2025 Cambridge IAS-Level Biology (9700) Practice Paper with Answers

Inhibitor X is a competitive inhibitor because increasing the substrate concentration allows the substrate to outcompete the inhibitor for the active site, meaning the maximum rate of reaction (\(V_{\max}\)) can still be reached. This means Inhibitor X binds directly to the active site. Inhibitor Y is a non-competitive inhibitor because it reduces the \(V_{\max}\) regardless of substrate concentration. This is because non-competitive inhibitors bind to an allosteric site, altering the shape of the active site so that the substrate can no longer bind or be converted to product.

[1 mark] - Correctly identifies that Inhibitor X binds to the active site and its effect can be overcome by increasing substrate concentration.

Denaturation by heating to \(70\ ^\circ\text{C}\) provides enough kinetic energy to break weak, non-covalent interactions like hydrogen bonds and ionic bonds which maintain the specific tertiary structure of the active site. Disulfide bonds, which are strong covalent bonds, are much more stable and are not broken at this temperature.

[1 mark] - Identifies that hydrogen and ionic bonds are broken, while disulfide bonds remain intact.

A mature xylem vessel element is a dead cell that lacks cytoplasm, has a lignified cell wall to withstand negative pressure, and is not associated with companion cells. A mature phloem sieve tube element is a living cell that contains peripheral cytoplasm (though it lacks a nucleus and many organelles), does not have lignin in its cell wall, and is closely associated with companion cells via plasmodesmata to maintain its metabolic activities.

[1 mark] - Correctly identifies cytoplasm, lignin, and companion cell presence/absence for both xylem vessel elements and phloem sieve tube elements.

In a herbaceous dicotyledonous stem, vascular bundles are located in a ring around the periphery, with the xylem tissue on the inner side (towards the pith) and phloem on the outer side (towards the cortex). In a herbaceous dicotyledonous root, there is a central vascular bundle (stele or vascular cylinder) where the xylem forms a central 'X' or star-like shape, and the phloem tissue is clustered in the regions between the arms of this star.

[1 mark] - Correctly identifies the anatomical distribution of xylem and phloem in both dicotyledonous stems and roots.

Malaria is caused by Plasmodium species, which are protoctists. It is transmitted via the vector of a female Anopheles mosquito and infects liver cells and red blood cells. Tuberculosis is caused by Mycobacterium tuberculosis (or Mycobacterium bovis), which is a bacterium. It is transmitted via the inhalation of aerosolized respiratory droplets and primarily infects the lungs.

[1 mark] - Correctly identifies the pathogen type, transmission method, and primary infection site for both malaria and tuberculosis.

Cholesterol sits between the hydrophobic tails of phospholipids, regulating membrane fluidity (preventing the membrane from becoming too fluid at high temperatures or too rigid/frozen at low temperatures) and contributing to mechanical stability. Glycoproteins extend outwards from the membrane surface, where they act as receptors for cell signaling molecules (like hormones) and act as cell surface markers (antigens) for cell-to-cell recognition.

[1 mark] - Identifies the correct physiological roles of cholesterol and glycoproteins in cell membranes.

The apoplast pathway consists of water moving through the cell walls and intercellular spaces. The Casparian strip is a band of suberin (a waterproof substance) located in the cell walls of endodermal cells. This blocks the apoplast pathway, forcing water and dissolved mineral ions to cross the selectively permeable cell surface membrane of endodermal cells to enter their cytoplasm (joining the symplast pathway). This allows the plant to regulate which ions enter the xylem.

[1 mark] - Correctly describes the apoplast pathway as cell-wall-based, the symplast pathway as cytoplasm-based, and that the Casparian strip forces apoplastic water to enter the symplast pathway at the endodermis.

According to the distribution of tissues in the gas exchange system:
- Trachea contains C-shaped rings of cartilage, smooth muscle, and goblet cells within its ciliated epithelium.
- Bronchioles do not have cartilage (which is absent to allow flexibility/constriction), but they do have smooth muscle to control airway diameter, and they do not have goblet cells (mucus in narrow bronchioles would block airflow).
- Alveoli have none of these; they consist only of a single layer of squamous epithelial cells and elastic fibres to maximize gas exchange efficiency.

[1 mark] - Correctly identifies the presence and absence of cartilage, smooth muscle, and goblet cells across the trachea, bronchioles, and alveoli.

Non-competitive inhibitors bind to an allosteric site (a site other than the active site) on the enzyme. This changes the shape of the active site, preventing the catalysis of the reaction even if the substrate binds. Consequently, the maximum rate of reaction (\(V_{\max}\)) is reduced. Increasing the substrate concentration does not overcome non-competitive inhibition, so the substrate concentration required to reach half of the new \(V_{\max}\) (the \(K_m\)) remains the same. Competitive inhibitors, on the other hand, increase the \(K_m\) but leave \(V_{\max}\) unchanged.

B is correct. 1 mark for identifying that a decrease in \(V_{\max}\) with an unchanged \(K_m\) is characteristic of a non-competitive inhibitor, which binds to a site other than the active site.

Mature xylem vessel elements are dead cells that lose their living contents, so cytoplasm is absent. Their cell walls are heavily thickened with waterproof lignin, and they do not have sieve plates. Mature phloem sieve tube elements are living cells containing a thin layer of peripheral cytoplasm, but lack a nucleus and most organelles. Their walls are not lignified (composed of cellulose), and they have perforated end walls called sieve plates to allow the transport of organic solutes.

A is correct. 1 mark for identifying the correct combination of features for both mature xylem vessel elements and mature phloem sieve tube elements.

Cholera is caused by the bacterium Vibrio cholerae. It is transmitted via the faecal-oral route through contaminated water or food. Once ingested, the bacteria colonise the lining of the small intestine, where they secrete cholera toxin, leading to severe watery diarrhoea.

