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Secondary 4 Pure Biology Preliminary Examination Paper 5
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TuitionGoWhere Preliminary Practice Paper - Pure Biology Secondary 4
TuitionGoWhere Secondary School (AI)
| Subject: | Pure Biology |
| Level: | Secondary 4 |
| Paper: | Preliminary Paper 2 – Structured & Free Response |
| Duration: | 1 hour 45 minutes |
| Total Marks: | 70 |
| Name: | ______________________________ |
| Class: | ______________________________ |
| Date: | ______________________________ |
| Version: | 5 of 5 |
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, graphs, or rough working.
- Do not use correction fluid.
- The number of marks is given in brackets [ ] at the end of each question or part question.
- The total mark for this paper is 70.
Section A: Multiple Choice Questions [10 marks]
Questions 1–10
Each question is worth 1 mark. Shade the correct option on the Optical Answer Sheet (OAS) provided. Use soft pencil (2B).
1. Which cell structure is responsible for the synthesis of proteins in both plant and animal cells?
A. Golgi body
B. Mitochondrion
C. Ribosome
D. Vacuole
2. A student observed a cell under an electron micrograph and noted the presence of a double membrane, cristae, and a matrix. Which organelle was the student observing?
A. Chloroplast
B. Endoplasmic reticulum
C. Mitochondrion
D. Nucleus
3. Which of the following is a function of the cell wall in plant cells?
A. Controls the entry and exit of substances
B. Provides shape and prevents the cell from bursting in a hypotonic solution
C. Stores genetic information
D. Synthesises proteins for export
4. An enzyme was added to a starch solution at 37 °C. After 10 minutes, iodine solution was added and the mixture remained brown. What conclusion can be drawn?
A. The enzyme denatured at 37 °C.
B. The starch was broken down into reducing sugars.
C. The enzyme is a protease.
D. The pH was unsuitable for the enzyme.
5. Which biomolecule yields the most energy per gram when completely oxidised in the body?
A. Carbohydrate
B. Fat
C. Protein
D. Vitamin
6. Red blood cells were placed in distilled water. After 30 minutes, the cells were observed to have burst. Which process caused water to enter the cells?
A. Active transport
B. Diffusion
C. Osmosis
D. Facilitated diffusion
7. Which of the following correctly describes the effect of increasing temperature on enzyme activity?
A. Activity increases continuously as temperature rises.
B. Activity increases to an optimum, then decreases sharply as the enzyme denatures.
C. Activity decreases continuously as temperature rises.
D. Activity is unaffected by temperature changes.
8. A plant cell is placed in a concentrated salt solution. Which diagram best represents the appearance of the cell after 30 minutes?
A. Cell membrane pressed against the cell wall
B. Cell membrane pulled away from the cell wall (plasmolysed)
C. Cell swollen and turgid
D. Cell unchanged in appearance
9. Which of the following is a reducing sugar?
A. Sucrose
B. Starch
C. Maltose
D. Glycogen
10. Which cell structure is found in plant cells but NOT in animal cells?
A. Cell membrane
B. Mitochondrion
C. Chloroplast
D. Ribosome
Section B: Structured Questions [40 marks]
Questions 11–18
Answer all questions in the spaces provided.
11. Fig. 11.1 shows two types of cells, P and Q, observed under a light microscope.
(Diagram description for generation: Cell P is a rectangular plant cell with a visible cell wall, large central vacuole, and chloroplasts. Cell Q is an irregularly shaped animal cell with a visible nucleus, cytoplasm, and cell membrane but no cell wall or chloroplasts.)
