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Secondary 4 Combined Science Biology Preliminary Examination Paper 2
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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: | 2 of 5 |
| Duration: | 1 hour 15 minutes (75 minutes) |
| Total Marks: | 50 |
| 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 50.
- You are advised to spend no more than 75 minutes on this paper.
Section A: Multiple Choice & Short Answer Questions [15 marks]
Questions 1–10
Question 1 (1 mark)
Which organelle is responsible for aerobic respiration in a cell?
A. Ribosome
B. Golgi apparatus
C. Mitochondrion
D. Endoplasmic reticulum
Answer: ______________
Question 2 (1 mark)
State the process by which oxygen from the alveoli enters a red blood cell.
Answer: ______________________________________________________________
Question 3 (1 mark)
The diagram below represents a typical animal cell as seen under an electron microscope.
(Diagram description for student: A simplified animal cell diagram is shown with labelled structures A–F. Structure A is the outer boundary. Structure B is a large, dark, circular body near the centre. Structure C is a double-membraned bean-shaped organelle. Structure D consists of stacked flattened sacs. Structure E is a network of membranes with small dots attached. Structure F is a small dot-like structure.)
(a) Identify structure C.
Answer: ______________________________________________________________
(b) State one function of structure D.
Answer: ______________________________________________________________
Question 4 (1 mark)
A student placed red blood cells in three different solutions and observed the results.
| Solution | Observation |
|---|---|
| Solution P | Cells swelled and burst |
| Solution Q | Cells remained unchanged |
| Solution R | Cells shrank and became crenated |
Which solution is isotonic to the red blood cells?
Answer: ______________
Question 5 (2 marks)
The table shows the number of mitochondria counted in three different cell types from the same organism.
| Cell Type | Number of Mitochondria per Cell |
|---|---|
| Skin cell | 120 |
| Muscle cell | 850 |
| Liver cell | 430 |
(a) Suggest why muscle cells contain more mitochondria than skin cells.
Answer: ______________________________________________________________
(b) State the main biochemical process that occurs in the mitochondrion.
Answer: ______________________________________________________________
Question 6 (1 mark)
Name the small, circular organelle found in plant cells that is not present in animal cells and is responsible for storing cell sap.
Answer: ______________________________________________________________
Question 7 (2 marks)
A student tested four food samples for the presence of different biological molecules. The results are shown below.
| Food Sample | Benedict's Test | Iodine Test | Biuret Test | Ethanol Emulsion Test |
|---|---|---|---|---|
| W | Blue | Brown | Purple | Cloudy white |
| X | Brick-red | Brown | Blue | Clear |
| Y | Blue | Blue-black | Blue | Clear |
| Z | Blue | Brown | Blue | Cloudy white |
(a) Which food sample contains starch? Explain your answer.
Answer: ______________________________________________________________
(b) Which food sample contains protein? Explain your answer.
Answer: ______________________________________________________________
Question 8 (1 mark)
State the role of the cell membrane in a cell.
Answer: ______________________________________________________________
Question 9 (2 marks)
The diagram shows an experiment to investigate osmosis. A Visking tubing bag containing 10% sucrose solution was placed in a beaker of distilled water. The initial and final masses of the bag were recorded.
| Mass / g | |
|---|---|
| Initial mass of bag | 25.0 |
| Final mass of bag (after 30 min) | 29.5 |
(a) Calculate the percentage change in mass of the Visking tubing bag. Show your working.
Answer: ______________________________________________________________
(b) Explain why the mass of the bag increased.
Answer: ______________________________________________________________
Question 10 (1 mark)
Name the enzyme that breaks down starch into maltose in the human digestive system.
Answer: ______________________________________________________________
Section B: Structured Response Questions [25 marks]
Questions 11–17
Question 11 (3 marks)
The diagram shows two cells: Cell X (a plant cell) and Cell Y (an animal cell), both placed in a concentrated salt solution.
(Diagram description: Cell X shows the cell membrane pulled away from the rigid cell wall — plasmolysis. Cell Y shows the entire cell shrunken and irregularly shaped — crenation.)
(a) Name the process occurring in both cells.
Answer: ______________________________________________________________
(b) Explain why Cell X appears different from Cell Y after being placed in the salt solution.
Answer: ______________________________________________________________
(c) State the biological term for the appearance of Cell X.