A is correct. 1 mark for correctly matching cholera with its bacterial pathogen, transmission via contaminated water or food, and the small intestine as its primary site of action.

Glycolipids project from the outer surface of the cell membrane and act as cell-surface markers or antigens involved in cell-to-cell recognition. Cholesterol regulates membrane fluidity and stability but does not act as a hormone receptor. Glycoproteins (not cholesterol) act as receptors for hormones. The hydrophobic core of the phospholipid bilayer acts as a barrier to polar ions, preventing their free diffusion.

B is correct. 1 mark for identifying that glycolipids function as antigens for cell-to-cell recognition.

The apoplast pathway is the movement of water through the cell walls and intercellular spaces. At the endodermis, the cell walls contain a band of suberin called the Casparian strip, which is impermeable to water. This blocks the apoplast pathway, forcing water and solutes to cross the cell membrane of the endodermal cells and enter the symplast pathway. The symplast pathway involves water moving through the cytoplasm via plasmodesmata. Water cannot be actively transported; it moves passively down a water potential gradient.

A is correct. 1 mark for correctly explaining that the Casparian strip blocks the apoplast pathway at the endodermis, redirecting water into the symplast pathway.

Bronchioles lack cartilage and goblet cells to ensure their walls are thin and flexible. However, they contain smooth muscle in their walls, which can contract or relax to adjust the diameter of the airway and regulate airflow to the alveoli.

A is correct. 1 mark for identifying that bronchioles do not have cartilage or goblet cells, but do possess smooth muscle in their walls.

Carbon dioxide reacts with water inside the red blood cell to form carbonic acid, a reaction catalyzed by the enzyme carbonic anhydrase. The carbonic acid then dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)). The \(\text{HCO}_3^-\) ions diffuse out of the red blood cell into the plasma. To maintain electrical neutrality, chloride ions (\(\text{Cl}^-\)) diffuse into the red blood cell from the plasma. This is known as the chloride shift.

A is correct. 1 mark for identifying carbonic anhydrase as the catalyst, hydrogencarbonate leaving the cell, and chloride entering the cell.

Increasing the temperature from \(20\ ^\circ\text{C}\) to \(40\ ^\circ\text{C}\) increases the kinetic energy of both the enzyme and substrate molecules. They move faster, which increases the rate of collisions and the proportion of collisions with energy equal to or greater than the activation energy, leading to a higher frequency of successful collisions. Breaking of hydrogen bonds causing denaturation typically occurs at higher temperatures (above the optimum).

B is correct. 1 mark for explaining that increasing temperature in this range increases kinetic energy, leading to a higher frequency of successful collisions.

In competitive inhibition, the inhibitor competes with the substrate for binding at the active site. Because this binding is reversible, increasing the substrate concentration increases the probability of substrate-enzyme collisions, which outcompetes the inhibitor. Consequently, the maximum rate of reaction (Vmax) remains unchanged. However, more substrate is needed to achieve the same rate of reaction, which increases the apparent Michaelis-Menten constant (Km).

1 mark for selecting option C. This is a recall and application of competitive inhibition kinetics. Correct identification of the unchanged Vmax and increased Km as diagnostic of competitive inhibition.

The temperature coefficient (Q10) is the factor by which the rate of reaction increases for every 10 degrees Celsius rise in temperature. The temperature increase from 15 to 35 degrees Celsius is a rise of 20 degrees Celsius, which represents two intervals of 10 degrees Celsius. Therefore, the rate of reaction doubles twice. First interval (15 to 25 degrees Celsius): 2.4 * 2 = 4.8. Second interval (25 to 35 degrees Celsius): 4.8 * 2 = 9.6 mmol per cubic decimeter per second.

1 mark for option C. Award 1 mark for calculating the two Q10 doubling steps to arrive at 9.6.

Phloroglucinol stains lignin red. In a plant stem, lignin is a major component of the secondary cell walls of support and water-transporting cells. Sclerenchyma fibres (which provide mechanical support) and xylem vessel elements (which transport water under tension) have highly lignified walls and will stain red. Companion cells and phloem sieve tube elements have unlignified cellulose cell walls and will not stain red.

1 mark for option A. Award for correctly identifying sclerenchyma fibres and vessel elements as containing lignin, and companion cells and sieve tube elements as lacking lignin.

Water potential (psi) is calculated using the formula: psi = solute potential (psi_s) + pressure potential (psi_p). For Cell X: psi_X = -600 kPa + 200 kPa = -400 kPa. For Cell Y: psi_Y = -800 kPa + 300 kPa = -500 kPa. Water always moves by osmosis from a region of higher water potential (less negative) to a region of lower water potential (more negative). Since -400 kPa is higher than -500 kPa, water will move from Cell X to Cell Y.

1 mark for option B. Award for correctly calculating both water potentials and applying the principle of water movement from higher to lower water potential.

Malaria is caused by protoctists of the genus Plasmodium and is transmitted by the bite of an infected female Anopheles mosquito. Cholera is caused by the bacterium Vibrio cholerae and is transmitted via the faecal-oral route through contaminated food and water. Tuberculosis is caused by the bacterium Mycobacterium tuberculosis (or M. bovis) and is transmitted via airborne droplets released during coughing or sneezing.

1 mark for option A. Award for correctly matching the causative pathogens and transmission routes for all three infectious diseases.

Glycoproteins (P) on the outer surface of the cell membrane act as receptor molecules for hormones and neurotransmitters, and serve as antigens in cell-to-cell recognition. Cholesterol (Q) regulates membrane fluidity but does not make it completely impermeable to water. Channel proteins (R) provide hydrophilic paths for facilitated diffusion, which is a passive process and does not use ATP (active transport pumps do). The phospholipid bilayer (S) is a barrier to large or charged ions, preventing their simple diffusion.

