2025 Cambridge IAL Biology (9700) Practice Paper with Answers

A competitive inhibitor competes directly with the substrate for binding at the active site. If the substrate concentration is increased to a very high level, the substrate outcompetes the inhibitor, and the reaction can still reach its normal maximum rate (\(V_{\text{max}}\)), meaning \(V_{\text{max}}\) remains unchanged. However, because the inhibitor reduces the affinity of the enzyme for the substrate, a higher concentration of substrate is required to reach half of \(V_{\text{max}}\). Therefore, the Michaelis-Menten constant (\(K_{\text{m}}\)) increases.

1 mark for identifying that \(V_{\text{max}}\) remains unchanged and \(K_{\text{m}}\) increases under the influence of a competitive inhibitor.

Enzymes function by lowering the activation energy of the transition state (statement 1) and providing an alternative, lower-energy pathway for the reaction to occur (statement 3). They do not alter the temperature-dependent kinetic energy of the substrate molecules (statement 2 is incorrect), nor do they change the thermodynamics of the reaction, meaning the overall free energy change (\(\Delta G\)) remains the same (statement 4 is incorrect).

1 mark for identifying statements 1 and 3 as correct, and statements 2 and 4 as incorrect.

Immobilizing an enzyme within a matrix (like alginate beads) actually creates a physical barrier that restricts the diffusion of substrate molecules to the active site (mass transfer limitation). Therefore, the rate of reaction is generally not higher due to increased diffusion. However, ease of recovery (option A), avoiding product contamination (option B), and improved thermal/chemical stability (option C) are all genuine, major advantages of enzyme immobilization.

1 mark for identifying option D as the statement that is not an advantage of using immobilized lactase.

The apoplast pathway involves water moving through the non-living cell walls and intercellular spaces. At the endodermis, this pathway is blocked by the Casparian strip, which is a band of waterproof suberin deposited in the cell walls. This forces water to cross the selectively permeable cell surface membrane and enter the symplast pathway, enabling the plant to regulate solute uptake.

1 mark for identifying the apoplast pathway, blocked by the Casparian strip composed of suberin.

Companion cells are highly metabolically active cells that support sieve tube elements. They contain a prominent nucleus, dense cytoplasm, and numerous mitochondria to generate ATP. Sieve tube elements are specialized for transport and have lost most of their organelles during development, lacking a nucleus, having only a thin peripheral layer of cytoplasm, and having modified end walls called sieve plates.

1 mark for correctly matching the cellular features of companion cells and sieve tube elements.

Active loading of sucrose into the companion cells and then into the sieve tube elements at the source lowers the water potential inside the sieve tube. As a result, water enters the sieve tube elements from the adjacent xylem by osmosis. This entry of water builds up hydrostatic (turgor) pressure at the source, driving the bulk flow of phloem sap towards the sink.

1 mark for correctly explaining that high hydrostatic pressure at the source is generated by water entering the phloem via osmosis due to the active loading of sucrose.

During the repolarisation phase of an action potential, voltage-gated sodium channels close (inactivate), preventing further entry of sodium ions. At the same time, voltage-gated potassium channels open, allowing potassium ions (\(\text{K}^+\)) to diffuse rapidly out of the axon down their electrochemical gradient. This loss of positive charge restores the negative resting potential inside the membrane.

1 mark for identifying that potassium channels open and potassium ions diffuse out of the axon during repolarisation.

When an action potential depolarises the pre-synaptic membrane, voltage-gated calcium channels open, allowing calcium ions (\(\text{Ca}^{2+}\)) to diffuse into the pre-synaptic knob down their concentration gradient. This influx of calcium ions stimulates synaptic vesicles containing acetylcholine to move towards and fuse with the pre-synaptic membrane, releasing the neurotransmitter into the cleft via exocytosis.

1 mark for identifying that calcium ions enter the pre-synaptic neurone via voltage-gated channels and trigger the exocytosis of acetylcholine.

Competitive inhibitors compete with the substrate for the active site, which increases the substrate concentration needed to reach half of the maximum velocity, thereby increasing the \(K_m\) value. However, the maximum velocity (\(V_{max}\)) can still be achieved at very high substrate concentrations. Non-competitive inhibitors bind to an allosteric site on the enzyme, preventing the catalytic reaction from occurring regardless of substrate concentration, which reduces \(V_{max}\) but does not change the affinity of the unaffected active sites (so \(K_m\) remains unchanged).

Award 1 mark for the correct option (A). Reject all other options.

Active loading of sucrose involves protons being actively pumped out of the companion cells into the apoplast (cell wall spaces). The resulting concentration gradient allows protons to flow back into companion cells down their electrochemical gradient via co-transporter proteins, bringing sucrose along with them. Finally, sucrose moves by diffusion down its concentration gradient from the companion cells into the sieve tube elements via plasmodesmata connections.

Award 1 mark for the correct option (D). Reject all other options.

The correct sequence begins with the arrival of an action potential which causes the opening of voltage-gated calcium channels (III), allowing calcium ions to enter the presynaptic neurone down their concentration gradient (I). The rising calcium concentration triggers vesicle fusion and the exocytosis of acetylcholine (II). Acetylcholine diffuses across the synaptic cleft and binds to receptors on ligand-gated sodium channels (V), opening them to allow sodium influx, which results in the depolarisation of the postsynaptic membrane (IV).

Award 1 mark for the correct option (A). Reject all other options.

The activation energy of the catalyzed reaction is the energy required to reach the transition state from the reactants' initial energy level: \(50\text{ kJ mol}^{-1} - 30\text{ kJ mol}^{-1} = +20\text{ kJ mol}^{-1}\). The overall enthalpy change (\(\Delta H\)) of the reaction is the energy of the products minus the energy of the reactants: \(10\text{ kJ mol}^{-1} - 30\text{ kJ mol}^{-1} = -20\text{ kJ mol}^{-1}\).

Award 1 mark for the correct option (A). Reject all other options.

All three statements are correct. Lignification provides mechanical support to resist the strong negative pressure (tension) in the xylem. The loss of protoplasm and end walls creates a continuous, hollow tube which reduces friction. Pits permit the lateral flow of water between adjacent vessels, allowing water to bypass any localized blockages (such as air bubbles).

Award 1 mark for the correct option (D). Reject all other options.

According to the sliding filament theory, during contraction, actin filaments slide over myosin filaments toward the M-line. The A-band represents the total length of the thick myosin filaments, which does not change. The I-band (consisting only of thin actin filaments) and the H-zone (consisting only of thick myosin filaments) both shorten as the degree of overlap increases.

Award 1 mark for the correct option (A). Reject all other options.

Immobilized enzymes are physically attached to or trapped within an inert matrix, which allows them to be separated easily from products and reused (1). This attachment often stabilizes their tertiary structure, making them more resilient to high temperatures and changes in pH (2). However, immobilization typically decreases or does not change the \(V_{max}\) due to mass-transfer limitations (substrates diffusing slower to the enzyme active sites) (3 is incorrect).

Award 1 mark for the correct option (A). Reject all other options.

At equilibrium, the water potential of the cell (\(\psi_{cell}\)) equals the water potential of the surrounding solution (\(\psi_{solution}\)), so \(\psi_{cell} = -500\text{ kPa}\). Using the formula for water potential: \(\psi = ψ_s + ψ_p\), we can substitute the known values: \(-500 = -800 + ψ_p\). Solving for \(ψ_p\) gives: \(ψ_p = -500 - (-800) = +300\text{ kPa}\).

Award 1 mark for the correct option (A). Reject all other options.

A competitive inhibitor competes with the substrate for binding at the active site. Increasing substrate concentration can overcome this inhibition, allowing the reaction to reach the same \(V_{max}\). However, a higher concentration of substrate is required to reach half \(V_{max}\), meaning the Michaelis-Menten constant (\(K_m\)) increases. Therefore, X is a competitive inhibitor binding to the active site.