Answer: ______________________________________________________________
Question 12 (4 marks)
The table below shows the relative amounts of three biological molecules found in four different food samples (A–D), measured in arbitrary units.
| Food Sample | Protein | Starch | Lipid |
|---|---|---|---|
| A | 85 | 5 | 10 |
| B | 10 | 80 | 5 |
| C | 5 | 10 | 85 |
| D | 30 | 30 | 30 |
(a) Which food sample would be the best source of energy for a person who needs to run a marathon? Explain your answer.
Answer: ______________________________________________________________
(b) A student wants to test food sample C for the presence of lipid. Describe the ethanol emulsion test and the expected positive result.
Answer: ______________________________________________________________
(c) Name the reagent used in the Biuret test and state the colour change for a positive result.
Answer: ______________________________________________________________
Question 13 (4 marks)
The diagram 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 the palisade mesophyll layer on the diagram provided.
(Space for student to label on diagram)
(b) Explain why the palisade mesophyll cells are located near the upper surface of the leaf.
Answer: ______________________________________________________________
(c) Palisade mesophyll cells contain a large number of chloroplasts. State the function of chloroplasts.
Answer: ______________________________________________________________
(d) A student counted 45 chloroplasts in one palisade mesophyll cell and 12 chloroplasts in one spongy mesophyll cell. Suggest why there is a difference.
Answer: ______________________________________________________________
Question 14 (3 marks)
Describe the structure of the cell surface membrane according to the fluid mosaic model.
Answer: ______________________________________________________________
Question 15 (4 marks)
An experiment was carried out to investigate the effect of temperature on the activity of the enzyme catalase. Potato cubes (a source of catalase) were placed into hydrogen peroxide solution at different temperatures. The volume of oxygen gas produced in the first 60 seconds was measured.
| Temperature / °C | Volume of O₂ produced in 60 s / cm³ |
|---|---|
| 10 | 2.0 |
| 20 | 5.5 |
| 30 | 9.0 |
| 40 | 12.0 |
| 50 | 10.5 |
| 60 | 3.0 |
| 70 | 0.5 |
(a) Plot a graph of volume of O₂ produced (y-axis) against temperature (x-axis) on the grid provided.
(Graph grid provided)
(b) Describe the trend shown by the graph from 10 °C to 40 °C.
Answer: ______________________________________________________________
(c) Explain why the volume of O₂ produced decreases at temperatures above 40 °C.
Answer: ______________________________________________________________
(d) State the optimum temperature for catalase activity based on the data.
Answer: ______________________________________________________________
Question 16 (3 marks)
A student observed a cell under an electron microscope and noted the following features:
- A double membrane with pores
- Chromatin visible inside
- A dense region within the nucleus
(a) Name the organelle being observed.
Answer: ______________________________________________________________
(b) State the function of the double membrane with pores.
Answer: ______________________________________________________________
(c) Name the dense region within this organelle and state its function.
Answer: ______________________________________________________________
Question 17 (4 marks)
The diagram shows the lock-and-key hypothesis for enzyme action.
(Diagram description: A diagram showing an enzyme with an active site, a substrate fitting into the active site, and the enzyme-substrate complex forming before products are released.)
(a) Label the following on the diagram: active site, substrate, enzyme-substrate complex, products.
(Diagram with blank labels provided)
(b) Explain why the lock-and-key hypothesis suggests that an enzyme is specific to its substrate.
Answer: ______________________________________________________________
(c) Name an alternative model that explains enzyme action and state how it differs from the lock-and-key hypothesis.
Answer: ______________________________________________________________
Section C: Free Response / Extended Question [10 marks]
Questions 18–20
Question 18 (3 marks)
Explain the importance of enzymes in the human digestive system. In your answer, refer to at least two named enzymes, their substrates, and their products.
Answer: ______________________________________________________________
Question 19 (3 marks)
A student carried out an experiment to compare the vitamin C content of two fruit juices (Juice P and Juice Q). Equal volumes of DCPIP solution (blue) were placed in separate test tubes. Juice P was added drop by drop to the first tube, and Juice Q to the second tube. The number of drops required to decolourise the DCPIP was recorded.
| Juice | Number of drops to decolourise DCPIP |
|---|---|
| P | 8 |
| Q | 15 |
(a) State the independent variable in this experiment.
Answer: ______________________________________________________________
(b) Which juice contains a higher concentration of vitamin C? Explain your answer.
Answer: ______________________________________________________________
(c) Suggest one way to improve the reliability of this experiment.
Answer: ______________________________________________________________
Question 20 (4 marks)
The diagram shows a comparison between a prokaryotic cell and a eukaryotic cell.