1 mark for option A. Award for correctly identifying the function of glycoproteins and rejecting the incorrect statements about cholesterol, channel proteins, and the phospholipid bilayer.

The trachea contains C-shaped cartilage rings, goblet cells in the ciliated epithelium, and smooth muscle. The bronchus contains irregular plates of cartilage, goblet cells, and smooth muscle. Bronchioles lack cartilage entirely and do not contain goblet cells (except in some larger bronchioles, but standard AS syllabus models them as absent/replaced by Clara cells/epithelial cells), but they do have smooth muscle, which allows them to constrict and dilate to regulate airflow.

1 mark for option A. Award for correctly identifying the distribution of cartilage, goblet cells, and smooth muscle in all three components of the respiratory tract.

In actively respiring tissues, carbon dioxide is released, increasing its concentration in the blood. Carbon dioxide reacts with water inside red blood cells to form carbonic acid, which dissociates into hydrogen ions and hydrogencarbonate ions, causing blood pH to decrease. The increased concentration of hydrogen ions (lower pH) decreases the affinity of haemoglobin for oxygen (the Bohr effect), causing haemoglobin to release oxygen more readily. This shifts the oxygen dissociation curve to the right.

1 mark for option A. Award for correctly linking active respiration to high CO2, low pH, decreased oxygen affinity, and a rightward shift of the oxygen dissociation curve.

The observation that the maximum rate of reaction (\(V_{\max}\)) is unchanged at high substrate concentrations indicates that the inhibitor is competitive. Competitive inhibitors have a molecular shape similar to the substrate and bind reversibly to the active site of the enzyme. At high substrate concentrations, substrate molecules outcompete the inhibitor molecules for the active sites, so the maximum rate of reaction can still be achieved.

[1 mark] B is the correct answer. Competitive inhibitors bind reversibly to the active site and their effect is overcome by high substrate concentrations.

Immobilizing an enzyme in alginate beads physically restricts the movement of the enzyme's protein structure. This restriction stabilizes the tertiary structure, making the hydrogen and ionic bonds within the enzyme more resistant to disruption by excess hydrogen (\(\text{H}^+\)) or hydroxide (\(\text{OH}^-\)) ions. Consequently, the enzyme is less susceptible to denaturation at extreme pH values.

[1 mark] B is the correct answer. Immobilization increases the stability of the tertiary structure against pH-induced denaturation.

Cell X is dead at maturity and has a heavily lignified cell wall with pits, which is characteristic of a xylem vessel element. Cell Y is living, contains minimal cytoplasm and organelles, and has a cell wall with sieve plates, which describes a phloem sieve tube element. Cell Z has active cytoplasm with numerous organelles (mitochondria, ribosomes) and is connected to other cells by plasmodesmata, which describes a companion cell.

[1 mark] C is the correct answer. X is a xylem vessel element, Y is a phloem sieve tube element, and Z is a companion cell.

Statement 1 is correct: Lignin provides structural support and high tensile strength, preventing the vessel elements from imploding under the negative pressure (tension) generated by transpiration. Statement 2 is correct: The removal of end walls creates a continuous, low-resistance column for water flow. Statement 3 is correct: Pits allow lateral movement of water to bypass blockages (e.g., air bubbles). Statement 4 is incorrect: Xylem vessel elements are dead cells containing no cytoplasm, so they cannot perform active transport of ions.

[1 mark] A is the correct answer. Statements 1, 2, and 3 are correct adaptations, while statement 4 is incorrect.

Statement 1 is correct: Anopheles mosquitoes can become resistant to insecticides (like DDT), reducing vector control effectiveness. Statement 2 is correct: Plasmodium parasites have developed resistance to common antimalarial drugs (like chloroquine). Statement 3 is correct: Plasmodium replicates inside hepatocytes (liver cells) and erythrocytes (red blood cells), hiding from antibodies in the blood plasma. Statement 4 is incorrect: Plasmodium is a eukaryotic protoctist, not a bacterium, and does not form bacterial endospores.

[1 mark] A is the correct answer. Statements 1, 2, and 3 are biological reasons for the difficulty of malaria control.

Glycoproteins consist of carbohydrate chains attached to proteins and project from the outer surface of the cell membrane, serving as antigens for cell-to-cell recognition. Option A is incorrect because cholesterol regulates membrane fluidity, not hormone reception. Option B is incorrect because cholesterol, not glycolipids, stabilizes membrane fluidity. Option D is incorrect because phospholipids form a barrier to polar molecules; active transport of ions is carried out by transmembrane protein pumps, not the phospholipid bilayer itself.

[1 mark] C is the correct answer as glycoproteins serve as cell identity markers/antigens.

The apoplast pathway refers to the movement of water and dissolved minerals through the non-living parts of the plant, specifically the cellulose cell walls and intercellular spaces. When water reaches the endodermis, it is blocked by the Casparian strip (composed of water-impermeable suberin), forcing it to cross the cell membrane into the cytoplasm (symplast pathway).

[1 mark] B is the correct answer. The apoplast pathway involves cell walls and is blocked at the endodermis by the Casparian strip.

In actively respiring tissues, carbon dioxide diffuses into red blood cells and reacts with water to form carbonic acid, which dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)). To prevent the buildup of free \(\text{H}^+\) ions from lowering the cytoplasmic pH, hemoglobin binds to these ions to form haemoglobinic acid (\(\text{HHb}\)), acting as an intracellular buffer.

[1 mark] B is the correct answer. Hemoglobin acts as a buffer by binding to hydrogen ions to form haemoglobinic acid.