(a) Identify cell P and cell Q. [2]
Cell P: ______________________________
Cell Q: ______________________________
(b) State two structural differences between cell P and cell Q that can be seen under a light microscope. [2]
(c) Name one organelle visible only under an electron microscope that would be found in both cell P and cell Q. State its function. [2]
Organelle: ______________________________
Function: ______________________________
(d) Cell P is a palisade mesophyll cell. Explain how its structure is adapted for its function in photosynthesis. [2]
[Total: 8 marks]
12. An experiment was carried out to investigate the effect of pH on the activity of enzyme catalase in potato tissue. Five test tubes were set up as shown in Table 12.1. Each test tube contained 5 cm³ of hydrogen peroxide solution and one small cube of potato tissue. The volume of oxygen gas collected in 2 minutes was measured.
| Test tube | pH of buffer solution | Volume of O₂ collected in 2 min / cm³ |
|---|---|---|
| A | 3 | 2.1 |
| B | 5 | 5.8 |
| C | 7 | 9.4 |
| D | 9 | 4.3 |
| E | 11 | 0.6 |
(a) State the aim of this experiment. [1]
(b) Identify the independent variable and the dependent variable. [2]
Independent: ______________________________
Dependent: ______________________________
(c) State two variables that should be kept constant in this experiment. [2]
(d) Describe the trend shown by the results. [2]
(e) Explain why very little oxygen was collected in test tube E. [2]
(f) Predict the volume of oxygen collected at pH 6. Justify your answer. [1]
[Total: 10 marks]
13. Fig. 13.1 shows the structure of a molecule of a biological polymer.
(Diagram description: A simplified diagram showing a chain of glucose monomers linked by glycosidic bonds, representing a portion of a starch molecule. Three glucose units are shown connected, with –OH groups and oxygen bridges labelled.)
(a) Name the polymer shown in Fig. 13.1. [1]
(b) Name the type of chemical bond labelled X between the monomers. [1]
(c) Name the chemical reaction that joins the monomers together. [1]
(d) Describe a chemical test that could be used to confirm the identity of this polymer. Include the reagent used, the procedure, and the positive result. [3]
Reagent: ______________________________
Procedure: ______________________________
Positive result: ______________________________
(e) State one function of this polymer in living organisms. [1]
[Total: 7 marks]
14. A student carried out food tests on a sample of food. The results are shown in Table 14.1.
| Test | Reagent used | Observation |
|---|---|---|
| Test 1 | Iodine solution | Solution turned blue-black |
| Test 2 | Benedict's solution (heated) | Orange-red precipitate formed |
| Test 3 | Biuret solution | Solution turned violet/purple |
| Test 4 | Ethanol emulsion test | Milky-white emulsion formed |
(a) State what each positive result indicates about the food sample. [4]
Test 1: ______________________________
Test 2: ______________________________
Test 3: ______________________________
Test 4: ______________________________
(b) For Test 2, explain why the mixture must be heated. [1]
(c) Name the biomolecule class detected in Test 4. [1]
[Total: 6 marks]
15. Fig. 15.1 shows a red blood cell placed in three different solutions, X, Y, and Z.
(Diagram description: Three diagrams side by side. In solution X, the red blood cell appears normal and biconcave. In solution Y, the cell appears swollen and spherical. In solution Z, the cell appears shrunken and crenated.)
(a) For each solution, state whether it is hypotonic, isotonic, or hypertonic relative to the cell's cytoplasm. [3]
Solution X: ______________________________
Solution Y: ______________________________
Solution Z: ______________________________
(b) Explain what happened to the red blood cell in solution Y in terms of osmosis. [3]
(c) Name the process by which water molecules move across the cell membrane in this experiment. [1]
[Total: 7 marks]
16. Fig. 16.1 shows the effect of temperature on the rate of an enzyme-catalysed reaction.
(Graph description: A bell-shaped curve with temperature (°C) on the x-axis from 0 to 70, and rate of reaction (arbitrary units) on the y-axis. The curve rises from 0 °C to a peak at 37 °C, then drops sharply to near zero at 65 °C.)
(a) Describe the effect of temperature on the rate of reaction from 0 °C to 37 °C. [2]
(b) Explain the increase in the rate of reaction between 0 °C and 37 °C. [2]
(c) Explain why the rate of reaction decreases sharply above 37 °C. [2]
(d) A student repeated the experiment using pepsin instead of the original enzyme. Predict the optimum temperature for pepsin and explain your reasoning. [2]
Prediction: ______________________________
Reason: _______________________________________________________________
[Total: 8 marks]
17. Fig. 17.1 shows a specialised cell from the human body.
(Diagram description: A long, thin nerve cell/neurone with a cell body containing a nucleus, many short dendrites extending from the cell body, and a long axon extending from the opposite end, surrounded by a myelin sheath with nodes of Ranvier visible.)