A non-competitive inhibitor binds to an allosteric site (a site other than the active site), changing the overall shape of the enzyme and reducing its catalytic activity. This lowers the maximum rate of reaction (\(V_{max}\)) regardless of how much substrate is added, but does not alter the affinity of the active enzymes for the substrate, meaning \(K_m\) remains unchanged. Therefore, Y is a non-competitive inhibitor binding to an allosteric site.

Correct option is A (1 mark).
- Reject B: Roles of X and Y are reversed.
- Reject C: Incorrectly states that competitive inhibitors bind to allosteric sites.
- Reject D: Incorrectly states that non-competitive inhibitors bind to active sites.

All three statements are correct.
1. Proton pumps in the companion cell membrane use ATP to actively pump hydrogen ions (\(\text{H}^+\)) out of the cytoplasm into the cell wall space (apoplast).
2. This active transport establishes a steep electrochemical gradient of hydrogen ions. The hydrogen ions then diffuse back down their gradient into the companion cell via co-transporter proteins, which simultaneously transport sucrose molecules with them.
3. Sucrose is loaded into the companion cells against its concentration gradient, powered by the entry of \(\text{H}^+\).

Correct option is A (1 mark) since all three statements correctly describe the mechanisms of active companion cell loading of sucrose.

The activation energy of a reaction is the minimum energy required to reach the transition state from the initial energy level of the reactants. For an enzyme-catalysed reaction, the transition state is lowered to energy level \(Z\). Therefore, the activation energy of the catalysed reaction is the difference between this level and the energy level of the reactants: \(Z - W\).
The overall energy change of the reaction (enthalpy change, \(\Delta H\)) is the difference between the final energy of the products and the initial energy of the reactants: \(X - W\).

Correct option is A (1 mark).
- Reject B: Incorrectly calculates activation energy as starting from products instead of reactants, and reverses the calculation of energy change.
- Reject C: Uses the uncatalysed transition state \(Y\) instead of \(Z\).
- Reject D: Subtracts the two transition states for activation energy, which is incorrect.

The correct physiological sequence is:
- First, the depolarization of the action potential arriving at the presynaptic knob causes voltage-gated calcium channels to open (3).
- Next, calcium ions diffuse down their electrochemical gradient into the presynaptic cytoplasm (1).
- This influx of calcium ions triggers the movement of synaptic vesicles, containing acetylcholine, which fuse with the presynaptic membrane to release the neurotransmitter into the cleft by exocytosis (4).
- Acetylcholine diffuses across the cleft and binds to specific receptor proteins on the postsynaptic membrane (2).
- Binding causes ligand-gated sodium channels to open, allowing sodium ions to enter and depolarise the postsynaptic neurone (5).
Thus, the correct sequence is 3 \(\rightarrow\) 1 \(\rightarrow\) 4 \(\rightarrow\) 2 \(\rightarrow\) 5.

Correct option is A (1 mark).
- Reject B: Synaptic vesicles must fuse (4) to release neurotransmitter before it can bind to receptors (2).
- Reject C and D: Calcium channels must open (3) before calcium ions can actually enter the presynaptic neurone (1).

During transpiration, liquid water in the leaf xylem leaves the vessels and moves through the apoplast (cell walls) or symplast (cytoplasm) of the mesophyll cells. It then evaporates from the damp surface of the mesophyll cell walls into the intercellular air spaces of the leaf as water vapour. From these spaces, the water vapour diffuses down its water vapour potential gradient, through the sub-stomatal cavity and the stomatal pore, and out into the drier surrounding atmosphere.

Correct option is A (1 mark).
- Reject B, C, and D: These paths fail to represent the correct sequence of evaporation (from cell walls to air spaces) and diffusion (from air spaces through stomatal pores to the outside atmosphere).

During muscle contraction, ATP performs crucial roles in the cross-bridge cycle:
1. When a new ATP molecule binds to the myosin head, it causes the myosin head to detach from the actin binding site.
2. This ATP is then hydrolysed by ATPase on the myosin head to ADP and inorganic phosphate. The energy released from this hydrolysis is used to cock the myosin head into its high-energy, ready position.
3. Statement 3 is incorrect because it is calcium ions (\(\text{Ca}^{2+}\)), not ATP, that bind to troponin to cause the conformational change in tropomyosin.

Correct option is A (1 mark).
- Reject B: Includes statement 3, which is false because calcium ions bind to troponin.
- Reject C: Includes statement 3 and misses statement 1.
- Reject D: Includes statement 3.

By definition, the Michaelis-Menten constant (\(K_m\)) is the substrate concentration at which the rate of the enzyme-catalysed reaction is exactly half of its maximum rate (\(1/2 V_{max}\)).

Given that \(V_{max} = 120\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}\):
\(\frac{1}{2} V_{max} = \frac{120}{2} = 60\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}\).

Looking at the experimental results in the table, the initial rate of reaction is \(60\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}\) when the substrate concentration is \(2.0\text{ mmol dm}^{-3}\).

Therefore, \(K_m = 2.0\text{ mmol dm}^{-3}\).

Correct option is A (1 mark).
- Reject B: This is the substrate concentration when the rate is \(80\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}\), which is not half of \(V_{max}\).
- Reject C: This is the rate of reaction at half \(V_{max}\), not the substrate concentration.
- Reject D: This is the value of \(V_{max}\).

A mature xylem vessel element is a dead cell that is completely empty to facilitate continuous water transport. It has a heavily lignified cell wall, but lacks cytoplasm and a nucleus.

A mature sieve tube element is a living cell, but has highly reduced contents to allow mass flow of phloem sap. It has a non-lignified cellulose cell wall, contains a thin peripheral layer of cytoplasm (lacking many organelles), and does not have a nucleus (it is controlled by its companion cell).

Correct option is A (1 mark).
- Reject B: Sieve tube elements do contain a thin layer of cytoplasm.
- Reject C: Xylem vessel elements are dead and do not have a nucleus.
- Reject D: Sieve tube elements do not have lignified cell walls or a nucleus.

1. The activation energy of the forward uncatalysed reaction is calculated as: \(E_{a} = \text{energy of uncatalysed transition state} - \text{energy of substrate} = 90 - 20 = 70\text{ kJ/mol}\).
2. The activation energy of the forward catalysed reaction is: \(E_{a} = 55 - 20 = 35\text{ kJ/mol}\).
3. The reduction in the forward activation energy is: \(70 - 35 = 35\text{ kJ/mol}\).
4. The activation energy for the reverse catalysed reaction is the energy difference between the catalysed transition state and the product: \(E_{a(\text{reverse})} = 55 - (-10) = 65\text{ kJ/mol}\).

Award 1 mark for the correct combination of the reduction in forward activation energy (35 kJ/mol) and the reverse catalysed activation energy (65 kJ/mol).

1. The temperature increases from 15 °C to 35 °C, which is an increase of 20 °C (two intervals of 10 °C).
2. With \(Q_{10} = 2.0\), each 10 °C rise doubles the rate.
3. At 25 °C, the rate is \(2 \times x = 2x\).
4. At 35 °C (without inhibitor), the rate is \(2 \times 2x = 4x\).
5. The competitive inhibitor reduces this uninhibited rate at 35 °C by 50%, so the final rate is \(0.50 \times 4x = 2x\).

Award 1 mark for identifying that the rate doubles twice to \(4x\) and is then halved to \(2x\).

Statement 1 is correct because the charged alginate matrix can alter the local hydrogen ion concentration around the immobilised enzyme, creating a microenvironment with a different pH than the bulk solution. Statement 2 is incorrect because immobilisation involves physical entrapment or weak bonding to a support, which does not alter the primary structure (the sequence of amino acids) of the enzyme. Statement 3 is correct because immobilisation restricts conformational changes, thereby making the tertiary structure more stable and less prone to unfolding/denaturation at extreme pH values.