A non-competitive inhibitor binds to an allosteric site on the enzyme, changing the shape of the active site so that substrates can no longer bind. This effectively decreases the concentration of active enzymes, which reduces the maximum rate of reaction (\(V_{\max}\)). Because the inhibitor does not compete with the substrate for the active site, the affinity of the remaining active enzyme molecules for the substrate is unaffected, meaning the Michaelis-Menten constant (\(K_m\)) remains unchanged.

Award 1 mark for identifying that a non-competitive inhibitor decreases \(V_{\max}\) but leaves \(K_m\) unchanged (Option B).

At low substrate concentrations (Point X), many active sites are free, so increasing the substrate concentration increases the rate of reaction. Therefore, substrate concentration is the limiting factor. At high substrate concentrations (Point Y), all enzyme active sites are fully saturated with substrate. Increasing the substrate concentration further has no effect on the rate of reaction because the rate is limited by how quickly the enzyme can process the substrate, which depends on the enzyme concentration. Therefore, enzyme concentration is the limiting factor at Point Y.

Award 1 mark for correctly identifying that substrate concentration is limiting at Point X and enzyme concentration is limiting at Point Y (Option A).

A mature xylem vessel element is a dead cell that completely lacks cytoplasm, has lignified secondary walls to prevent collapse under tension, and is not associated with companion cells. A mature sieve tube element is a living cell that retains a thin layer of peripheral cytoplasm (but no nucleus, ribosomes, or vacuole), has non-lignified walls (cellulose), and is always situated next to one or more companion cells, which carry out metabolic functions for it.

Award 1 mark for the correct set of characteristics for both xylem vessel elements and sieve tube elements (Option A).

Lignin is a strong, waterproof polymer deposited in the walls of xylem vessel elements (to support the vessel and prevent collapse) and sclerenchyma fibres (for mechanical support of the stem). Callose is a carbohydrate polymer deposited at the sieve plates of phloem sieve tube elements, particularly in response to damage or during dormancy.

Award 1 mark for identifying that lignin is found in xylem vessel elements and sclerenchyma fibres, and callose is found in sieve tube elements (Option B).

Cholera is caused by the bacterium *Vibrio cholerae*, which is transmitted via the faecal-oral route through contaminated food or water. Therefore, treating drinking water with chlorine (1) and proper sewage disposal (2) are highly effective in preventing its transmission. Tuberculosis is caused by *Mycobacterium tuberculosis*, which is transmitted through the air in droplets produced by coughing or sneezing; sewage treatment and water chlorination have no impact on its spread. Antibiotics (3) are used to treat both diseases, and the BCG vaccine (4) is used to protect against tuberculosis, not cholera.

Award 1 mark for selecting 1 and 2 only as water sanitation measures that prevent cholera but do not affect airborne tuberculosis transmission (Option A).

Glycoproteins on the outer surface of the cell membrane function as receptors for hormones/neurotransmitters (cell-signalling) and as cell-marker antigens (allowing the immune system to recognize the cell as 'self'). Option A is incorrect because cholesterol decreases membrane fluidity at high temperatures. Option C is incorrect because phosphate heads are hydrophilic, while fatty acid tails are hydrophobic. Option D is incorrect because carrier proteins transport polar or charged substances, not non-polar ones.

Award 1 mark for the correct pairing of glycoprotein with its function in cell signalling and cell recognition (Option B).

The apoplast pathway involves water moving through the cell walls and intercellular spaces (the non-living parts of the plant). When the water reaches the endodermis, its progress is blocked by the suberin-rich, waterproof Casparian strip. This forces water to cross the cell membrane of the endodermal cells and enter the symplast pathway. Option A describes the symplast pathway, and Option C describes the vacuolar pathway. Option D is incorrect because the Casparian strip blocks water rather than facilitating its free movement.

Award 1 mark for correctly identifying that the apoplast pathway goes through cell walls and intercellular spaces and is blocked by the Casparian strip (Option B).

Actively respiring tissues produce carbon dioxide, which diffuses into red blood cells. Carbonic anhydrase catalyses the conversion of carbon dioxide and water to carbonic acid, which dissociates into hydrogen ions (\(H^+\)) and hydrogencarbonate ions. The binding of \(H^+\) ions to oxyhaemoglobin causes a conformational change that reduces its affinity for oxygen, shifting the oxygen dissociation curve to the right (the Bohr shift). This physiological mechanism ensures that more oxygen is released to tissues with high metabolic activity.

Award 1 mark for explaining that high carbon dioxide production increases hydrogen ion concentration, decreasing haemoglobin's affinity for oxygen to enhance unloading (Option C).

(a) Activation energy is the minimum energy needed for a chemical reaction to occur. An enzyme lowers this threshold by providing an alternative pathway with a lower activation energy. When lactose binds to the complementary active site of lactase, it forms an enzyme-substrate complex. This binding puts structural tension on the glycosidic bond, destabilizing it so that less thermal energy is required to break it. (b) Immobilization of the lactase enzyme physically traps it within an alginate gel lattice. This matrix restricts the thermal movement and vibration of the polypeptide chains at elevated temperatures. Consequently, the weak interactions (hydrogen and ionic bonds) maintaining the specific tertiary conformation of the active site are less easily disrupted, extending the operational stability and optimum temperature of the enzyme. (c) Using immobilized enzymes is highly cost-effective because the enzymes can be recovered from the reaction mixture and reused multiple times. Additionally, because the enzyme remains bound in the matrix, the product does not require expensive downstream purification to remove the enzyme. This setup also allows the reaction to run as a continuous flow process rather than in batches.