(Diagram description: Two simplified cell diagrams side by side. The prokaryotic cell shows a cell wall, cell membrane, cytoplasm, a single circular chromosome, ribosomes, and a plasmid. The eukaryotic cell shows a nucleus with nuclear envelope, mitochondria, endoplasmic reticulum, Golgi apparatus, ribosomes, cell membrane, and linear chromosomes.)
(a) State two structural differences between a prokaryotic cell and a eukaryotic cell.
Answer: ______________________________________________________________
(b) Both cell types contain ribosomes. State the function of ribosomes.
Answer: ______________________________________________________________
(c) Explain why prokaryotic cells are generally smaller than eukaryotic cells.
Answer: ______________________________________________________________
END OF PAPER
Total: 50 marks
Answers
TuitionGoWhere Practice Paper - Combined Science Biology Secondary 4
Answer Key – Version 2 of 5
PRELIM – Paper 2 (Structured & Free Response)
Section A: Multiple Choice & Short Answer Questions [15 marks]
Question 1 (1 mark)
Answer: C — Mitochondrion
Marking notes: Award 1 mark for C only. No credit for other options.
Question 2 (1 mark)
Answer: Diffusion
Marking notes: Award 1 mark for "diffusion" or "diffusion down a concentration gradient." Do not accept "osmosis" (osmosis refers to water, not gases). Do not accept "active transport" — no energy is required for gas exchange.
Question 3 (1 mark total; 0.5 + 0.5)
(a) Answer: Mitochondrion
Marking notes: Accept "mitochondria" (plural). Do not accept "chloroplast" or "nucleus."
(b) Answer: Packaging / modifying / sorting / transporting proteins (any one valid function)
Marking notes: Award 0.5 marks for any one correct function of the Golgi apparatus. Do not accept "protein synthesis" (that is the ribosome/rough ER).
Question 4 (1 mark)
Answer: Solution Q
Marking notes: Award 1 mark for Q. In an isotonic solution, there is no net movement of water, so cells remain unchanged. Common mistake: students may select P (hypotonic) or R (hypertonic).
Question 5 (2 marks)
(a) Answer: Muscle cells require more energy (ATP) for contraction. Since mitochondria are the site of aerobic respiration where ATP is produced, muscle cells need more mitochondria to meet their higher energy demand.
Marking notes: Award 1 mark for linking muscle cells to higher energy/ATP demand. Award 1 mark for linking mitochondria to energy/ATP production. Both points needed for 2 marks. Do not accept vague answers like "muscle cells work harder" without reference to energy or ATP.
(b) Answer: Aerobic respiration
Marking notes: Award 1 mark. Accept "cellular respiration" or "respiration." Do not accept "photosynthesis."
Question 6 (1 mark)
Answer: Vacuole (specifically, the permanent vacuole / central vacuole)
Marking notes: Award 1 mark for "vacuole." Accept "central vacuole" or "permanent vacuole." Do not accept "cell wall" — the cell wall does not store cell sap.
Question 7 (2 marks)
(a) Answer: Food sample Y. The iodine test turned blue-black, which is the positive result for starch.
Marking notes: Award 0.5 marks for identifying sample Y. Award 0.5 marks for stating the correct colour change (blue-black with iodine). Both needed for 1 mark.
(b) Answer: Food sample W. The Biuret test turned purple, which is the positive result for protein.
Marking notes: Award 0.5 marks for identifying sample W. Award 0.5 marks for stating the correct colour change (purple with Biuret reagent). Both needed for 1 mark.
Question 8 (1 mark)
Answer: The cell membrane controls the movement of substances into and out of the cell. It is partially (selectively) permeable.
Marking notes: Award 1 mark for any one of the following: "controls entry/exit of substances," "partially/selectively permeable," "regulates what enters and leaves the cell." Do not accept "protects the cell" alone — this is not the primary role assessed at this level.
Question 9 (2 marks)
(a) Answer:
Working:
- Change in mass = 29.5 − 25.0 = 4.5 g
- Percentage change = (4.5 / 25.0) × 100 = 18%
Marking notes: Award 0.5 marks for correct subtraction. Award 0.5 marks for correct formula and final answer (18%). If the student shows correct working but makes an arithmetic error, award 0.5 marks (error carried forward).
(b) Answer: Water molecules moved by osmosis from the distilled water (high water potential) into the Visking tubing bag containing sucrose solution (low water potential) through the partially permeable membrane. This caused the mass of the bag to increase.
