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Secondary 4 Combined Science Biology Preliminary Examination Paper 4
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Questions
TuitionGoWhere Practice Paper - Combined Science Biology Secondary 4
TuitionGoWhere Secondary School (AI)
| Subject: | Combined Science (Biology) |
| Level: | Secondary 4 |
| Paper: | PRELIM – Paper 2 (Structured & Free Response) |
| Version: | 4 of 5 |
| Duration: | 1 hour 15 minutes (75 minutes) |
| Total Marks: | 60 |
| Name: | ______________________________ |
| Class: | ______________________________ |
| Date: | ______________________________ |
Instructions to Candidates
- Write your name, class, and date in the spaces provided above.
- Answer ALL questions in the spaces provided.
- Write in dark blue or black pen.
- You may use a pencil for any diagrams or graphs.
- Do not use correction fluid.
- The number of marks for each question or part-question is shown in brackets [ ].
- The total mark for this paper is 60.
- You are advised to spend no more than 75 minutes on this paper.
Section A: Multiple Choice & Short Answer [20 marks]
Questions 1–10
Answer ALL questions. Each question carries 2 marks unless otherwise stated.
1. The diagram below represents a typical animal cell as seen under an electron microscope.
(Diagram description for student: A labelled animal cell showing structures A (nucleus), B (mitochondrion), C (rough endoplasmic reticulum), D (cell membrane), E (Golgi apparatus))
Which labelled structure is the site of aerobic respiration?
A. A B. B C. C D. D
[2]
2. A student placed red blood cells into three different solutions and observed the results after 30 minutes.
| Solution | Observation |
|---|---|
| Solution P | Cells appeared swollen and some had burst |
| Solution Q | Cells appeared normal |
| Solution R | Cells appeared shrunken and crenated |
Which statement correctly describes the concentration of each solution compared to the cytoplasm of the red blood cells?
A. Solution P is hypertonic, Solution Q is isotonic, Solution R is hypotonic B. Solution P is hypotonic, Solution Q is isotonic, Solution R is hypertonic C. Solution P is isotonic, Solution Q is hypotonic, Solution R is hypertonic D. Solution P is hypotonic, Solution Q is hypertonic, Solution R is isotonic
[2]
3. State the process by which oxygen molecules in the alveoli of the lung reach a red blood cell in a surrounding capillary.
[2]
4. The table shows the results of food tests carried out on a food sample.
| Test | Reagent Used | Observation |
|---|---|---|
| Test 1 | Iodine solution | Blue-black colour |
| Test 2 | Benedict's solution (heated) | Orange-red precipitate |
| Test 3 | Biuret solution | Violet colour |
(a) Identify the food substances present in the sample.
[2]
(b) State the colour change observed in Test 2 if the food sample had contained only a non-reducing sugar.
[1]
5. The diagram shows an enzyme-catalysed reaction at two different temperatures.
(Graph description: Reaction rate (y-axis) vs. Temperature in °C (x-axis). Curve A peaks at 37°C. Curve B peaks at 60°C.)
(a) Which curve, A or B, is more likely to represent a human enzyme? Explain your answer.
[2]
(b) Explain why the reaction rate decreases above the optimum temperature for Curve A.
[2]
6. A student investigated the effect of pH on the activity of enzyme X. The results are shown in the table below.
| pH | Rate of reaction (arbitrary units) |
|---|---|
| 2 | 5 |
| 4 | 28 |
| 6 | 52 |
| 7 | 60 |
| 8 | 35 |
| 10 | 8 |
| 12 | 2 |
(a) State the optimum pH for enzyme X.
[1]
(b) Suggest what would happen to the rate of reaction if the enzyme was placed in pH 1 for 10 minutes and then returned to pH 7. Explain your answer.
[2]
7. The diagram shows a plant cell placed in a concentrated sucrose solution.
(Diagram description: A plant cell showing the cell membrane pulled away from the cell wall; the large central vacuole appears shrunken.)
(a) Name the process that has caused the change observed.
[1]
(b) Name the condition of the cell shown in the diagram.
