpiration (e.g., blocking stomata, removing leaves) and explain how the results would verify the conclusion. Award 1 mark for the method and 1 mark for the expected outcome/explanation.
13. (a) State the function of the root hair labelled E. [1 mark]
Answer: To absorb water and dissolved mineral ions from the soil. [1 mark]
Marking note: Must mention both water and mineral ions. Accept "absorption of water and minerals."
(b) Explain how the structure of the root hair cell is adapted for its function. [2 marks]
Answer:
- The root hair cell has a long, narrow extension (the root hair) that increases the surface area to volume ratio [1 mark], maximising the rate of absorption of water and mineral ions.
- The cell membrane contains many carrier proteins/transport proteins for active transport of mineral ions [1 mark]. / The cell has a large vacuole with a lower water potential to facilitate water uptake by osmosis.
Marking note: Award 1 mark for surface area adaptation and 1 mark for any other valid adaptation (e.g., thin cell wall, many mitochondria for active transport, large vacuole).
(c) Describe the pathway taken by water from the root hair to the xylem vessel labelled H. [3 marks]
Answer: Water enters the root hair cell by osmosis [1 mark]. It then moves across the cortex cells (through the cell walls/apoplast pathway, cytoplasm/symplast pathway, or vacuolar pathway) [1 mark] until it reaches the endodermis. The Casparian strip in the endodermis blocks the apoplast pathway, forcing water into the symplast pathway. Water then moves into the xylem vessels [1 mark].
Marking note: Must mention osmosis, movement through cortex, and entry into xylem. Award 1 mark for each key stage. Accept reference to apoplast/symplast/vacuolar pathways.
14. (a) Describe the trend shown in the graph. [2 marks]
Answer: The rate of photosynthesis increases as temperature rises from 0°C to 25°C [1 mark], reaching a maximum at 25°C. Beyond 25°C, the rate decreases sharply, falling to zero at around 50°C [1 mark].
Marking note: Must describe both the increase and the decrease. Award 1 mark for each phase.
(b) Explain why the rate of photosynthesis increases as temperature rises from 0°C to 25°C. [2 marks]
Answer: As temperature increases, the kinetic energy of the enzyme and substrate molecules increases [1 mark], leading to more frequent and successful collisions between enzymes and substrates. This increases the rate of the light-independent reactions (Calvin cycle), thus increasing the overall rate of photosynthesis [1 mark].
Marking note: Must refer to enzyme activity/kinetic energy and link to increased reaction rate.
(c) Explain why the rate of photosynthesis decreases rapidly at temperatures above 30°C. [2 marks]
Answer: At temperatures above 30°C, the enzymes involved in photosynthesis (e.g., Rubisco) begin to denature [1 mark]. The active site of the enzyme loses its specific shape, so the substrate can no longer bind, and the rate of reaction decreases sharply [1 mark].
Marking note: Must mention enzyme denaturation and loss of active site shape/function. Accept reference to photorespiration increasing at high temperatures.
15. (a) State the colour change observed when iodine solution is added to a part of the leaf that contains starch. [1 mark]
Answer: The iodine solution changes from brown/yellow to blue-black. [1 mark]
Marking note: Must state the colour change clearly. Accept "turns blue-black."
(b) Predict and explain the results of the iodine test on the green and white patches of the leaf. [3 marks]
Answer:
- The green patches will turn blue-black with iodine solution [1 mark] because they contain chlorophyll, which enabled photosynthesis to occur, producing starch [1 mark].
- The white patches will remain brown/yellow (no colour change) [1 mark] because they lack chlorophyll, so photosynthesis could not occur, and no starch was produced.
Marking note: Award 1 mark for each prediction and 1 mark for the explanation linking chlorophyll to photosynthesis and starch production.
(c) Explain why the leaf was destarched before the experiment. [2 marks]
Answer: Destarching removes any starch that was already present in the leaf before the experiment [1 mark]. This ensures that any starch detected after the experiment was produced during the experimental period, making the results valid [1 mark].
Marking note: Must mention removing pre-existing starch and ensuring validity/accuracy of results.
Section D: Extended Response (10 marks)
16. Describe the process of water uptake and transport from the soil to the leaves of a flowering plant. In your answer, explain the roles of root pressure, capillary action, and transpiration pull. [6 marks]
Answer:
- Water uptake: Water enters the root hair cells from the soil by osmosis, down a water potential gradient. Mineral ions are actively transported into the root hair cells, lowering the water potential and facilitating water uptake. Water then moves across the cortex to the xylem via the apoplast, symplast, and vacuolar pathways.
- Root pressure: The active transport of mineral ions into the xylem vessels in the roots lowers the water potential in the xylem. Water enters the xylem by osmosis, creating a positive hydrostatic pressure called root pressure. This pushes water up the xylem, but its effect is limited and mainly contributes to transport in short plants or at night.