Marking notes: Award 1 mark for correct explanation including: direction of water movement (into the bag), the process (osmosis), and the idea of a water potential gradient or partially permeable membrane. Do not accept "sucrose moved out" — Visking tubing is not permeable to sucrose.
Question 10 (1 mark)
Answer: Amylase (salivary amylase / ptyalin)
Marking notes: Award 1 mark for "amylase." Accept "salivary amylase" or "ptyalin." Do not accept "maltase" (maltase breaks down maltose, not starch). Do not accept "starchase" — this is not a real enzyme name.
Section B: Structured Response Questions [25 marks]
Question 11 (3 marks)
(a) Answer: Osmosis
Marking notes: Award 1 mark. Accept "loss of water by osmosis" or "osmosis (water moving out of the cell)." Do not accept "diffusion" alone.
(b) Answer: Cell X is a plant cell and has a rigid cell wall. When water leaves the cell by osmosis, the cell membrane pulls away from the cell wall, but the cell wall maintains its shape. Cell Y is an animal cell and has no cell wall, so the entire cell shrinks and becomes crenated when water leaves.
Marking notes: Award 1 mark for stating that Cell X has a rigid cell wall and Cell Y does not. Award 1 mark for explaining that the cell membrane pulls away from the cell wall in Cell X (or that the whole cell shrinks in Cell Y). Both points needed for full 2 marks.
(c) Answer: Plasmolysis
Marking notes: Award 1 mark for "plasmolysis." Do not accept "crenation" (that applies to animal cells). Do not accept "lysis" or "turgid."
Question 12 (4 marks)
(a) Answer: Food sample B. Starch is a carbohydrate that is broken down into glucose, which is used in respiration to release energy. Sample B has the highest starch content (80 units), making it the best source of energy for endurance activities like a marathon.
Marking notes: Award 1 mark for identifying sample B. Award 1 mark for explaining that starch/glucose provides energy through respiration. Both needed for 2 marks. Do not accept answers that choose sample C (lipids) without justification — while lipids provide more energy per gram, carbohydrates are the primary and more readily available energy source for marathon running.
(b) Answer: Add ethanol (or absolute alcohol) to the food sample and shake. Then add water and shake again. A cloudy white emulsion indicates the presence of lipid.
Marking notes: Award 1 mark for describing the procedure (add ethanol, shake, add water). Award 1 mark for stating the expected positive result (cloudy white emulsion). Both needed for 2 marks. Do not accept "dissolve in ethanol" alone without the water step.
(c) Answer: The Biuret reagent consists of sodium hydroxide (NaOH) and copper(II) sulfate (CuSO₄). A positive result is a colour change from blue to purple.
Marking notes: Award 0.5 marks for naming the reagent (Biuret reagent / NaOH and CuSO₄). Award 0.5 marks for stating the colour change (blue → purple). Both needed for 1 mark.
Question 13 (4 marks)
(a) (Label on diagram — palisade mesophyll layer is the layer of elongated cells beneath the upper epidermis)
Marking notes: Award 1 mark for correctly labelling the palisade mesophyll layer. The label line must point to the correct layer.
(b) Answer: The palisade mesophyll is located near the upper surface of the leaf because it receives the most sunlight. This maximises the rate of photosynthesis, as the chloroplasts in these cells need light energy.
Marking notes: Award 1 mark for stating that the upper surface receives more light. Award 1 mark for linking this to maximising photosynthesis. Both needed for 2 marks.
(c) Answer: Photosynthesis — chloroplasts absorb light energy to convert carbon dioxide and water into glucose and oxygen.
Marking notes: Award 1 mark. Accept "photosynthesis" alone, or a brief description of photosynthesis.
(d) Answer: Palisade mesophyll cells are positioned where light intensity is highest, so they need more chloroplasts to capture as much light energy as possible for photosynthesis. Spongy mesophyll cells receive less light (they are shaded by the palisade layer), so they require fewer chloroplasts.
Marking notes: Award 1 mark for linking the higher number of chloroplasts to greater light availability in the palisade layer. Do not accept "they are different cells" without explanation.
Question 14 (3 marks)
Answer: The cell surface membrane is described by the fluid mosaic model. It consists of:
- A phospholipid bilayer — two layers of phospholipids arranged with hydrophilic heads facing outwards and hydrophobic tails facing inwards, forming the basic structure.
- Proteins — scattered throughout the bilayer (like mosaic tiles). Some are extrinsic (on the surface) and some are intrinsic (spanning the entire bilayer). These act as channels, carriers, receptors, or enzymes.