[1]
8. Explain why a muscle cell contains a large number of mitochondria.
[2]
9. Distinguish between diffusion and active transport. Give one key difference.
[2]
10. A piece of potato of known mass was placed in distilled water for 2 hours. After removal, the potato piece was reweighed and found to have increased in mass by 8%.
(a) Explain, in terms of water potential, why the potato piece gained mass.
[2]
(b) Predict and explain what would happen to the mass of an identical potato piece placed in a concentrated salt solution for 2 hours.
[2]
Section B: Structured Response [25 marks]
Questions 11–15
Answer ALL questions. Write your answers in the spaces provided.
11. Figure 11.1 shows a section through a leaf as seen under a light microscope.
(Diagram description: A cross-section of a leaf showing upper epidermis, palisade mesophyll layer, spongy mesophyll layer, lower epidermis with stomata, and vascular bundles.)
(a) Label structures A (palisade mesophyll), B (spongy mesophyll), and C (stoma) on Figure 11.1.
[3]
(b) Explain how the structure of the palisade mesophyll cells is adapted for photosynthesis.
[3]
(c) State two ways in which the structure of a palisade mesophyll cell differs from that of a red blood cell.
[2]
[Total: 8 marks]
12. An experiment was carried out to investigate the effect of temperature on the permeability of beetroot cell membranes. Beetroot discs of equal size were placed in test tubes containing 10 cm³ of distilled water at different temperatures for 15 minutes. The colour intensity of the solution was then measured using a colorimeter.
The results are shown in the table below.
| Temperature (°C) | Colorimeter reading (absorbance units) |
|---|---|
| 10 | 0.05 |
| 20 | 0.08 |
| 30 | 0.12 |
| 40 | 0.25 |
| 50 | 0.55 |
| 60 | 0.82 |
| 70 | 0.85 |
(a) Describe the relationship between temperature and membrane permeability as shown by the results.
[2]
(b) Explain the results obtained at 60 °C and 70 °C.
[3]
(c) Suggest why the beetroot discs were rinsed with distilled water before being placed in the test tubes.
[1]
(d) State one variable that should be kept constant in this experiment to ensure a fair test.
[1]
[Total: 7 marks]
13. The diagram shows a section through a villus in the small intestine.
(Diagram description: A villus showing the epithelium, capillary network, and a central lacteal.)
(a) Name the process by which glucose molecules are absorbed from the lumen of the small intestine into the capillary network of the villus.
[1]
(b) Explain how the structure of the villus is adapted for efficient absorption of digested food.
[3]
(c) A student claims that amino acids are absorbed into the lacteal. Evaluate this claim.
[2]
[Total: 6 marks]
14. A student carried out an experiment to test the effect of enzyme concentration on the rate of breakdown of starch by amylase. The results are shown in the graph below.
(Graph description: Rate of reaction (y-axis) vs. Amylase concentration in % (x-axis). The curve rises steeply from 0% to 4% amylase, then levels off between 6% and 10%.)
(a) Describe the trend shown by the graph.
[2]
(b) Explain why the rate of reaction levels off at higher enzyme concentrations.
[2]
(c) State two conditions, other than enzyme concentration, that should be kept constant in this experiment.
[2]
[Total: 6 marks]
15. Explain why enzymes are described as biological catalysts. In your answer, include how enzymes affect the activation energy of a reaction and what happens to the enzyme after the reaction.
[3]
[Total: 3 marks]
Section C: Free Response & Data-Based Questions [15 marks]
Questions 16–20
Answer ALL questions. Write your answers in the spaces provided.
16. A student investigated the effect of different sugar solutions on the mass of potato cylinders. Five potato cylinders of equal length and diameter were weighed and placed in separate sugar solutions of different concentrations for 4 hours. The percentage change in mass was calculated.
The results are shown in the table.
| Sugar concentration (mol/dm³) | Percentage change in mass (%) |
|---|---|
| 0.0 | +12.0 |
| 0.2 | +5.5 |
| 0.4 | −2.0 |
| 0.6 | −8.5 |
| 0.8 | −14.0 |
(a) Plot a graph of percentage change in mass (y-axis) against sugar concentration (x-axis) on the grid provided.