Part (a): [Max 3 marks] 1. Definition of activation energy: the minimum energy required for a chemical reaction to occur / for substrates to react; 2. Lactose binds to the active site to form an enzyme-substrate (ES) complex; 3. This puts physical strain on the glycosidic bond, lowering the energy needed to break it / aligns the substrate in an optimal orientation. Part (b): [Max 4 marks] 1. Alginate matrix restricts the movement / kinetic energy of enzyme molecules; 2. Prevents unfolding / denaturation of the tertiary structure of the enzyme; 3. Protects/stabilizes hydrogen bonds, ionic bonds, or hydrophobic interactions; 4. The active site remains complementary in shape to the substrate at higher temperatures. Part (c): [Max 3 marks] 1. Enzyme can be easily recovered/recycled, reducing overall operational costs; 2. Product is not contaminated with the enzyme (no downstream purification needed); 3. Allows the process to be run continuously (continuous flow process); 4. Increases stability over a wider range of pH/temperatures.

(a) Competitive inhibitors compete directly with the substrate for the active site. At infinitely high substrate concentrations, the substrate completely outcompetes the inhibitor, meaning the theoretical maximum rate (\(V_{\max}\)) remains unchanged, but a higher substrate concentration is required to reach half-maximal velocity, raising the \(K_{\text{m}}\) value. Non-competitive inhibitors bind to an allosteric site, altering the shape of the active site so that the enzyme cannot function. This permanently reduces the number of functional enzymes, decreasing the \(V_{\max}\). However, the remaining unaffected enzyme molecules still have normal affinity for the substrate, meaning the \(K_{\text{m}}\) value remains unchanged. (b) Competitive inhibitors like orthovanadate have a molecular shape that mimics the substrate. This allows them to bind to the active site, temporarily preventing substrate molecules from entering and forming enzyme-substrate complexes. By increasing substrate concentration significantly, the ratio of substrate to inhibitor molecules increases, making it much more likely for a substrate molecule to bind to an empty active site than an inhibitor molecule, thus overcoming the inhibition. (c) The \(K_{\text{m}}\) is defined as the substrate concentration at which the reaction rate is half of \(V_{\max}\). A low \(K_{\text{m}}\) indicates that the enzyme reaches half-maximal speed at a very low substrate concentration, representing high substrate affinity. A high \(K_{\text{m}}\) means high substrate concentrations are needed, representing low affinity.

Part (a): [Max 4 marks, 1 mark per point] 1. Competitive inhibitor: \(V_{\max}\) remains unchanged / is achieved at high substrate concentrations; 2. Competitive inhibitor: \(K_{\text{m}}\) increases; 3. Non-competitive inhibitor: \(V_{\max}\) decreases; 4. Non-competitive inhibitor: \(K_{\text{m}}\) remains unchanged. Part (b): [Max 4 marks] 1. Inhibitor has a shape complementary to the active site / similar molecular structure to the substrate; 2. Binds temporarily / reversibly to the active site; 3. Blocks the active site, preventing the substrate from entering / forming enzyme-substrate complexes; 4. Can be overcome by increasing substrate concentration; 5. High substrate concentration increases the probability of substrate collision with active sites compared to the inhibitor. Part (c): [Max 2 marks] 1. A low \(K_{\text{m}}\) value indicates high affinity of the enzyme for its substrate; 2. A high \(K_{\text{m}}\) value indicates low affinity of the enzyme for its substrate; 3. Definition: \(K_{\text{m}}\) is the substrate concentration at which the reaction velocity is half of \(V_{\max}\).

(a) In a transverse section of a dicotyledonous root, the vascular tissue is concentrated in the center, known as the vascular cylinder or stele. Here, xylem forms a star or cross shape, with clusters of phloem located in the spaces between the arms of the star. In contrast, the stem contains discrete vascular bundles arranged in a ring near the cortex. Within each of these stem bundles, xylem is oriented towards the center (inner side) and phloem towards the epidermis (outer side). (b) Mature xylem vessel elements are dead cells that have lost their nuclei, cytoplasm, and all organelles. This creates a clear, unobstructed lumen that minimizes resistance to water flow. The end walls of these cells are completely broken down, forming continuous hollow conduits that facilitate the uninterrupted mass flow of water. The thick secondary walls are reinforced with lignin, which provides high mechanical strength to prevent the vessels from collapsing inward under the high tension (negative pressure) generated by transpiration. Furthermore, unlignified regions called pits are present in the cell walls, allowing lateral movement of water to adjacent cells or vessels. (c) Lignin is a highly hydrophobic polymer. By lining the interior of the cellulose wall, it waterproofs the vessel, ensuring that water does not leak out laterally before reaching the leaves. Secondly, the lignified walls provide structural rigidity to the stem, helping the plant stand upright and support its own weight.

Part (a): [Max 3 marks] 1. Root vascular tissues are located in the center (vascular cylinder/stele) whereas stem vascular tissues are in discrete bundles; 2. Root xylem forms a star/cross shape in the center with phloem in between the arms; 3. Stem vascular bundles are arranged in a ring; 4. In stem bundles, xylem is on the inside and phloem is on the outside. Part (b): [Max 5 marks] 1. No cytoplasm / no organelles / dead cells, providing an empty lumen that reduces resistance to water flow; 2. End walls are broken down / absent, creating continuous tubes for mass flow; 3. Walls are thickened with lignin to prevent collapse under negative pressure / tension; 4. Pits (unlignified areas) allow lateral movement of water between vessels / to surrounding tissues; 5. Spiral / annular / reticulate patterns of lignin allow stretching and growth of the stem; 6. Cellulose cell wall has hydrophilic properties to promote adhesion of water molecules. Part (c): [Max 2 marks] 1. Waterproofing the cell wall to prevent water loss / keep water inside the vessels; 2. Structural support / mechanical rigidity for the entire stem / plant; 3. Protection against decay / pathogen entry.