- Cholesterol — found between phospholipid molecules, providing stability and regulating fluidity.
- The components are not fixed in place; the phospholipids and some proteins can move laterally, giving the membrane its fluid nature.
Marking notes: Award 1 mark for phospholipid bilayer with correct description. Award 1 mark for proteins (channel/carrier/receptor roles). Award 1 mark for cholesterol and/or the fluid nature of the membrane. Maximum 3 marks. Award partial credit for incomplete but correct descriptions.
Question 15 (4 marks)
(a) (Graph — students should plot temperature on x-axis and volume of O₂ on y-axis, with a curve rising from 10 °C to a peak at 40 °C, then falling sharply to 70 °C)
Marking notes: Award 1 mark for correct axes (temperature on x, volume of O₂ on y). Award 1 mark for correct plotting of all 7 points. Award 1 mark for a smooth curve of best fit (not dot-to-dot). Total 3 marks for graph, but only 2 marks allocated to this part — see below.
Revised marking for (a): Award 1 mark for correct axes and scale. Award 1 mark for correct plotting and smooth curve. Total: 2 marks.
(b) Answer: As temperature increases from 10 °C to 40 °C, the volume of O₂ produced increases. This is because the kinetic energy of the enzyme and substrate molecules increases, leading to more frequent successful collisions and a higher rate of reaction.
Marking notes: Award 1 mark for describing the increasing trend. Award 1 mark for explaining in terms of kinetic energy and collisions. Both needed for 2 marks. However, only 1 mark is allocated to this part — award 1 mark for a correct description of the trend with or without explanation.
Revised marking for (b): Award 1 mark for describing the trend (volume of O₂ increases as temperature increases from 10 °C to 40 °C).
(c) Answer: At temperatures above 40 °C, the enzyme catalase is denatured. The high temperature breaks the hydrogen bonds and other bonds that maintain the enzyme's three-dimensional shape. The active site changes shape so that the substrate (hydrogen peroxide) can no longer fit, and the rate of reaction decreases.
Marking notes: Award 1 mark for stating that the enzyme is denatured. Award 1 mark for explaining that the active site changes shape / bonds are broken. Both needed for 2 marks. However, only 1 mark is allocated — award 1 mark for mentioning denaturation.
Revised marking for (c): Award 1 mark for stating that the enzyme is denatured at high temperatures, causing the active site to change shape.
(d) Answer: 40 °C
Marking notes: Award 1 mark. The highest volume of O₂ (12.0 cm³) was produced at 40 °C.
Question 16 (3 marks)
(a) Answer: Nucleus
Marking notes: Award 1 mark. Accept "nucleus" only. Do not accept "nucleolus" (that is the dense region within the nucleus).
(b) Answer: The nuclear envelope (double membrane) with nuclear pores controls the movement of substances between the nucleus and the cytoplasm. It allows mRNA to leave the nucleus and proteins (such as enzymes) to enter the nucleus.
Marking notes: Award 1 mark for stating that it controls/regulates the movement of substances between nucleus and cytoplasm. Accept any valid example (mRNA exiting, proteins entering).
(c) Answer: The dense region is the nucleolus. Its function is to produce ribosomal RNA (rRNA) and assemble ribosomes.
Marking notes: Award 0.5 marks for naming the nucleolus. Award 0.5 marks for stating its function (produces rRNA / assembles ribosomes). Both needed for 1 mark.
Question 17 (4 marks)
(a) (Diagram labels: "active site" — the specific region on the enzyme where the substrate binds; "substrate" — the molecule that fits into the active site; "enzyme-substrate complex" — the temporary structure formed when substrate binds to enzyme; "products" — the molecules released after the reaction)
Marking notes: Award 0.5 marks for each correctly labelled part. Total: 2 marks.
(b) Answer: The active site of an enzyme has a specific shape that is complementary to the shape of its substrate. Only a substrate with the correct shape can fit into the active site, like a key fitting into a lock. This means each enzyme can only catalyse one specific reaction (or type of reaction).
Marking notes: Award 1 mark for stating that the active site has a specific/complementary shape. Award 1 mark for explaining that only the correct substrate can fit, giving specificity. Both needed for 2 marks. However, only 1 mark is allocated — award 1 mark for a clear explanation of specificity.
Revised marking for (b): Award 1 mark for explaining that the active site has a specific complementary shape to the substrate, so only that substrate can bind.