[3]
(b) Use your graph to determine the sugar concentration at which there is no net change in mass of the potato cylinder. Explain what this indicates about the water potential of the potato cells.
[3]
(c) Explain the results obtained at 0.0 mol/dm³ sugar concentration in terms of osmosis.
[2]
[Total: 8 marks]
17. The diagram shows two cells, Cell X and Cell Y, as seen under an electron microscope.
(Diagram description: Cell X shows a cell wall, large central vacuole, chloroplasts, and a nucleus. Cell Y shows no cell wall, no chloroplasts, many small vacuoles, and a nucleus.)
(a) Identify Cell X and Cell Y as either a plant cell or an animal cell. Give two reasons for each identification.
[4]
(b) Both cells contain ribosomes. State the function of ribosomes.
[1]
(c) Suggest why Cell X contains more mitochondria than a typical plant cell.
[2]
[Total: 7 marks]
18. A student tested four different food samples (A, B, C, and D) for the presence of starch, reducing sugar, protein, and fats. The results are shown in the table.
| Food Sample | Iodine test | Benedict's test | Biuret test | Ethanol emulsion test |
|---|---|---|---|---|
| A | Blue-black | Blue (no change) | Violet | Cloudy white |
| B | Brown-yellow | Orange-red | Pale blue | Clear |
| C | Brown-yellow | Blue (no change) | Violet | Clear |
| D | Blue-black | Orange-red | Pale blue | Cloudy white |
(a) Which food sample contains starch but no reducing sugar? Explain your answer.
[2]
(b) Which food sample contains protein but no fats? Explain your answer.
[2]
(c) A student concluded that food sample B contains only reducing sugar. Suggest one additional test the student could carry out to confirm that no non-reducing sugar is present.
[2]
[Total: 6 marks]
19. The graph shows the effect of substrate concentration on the rate of an enzyme-catalysed reaction in the presence and absence of an inhibitor.
(Graph description: Two curves. Curve 1 (no inhibitor) rises steeply and plateaus at a high rate. Curve 2 (with inhibitor) rises more gradually and plateaus at a lower rate. Both curves start from the origin.)
(a) Describe and explain the shape of Curve 1.
[3]
(b) Suggest what type of inhibitor is present in Curve 2. Explain your answer.
[2]
(c) State one factor, other than substrate concentration and inhibitor presence, that would affect the rate of this reaction.
[1]
[Total: 6 marks]
20. A student placed a strip of dialysis tubing filled with 10% starch solution and 10% glucose solution into a beaker of distilled water. After 30 minutes, the student tested the water outside the tubing for the presence of starch and glucose.
(a) Predict and explain the results for the test for starch in the water outside the tubing.
[2]
(b) Predict and explain the results for the test for glucose in the water outside the tubing.
[2]
(c) Explain why this experiment models the function of the small intestine in the human digestive system.
[2]
[Total: 6 marks]
END OF PAPER
Total: 60 marks
Answers
TuitionGoWhere Practice Paper - Combined Science Biology Secondary 4
Answer Key – Version 4 of 5
PRELIM – Paper 2 (Structured & Free Response)
Section A: Multiple Choice & Short Answer [20 marks]
1. B [2]
Reasoning: Mitochondria (structure B) are the organelles responsible for aerobic respiration, where glucose is broken down in the presence of oxygen to release energy in the form of ATP. The nucleus (A) contains genetic material, rough ER (C) is involved in protein synthesis, and the cell membrane (D) controls entry and exit of substances.
Marking: 1 mark for correct answer. No marks for incorrect or ambiguous response.
2. B [2]
Reasoning: In a hypotonic solution (lower solute concentration than the cell cytoplasm), water enters the cell by osmosis, causing it to swell and potentially burst (Solution P). In an isotonic solution, there is no net movement of water and cells appear normal (Solution Q). In a hypertonic solution (higher solute concentration), water leaves the cell, causing it to shrink and become crenated (Solution R).
Marking: 1 mark for correct answer.
3. Diffusion [2]
Reasoning: Oxygen moves from a region of high concentration (in the alveoli) to a region of low concentration (in the red blood cell) down a concentration gradient. This passive process is called diffusion. No energy is required.