(a) Phloem sieve tube elements are living cells, but they have a highly reduced internal structure. They lack a nucleus, ribosomes, and a large vacuole, and have only a very thin layer of cytoplasm aligned against the cell wall. This configuration dramatically reduces the internal resistance to the movement of phloem sap. At their junction ends, they have specialized sieve plates with large pores, which act as sieves and allow the mass flow of assimilates. They are also linked to companion cells via cytoplasmic channels called plasmodesmata, which facilitate the entry of sucrose. (b) Companion cells carry out the metabolic processes needed to load sucrose into the sieve tube elements. They are packed with mitochondria, which produce ATP via aerobic respiration. This ATP is used to power active proton pumps in the companion cell membrane, which pump hydrogen ions (\(\text{H}^+\)) out of the cell into the cell wall (apoplastic space), establishing an electrochemical gradient. Sucrose is then loaded into the companion cell against its concentration gradient by co-transporter proteins, which bring sucrose in alongside hydrogen ions as they flow back down their concentration gradient. (c) Besides sucrose, the phloem transports other essential organic and inorganic molecules. These include amino acids (used for protein synthesis throughout the plant), plant hormones (such as auxins, cytokinins, and gibberellins), and various mineral ions like potassium, phosphate, and chloride.

Part (a): [Max 4 marks] 1. Peripheral cytoplasm / lack of nucleus / ribosomes / vacuole, creating less resistance to flow of sap; 2. Sieve plates have large pores to allow uninterrupted mass flow; 3. Plasmodesmata connect sieve tube elements to companion cells to allow movement of solutes; 4. Thin cell walls to assist rapid diffusion/exchange; 5. Cell membrane contains carrier/transport proteins to maintain solute concentrations. Part (b): [Max 4 marks] 1. Many mitochondria to generate ATP for active transport; 2. Proton pumps (\(\text{H}^+\)-ATPase) in plasma membrane to pump hydrogen ions into the cell wall / apoplast; 3. Co-transporter proteins in membrane to transport sucrose into the cell alongside hydrogen ions; 4. Highly folded cell surface membrane (transfer cell structure) to increase surface area for transport; 5. Abundant ribosomes / rough ER to synthesize transporter proteins. Part (c): [Max 2 marks, 1 mark per correct answer] Accept any two of: 1. Amino acids; 2. Plant growth regulators / hormones (e.g., auxin / gibberellins); 3. Inorganic ions / minerals (e.g., potassium / phosphate); 4. Water; 5. ATP / proteins / RNA.

(a) Tuberculosis is caused by the bacterium *Mycobacterium tuberculosis* (or *Mycobacterium bovis* in cattle and sometimes humans) and is transmitted via droplet infection, where healthy individuals inhale airborne droplets sneezed or coughed by infected individuals. Cholera is caused by the bacterium *Vibrio cholerae* and is primarily waterborne, transmitted when individuals ingest water or food contaminated with human faeces containing the pathogen. (b) In dense populations without safe sanitation, untreated faecal waste from cholera patients can easily enter and contaminate local drinking water and food supplies. Since the pathogen is transmitted via the faecal-oral route, ingestion of this contaminated water spreads the disease rapidly. Prevention requires constructing modern sewage treatment facilities to separate human waste from water sources, purifying public drinking water with chlorine, teaching proper hand hygiene, and implementing oral cholera vaccination campaigns. (c) Tuberculosis is highly challenging to control because the bacteria can survive intracellularly inside macrophages, where they are protected from antibiotics and host immune responses. Furthermore, curing TB requires a multi-drug regimen lasting at least 6 to 9 months. Many patients stop taking their medication once symptoms improve, which leads to treatment failure and has driven the selection of multi-drug resistant (MDR-TB) and extensively drug-resistant (XDR-TB) strains.

Part (a): [4 marks, 1 mark per point] (i) *Mycobacterium tuberculosis* / *Mycobacterium bovis* (genus capitalized, species lowercase, must be italicized or underlined); (ii) *Vibrio cholerae* (genus capitalized, species lowercase, must be italicized or underlined); (iii) Airborne droplets / inhalation of aerosol droplets; (iv) Ingestion of contaminated water / food / faecal-oral route. Part (b): [Max 4 marks] 1. Inadequate sanitation allows human faeces / sewage containing the pathogen to contaminate drinking water; 2. High population density increases the speed of transmission and exposure to infected material; 3. Prevention: providing safe, clean, chlorinated drinking water; 4. Prevention: proper sewage disposal / treatment to prevent water contamination; 5. Prevention: public health education on hand washing / food hygiene; 6. Prevention: use of oral cholera vaccine. Part (c): [Max 2 marks] 1. Pathogen lives intracellularly (inside macrophages), evading the immune response and antibiotics; 2. Very long treatment course (6-9 months) leads to poor patient compliance / incomplete courses; 3. Emergence of multi-drug resistant (MDR) and extensively drug-resistant (XDR) bacterial strains; 4. Latent TB infections can remain dormant for years and reactivate later.

(a) The cell membrane is called 'fluid' because both phospholipid molecules and many membrane proteins are free to diffuse laterally within their respective monolayers. It is a 'mosaic' because the various protein molecules are embedded and scattered randomly throughout the phospholipid bilayer, resembling individual colored tiles in a mosaic pattern. (b) Cholesterol plays a vital role as a membrane buffer. At high temperatures, the increase in thermal energy makes phospholipids move more dynamically; cholesterol binds to the hydrophobic tails, restricting their lateral movement and preventing the membrane from becoming too fluid or losing structure. At low temperatures, cholesterol prevents the fatty acid tails from packing tightly together and freezing into a rigid, non-functional gel, thereby maintaining membrane fluidity. (c) The signaling pathway begins when a extracellular ligand binds specifically to a complementary receptor on the membrane's outer surface. This binding induces a conformational change in the intracellular domain of the receptor protein. This shape change activates a G-protein or an enzyme on the inner surface of the membrane. The activated G-protein/enzyme triggers the production of small, fast-diffusing second messengers (e.g., cyclic AMP). These second messengers initiate a phosphorylation cascade, activating a series of intracellular enzymes (kinases) which amplify the signal, ultimately leading to a specific cellular response such as gene transcription or a change in metabolism.