(a) Name the cell shown in Fig. 17.1. [1]
(b) State the function of this cell. [1]
(c) Explain how two structural features of this cell are adapted to its function. [4]
Feature 1: ______________________________
Explanation: _______________________________________________________________
Feature 2: ______________________________
Explanation: _______________________________________________________________
[Total: 6 marks]
18. A piece of fresh potato of mass 5.0 g was placed in a sucrose solution of concentration 0.4 mol/dm³ for 30 minutes. After removal and blotting dry, the mass was found to be 4.6 g.
(a) Calculate the percentage change in mass of the potato. Show your working. [2]
(b) Explain why the potato lost mass in terms of water potential and osmosis. [3]
(c) Predict and explain what would happen to the mass of a similar potato piece if it were placed in distilled water for 30 minutes. [2]
[Total: 7 marks]
Section C: Free Response Question [20 marks]
Questions 19–20
Answer all questions. Write your answers in the spaces provided. You are advised to organise your answers clearly and use appropriate biological terminology.
19. Enzymes are biological catalysts that play a vital role in all living organisms.
(a) Explain what is meant by the term biological catalyst. [2]
(b) Describe the 'lock and key' hypothesis of enzyme action. In your answer, refer to the enzyme, substrate, active site, and enzyme-substrate complex. [4]
(c) Explain how both temperature and pH affect enzyme activity. Include reference to the active site in your answer. [6]
Effect of temperature:
Effect of pH:
(d) A student investigated the effect of enzyme concentration on the rate of an enzyme-catalysed reaction, keeping all other variables constant. Sketch a graph to show the expected results. Label both axes clearly. [3]
(Grid/graph space provided)
(e) Explain the shape of the graph you have drawn in (d). [3]
(f) Give one commercial or industrial application of enzymes. [2]
[Total: 20 marks]
20. Cell specialisation is a key concept in multicellular organisms.
(a) Define the term cell specialisation. [1]
(b) Describe how each of the following cells is specialised for its function. In each case, refer to specific structural features. [12]
(i) Sperm cell [4]
(ii) Red blood cell [4]
(iii) Root hair cell [4]
(c) Explain why cell specialisation is important for the survival of a multicellular organism. [3]
[Total: 16 marks]
END OF PAPER
Total: 70 marks
Answers
TuitionGoWhere Preliminary Practice Paper - Pure Biology Secondary 4
Answer Key – Version 5 of 5
Section A: Multiple Choice Questions [10 marks]
| Question | Answer | Marks | Notes |
|---|---|---|---|
| 1 | C – Ribosome | [1] | Ribosomes are the site of protein synthesis in all cells. Golgi body packages proteins; mitochondrion is for aerobic respiration; vacuole is for storage. |
| 2 | C – Mitochondrion | [1] | The double membrane, cristae, and matrix are distinctive features of mitochondria. Chloroplasts have thylakoids/grana; ER does not have cristae; nucleus has nuclear envelope but no cristae. |
| 3 | B – Provides shape and prevents the cell from bursting in a hypotonic solution | [1] | The cell wall is rigid and fully permeable. A is the function of the cell membrane; C is the nucleus; D is the ribosome/rough ER. |
| 4 | B – The starch was broken down into reducing sugars | [1] | Iodine remaining brown (not blue-black) indicates starch is no longer present, meaning it has been digested by the enzyme. |
| 5 | B – Fat | [1] | Fats yield approximately 37 kJ/g, compared to carbohydrates (~17 kJ/g) and proteins (~17 kJ/g). Vitamins are not energy sources. |
| 6 | C – Osmosis | [1] | Osmosis is the net movement of water molecules from a region of higher water potential (distilled water) to a region of lower water potential (cell cytoplasm) across a partially permeable membrane. |
| 7 | B – Activity increases to an optimum, then decreases sharply as the enzyme denatures | [1] | This is the classic bell-shaped curve. Enzymes have an optimum temperature; beyond it, the active site is permanently altered. |