- Capillary action: Xylem vessels are narrow, and water molecules exhibit cohesion (attraction to each other) and adhesion (attraction to the xylem walls). These forces cause capillary action, which helps draw water up the narrow xylem tubes. However, capillary action alone cannot account for water transport to the tops of tall trees.
- Transpiration pull: The main driving force is transpiration pull. Water evaporates from the mesophyll cells into the air spaces of the leaf and diffuses out through the stomata (transpiration). This loss of water lowers the water potential in the leaf cells, causing water to move from the xylem into the leaf cells by osmosis. Due to the cohesive property of water, the column of water in the xylem is pulled up as a continuous stream under tension. This transpiration pull can draw water up to great heights.
Marking note: Award marks for: (1) describing water uptake by osmosis in root hairs; (2) explaining root pressure (active transport, osmotic water entry, limited role); (3) explaining capillary action (cohesion, adhesion, narrow vessels); (4) explaining transpiration pull (evaporation, cohesion-tension, main driving force); (5) logical sequence and integration of processes; (6) use of correct terminology. Maximum 6 marks.
17. (a) Explain the colour change observed in test tube 1. [2 marks]
Answer: The indicator turned purple, indicating a low carbon dioxide concentration [1 mark]. In bright light, the leaf carries out photosynthesis at a higher rate than respiration, so it absorbs more carbon dioxide from the surrounding air than it releases, reducing the CO₂ concentration [1 mark].
Marking note: Must link purple colour to low CO₂ and explain that photosynthesis exceeds respiration.
(b) Explain the colour change observed in test tube 2. [2 marks]
Answer: The indicator turned yellow, indicating a high carbon dioxide concentration [1 mark]. In darkness, the leaf cannot carry out photosynthesis but continues to respire, releasing carbon dioxide. This increases the CO₂ concentration in the test tube [1 mark].
Marking note: Must link yellow colour to high CO₂ and explain that respiration occurs without photosynthesis.
(c) State the purpose of test tube 3 in this experiment. [1 mark]
Answer: Test tube 3 acts as a control [1 mark] to show that any colour change in the other test tubes is due to the presence of the leaf and not due to other factors (e.g., the effect of light on the indicator itself).
Marking note: Must mention "control" or equivalent explanation.
(d) Suggest why the indicator in test tube 4 turned orange-red instead of purple or yellow. [1 mark]
Answer: In dim light, the rate of photosynthesis is approximately equal to the rate of respiration [1 mark], so there is no net change in carbon dioxide concentration, and the indicator remains near its equilibrium colour (red/orange-red).
Marking note: Accept "photosynthesis rate equals respiration rate" or "compensation point."
18. Compare the structure and function of xylem and phloem tissues in flowering plants. [4 marks]
Answer:
- Xylem:
- Structure: Composed of dead cells with no cytoplasm or end walls, forming hollow tubes. Walls are thickened with lignin, which provides structural support and makes them waterproof.
- Function: Transports water and dissolved mineral ions from the roots to the stems and leaves. Also provides mechanical support to the plant.
- Phloem:
- Structure: Composed of living cells (sieve tube elements and companion cells). Sieve tube elements have sieve plates with pores, allowing cytoplasmic connections. Companion cells have many mitochondria to provide energy for active transport.
- Function: Transports sucrose and amino acids (products of photosynthesis) from the leaves to other parts of the plant (translocation).
Marking note: Award marks for: (1) xylem structure (dead, hollow, lignin); (2) xylem function (water/mineral transport, support); (3) phloem structure (living, sieve plates, companion cells); (4) phloem function (translocation of sucrose/amino acids). Maximum 4 marks.
19. A farmer noticed that plants in a waterlogged field were wilting even though the soil was saturated with water. Explain why the plants were wilting. [2 marks]
Answer: Waterlogged soil lacks oxygen [1 mark]. Root hair cells require oxygen for aerobic respiration to provide energy for the active transport of mineral ions into the root. Without oxygen, active transport stops, so the water potential in the root cells does not decrease sufficiently, and water uptake by osmosis is reduced. The plant loses water by transpiration faster than it can absorb water, leading to wilting [1 mark].
Marking note: Must mention lack of oxygen and link to reduced active transport/respiration and water uptake. Award 1 mark for identifying oxygen deficiency and 1 mark for explaining the consequence on water uptake.
20. Explain the importance of nitrate ions and magnesium ions for healthy plant growth. [2 marks]
Answer:
- Nitrate ions: Required for the synthesis of amino acids and proteins [1 mark], which are essential for cell growth and the formation of new tissues. Deficiency leads to stunted growth and yellowing of leaves.
- Magnesium ions: Required for the synthesis of chlorophyll [1 mark], the pigment that absorbs light for photosynthesis. Deficiency leads to chlorosis (yellowing of leaves) and reduced photosynthesis.
Marking note: Award 1 mark for each ion with its correct role. Must link nitrate to proteins/amino acids and magnesium to chlorophyll.
END OF ANSWER KEY