Part (a): [Max 3 marks] 1. Fluid: phospholipids (and proteins) can move laterally / diffuse within their monolayer; 2. Mosaic: protein molecules are scattered / randomly distributed throughout the bilayer; 3. Membrane consists of a bilayer of phospholipids with proteins embedded within it. Part (b): [Max 3 marks] 1. At high temperatures: cholesterol restricts movement of fatty acid tails, reducing membrane fluidity; 2. At low temperatures: cholesterol prevents close packing of fatty acid tails, preventing crystallization / maintaining fluidity; 3. Stability: increases mechanical stability / prevents leakage of polar substances. Part (c): [Max 4 marks] 1. Ligand binds to a specific / complementary cell surface receptor; 2. Causes a conformational change in the receptor protein; 3. Activates a G-protein / enzyme on the cytoplasmic side of the membrane; 4. Stimulates release of second messengers (e.g., cAMP / calcium ions); 5. Triggers a phosphorylation cascade / series of enzyme activations; 6. Leads to a final cellular response (e.g., gene transcription / metabolic changes); 7. Mention of signal amplification at any stage.

**(a) (i)** To prepare a 2-fold serial dilution, half of the volume of the preceding concentration is transferred and diluted with an equal volume of distilled water:
* **For \(1.0\%\) amylase**: Take \(5.0\text{ cm}^3\) of \(2.0\%\) stock **E**, add \(5.0\text{ cm}^3\) of distilled water.
* **For \(0.5\%\) amylase**: Take \(5.0\text{ cm}^3\) of the prepared \(1.0\%\) solution, add \(5.0\text{ cm}^3\) of distilled water.
* **For \(0.25\%\) amylase**: Take \(5.0\text{ cm}^3\) of the prepared \(0.5\%\) solution, add \(5.0\text{ cm}^3\) of distilled water.
* **For \(0.125\%\) amylase**: Take \(5.0\text{ cm}^3\) of the prepared \(0.25\%\) solution, add \(5.0\text{ cm}^3\) of distilled water.

**(a) (ii)** Rate calculations:
* \(1000 / 240\text{ s} = 4.17\)
* \(1000 / 120\text{ s} = 8.33\)
* \(1000 / 60\text{ s} = 16.67\)
* \(1000 / 30\text{ s} = 33.33\)
* \(1000 / 15\text{ s} = 66.67\)

The table should feature borders, clear column headers (with slashes separating headings and units), and consistent significant figures (e.g., all rates written to 2 decimal places).

**Table of Results**
| Concentration of amylase / % | Time taken for complete hydrolysis / s | Rate of reaction / arbitrary units |
| :---: | :---: | :---: |
| 0.125 | 240 | 4.17 |
| 0.250 | 120 | 8.33 |
| 0.500 | 60 | 16.67 |
| 1.000 | 30 | 33.33 |
| 2.000 | 15 | 66.67 |

**(b) (i)**
* **Axes**: x-axis "Concentration of amylase / %" (scale 0 to 2.0, with 0.2 per large division); y-axis "Rate of reaction / arbitrary units" (scale 0 to 70, with 10 per large division).
* **Points**: Plotted as neat crosses or dots in circles at exactly the calculated coordinates.
* **Line**: A straight, clean line of best fit passing directly through the origin (0,0).

**(b) (ii)** At \(0.75\%\) concentration, the line of best fit gives a rate of exactly \(25.0\) arbitrary units. The candidate should draw a dashed vertical line from \(0.75\%\) on the x-axis to the line of best fit, and then a dashed horizontal line across to \(25.0\) on the y-axis.

**(c)**
* The rate of starch hydrolysis is directly proportional to the amylase concentration.
* At higher amylase concentrations, there are more enzyme molecules and therefore more active sites available.
* This increases the frequency of successful collisions between starch molecules and the active sites.
* As a result, more enzyme-substrate complexes (ESCs) form per unit time.
* Since starch (substrate) is in excess, the rate continues to increase linearly without leveling off.

**(d)**
* **Source of error 1**: Subjective nature of identifying the endpoint (determining precisely when the blue-black colour transitions completely to yellow-brown).
* **Source of error 2**: Large sampling time intervals (every 15 seconds) mean the actual end-point likely occurred between sampling times (causing an overestimation of time and underestimation of rate).
* **Improvement**: Use a colorimeter to measure absorbance of the starch-iodine mixture quantitatively at each interval, OR perform tests at shorter intervals (e.g., every 5 seconds).

**(e)**
* **Control**: Set up a reaction tube containing starch solution and boiled, cooled amylase (or replace amylase with distilled water).
* **Explanation**: Boiled amylase is denatured and inactive. If the iodine remains blue-black (no hydrolysis), it confirms that the hydrolysis of starch is catalyzed specifically by active amylase, rather than occurring spontaneously.

**(f)**
* Set up water baths at at least five different temperatures (e.g., \(20^\circ\text{C}\), \(30^\circ\text{C}\), \(40^\circ\text{C}\), \(50^\circ\text{C}\), \(60^\circ\text{C}\)).
* Keep the concentration of amylase (e.g., \(1.0\%\)) and starch (e.g., \(1.0\%\)) constant, and pre-incubate the separate starch and amylase solutions at each target temperature for 5 minutes before mixing.

**(a)(i) [3 marks]**
* Award 1 mark for correct volumes of amylase used (\(5.0\text{ cm}^3\) for all four dilutions).
* Award 1 mark for correct source of amylase (the previous dilution tube is used for the next step: \(1.0\%\) derived from \(2.0\%\), \(0.5\%\) from \(1.0\%\), etc.).
* Award 1 mark for correct volumes of distilled water (\(5.0\text{ cm}^3\) for all four dilutions to achieve a total volume of \(10.0\text{ cm}^3\)).