Award 1 mark for identifying statements 1 and 3 as correct and statement 2 as incorrect.

1. The initial water potential of the plant cell is calculated as: \(\psi_{\text{cell}} = \psi_s + \psi_p = -0.8 + 0.3 = -0.5\text{ MPa}\).
2. The water potential of the surrounding solution is \(-0.7\text{ MPa}\).
3. Since water moves from a region of higher water potential (less negative) to a region of lower water potential (more negative), water will move out of the cell (\(-0.5\text{ MPa} \rightarrow -0.7\text{ MPa}\)).
4. As water leaves, the volume of the vacuole and cytoplasm decreases, causing the protoplast to shrink.

Award 1 mark for calculating the correct initial cell water potential and deducing that water leaves the cell, causing the protoplast to shrink.

A mature xylem vessel element is a dead cell; it completely lacks cytoplasm, has heavily lignified walls to withstand high tension, and does not have companion cells. A mature phloem sieve tube element is a living cell; it contains a thin layer of peripheral cytoplasm (though it lacks a nucleus and many organelles), has non-lignified cellulose walls, and is structurally and metabolically associated with companion cells.

Award 1 mark for the row showing: Xylem (Cytoplasm absent, Lignified walls present, Companion cells absent); Phloem (Cytoplasm present, Lignified walls absent, Companion cells present).

Active loading of sucrose involves the active pumping of hydrogen ions out of the companion cells (using ATP), creating an electrochemical gradient. Protons then flow back down their gradient via a co-transporter protein, bringing sucrose into the companion cells against its concentration gradient (secondary active transport). Once inside the companion cells, sucrose accumulates to a high concentration and moves passively down its concentration gradient into the sieve tube elements via connecting plasmodesmata.

Award 1 mark for identifying the diffusion of sucrose through plasmodesmata as the passive process.

Myelin acts as an electrical insulator, preventing depolarisation from occurring along the myelinated segments of the axon membrane. Voltage-gated sodium channels are highly concentrated at the uninsulated nodes of Ranvier. Consequently, depolarisation is restricted to the nodes, forcing local electrical currents to jump from one node to the next, which significantly increases conduction speed compared to continuous propagation.

Award 1 mark for identifying that depolarisation is restricted to the nodes of Ranvier and that local circuits jump between nodes.

1. Arrival of the action potential depolarises the presynaptic membrane, causing voltage-gated calcium channels to open (5).
2. Calcium ions diffuse down their electrochemical gradient into the presynaptic neurone (1).
3. This influx causes synaptic vesicles to fuse with the presynaptic membrane, releasing acetylcholine by exocytosis (3).
4. Acetylcholine diffuses across the cleft and binds to specific ligand-gated receptors on the postsynaptic membrane (2).
5. This binding causes ligand-gated sodium channels to open, resulting in sodium influx and depolarisation of the postsynaptic membrane (4).
Hence, the correct order is 5 \(\rightarrow\) 1 \(\rightarrow\) 3 \(\rightarrow\) 2 \(\rightarrow\) 4.

Award 1 mark for the correct chronological sequence of synaptic transmission events.

A competitive inhibitor has a molecular shape similar to that of the substrate and binds reversibly to the active site of the enzyme. Increasing the substrate concentration increases the probability of a substrate molecule, rather than an inhibitor molecule, colliding with and binding to an active site. Therefore, at high substrate concentrations, the inhibitor is outcompeted, and the rate of reaction can reach the same maximum velocity (\(V_{\max}\)) as without the inhibitor. The Michaelis-Menten constant (\(K_{\text{m}}\)) increases (indicating a lower apparent affinity), rather than decreases.

1 mark for identifying that increasing substrate concentration enables substrate molecules to outcompete competitive inhibitor molecules for the active sites.

By definition, the Michaelis-Menten constant (\(K_{\text{m}}\)) is the substrate concentration at which the rate of reaction is half of the maximum velocity (\(V_{\max}\)). For enzyme X, \(K_{\text{m}}\) is \(2.5\text{ mmol dm}^{-3}\), which means that at this substrate concentration, the rate of reaction is half of \(V_{\max}\) (i.e., \(60\text{ arbitrary units}\)). Option A and B are incorrect because a lower \(K_{\text{m}}\) indicates a higher affinity. Option D is incorrect because at its \(K_{\text{m}}\) of \(8.0\text{ mmol dm}^{-3}\), enzyme Y operates at half of its \(V_{\max}\) (\(125\text{ arbitrary units}\)), not its full \(V_{\max}\).

1 mark for correctly applying the definition of \(K_{\text{m}}\) to determine that enzyme X operates at half of its maximum velocity when substrate concentration is \(2.5\text{ mmol dm}^{-3}\).

As the temperature is raised, both the enzyme and substrate molecules gain kinetic energy. They move faster, resulting in more frequent collisions and a higher probability of successful collisions that form enzyme-substrate complexes, thereby increasing the rate of reaction. Enzymes lower activation energy, but temperature does not change this activation energy barrier itself.

1 mark for identifying that increased kinetic energy of molecules leads to a higher frequency of successful collisions.

The absence of cytoplasm and organelles in mature xylem vessel elements leaves an empty, hollow tube, which greatly reduces resistance to the longitudinal flow of water. Option A is incorrect because cohesion between water molecules, not lignification, prevents the water column from breaking. Option B is incorrect because lignified cell walls (not pits) prevent the collapse of vessels under tension. Option D is incorrect because absent end walls facilitate longitudinal flow, whereas pits allow lateral movement.

1 mark for correctly matching the absence of cytoplasm with the reduced resistance to longitudinal water flow.

Active loading begins when protons are actively pumped out of the companion cell into the cell wall using ATP (2), establishing a high concentration of protons in the cell wall space (4). Protons then diffuse back into the companion cell via co-transporter proteins, carrying sucrose against its concentration gradient (1). Finally, sucrose diffuses from the companion cell down its concentration gradient into the sieve tube element via plasmodesmata (3). Therefore, the correct sequence is 2 \(\rightarrow\) 4 \(\rightarrow\) 1 \(\rightarrow\) 3.

1 mark for identifying the correct sequential order of active sucrose loading (2 \(\rightarrow\) 4 \(\rightarrow\) 1 \(\rightarrow\) 3).

Water always moves down a water potential gradient (from a less negative/higher value to a more negative/lower value). Here, the gradient runs from the root hair cell (\(-150\text{ kPa}\)) to the cortex cell (\(-250\text{ kPa}\)) and then to the endodermal cell (\(-350\text{ kPa}\)). The symplast pathway involves water moving from cytoplasm to cytoplasm via plasmodesmata. Option A is incorrect because water moves from root hair to endodermal cell, not the reverse. Option B is incorrect because the Casparian strip blocks the apoplast pathway. Option D is incorrect because water movement is entirely passive by osmosis.

1 mark for correctly identifying that water moves down the water potential gradient from the root hair cell to the endodermal cell via plasmodesmata in the symplast pathway.

The arrival of an action potential depolarises the presynaptic membrane, opening voltage-gated calcium channels. Calcium ions diffuse in (1) and stimulate synaptic vesicles to fuse with the membrane, releasing acetylcholine via exocytosis into the cleft (4). Acetylcholine diffuses across and binds to specific receptor proteins on the postsynaptic membrane (2), which opens ligand-gated sodium channels, allowing sodium ions to enter and depolarise the postsynaptic neurone (3). Finally, acetylcholinesterase hydrolyses acetylcholine in the synaptic cleft (5) to reset the system. Hence, the correct sequence is 1 \(\rightarrow\) 4 \(\rightarrow\) 2 \(\rightarrow\) 3 \(\rightarrow\) 5.