| 8 | B – Cell membrane pulled away from the cell wall (plasmolysed) | [1] | In a hypertonic solution, water leaves the cell by osmosis. The cell membrane shrinks away from the rigid cell wall – this is plasmolysis. |
| 9 | C – Maltose | [1] | Maltose is a reducing sugar (a disaccharide with a free aldehyde group). Sucrose is a non-reducing sugar. Starch and glycogen are polysaccharides and do not act as reducing sugars in the standard Benedict's test. |
| 10 | C – Chloroplast | [1] | Chloroplasts are found only in plant cells (and some protists). Cell membrane, mitochondrion, and ribosome are found in both plant and animal cells. |
Section B: Structured Questions [40 marks]
Question 11 [8 marks]
(a) Identify cell P and cell Q. [2]
- Cell P: Plant cell (or palisade mesophyll cell) [1]
- Cell Q: Animal cell [1]
Marking note: Accept any specific plant cell type (e.g., palisade cell, epidermal cell) for P. Accept any specific animal cell type for Q.
(b) Two structural differences visible under a light microscope. [2]
- Cell P has a cell wall / Cell Q does not have a cell wall [1]
- Cell P has a large central vacuole / Cell Q has small or no vacuoles [1]
Also accept: Cell P has chloroplasts / Cell Q does not. Cell P is rectangular/fixed shape / Cell Q is irregular shape.
(c) One organelle visible only under an electron microscope found in both cells, and its function. [2]
- Organelle: Mitochondrion (also accept: endoplasmic reticulum, Golgi body, ribosome) [1]
- Function: Site of aerobic respiration / produces ATP (energy) [1]
If ribosome is named: Function = site of protein synthesis. If rough ER: synthesis and transport of proteins. If Golgi body: modifies, packages, and secretes proteins.
(d) How the palisade mesophyll cell is adapted for photosynthesis. [2]
- Contains many chloroplasts to absorb light energy for photosynthesis [1]
- Positioned near the upper surface of the leaf to receive maximum light / cells are elongated and closely packed to maximise light absorption [1]
Marking note: Must link structure to function for full marks. Simply stating "has chloroplasts" without linking to photosynthesis = 1 mark only.
Question 12 [10 marks]
(a) Aim of the experiment. [1]
- To investigate the effect of pH on the activity of catalase (in potato tissue) [1]
Marking note: Must mention both pH (independent variable) and catalase activity (dependent variable).
(b) Independent and dependent variables. [2]
- Independent: pH of the buffer solution [1]
- Dependent: Volume of oxygen gas collected in 2 minutes (or rate of reaction / catalase activity) [1]
(c) Two variables to be kept constant. [2]
- Temperature [1]
- Volume/concentration of hydrogen peroxide solution (also accept: size/mass of potato cube, time of reaction) [1]
Marking note: Do not accept "type of enzyme" or "type of potato" as these are inherent to the setup.
(d) Trend shown by the results. [2]
- As pH increases from 3 to 7, the volume of oxygen collected increases (catalase activity increases) [1]
- As pH increases from 7 to 11, the volume of oxygen collected decreases (catalase activity decreases) / The optimum pH is pH 7 [1]
Marking note: Must describe both the increase and decrease, or state the optimum pH with supporting data.
(e) Explanation for very little oxygen in test tube E (pH 11). [2]
- At pH 11, the enzyme catalase is denatured [1]
- The shape of the active site is altered so the substrate (hydrogen peroxide) can no longer fit into it / enzyme-substrate complexes cannot form [1]
Marking note: "Denatured" alone = 1 mark. Must explain what denaturation means at the molecular level for the second mark.
(f) Prediction for pH 6. [1]
- Volume of oxygen collected would be between 5.8 cm³ and 9.4 cm³ (accept any value in this range, e.g., ~7–8 cm³) [1]
- Justification: pH 6 is between pH 5 and pH 7, and since activity increases from pH 5 to pH 7, the value should be between the values at those two pH values [1]
Marking note: The justification mark is conditional on a reasonable predicted value.