**(a)(ii) [3 marks]**
* Award 1 mark for constructing a table with clear borders and appropriate headers with units: "Concentration of amylase / %", "Time taken for complete hydrolysis / s" (or "Time / s"), and "Rate of reaction / arbitrary units" (or "\(\text{s}^{-1}\)").
* Award 1 mark for correct calculations of all 5 rates based on the student's raw data.
* Award 1 mark for recording all rates to a consistent degree of precision (either all to 1 d.p. or all to 2 d.p.).

**(b)(i) [4 marks]**
* **A (Axes)**: Axes correctly labeled with units: x-axis "Concentration of amylase / %", y-axis "Rate of reaction / arbitrary units" [1].
* **S (Scale)**: Scales on both axes must be linear and allow the plotted points to occupy more than half of the grid in both dimensions [1].
* **P (Plotting)**: All five points plotted accurately to within half a small square of the grid [1].
* **L (Line)**: Clean, straight line of best fit drawn with a ruler, starting at the origin \((0,0)\) [1].

**(b)(ii) [2 marks]**
* Award 1 mark for reading the correct rate at \(0.75\%\) concentration from their graph (\(25.0 \pm 1.0\) arbitrary units).
* Award 1 mark for showing construction lines (vertical and horizontal dashed lines) clearly on the graph.

**(c) [3 marks]**
* Award 1 mark for identifying that rate of reaction is directly proportional to amylase concentration.
* Award 1 mark for explaining that higher enzyme concentration results in more active sites, leading to a higher frequency of successful collisions between amylase active sites and starch molecules.
* Award 1 mark for stating that this leads to more enzyme-substrate complexes (ESCs) forming per unit time.

**(d) [3 marks]**
* Award up to 2 marks for stating two distinct sources of error (e.g., subjective endpoint color, large 15 s intervals, lack of temperature control).
* Award 1 mark for a matching, valid improvement (e.g., use a colorimeter, use 5 s intervals, use a thermostatically controlled water bath).

**(e) [2 marks]**
* Award 1 mark for describing the control (using boiled and cooled amylase OR replacing amylase with the same volume of distilled water).
* Award 1 mark for explaining that it shows the hydrolysis of starch is due to active amylase and does not occur spontaneously.

**(f) [2 marks]**
* Award 1 mark for stating that at least 5 different temperatures must be tested (using water baths).
* Award 1 mark for specifying that the concentration of amylase and starch must remain constant, AND that both solutions must be pre-incubated at the test temperature before mixing.

a(i) To calculate magnification: Convert 48 mm to micrometres: 48 * 1000 = 48,000 \(\mu\text{m}\). Magnification = Image Size / Actual Size = 48,000 / 320 = \(\times 150\). a(ii) Actual width = 18 epu * 2.5 \(\mu\text{m}\)/epu = 45 \(\mu\text{m}\). b(i) Radius (r) = diameter / 2 = 0.45 mm / 2 = 0.225 mm. Area = \(\pi r^2\) = 3.142 * (0.225)^2 = 3.142 * 0.050625 = 0.159 \(\text{mm}^2\) (or 0.16 \(\text{mm}^2\)). b(ii) Mean number of stomata = (14 + 11 + 15 + 12 + 13) / 5 = 65 / 5 = 13. b(iii) Mean stomatal density = Mean count / Area of field of view = 13 / 0.159 = 81.76. To the nearest whole number, this is 82 stomata per \(\text{mm}^2\) (or 81 if using 0.16 \(\text{mm}^2\) as the area). c) Prepare an epidermal peel by using clear nail polish or adhesive tape, or by peeling the epidermis carefully with fine forceps. Mount the peel in water on a slide, place a coverslip on top, and observe under the light microscope. A typical xerophytic leaf has no stomata on the upper epidermis because the upper surface is exposed to direct sunlight and high temperatures, which would lead to excessive water loss by transpiration. d) Stomatal crypts trap moist, humid air next to the stomata, which reduces the water vapour potential gradient between the inside of the leaf and the external environment, thereby reducing the rate of transpiration.

a(i) [4 marks] 1. Correct conversion of 48 mm to 48,000 \(\mu\text{m}\) or 320 \(\mu\text{m}\) to 0.32 mm. 2. Formula stated: Magnification = Image size / Actual size. 3. Correct substitution: 48,000 / 320. 4. Correct answer: \(\times 150\) (accept 150). a(ii) [2 marks] 1. Correct multiplication: 18 * 2.5. 2. Correct answer with units: 45 \(\mu\text{m}\) (accept micrometres). b(i) [2 marks] 1. Correct calculation of radius: 0.225 mm. 2. Correct area: 0.159 \(\text{mm}^2\) (accept 0.16 \(\text{mm}^2\)). b(ii) [1 mark] 1. Correct mean: 13. b(iii) [3 marks] 1. Correct division of mean count by area: 13 / 0.159 (or 13 / 0.16). 2. Correct calculation: 81.76 (or 81.25). 3. Rounded correctly to the nearest whole number: 82 (or 81) stomata per \(\text{mm}^2\). c) [4 marks] 1. Describe preparation of peel using nail polish/tape or forceps. 2. Describe mounting in water/glycerol under a coverslip. 3. Explain that upper epidermis is exposed to direct sunlight/heat/wind, leading to excessive transpiration if stomata were present. 4. Mention presence of thick cuticle on upper surface to prevent water loss. d) [2 marks] 1. Crypts trap water vapour / humid air, creating a local microenvironment of high humidity. 2. This reduces the water vapour potential gradient, reducing the rate of diffusion/transpiration.