Marking: 1 mark for "diffusion" or "diffusion down a concentration gradient". Do not accept "osmosis" or "active transport".
4. (a) Starch, reducing sugar, and protein are present. [2]
Reasoning:
- Test 1: Iodine solution turns blue-black → starch is present.
- Test 2: Benedict's solution heated gives orange-red precipitate → reducing sugar is present.
- Test 3: Biuret solution turns violet → protein is present.
Marking: 1 mark for each correct food substance identified (max 2 marks; all three must be named for full marks).
(b) The solution would remain blue (no colour change). [1]
Reasoning: Benedict's test specifically detects reducing sugars. A non-reducing sugar (e.g., sucrose) would not react with Benedict's solution, so the solution would stay blue.
Marking: 1 mark for "remains blue" or "no change" or "stays blue".
5. (a) Curve A. Human enzymes have an optimum temperature of approximately 37 °C, which is the normal human body temperature. [2]
Reasoning: Human enzymes function best at body temperature (~37 °C). Curve A peaks at 37 °C, while Curve B peaks at 60 °C, which is too high for most human enzymes.
Marking: 1 mark for identifying Curve A. 1 mark for correct explanation linking to body temperature.
(b) Above the optimum temperature, the enzyme's active site becomes denatured. The shape of the active site changes due to the breaking of bonds (e.g., hydrogen bonds) that maintain the enzyme's tertiary structure. The substrate can no longer fit into the active site, so the rate of reaction decreases. [2]
Reasoning: High temperatures cause the enzyme to lose its three-dimensional shape (denaturation). This is irreversible for most enzymes.
Marking: 1 mark for stating "denaturation" or "active site changes shape". 1 mark for explaining that the substrate can no longer bind/fit.
6. (a) pH 7 [1]
Reasoning: The highest rate of reaction (60 arbitrary units) occurs at pH 7.
Marking: 1 mark for pH 7.
(b) The rate of reaction would remain very low (close to zero). At pH 1, the enzyme would be denatured (the active site would be permanently altered). Returning to pH 7 would not restore the enzyme's original shape, so the enzyme would not function. [2]
Reasoning: Extremely low pH causes irreversible denaturation of the enzyme. The enzyme cannot recover its original shape even when returned to the optimum pH.
Marking: 1 mark for stating the enzyme is denatured at pH 1. 1 mark for stating the effect is irreversible / enzyme does not regain function.
7. (a) Osmosis [1]
Reasoning: Water moved out of the plant cell by osmosis (from a region of higher water potential inside the cell to a region of lower water potential in the concentrated sucrose solution).
Marking: 1 mark for "osmosis".
(b) Plasmolysis (or plasmolysed) [1]
Reasoning: The cell membrane has pulled away from the cell wall due to water loss. This condition is called plasmolysis.
Marking: 1 mark for "plasmolysis" or "plasmolysed".
8. Muscle cells require large amounts of energy (ATP) for contraction. Mitochondria are the sites of aerobic respiration, where ATP is produced. Therefore, muscle cells contain many mitochondria to meet their high energy demand. [2]
Reasoning: The link between energy demand, mitochondria, and ATP production must be made clear.
Marking: 1 mark for linking muscle cells to high energy demand. 1 mark for linking mitochondria to ATP/energy production.
9. Diffusion is the passive movement of particles from a region of high concentration to a region of low concentration (down the concentration gradient) and does not require energy. Active transport is the movement of particles from a region of low concentration to a region of high concentration (against the concentration gradient) and requires energy (ATP). [2]
Reasoning: Any one valid difference is acceptable: direction of movement, energy requirement, or involvement of carrier proteins.
Marking: 1 mark for stating diffusion is passive / does not require energy / moves down the gradient. 1 mark for stating active transport requires energy / moves against the gradient.
10. (a) The water potential of the distilled water is higher (less negative) than the water potential of the potato cells. Water molecules move by osmosis from the distilled water (higher water potential) into the potato cells (lower water potential), causing the potato to gain mass. [2]
Reasoning: Osmosis is the net movement of water molecules from a region of higher water potential to a region of lower water potential across a partially permeable membrane.