Question 13 [7 marks]
(a) Name the polymer. [1]
- Starch (also accept: amylose or amylopectin) [1]
(b) Type of bond labelled X. [1]
- Glycosidic bond [1]
(c) Chemical reaction that joins monomers. [1]
- Condensation (or condensation reaction / dehydration synthesis) [1]
(d) Chemical test for this polymer. [3]
- Reagent: Iodine solution (or iodine/potassium iodide solution) [1]
- Procedure: Add a few drops of iodine solution to the sample [1]
- Positive result: Solution turns blue-black [1]
Marking note: Must include all three components for full marks. "Add iodine" alone without stating the colour change = 2 marks.
(e) One function in living organisms. [1]
- Energy storage (in plants) [1]
Also accept: stores glucose / source of energy when hydrolysed.
Question 14 [6 marks]
(a) What each positive result indicates. [4]
- Test 1: Starch is present [1]
- Test 2: Reducing sugar is present [1]
- Test 3: Protein is present [1]
- Test 4: Fat/lipid is present [1]
(b) Why the mixture must be heated in Test 2. [1]
- Heating is required for Benedict's solution to react with reducing sugars / to produce a coloured precipitate [1]
Marking note: Accept "to speed up the reaction" or "to provide activation energy for the redox reaction between copper(II) sulfate and the reducing sugar."
(c) Biomolecule class detected in Test 4. [1]
- Lipid (or fat) [1]
Question 15 [7 marks]
(a) Nature of each solution. [3]
- Solution X: Isotonic [1]
- Solution Y: Hypotonic [1]
- Solution Z: Hypertonic [1]
(b) Explanation for what happened in solution Y. [3]
- Solution Y has a higher water potential than the cytoplasm of the red blood cell [1]
- Water molecules move into the cell by osmosis (from a region of higher water potential to a region of lower water potential) [1]
- The cell swells and may burst (haemolysis) because the cell membrane cannot withstand the increasing internal pressure [1]
Marking note: Must use the term "osmosis" and refer to water potential or concentration gradient. Award 1 mark for each valid point.
(c) Name of the process. [1]
- Osmosis [1]
Question 16 [8 marks]
(a) Effect of temperature from 0 °C to 37 °C. [2]
- As temperature increases from 0 °C to 37 °C, the rate of reaction increases [1]
- The rate is highest at 37 °C (the optimum temperature) [1]
Marking note: Must describe the trend, not just state "it increases." Award 1 mark for stating the trend and 1 mark for identifying the optimum.
(b) Explanation for the increase (0 °C to 37 °C). [2]
- As temperature increases, the kinetic energy of both enzyme and substrate molecules increases [1]
- This leads to more frequent successful collisions / more enzyme-substrate complexes are formed per unit time, so the rate of reaction increases [1]
Marking note: Must refer to kinetic energy and collisions/complex formation.
(c) Explanation for the decrease above 37 °C. [2]
- Above 37 °C, the enzyme denatures [1]
- The shape of the active site changes so that substrate molecules can no longer bind to it / enzyme-substrate complexes can no longer form [1]
Marking note: "Denatured" alone = 1 mark. Must explain the effect on the active site for the second mark.
(d) Prediction for pepsin. [2]
- Prediction: Optimum temperature of approximately 37 °C [1]
- Reason: Pepsin is an enzyme found in the human body (stomach), and human body temperature is approximately 37 °C, so the enzyme has evolved to work best at this temperature [1]
Marking note: Accept any temperature in the range 35–40 °C. The reason must link to human body temperature.