1 mark for identifying the correct chronological order of synaptic transmission events (1 \(\rightarrow\) 4 \(\rightarrow\) 2 \(\rightarrow\) 3 \(\rightarrow\) 5).

According to the sliding filament model of muscle contraction, thick (myosin) and thin (actin) filaments slide past one another. The A-band represents the entire length of the myosin filaments, which do not shorten; thus, the A-band length remains constant. The I-band (actin only) and H-zone (myosin only) both decrease in width as actin filaments slide into the H-zone and overlap more with myosin. The overall sarcomere shortens, meaning the distance between the Z-lines decreases. Therefore, row A is correct.

1 mark for correctly identifying the change (or lack thereof) in the A-band, I-band, H-zone, and the distance between Z-lines during skeletal muscle contraction.

Detailed explanation of the answers: (a) The hydrolysis of urea produces ammonia, which is alkaline and raises pH. This can be measured using a pH meter over time. Alternatively, carbon dioxide gas can be collected using a gas syringe. (b)(i) At extremely high substrate concentrations, the inhibitor has a negligible effect because substrate molecules are far more likely to bind to the active sites, allowing the reaction to reach the same Vmax as the uninhibited reaction. (ii) A competitive inhibitor increases Km because more substrate is needed to achieve half the maximum velocity of the enzyme-catalysed reaction. (c) Urease is a globular protein. Its secondary structure involves folding into alpha-helices and beta-sheets. Its tertiary structure involves further complex folding into a specific 3D conformation. Hydrophilic R-groups are oriented towards the exterior, allowing it to interact with water and remain soluble, while hydrophobic R-groups are buried in the core.

(a) Max 2 marks: 1 mark for measuring increase in pH / alkalinity / electrical conductivity over time; 1 mark for measuring the volume of carbon dioxide gas produced per unit time. (b)(i) Max 2 marks: 1 mark for stating Vmax is unchanged; 1 mark for explaining that high substrate concentration outcompetes / overcomes the effect of the inhibitor. (b)(ii) Max 2 marks: 1 mark for stating Km increases; 1 mark for explaining that a higher substrate concentration is needed to reach half Vmax. (c) Max 4 marks: 1 mark for secondary structure consisting of alpha-helices and beta-pleated sheets held by hydrogen bonds; 1 mark for tertiary structure being the overall 3D folding stabilized by ionic / disulfide / hydrophobic / hydrogen interactions; 1 mark for hydrophilic R-groups on the outside of the molecule to allow solubility; 1 mark for hydrophobic R-groups clustered on the inside.

Detailed explanation of the answers: (a) Increased temperature provides excess kinetic energy, causing the atoms in human DNA polymerase to vibrate violently. This breaks weaker chemical bonds (hydrogen and ionic bonds) that maintain its tertiary structure. The active site loses its specific 3D shape, preventing complementary substrates from binding and forming enzyme-substrate complexes. (b) Thermophilic enzymes like Taq polymerase have evolved structural stability, often containing a higher proportion of covalent disulfide bridges, dense internal hydrogen bonding, and tightly packed hydrophobic cores that resist denaturation at high temperatures. (c) Activation energy is the energy barrier that reactants must overcome to undergo a chemical reaction. DNA polymerase lowers this barrier by correctly aligning the incoming nucleotide with the template strand and holding them in a position that facilitates covalent phosphodiester bond formation.

(a) Max 4 marks: 1 mark for heat increasing kinetic energy of the enzyme / increased vibration of atoms; 1 mark for breaking of hydrogen bonds and ionic bonds; 1 mark for loss of tertiary structure / change in shape of active site; 1 mark for substrate no longer complementary / no enzyme-substrate complexes formed / denaturation. (b) Max 3 marks: 1 mark for greater density of hydrogen bonds / ionic bonds; 1 mark for presence of more covalent disulfide bridges; 1 mark for highly stable / tightly packed hydrophobic core resisting thermal motion. (c) Max 3 marks: 1 mark for defining activation energy as the minimum energy required to initiate a chemical reaction / reach transition state; 1 mark for bringing substrates into close proximity / correct orientation; 1 mark for putting strain on chemical bonds of substrates to facilitate reaction.

Detailed explanation of the answers: (a) Xylem vessels are hollow tubes formed from dead cells with no cytoplasm or end walls, minimizing resistance to the mass flow of water. Their walls are impregnated with lignin, which is waterproof and provides high mechanical strength to prevent the vessel from collapsing under the negative pressure (tension) generated by transpiration. (b)(i) Active loading of sucrose requires a high amount of energy. Companion cells contain many mitochondria to generate ATP via aerobic respiration. This ATP powers proton pumps that transport H+ ions out of the cell. Companion cells also have co-transporter proteins to transport H+ back into the cell down their gradient, carrying sucrose with them. (ii) Plasmodesmata are direct physical connections between the cytoplasms of adjacent companion cells and sieve tube elements. Once sucrose is actively loaded into the companion cell, it diffuses through plasmodesmata into the sieve tube element.

(a) Max 4 marks: 1 mark for lignified cell walls preventing collapse / resisting negative pressure; 1 mark for lignin providing physical / mechanical support to the plant; 1 mark for hollow lumen / lack of cytoplasm / loss of end walls to form a continuous tube with low resistance to water flow; 1 mark for pits in walls to allow lateral movement of water. (b)(i) Max 4 marks: 1 mark for large numbers of mitochondria to produce ATP; 1 mark for active transport / pumping of hydrogen ions (H+) out of the cell; 1 mark for presence of sucrose-H+ co-transporter proteins in the cell membrane; 1 mark for folded cell membrane / microvilli to increase surface area for transport. (b)(ii) Max 2 marks: 1 mark for plasmodesmata acting as cytoplasmic channels / connections; 1 mark for allowing passive movement / diffusion / mass flow of sucrose from companion cells into sieve tube elements down a concentration gradient.

Detailed explanation of the answers: (a) To ensure a valid measurement of water uptake, which closely matches transpiration, no air bubbles must enter the xylem vessels. Cutting and assembling the apparatus under water ensures this. Seal all connections with vaseline to maintain an airtight system. Dry the leaves to prevent a water layer from blocking stomata. Measure the rate by timing the air bubble across a scale, repeating for reliability. (b) In still air, transpired water vapour builds up near stomata, creating a highly humid boundary layer that reduces the water potential gradient. Wind sweeps this water vapour away, increasing the steepness of the water potential gradient and increasing the diffusion rate of water vapour out of the leaf. (c) Apoplast: water moves by mass flow through the cellulose cell walls and intercellular spaces. Symplast: water moves by osmosis from cell to cell through the cytoplasm, connected via plasmodesmata.

(a) Max 4 marks: 1 mark for cutting shoot under water to prevent air bubbles / embolism in the xylem; 1 mark for assembling potometer under water / using vaseline to ensure airtight seals; 1 mark for drying the leaves before taking measurements; 1 mark for timing the movement of the air bubble / repeating the experiment and calculating a mean. (b) Max 3 marks: 1 mark for water vapour accumulating around stomata in still air to create a humid boundary layer; 1 mark for wind blowing away / removing this boundary layer of water vapour; 1 mark for maintaining / steepening the water potential gradient between the leaf interior and the outside air. (c) Max 3 marks: 1 mark for apoplast pathway involving movement of water through cell walls / intercellular spaces; 1 mark for symplast pathway involving movement through cytoplasm / plasmodesmata; 1 mark for water moving by mass flow in apoplast and by osmosis in symplast.