Marking: 1 mark for correct comparison of water potentials. 1 mark for stating osmosis / water moves into the potato.
(b) The potato piece would lose mass. The concentrated salt solution has a lower (more negative) water potential than the potato cells. Water molecules would move out of the potato cells into the solution by osmosis, causing the potato to decrease in mass. [2]
Reasoning: Water moves from high water potential (potato) to low water potential (concentrated salt solution).
Marking: 1 mark for predicting mass loss. 1 mark for correct explanation in terms of water potential / osmosis.
Section B: Structured Response [25 marks]
11. (a) A = Palisade mesophyll; B = Spongy mesophyll; C = Stoma (plural: stomata) [3]
Marking: 1 mark for each correct label.
(b) Palisade mesophyll cells are adapted for photosynthesis in the following ways:
- They contain many chloroplasts, which contain chlorophyll to absorb light energy.
- They are located near the upper surface of the leaf, where they receive the most sunlight.
- They are elongated and closely packed, maximising the number of cells (and chloroplasts) exposed to light.
- They have a large surface area to volume ratio for efficient gas exchange. [3]
Marking: 1 mark for each valid adaptation (max 3 marks). Must link structure to function for each mark.
(c) Two differences:
- Palisade mesophyll cells have a cell wall; red blood cells do not have a cell wall.
- Palisade mesophyll cells contain chloroplasts; red blood cells do not contain chloroplasts. (Also accept: palisade cells have a large central vacuole; RBCs do not. Palisade cells are rectangular/columnar; RBCs are biconcave disc shape.) [2]
Marking: 1 mark for each valid difference (max 2 marks).
12. (a) As temperature increases, the colorimeter reading (absorbance) increases, indicating that membrane permeability increases. The increase is gradual from 10 °C to 40 °C, and becomes much more rapid from 40 °C to 60 °C. [2]
Marking: 1 mark for stating that permeability increases with temperature. 1 mark for describing the pattern (gradual then rapid / non-linear increase).
(b) At 60 °C and 70 °C, the high temperature causes the phospholipid bilayer of the cell membrane to break down / become more fluid. The membrane proteins become denatured. This increases membrane permeability significantly, allowing more pigment (betacyanin) to leak out of the cells into the surrounding water. [3]
Marking: 1 mark for stating high temperature disrupts/damages the membrane. 1 mark for mentioning denaturation of membrane proteins or increased fluidity of phospholipids. 1 mark for linking to pigment leaking out / increased permeability.
(c) To remove any pigment that may have been released from damaged cells on the surface of the beetroot during cutting, which would otherwise contaminate the distilled water and give a false reading. [1]
Marking: 1 mark for stating removal of surface pigment / to ensure a fair test / to prevent contamination.
(d) Any one of the following:
- Size/mass/volume of beetroot discs
- Time the discs are left in the water
- Volume of distilled water used
- Same type/source of beetroot [1]
Marking: 1 mark for any valid controlled variable.
13. (a) Active transport [1]
Reasoning: Glucose is absorbed against a concentration gradient (from low concentration in the lumen to higher concentration in the capillary) using energy from ATP. This is active transport. (Note: In some contexts, facilitated diffusion via carrier proteins is also accepted at this level, but active transport is the primary mechanism for glucose absorption in the villus.)
Marking: 1 mark for "active transport".
(b) The villus is adapted for efficient absorption in the following ways:
- It has a large surface area due to its finger-like projection and the presence of microvilli on the epithelial cells, increasing the rate of absorption.
- The epithelium is only one cell thick, providing a short diffusion distance for digested food to pass through.
- It has a dense network of capillaries inside, which rapidly transport absorbed nutrients away, maintaining a steep concentration gradient.
- It contains a lacteal for the absorption of fatty acids and glycerol. [3]
Marking: 1 mark for each valid adaptation (max 3 marks). Must link structure to function.