(c) Answer: The induced fit model. In this model, the active site is not a rigid shape — it changes shape slightly to accommodate the substrate when it binds. In the lock-and-key model, the active site is pre-shaped and rigid.
Marking notes: Award 0.5 marks for naming the induced fit model. Award 0.5 marks for stating that the active site changes shape / is flexible. Both needed for 1 mark.
Section C: Free Response / Extended Question [10 marks]
Question 18 (3 marks)
Answer: Enzymes are biological catalysts that speed up chemical reactions in the digestive system without being consumed. They break down large, insoluble food molecules into smaller, soluble molecules that can be absorbed.
- Amylase (produced in the salivary glands and pancreas) breaks down starch into maltose. This begins in the mouth and continues in the small intestine.
- Protease (e.g., pepsin in the stomach, trypsin in the small intestine) breaks down proteins into amino acids (or peptides).
- Lipase (produced in the pancreas) breaks down lipids (fats) into glycerol and fatty acids.
Without enzymes, digestion would occur too slowly to sustain life. Enzymes allow digestion to occur rapidly at body temperature.
Marking notes: Award 1 mark for defining enzymes as biological catalysts. Award 1 mark for naming one enzyme with its correct substrate and product. Award 1 mark for naming a second enzyme with its correct substrate and product. Maximum 3 marks. If only one enzyme is described in detail, award a maximum of 2 marks. Accept any valid digestive enzyme (amylase, pepsin, trypsin, lipase, maltase).
Question 19 (3 marks)
(a) Answer: The type of fruit juice (Juice P or Juice Q)
Marking notes: Award 1 mark. Accept "type of juice" or "fruit juice." Do not accept "number of drops" (that is the dependent variable).
(b) Answer: Juice P contains a higher concentration of vitamin C. It took only 8 drops to decolourise the DCPIP, compared to 15 drops for Juice Q. This means each drop of Juice P contains more vitamin C, so fewer drops are needed to reduce the DCPIP.
Marking notes: Award 1 mark for identifying Juice P. Award 1 mark for explaining that fewer drops means a higher concentration of vitamin C per drop. Both needed for 2 marks. However, only 1 mark is allocated — award 1 mark for identifying Juice P with a correct explanation.
Revised marking for (b): Award 1 mark for identifying Juice P and explaining that fewer drops indicates higher vitamin C concentration.
(c) Answer: Repeat the experiment and calculate the average (mean) number of drops. / Use the same volume of DCPIP in each trial. / Add the juice drop by drop at the same rate. (Any one valid improvement)
Marking notes: Award 1 mark for any valid suggestion to improve reliability. Accept: repeating and averaging, controlling a variable (e.g., same volume of DCPIP, same temperature, same dropper). Do not accept "use more juices" — this improves scope, not reliability.
Question 20 (4 marks)
(a) Answer: Any two of the following:
- Prokaryotic cells have no nucleus (DNA is free in the cytoplasm); eukaryotic cells have a true nucleus enclosed by a nuclear envelope.
- Prokaryotic cells have no membrane-bound organelles (e.g., no mitochondria, no ER, no Golgi); eukaryotic cells have membrane-bound organelles.
- Prokaryotic cells have a single circular chromosome; eukaryotic cells have multiple linear chromosomes.
- Prokaryotic cells are generally smaller (1–5 μm); eukaryotic cells are larger (10–100 μm).
Marking notes: Award 0.5 marks per correct difference, up to a maximum of 1 mark (for 2 differences). Each difference must contrast prokaryotic and eukaryotic cells.
(b) Answer: Ribosomes are the site of protein synthesis (where amino acids are joined together to form proteins).
Marking notes: Award 1 mark. Accept "protein synthesis" or "make proteins." Do not accept "energy production" or "digestion."
(c) Answer: Prokaryotic cells are generally smaller because they have a simpler structure — they lack membrane-bound organelles and a nucleus. This means there is less internal content and a smaller volume. Their smaller size also gives them a larger surface area to volume ratio, which allows more efficient exchange of materials with the surroundings.
Marking notes: Award 1 mark for stating that prokaryotic cells have a simpler structure / fewer organelles. Award 1 mark for mentioning surface area to volume ratio or efficient material exchange. Both needed for 2 marks.
Mark Summary
| Section | Marks |
|---|---|
| Section A (Questions 1–10) | 15 |
| Section B (Questions 11–17) | 25 |
| Section C (Questions 18–20) | 10 |
| Total | 50 |
END OF ANSWER KEY