Detailed explanation of the answers: (a) The depolarization of the presynaptic knob opens voltage-gated calcium channels. Calcium ions diffuse rapidly down their electrochemical gradient into the presynaptic cytoplasm. This influx triggers synaptic vesicles containing acetylcholine to move towards and fuse with the presynaptic membrane, releasing the neurotransmitter into the cleft via exocytosis. (b) Acetylcholine molecules diffuse across the 20 nm synaptic cleft. They bind to complementary receptor sites on ligand-gated sodium channels in the postsynaptic membrane. This binding causes a conformational change that opens the channels, allowing sodium ions to diffuse into the postsynaptic cell, causing a local depolarization. (c)(i) Acetylcholinesterase breaks down acetylcholine into choline and ethanoic acid (acetate). (ii) If acetylcholine remained in the cleft, it would keep the ligand-gated sodium channels open, causing constant, uncontrolled firing of action potentials. Hydrolysis stops the signal and allows choline to be actively reabsorbed and recycled.

(a) Max 4 marks: 1 mark for depolarisation of the presynaptic membrane opening voltage-gated calcium channels; 1 mark for calcium ions diffusing into the presynaptic neurone down an electrochemical gradient; 1 mark for calcium ions causing synaptic vesicles to move towards and fuse with the presynaptic membrane; 1 mark for release of acetylcholine into the synaptic cleft by exocytosis. (b) Max 3 marks: 1 mark for acetylcholine diffusing across the synaptic cleft; 1 mark for binding to specific receptor proteins on the postsynaptic membrane; 1 mark for opening ligand-gated sodium channels / allowing sodium ions (Na+) to diffuse into the postsynaptic neurone causing depolarisation. (c)(i) 1 mark for choline and ethanoic acid / acetate. (c)(ii) Max 2 marks: 1 mark for preventing continuous depolarisation / continuous firing of action potentials in the postsynaptic neurone; 1 mark for recycling acetylcholine components / clearing receptor sites so they can respond to new signals.

Detailed explanation of the answers: (a)(i) Tropomyosin is a fibrous protein that wraps around actin filaments. In relaxed muscle, it covers the myosin-binding sites on actin, physically preventing the myosin heads from forming cross-bridges. (ii) Troponin is a globular regulatory protein complex bound to tropomyosin. When calcium ions bind to troponin, they cause a conformational shape change that pulls tropomyosin away from the active binding sites. (b) ATP has two major roles in the contraction cycle. First, the binding of ATP to the myosin head causes the head to detach from actin. Second, ATP is hydrolysed into ADP and inorganic phosphate by ATPase in the myosin head. This release of energy 'cocks' the myosin head into its high-energy state so it can form a new cross-bridge further along the actin filament. (c) If calcium release is blocked, troponin cannot change shape, and tropomyosin remains in its blocking position. Cross-bridges cannot be formed, so the sliding of filaments cannot occur and the muscle remains flaccid / cannot contract.

(a)(i) Max 2 marks: 1 mark for covering / blocking myosin-binding sites on the actin filament in relaxed muscle; 1 mark for preventing cross-bridge formation / preventing myosin heads from binding to actin. (a)(ii) Max 2 marks: 1 mark for binding to calcium ions; 1 mark for changing shape / undergoing conformational change to pull tropomyosin away from myosin-binding sites on actin. (b) Max 4 marks: 1 mark for ATP binding to the myosin head causing its detachment from actin; 1 mark for hydrolysis of ATP to ADP and inorganic phosphate; 1 mark for energy from hydrolysis cocking / returning the myosin head to its high-energy state; 1 mark for release of ADP and phosphate allowing the power stroke / pulling of actin filament. (c) Max 2 marks: 1 mark for troponin not changing shape / tropomyosin remaining over binding sites on actin; 1 mark for no actin-myosin cross-bridges being formed / muscle remaining relaxed / unable to contract.

(a) (i) Proportional dilution preparation:
To prepare 10 cm³ of each concentration, dilute the 1.0% stock inhibitor solution with distilled water as follows:
- 0.8% inhibitor: 8.0 cm³ stock + 2.0 cm³ distilled water.
- 0.6% inhibitor: 6.0 cm³ stock + 4.0 cm³ distilled water.
- 0.4% inhibitor: 4.0 cm³ stock + 6.0 cm³ distilled water.
- 0.2% inhibitor: 2.0 cm³ stock + 8.0 cm³ distilled water.
(Measure all volumes using a graduated 10 cm³ syringe or pipette).

(a) (ii) Controlled variable: Volume / concentration of amylase solution (or volume / concentration of starch solution, or pH).
Method of control: Use a graduated pipette/syringe to measure exactly 2.0 cm³ of 1.0% amylase for every trial (or use a buffer solution of pH 6.8 to maintain a constant pH).

(a) (iii) Table of results:

| Inhibitor Concentration / % | Trial 1 / s | Trial 2 / s | Trial 3 / s | Mean Time (\(t\)) / s | Mean Rate of Reaction / arbitrary units (3 s.f.) |
|:---:|:---:|:---:|:---:|:---:|:---:|
| 0.0 | 60 | 90 | 60 | 70.0 | 14.3 |
| 0.2 | 90 | 120 | 90 | 100.0 | 10.0 |
| 0.4 | 150 | 120 | 150 | 140.0 | 7.14 |
| 0.6 | 210 | 180 | 210 | 200.0 | 5.00 |
| 0.8 | 300 | 270 | 330 | 300.0 | 3.33 |

(b) Graph plotting requirements:
- Axes correctly labeled with units: x-axis: 'Inhibitor concentration / %', y-axis: 'Mean rate of reaction / arbitrary units' (or '\(s^{-1}\)').
- Scales linear and occupying more than half of the grid in both directions.
- All 5 points plotted accurately using a small, clean cross (x) or circled dot.
- Points joined using straight, ruled lines from point to point, with no extrapolation beyond the plotted points.

(c) (i) Description: As the concentration of the plant extract (inhibitor) increases, the mean rate of reaction decreases (or time taken for starch hydrolysis increases).
Explanation: The inhibitor binds to the amylase enzyme, which alters or blocks the active site. This prevents substrate (starch) molecules from binding, resulting in fewer successful collisions and fewer enzyme-substrate complexes (ESCs) forming per unit time.

(c) (ii) Modification:
- Keep the concentration of the inhibitor constant.
- Vary the concentration of the substrate (starch) over a wide range.
- Measure the rate of reaction for each substrate concentration with and without the inhibitor.
- Plot a graph of rate of reaction against substrate concentration. If the inhibitor is competitive, high substrate concentrations will overcome the inhibition and reach the same maximum rate (\(V_{\max}\)) as the control. If non-competitive, the maximum rate (\(V_{\max}\)) will remain significantly lower.

Part (a)(i) [3 marks max]:
- 1 mark: Correctly states volumes of stock solution and distilled water for all four dilutions (0.8%, 0.6%, 0.4%, 0.2%) to make a total volume of 10 cm³.
- 1 mark: Explicitly states that the total volume of each solution is 10 cm³.
- 1 mark: Describes using a syringe or graduated pipette to measure the volumes.

Part (a)(ii) [2 marks max]:
- 1 mark: Identifies a suitable controlled variable (e.g., volume/concentration of starch, volume/concentration of amylase, or pH).
- 1 mark: Describes a realistic method of controlling this variable (e.g., using a syringe to measure a specific volume, or adding a specific volume of buffer solution).

Part (a)(iii) [5 marks max]:
- 1 mark: Table is fully enclosed with ruled borders and contains clear headings with appropriate units (Inhibitor Concentration / %, Trial times / s, Mean Time / s, Mean Rate / arbitrary units or s⁻¹).
- 1 mark: Correctly calculates the mean time for all five concentrations (70, 100, 140, 200, 300).
- 1 mark: Correctly calculates the rate of reaction using the formula 1000/mean time (14.3, 10.0, 7.14, 5.00, 3.33).
- 1 mark: All calculated rate values are rounded consistently to 3 significant figures.
- 1 mark: Times and concentrations are arranged in a logical, ascending sequence.