Question 17 [6 marks]
(a) Name the cell. [1]
- Neurone (or nerve cell) [1]
(b) Function of this cell. [1]
- To transmit electrical impulses (nerve impulses / action potentials) from one part of the body to another [1]
(c) Two structural adaptations and explanations. [4]
Feature 1: Long axon [1]
- Explanation: The long axon allows the cell to transmit impulses over long distances without loss of signal [1]
Feature 2: Myelin sheath [1]
- Explanation: The myelin sheath insulates the axon and speeds up the transmission of nerve impulses (by saltatory conduction) [1]
Also accept:
- Many dendrites – to receive impulses from many other neurones / increase surface area for connections
- Nodes of Ranvier – allow the impulse to jump from node to node, speeding up transmission
- Synaptic terminals – release neurotransmitters to pass the signal to the next neurone or effector
Marking note: 1 mark for naming the feature, 1 mark for linking it to function. Award a maximum of 2 features × 2 marks = 4 marks.
Question 18 [7 marks]
(a) Calculate the percentage change in mass. [2]
Working:
Change in mass = 5.0 g − 4.6 g = 0.4 g [1]
Percentage change = (0.4 / 5.0) × 100 = 8% [1]
Marking note: Award 1 mark for correct working (subtraction and formula) and 1 mark for correct final answer with unit (%). Accept 8.0%.
(b) Explanation in terms of water potential and osmosis. [3]
- The sucrose solution has a lower water potential than the potato cells [1]
- Water moves out of the potato cells by osmosis (from a region of higher water potential in the potato cells to a region of lower water potential in the sucrose solution) [1]
- The net loss of water causes the potato to lose mass [1]
Marking note: Must use the term "osmosis" and correctly describe the water potential gradient. Award 1 mark per valid point.
(c) Prediction for distilled water. [2]
- The potato would gain mass [1]
- Distilled water has a higher water potential than the potato cells, so water moves into the potato cells by osmosis, causing the potato to gain mass [1]
Marking note: The explanation mark is conditional on a correct prediction. Must mention osmosis and water potential.
Section C: Free Response Question [20 marks]
Question 19 [20 marks]
(a) Explain what is meant by "biological catalyst." [2]
- A biological catalyst is an enzyme [1]
- It is a protein that speeds up the rate of a chemical reaction without being used up / without being permanently changed, and it is produced by living organisms [1]
Marking note: Must state that it is an enzyme AND that it speeds up reactions without being consumed. Award 1 mark for each point.
(b) Describe the 'lock and key' hypothesis. [4]
- The enzyme has a specific region called the active site [1]
- The substrate has a shape that is complementary to the shape of the active site (like a key fits into a lock) [1]
- The substrate binds to the active site of the enzyme, forming an enzyme-substrate complex [1]
- The reaction takes place / the substrate is converted into products, which then leave the active site, and the enzyme is unchanged and can be reused [1]
Marking note: Must include all four key terms (enzyme, substrate, active site, enzyme-substrate complex) and describe the process. Award 1 mark per valid point, maximum 4 marks.
(c) Effect of temperature and pH on enzyme activity. [6]
Effect of temperature: [3]
- At low temperatures, enzyme activity is low because molecules have low kinetic energy, so there are fewer successful collisions between enzyme and substrate [1]
- As temperature increases, kinetic energy increases, so the rate of reaction increases up to the optimum temperature [1]
- Above the optimum temperature, the enzyme denatures – the shape of the active site is permanently altered so the substrate can no longer fit, and the enzyme loses its function [1]
Effect of pH: [3]
- Each enzyme has an optimum pH at which it works best [1]
- At the optimum pH, the shape of the active site is correct for the substrate to bind [1]
- At pH values above or below the optimum, the enzyme denatures – the ionic bonds holding the enzyme's shape are disrupted, altering the shape of the active site so the substrate can no longer bind [1]
Marking note: Award up to 3 marks for temperature and 3 marks for pH. Must refer to the active site in each explanation for full marks.
(d) Sketch graph of enzyme concentration vs. rate of reaction. [3]
- x-axis: Enzyme concentration (arbitrary units) [1]
- y-axis: Rate of reaction (arbitrary units) [1]
- Shape: A straight line passing through the origin (linear relationship) – as enzyme concentration increases, rate of reaction increases proportionally [1]
Marking note: Award 1 mark for each correctly labelled axis and 1 mark for the correct shape (straight line through origin). If the graph curves/plateaus, award only 1 mark for axes (substrate becomes limiting – but the question states all other variables are constant, so a linear relationship is expected).
