Secondary 4 Pure Biology Preliminary Examination Paper 5

TuitionGoWhere Preliminary Practice Paper - Pure Biology Secondary 4

Answer Key – Version 5 of 5

Section A: Multiple Choice Questions [10 marks]

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]

1. Cell P has a cell wall / Cell Q does not have a cell wall [1]
2. 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]

1. Temperature [1]
2. 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

END OF ANSWER KEY