Part (b) [4 marks max]:
- 1 mark: X-axis is 'Inhibitor concentration / %' and Y-axis is 'Mean rate of reaction / arbitrary units'.
- 1 mark: Linear scales chosen so that data points cover at least 50% of the grid height and width.
- 1 mark: All points plotted accurately within half a small square.
- 1 mark: Points joined point-to-point with straight, sharp, ruled lines, or a smooth line of best fit with no extrapolation.

Part (c)(i) [3 marks max]:
- 1 mark: States that increasing inhibitor concentration decreases the rate of starch hydrolysis.
- 1 mark: Explains that the inhibitor binds to the enzyme / prevents substrate binding to the active site.
- 1 mark: States this results in fewer successful enzyme-substrate complexes (ESCs) forming per unit time.

Part (c)(ii) [3 marks max]:
- 1 mark: Suggests carrying out the reaction at a fixed concentration of inhibitor while varying the substrate (starch) concentration.
- 1 mark: Describes comparing the resulting curves or determining the maximum rate of reaction (\(V_{\max}\)).
- 1 mark: Explains that a competitive inhibitor's effect is overcome at high substrate concentrations (reaching the original \(V_{\max}\)), whereas a non-competitive inhibitor's effect is not overcome (lower \(V_{\max}\)).

(a) (i) Simple dilution preparation to make 10 cm³ of each solution:
- For 0.2 mol dm⁻³: Use a syringe to measure 2.0 cm³ of 1.0 mol dm⁻³ stock solution and add 8.0 cm³ of distilled water.
- For 0.4 mol dm⁻³: Measure 4.0 cm³ of stock and add 6.0 cm³ of distilled water.
- For 0.6 mol dm⁻³: Measure 6.0 cm³ of stock and add 4.0 cm³ of distilled water.
- For 0.8 mol dm⁻³: Measure 8.0 cm³ of stock and add 2.0 cm³ of distilled water.
(Control: Total volume is kept constant at 10.0 cm³).

(a) (ii) Table of results:

| Concentration of Sucrose Solution / mol dm⁻³ | Initial Length / mm | Final Length / mm | Change in Length / mm | Percentage Change in Length / % |
|:---:|:---:|:---:|:---:|:---:|
| 0.0 | 50.0 | 53.5 | +3.5 | +7.0 |
| 0.2 | 50.0 | 51.5 | +1.5 | +3.0 |
| 0.4 | 50.0 | 49.0 | -1.0 | -2.0 |
| 0.6 | 50.0 | 47.5 | -2.5 | -5.0 |
| 0.8 | 50.0 | 46.0 | -4.0 | -8.0 |

(Note: Change in length = Final length - Initial length. Percentage change = (Change in length / Initial length) * 100. Positive and negative signs indicate gain or loss in length, respectively).

(b) (i) Graph plotting requirements:
- Axes labeled with units: x-axis: 'Concentration of sucrose solution / mol dm⁻³', y-axis: 'Percentage change in length / %'.
- The origin of the y-axis must allow for both positive (above 0) and negative (below 0) values, with linear scales.
- Points plotted accurately: (0.0, 7.0), (0.2, 3.0), (0.4, -2.0), (0.6, -5.0), (0.8, -8.0).
- A smooth line of best fit drawn through the points, showing a clear intersection with the x-axis.

(b) (ii) Determination:
- The concentration of sucrose solution is 0.32 mol dm⁻³ (accept values between 0.31 and 0.33 mol dm⁻³ depending on the line of best fit).
- Explanation: This is the concentration where the line of best fit crosses the x-axis (percentage change in length is 0%). At this point, there is no net movement of water by osmosis because the water potential of the sucrose solution is equal to the water potential of the potato tissue.

(c) (i) Sources of error and improvements:
- Error 1: Surface liquid remains on the potato cylinders, affecting measurement of final length / thickness.
- Improvement 1: Gently blot each potato cylinder with a paper towel using a standardized procedure before measuring.
- Error 2: Evaporation of water from the tubes during the 24-hour immersion, which increases the concentration of the sucrose solutions.
- Improvement 2: Cover each tube with a rubber bung, cap, or Parafilm.
- Error 3: Cylinders cut from different parts of the same potato tuber might have different initial cell densities or sugar concentrations.
- Improvement 3: Cut all cylinders from the same region (e.g., central parenchyma) of a single potato tuber, or mix and randomize cylinders across the treatments.

(c) (ii) Adaptation for comparison:
- Cut identical cylinders (e.g., 50.0 mm length, same diameter using the same cork borer) from a sweet potato and subject them to the exact same range of sucrose concentrations for 24 hours.
- Fairness / Control: Ensure both sets of cylinders are kept at the same constant ambient temperature during the 24-hour period, and ensure the same volumes of sucrose solutions (10 cm³) are used.

Part (a)(i) [3 marks max]:
- 1 mark: Correctly identifies the volumes of 1.0 mol dm⁻³ sucrose stock and distilled water needed to prepare all four concentrations.
- 1 mark: Explains that the total final volume of each solution is maintained at exactly 10 cm³.
- 1 mark: Specifies the use of a measuring device such as a syringe or graduated pipette.

Part (a)(ii) [5 marks max]:
- 1 mark: Table is completely enclosed with clear headings and appropriate slash-separated units (e.g., Concentration of sucrose solution / mol dm⁻³, Length / mm, Percentage change / %).
- 1 mark: All five initial and final lengths are recorded to 1 decimal place (e.g., 50.0, 53.5).
- 1 mark: Correctly calculates change in length for all concentrations with '+' or '-' signs clearly shown.
- 1 mark: Correctly calculates percentage change in length (7.0, 3.0, -2.0, -5.0, -8.0).
- 1 mark: All percentage values are reported consistently to 1 decimal place.

Part (b)(i) [4 marks max]:
- 1 mark: X-axis labeled 'Concentration of sucrose solution / mol dm⁻³' and Y-axis labeled 'Percentage change in length / %'.
- 1 mark: The scale is linear and positioned to accommodate both positive and negative percentage changes, occupying more than 50% of the grid.
- 1 mark: Points are plotted accurately to within half a small square.
- 1 mark: A clean, smooth line of best fit is drawn, passing through all points without double-lining.

Part (b)(ii) [2 marks max]:
- 1 mark: Correctly reads the concentration where the percentage change in length is 0% (0.32 mol dm⁻³, accept 0.31 to 0.33 mol dm⁻³).
- 1 mark: Explains that at this concentration, there is no net movement of water / water potentials are equal / no change in mass or length.

Part (c)(i) [4 marks max]:
- 2 marks: Identifies two distinct and plausible errors (e.g., surface moisture, evaporation, temperature fluctuations, variations in cork borer diameter).
- 2 marks: Proposes a scientifically sound, corresponding improvement for each identified error.

Part (c)(ii) [2 marks max]:
- 1 mark: Explains that the procedure must be replicated identically using sweet potato tissue.
- 1 mark: Identifies at least two key variables to keep constant to ensure a fair comparison (e.g., temperature, surface area to volume ratio, volume of solution, duration of immersion).

(a) The primary structure is the unique sequence of amino acids joined by peptide bonds. This sequence determines the positioning of R-groups, which interact to form hydrogen, ionic, and disulfide bonds, as well as hydrophobic interactions. These interactions fold the polypeptide into a specific tertiary structure, creating a highly specific active site that is complementary to the substrate. (b)(i) Taq polymerase will have a much higher optimum temperature (around 72 to 75 degrees C) and higher thermal stability than human DNA polymerase (optimum around 37 degrees C). This is because Taq polymerase has evolved in hot springs and its tertiary structure is stabilised by a higher density of stronger bonds, such as disulfide bridges and ionic bonds, preventing denaturation at high temperatures. In contrast, human DNA polymerase denatures at temperatures above 40 to 45 degrees C as hydrogen bonds break, changing the shape of the active site. (ii) As temperature increases, the kinetic energy of both enzyme and substrate molecules increases, causing them to move faster. This increases the frequency of collisions between substrate molecules and the active site of the enzyme, resulting in a higher rate of successful enzyme-substrate complex formation.