(c) The student's claim is incorrect. Amino acids are absorbed into the capillary network (blood capillaries) of the villus, not the lacteal. The lacteal absorbs fatty acids and glycerol (products of fat digestion). [2]
Marking: 1 mark for stating the claim is incorrect. 1 mark for correctly identifying where amino acids are absorbed (capillaries) and/or what the lacteal absorbs (fatty acids and glycerol).
14. (a) As the amylase concentration increases from 0% to 4%, the rate of reaction increases rapidly. Beyond 4%, the rate of reaction levels off and remains relatively constant despite further increases in enzyme concentration. [2]
Marking: 1 mark for describing the initial increase. 1 mark for describing the levelling off.
(b) At higher enzyme concentrations, all the substrate (starch) molecules are already bound to enzyme active sites. The substrate concentration becomes the limiting factor. Adding more enzyme does not increase the rate because there is no additional substrate available for the extra enzyme molecules to act on. [2]
Marking: 1 mark for stating substrate is the limiting factor / all substrate is being used. 1 mark for explaining that extra enzyme has no substrate to act on.
(c) Any two of the following:
- Temperature
- pH of the solution
- Concentration of starch (substrate)
- Volume of starch solution
- Time allowed for the reaction [2]
Marking: 1 mark for each valid condition (max 2 marks).
15. Enzymes are biological catalysts because they speed up chemical reactions in living organisms without being used up or permanently changed in the process. Enzymes lower the activation energy required for a reaction to proceed, allowing the reaction to occur more quickly at body temperature. After the reaction, the enzyme remains unchanged and can be reused to catalyse further reactions. [3]
Marking: 1 mark for stating enzymes speed up reactions without being used up. 1 mark for stating enzymes lower activation energy. 1 mark for stating enzymes remain unchanged / can be reused after the reaction.
Section C: Free Response & Data-Based Questions [15 marks]
16. (a) Graph plotting: [3]
Marking scheme for graph:
- 1 mark for correct labelling of axes (x-axis: Sugar concentration in mol/dm³; y-axis: Percentage change in mass (%))
- 1 mark for correct plotting of all 5 points
- 1 mark for drawing a smooth curve or line of best fit through the points
(Expected graph: A downward-sloping curve/line crossing the x-axis between 0.2 and 0.4 mol/dm³.)
(b) The sugar concentration at which there is no net change in mass is approximately 0.3 mol/dm³ (accept 0.28–0.32 mol/dm³). This indicates that the water potential of the potato cells is equal to the water potential of the sugar solution at this concentration. There is no net movement of water into or out of the potato cells. [3]
Marking: 1 mark for reading the correct concentration from the graph (0.28–0.32 mol/dm³). 1 mark for stating that water potentials are equal. 1 mark for stating no net movement of water / equilibrium.
(c) At 0.0 mol/dm³ (distilled water), the water potential of the solution is higher (less negative) than the water potential inside the potato cells. Water molecules move by osmosis from the distilled water (higher water potential) into the potato cells (lower water potential) across the partially permeable cell membrane. This causes the potato cylinder to gain mass. [2]
Marking: 1 mark for correct comparison of water potentials. 1 mark for stating osmosis / water moves into the potato cells.
17. (a) Cell X is a plant cell. Reasons:
- It has a cell wall (animal cells do not have a cell wall).
- It has a large central vacuole (animal cells have small or no vacuoles).
- It contains chloroplasts (animal cells do not have chloroplasts).
Cell Y is an animal cell. Reasons:
- It does not have a cell wall (plant cells have a cell wall).
- It does not contain chloroplasts (plant cells have chloroplasts).
- It has small vacuoles instead of a large central vacuole. [4]
Marking: 1 mark for correctly identifying Cell X as a plant cell. 1 mark for two valid reasons for Cell X. 1 mark for correctly identifying Cell Y as an animal cell. 1 mark for two valid reasons for Cell Y.
(b) Ribosomes are the site of protein synthesis (where amino acids are assembled into proteins). [1]
Marking: 1 mark for "protein synthesis" or "making proteins".