(e) Explanation of the graph shape. [3]
- As enzyme concentration increases, there are more enzyme molecules available [1]
- This means more active sites are available for substrate molecules to bind to [1]
- Therefore, more enzyme-substrate complexes are formed per unit time, so the rate of reaction increases proportionally [1]
Marking note: Award 1 mark per valid point. Must explain the relationship, not just describe the graph.
(f) One commercial or industrial application of enzymes. [2]
Accept any one of the following (or other valid examples):
- Biological detergents – proteases and lipases are added to break down protein and fat stains on clothes [2]
- Food industry – pepsin or rennet is used in cheese-making to curdle milk [2]
- Medical diagnostics – glucose oxidase is used in blood glucose test strips to measure blood sugar levels [2]
- Textile industry – cellulases are used to soften denim fabric (stone-washing effect) [2]
Marking note: Award 1 mark for naming the application and 1 mark for explaining how the enzyme is used. A named enzyme must be included for full marks.
Question 20 [16 marks]
(a) Define cell specialisation. [1]
- Cell specialisation is the process by which cells develop specific structures (or become differentiated) to carry out specific functions [1]
Marking note: Must convey that cells become different/adapted for particular roles.
(b) How each cell is specialised. [12]
(i) Sperm cell [4]
- Has a tail (flagellum) for swimming/motility towards the egg [1]
- Contains many mitochondria in the mid-piece to provide ATP (energy) for movement [1]
- Has an acrosome (enzyme-filled cap) at the head that releases enzymes to digest the egg's outer layer for penetration [1]
- Has a haploid nucleus (half the number of chromosomes) so that upon fertilisation, the diploid number is restored [1]
Also accept: streamlined/compact head for reduced resistance during swimming; small size for ease of movement.
Marking note: Award 1 mark per valid structure-function link, maximum 4 marks.
(ii) Red blood cell [4]
- Biconcave disc shape provides a large surface area to volume ratio for efficient gas exchange (oxygen uptake and carbon dioxide release) [1]
- No nucleus – provides more space for haemoglobin to carry more oxygen [1]
- Contains haemoglobin which binds to oxygen (forming oxyhaemoglobin) for oxygen transport [1]
- Small and flexible – can squeeze through narrow capillaries to deliver oxygen to all body tissues [1]
Also accept: thin cell membrane for short diffusion distance; no mitochondria so that oxygen is not used up by the cell itself.
Marking note: Award 1 mark per valid structure-function link, maximum 4 marks.
(iii) Root hair cell [4]
- Has a long, thin extension (root hair) that increases the surface area for absorption of water and mineral ions from the soil [1]
- Has a thin cell wall to allow easy passage of water and dissolved minerals [1]
- Contains a large vacuole with cell sap (concentrated solution) that maintains a water potential gradient for water uptake by osmosis [1]
- Has no chloroplasts (not needed as the cell is underground) – energy is obtained from the shoot / has mitochondria for active transport of mineral ions [1]
Also accept: numerous mitochondria to provide ATP for active transport of mineral ions against a concentration gradient.
Marking note: Award 1 mark per valid structure-function link, maximum 4 marks.
(c) Why cell specialisation is important for multicellular organisms. [3]
- Different cells perform different functions, allowing the organism to carry out many different processes simultaneously [1]
- This increases efficiency – specialised cells are better at their specific function than a generalised cell would be [1]
- It allows for the division of labour among cells, enabling complex processes (e.g., digestion, transport, reproduction) to occur in a coordinated manner, which is essential for the survival of a multicellular organism [1]
Marking note: Award 1 mark per valid point, maximum 3 marks. Answers must go beyond simply restating the definition of specialisation.
Mark Summary
| Section | Marks |
|---|---|
| A: Multiple Choice (Q1–10) | 10 |
| B: Structured (Q11–18) | 40 |
| C: Free Response (Q19–20) | 20 |
| Total | 70 |
END OF ANSWER KEY