Part (a): [Max 3 marks] 1. Primary structure is sequence of amino acids; 2. Determines the position of R-groups and the types of bonds formed (hydrogen, ionic, disulfide, hydrophobic); 3. Determines the tertiary structure / 3D conformation; 4. Results in a specific active site shape complementary to the substrate. Part (b)(i): [Max 4 marks] 1. Taq polymerase has a higher optimum temperature (70 to 80 degrees C) than human DNA polymerase (37 degrees C); 2. Taq polymerase is thermostable / does not denature at high temperatures; 3. Taq polymerase has more or stronger bonds (e.g. disulfide bridges, ionic bonds, hydrophobic interactions) in its tertiary structure; 4. Human DNA polymerase denatures at high temperatures because weak hydrogen bonds break, changing the active site shape so substrate can no longer bind. Part (b)(ii): [Max 3 marks] 1. Increasing temperature increases the kinetic energy of molecules; 2. Molecules move faster / have more successful collisions per unit time; 3. More enzyme-substrate complexes (ESCs) formed per second.

(a) Vmax is the maximum rate of an enzyme-catalysed reaction when the enzyme is fully saturated with substrate. The Michaelis-Menten constant (Km) is the substrate concentration at which the reaction rate is half of its maximum velocity, representing the affinity of the enzyme for its substrate. (b)(i) A competitive inhibitor increases the value of Km but has no effect on the value of Vmax. (ii) At low substrate concentrations, the competitive inhibitor competes with the substrate for the active site because it has a complementary shape, blocking substrate binding and reducing the rate of reaction. This can be overcome by increasing the substrate concentration significantly, making it much more likely for a substrate molecule to bind to the active site rather than an inhibitor molecule. (c) The active site is where the substrate binds and where the catalytic reaction occurs, whereas an allosteric site is a distinct regulatory site away from the active site where non-competitive inhibitors or activators bind to change the enzyme's conformation.

Part (a): [Max 3 marks] 1. Vmax is the maximum velocity / rate of reaction; 2. Occurs when all enzyme active sites are saturated with substrate; 3. Km is the substrate concentration at half Vmax; 4. Km is an indicator of affinity (high Km = low affinity, low Km = high affinity). Part (b)(i): [Max 2 marks] 1. Km increases; 2. Vmax remains unchanged. Part (b)(ii): [Max 3 marks] 1. Competitive inhibitor has a similar shape to the substrate and binds to the active site; 2. Prevents substrate binding / fewer enzyme-substrate complexes formed; 3. Increasing substrate concentration overcomes inhibition as substrate outcompetes the inhibitor for the active sites. Part (c): [Max 2 marks] 1. Active site binds the substrate / undergoes catalysis; 2. Allosteric site is a separate site where regulatory molecules / non-competitive inhibitors bind, altering the shape of the active site.

(a) Advantages of using immobilised lactase include: the enzyme can be easily recovered and reused, reducing costs; the final milk product is not contaminated with the enzyme, avoiding downstream purification; and the enzyme is more stable at wider ranges of temperature and pH due to the protective matrix. (b) The calcium alginate beads containing the immobilised lactase are packed into a vertical column. Milk containing lactose is poured into the top of the column and flows downward through the beads under gravity. As it passes over the beads, lactose is hydrolysed into glucose and galactose, and the lactose-free milk is collected at the bottom. (c) The reaction rate may be lower because of diffusion limitations: the substrate (lactose) must diffuse through the alginate matrix to reach the active sites of the trapped enzymes, and the products must diffuse out. This slows down the rate of reaction compared to free enzyme where reactants are in direct, rapid contact. Smaller bead sizes increase the surface-area-to-volume ratio, reducing the diffusion pathway and thereby increasing the reaction rate.

Part (a): [Max 3 marks] 1. Enzyme can be easily recovered / reused; 2. Product is free of enzyme contamination (no need for purification); 3. Increased thermal / pH stability of the enzyme; 4. Allows continuous production process. Part (b): [Max 3 marks] 1. Alginate beads packed into a column/tube; 2. Milk poured in at the top and trickles down through the column; 3. Hydrolysis of lactose occurs as milk contacts beads; 4. Lactose-free milk (containing glucose and galactose) is collected at the bottom. Part (c): [Max 4 marks] 1. Diffusion barrier / substrate must diffuse into the gel matrix to reach the active site; 2. Product must diffuse out of the bead; 3. Access to active sites is restricted compared to free enzyme; 4. Smaller bead size increases surface area to volume ratio; 5. Smaller bead size reduces diffusion distance, increasing the reaction rate.

(a) Mature xylem vessel elements are dead cells with no cytoplasm, organelles, or end-walls, forming a continuous, hollow tube that minimizes resistance to water flow. The cell walls are thickened with lignin, which provides mechanical strength and prevents the vessels from collapsing under the negative pressure of transpiration. Pit structures in the lignified walls allow the lateral movement of water between adjacent vessels. (b)(i) Lignin can be deposited in spiral, annular (ring-like), or reticulate (network) patterns. This patterning allows the vessel to remain flexible and stretch during plant growth. Lignin provides exceptional compressive strength to the cell walls, preventing them from collapsing inward when water is pulled up under extreme tension (negative pressure) generated by transpiration. (ii) Lignin is hydrophobic, which prevents water from penetrating or being absorbed into the secondary cell wall. This keeps water confined within the lumen of the xylem vessel. Furthermore, the waterproofing smooths the inner walls, reducing friction/drag as the water column moves upward, facilitating an uninterrupted transpiration stream.

Part (a): [Max 4 marks] 1. Cells are dead / contain no cytoplasm / no organelles, creating a hollow lumen with low resistance to water flow; 2. No end walls / perforated end walls, allowing continuous column of water; 3. Lignified walls provide strength / prevent collapse under tension; 4. Pits in walls allow lateral movement of water to bypass blockages. Part (b)(i): [Max 3 marks] 1. Spiral / annular / reticulate / pitted pattern; 2. Allows elongation / stretching / flexibility of the stem; 3. Lignin resists high tension / negative pressure / prevents collapse of vessels. Part (b)(ii): [Max 3 marks] 1. Lignin is hydrophobic / waterproof; 2. Prevents water escaping laterally from the vessel lumen; 3. Reduces friction / adhesion forces between water and the wall, facilitating faster mass flow.

(a) Proton pumps (H+ ATPases) in the cell surface membrane of the companion cell actively pump hydrogen ions (H+) out of the companion cell into the cell wall spaces, using energy from ATP hydrolysis. This creates a high electrochemical gradient of H+ outside the cell. H+ ions then diffuse back down their concentration gradient into the companion cell through co-transporter proteins. This facilitated diffusion of H+ is coupled with the transport of sucrose against its concentration gradient into the companion cell. Sucrose then diffuses from the companion cell into the sieve tube element via plasmodesmata. (b) The accumulation of sucrose in the sieve tube element lowers its solute potential, and therefore lowers its water potential. Water enters the sieve tube element from the adjacent xylem by osmosis down a water potential gradient. This increases the hydrostatic pressure within the sieve tube at the source, creating a pressure gradient that drives the mass flow of phloem sap towards the sink (where pressure is lower). (c) Sieve tube elements have no nucleus, very few organelles, and have sieve plates at their end-walls, whereas companion cells possess a prominent nucleus, numerous mitochondria, and lack sieve plates.