(c) Cell X may be a cell that is very metabolically active (e.g., a palisade mesophyll cell carrying out photosynthesis or a root hair cell carrying out active transport), requiring more energy (ATP) than a typical plant cell. Since mitochondria produce ATP through aerobic respiration, more mitochondria are needed to meet the higher energy demand. [2]
Marking: 1 mark for linking to high energy demand / metabolic activity. 1 mark for linking mitochondria to ATP production.
18. (a) Food sample A contains starch but no reducing sugar. The iodine test turned blue-black, indicating starch is present. The Benedict's test showed no colour change (remained blue), indicating no reducing sugar is present. [2]
Marking: 1 mark for identifying Sample A. 1 mark for correct explanation referencing both test results.
(b) Food sample C contains protein but no fats. The Biuret test turned violet, indicating protein is present. The ethanol emulsion test was clear, indicating no fats are present. [2]
Marking: 1 mark for identifying Sample C. 1 mark for correct explanation referencing both test results.
(c) The student could first hydrolyse the sample by heating it with dilute hydrochloric acid (HCl), then neutralise with sodium hydrogacid solution, and then re-test with Benedict's solution. If the Benedict's test now gives an orange-red precipitate, this confirms the presence of a non-reducing sugar (which was hydrolysed to reducing sugars by the acid). [2]
Marking: 1 mark for stating hydrolysis with dilute acid. 1 mark for re-testing with Benedict's solution and expecting a positive result if non-reducing sugar is present.
19. (a) As substrate concentration increases, the rate of reaction increases because more substrate molecules are available to bind with enzyme active sites, forming more enzyme-substrate complexes. At higher substrate concentrations, all enzyme active sites become saturated (occupied), so the rate of reaction reaches a maximum and levels off. Adding more substrate does not increase the rate further because there are no free active sites available. [3]
Marking: 1 mark for stating rate increases with substrate concentration. 1 mark for explaining more enzyme-substrate complexes form. 1 mark for stating active sites become saturated / maximum rate reached.
(b) The inhibitor is a non-competitive inhibitor. The maximum rate of reaction (Vmax) is lower in the presence of the inhibitor, but the shape of the curve is similar. A non-competitive inhibitor binds to a site other than the active site (allosteric site), changing the shape of the active site so that the substrate cannot bind effectively. This reduces the overall rate of reaction regardless of substrate concentration. [2]
Marking: 1 mark for identifying non-competitive inhibitor. 1 mark for correct explanation (binds to allosteric site / changes active site shape / reduces Vmax).
(Note: If the graph showed the same Vmax but higher Km, the answer would be competitive inhibitor. The description here assumes a lower Vmax for Curve 2.)
(c) Any one of the following:
- Temperature
- pH
- Enzyme concentration [1]
Marking: 1 mark for any valid factor.
20. (a) The test for starch in the water outside the tubing will be negative (iodine solution remains brown-yellow). Starch molecules are too large to pass through the pores of the dialysis tubing (partially permeable membrane). Therefore, starch remains inside the tubing and does not diffuse into the surrounding water. [2]
Marking: 1 mark for predicting a negative result. 1 mark for explaining that starch molecules are too large to pass through the membrane.
(b) The test for glucose in the water outside the tubing will be positive (Benedict's test gives orange-red precipitate after heating). Glucose molecules are small enough to pass through the pores of the dialysis tubing. Glucose diffuses out of the tubing (from high concentration inside to low concentration outside) into the surrounding water. [2]
Marking: 1 mark for predicting a positive result. 1 mark for explaining that glucose molecules are small enough to pass through / glucose diffuses out.
(c) The dialysis tubing acts as a model of the small intestine wall. The partially permeable membrane of the tubing represents the cell membrane of the intestinal epithelium. Small molecules (like glucose) can pass through the membrane, while large molecules (like starch) cannot. This models how only small, digested food molecules (e.g., glucose, amino acids) are absorbed through the intestinal wall into the blood, while large undigested molecules (like starch) are too large to be absorbed and must first be broken down by enzymes. [2]
Marking: 1 mark for linking the dialysis tubing membrane to the intestinal wall / cell membrane. 1 mark for linking the selective permeability to absorption of small molecules only / need for digestion of large molecules.
END OF ANSWER KEY
Total: 60 marks