Part (a): [Max 5 marks] 1. Hydrogen ions (protons / H+) are actively pumped out of companion cells; 2. Uses ATP / energy; 3. Creates a proton / electrochemical / concentration gradient across the membrane; 4. Protons diffuse back into companion cells through co-transporter proteins; 5. Sucrose is co-transported / carried in with protons against its concentration gradient; 6. Sucrose moves from companion cells into sieve tube elements through plasmodesmata. Part (b): [Max 3 marks] 1. High sucrose concentration lowers water potential in the sieve tube; 2. Water enters sieve tube from xylem by osmosis; 3. Increases hydrostatic pressure at the source, creating a pressure gradient for mass flow. Part (c): [Max 2 marks] 1. Companion cell has a nucleus / sieve tube element does not; 2. Companion cell has many mitochondria / sieve tube element has very few; 3. Sieve tube element has sieve plates / companion cell does not.

(a) Under water stress, the plant hormone abscisic acid (ABA) binds to specific receptors on the cell surface membrane of guard cells. This binding stimulates the opening of calcium ion channels, allowing calcium ions to enter the cytoplasm. The rise in intracellular calcium acts as a second messenger, triggering the efflux of potassium ions (K+) and chloride ions out of the guard cells. This loss of solutes raises the water potential inside the guard cells. Water then leaves the guard cells by osmosis, causing them to lose turgor and become flaccid, which closes the stomatal pore. (b)(i) Two features are: a thick waxy cuticle, and sunken stomata in pits. (ii) Sunken stomata trap moist air in the pit, reducing the water vapour potential gradient between the inside of the leaf and the atmosphere outside the pit, which significantly reduces the rate of transpiration. (c) The symplastic pathway involves the movement of water through the living cytoplasm and plasmodesmata of adjacent cells, whereas the apoplastic pathway involves water movement through the non-living cell walls and intercellular spaces without crossing any plasma membranes.

Part (a): [Max 4 marks] 1. ABA binds to receptors on guard cell membrane; 2. Stimulates influx of calcium ions (Ca2+) into the cytoplasm; 3. Promotes efflux of potassium ions (K+) / anions (Cl-) from guard cells; 4. Increases water potential inside guard cells; 5. Water leaves guard cells by osmosis, guard cells lose turgidity / become flaccid, closing stomata. Part (b)(i): [Max 2 marks] 1. Thick waxy cuticle; 2. Sunken stomata / stomatal pits; 3. Rolled leaves / hairs on leaf surface. (Accept any two) Part (b)(ii): [Max 2 marks] 1. Sunken stomata / hairs / rolled leaves trap a layer of humid / moist air; 2. Reduces the water vapour potential gradient (between inside and outside leaf); OR 3. Thick cuticle is impermeable to water; 4. Reduces cuticular transpiration. Part (c): [Max 2 marks] 1. Symplastic: moves through cytoplasm / plasmodesmata; 2. Apoplast: moves through cell walls / intercellular spaces.

(a) The resting potential is maintained by the active transport of ions by the sodium-potassium pump, which pumps three sodium ions (Na+) out of the axon for every two potassium ions (K+) pumped in, using energy from ATP hydrolysis. Additionally, the axon membrane is much more permeable to potassium ions than to sodium ions due to the presence of open potassium leak channels, allowing K+ to diffuse down its concentration gradient out of the cell. Large, negatively charged organic anions remain inside the cytoplasm, contributing to the overall negative interior charge. (b) Myelin is an electrical insulator provided by Schwann cells wrapping around the axon. Depolarisation cannot occur where myelin is present; it can only occur at the unmyelinated gaps called nodes of Ranvier, where sodium channels are highly concentrated. When an action potential occurs at one node, local circuits of current flow to the adjacent node, depolarising it to threshold. This causes the action potential to jump from node to node, a process called saltatory conduction, which greatly increases transmission speed. (c) Without myelin, the electrical insulation of the axon is lost, allowing ions to leak across the membrane along the entire length of the axon. This disrupts the local currents, making it difficult to depolarise the next region of the membrane. Consequently, saltatory conduction cannot occur, resulting in significantly slower impulse transmission or a complete failure of the action potential to reach its target destination.

Part (a): [Max 3 marks] 1. Sodium-potassium pump actively transports 3 Na+ out and 2 K+ in (using ATP); 2. Membrane is more permeable to K+ than Na+ / K+ leaks out down concentration gradient; 3. Presence of organic anions inside cell; 4. Results in net negative charge inside relative to outside. Part (b): [Max 4 marks] 1. Schwann cells wrap around axon to form myelin sheath which acts as an electrical insulator; 2. Depolarisation / ion movement can only occur at nodes of Ranvier; 3. Local circuits set up between adjacent nodes; 4. Action potential jumps from node to node (saltatory conduction); 5. Increases speed of transmission compared to unmyelinated axons. Part (c): [Max 3 marks] 1. Loss of insulation leads to leakage of ions / current; 2. Local circuits are weakened / cannot depolarise adjacent nodes; 3. Saltatory conduction is prevented / speed of transmission is reduced; 4. Impulses may block / fail to reach effector.

(a) When an action potential reaches the presynaptic membrane, it stimulates the opening of voltage-gated calcium ion channels, and calcium ions (Ca2+) diffuse into the presynaptic neurone down their concentration gradient. This influx of calcium ions causes synaptic vesicles containing acetylcholine (ACh) to move towards and fuse with the presynaptic membrane, releasing ACh into the synaptic cleft via exocytosis. ACh diffuses across the synaptic cleft and binds to specific receptor proteins on the postsynaptic membrane. This binding causes ligand-gated sodium channels to open, allowing sodium ions (Na+) to diffuse into the postsynaptic neurone, depolarising the membrane. (b)(i) Acetylcholinesterase normally hydrolyses acetylcholine into choline and ethanoic acid, terminating the signal. If this enzyme is inhibited, acetylcholine remains bound to the receptors on the postsynaptic membrane, causing ligand-gated sodium channels to stay open. This results in continuous depolarisation of the postsynaptic membrane and the continuous generation of action potentials. (ii) Continuous action potentials travel down motor neurones to neuromuscular junctions, causing continuous stimulation of muscle fibres. This leads to involuntary muscle contractions (twitching and spasms). Eventually, the muscles (including the diaphragm and intercostal muscles involved in breathing) remain in a state of permanent contraction and cannot relax, causing respiratory paralysis.

Part (a): [Max 5 marks] 1. Action potential causes opening of voltage-gated Ca2+ channels in presynaptic membrane; 2. Ca2+ ions enter presynaptic knob; 3. Vesicles move and fuse with presynaptic membrane, releasing acetylcholine (ACh) by exocytosis; 4. ACh diffuses across synaptic cleft; 5. ACh binds to receptors on postsynaptic membrane; 6. Ligand-gated sodium channels open, Na+ enters, causing depolarisation / generator potential. Part (b)(i): [Max 3 marks] 1. Acetylcholine is not broken down / remains in synaptic cleft; 2. ACh remains bound to postsynaptic receptors; 3. Sodium channels remain open / postsynaptic membrane remains depolarised; 4. Continuous action potentials are generated. Part (b)(ii): [Max 2 marks] 1. Continuous stimulation of muscles / neuromuscular junctions leads to muscle spasms / contraction; 2. Inability of respiratory muscles (diaphragm / intercostal muscles) to relax leads to asphyxiation / paralysis.

(a)
1. Myelin is an electrical insulator composed of lipid-rich Schwann cells wrapped repeatedly around the axon membrane.
2. This prevents the movement of ions (depolarization/repolarization) through the membrane in myelinated regions.
3. Voltage-gated sodium and potassium channels are concentrated almost entirely at the gaps between Schwann cells, known as the nodes of Ranvier.
4. Therefore, local currents are extended to the next node, causing the action potential to "jump" from one node of Ranvier to the next in a rapid process called saltatory conduction.

(b)
1. Without myelin insulation, depolarization must occur sequentially along the entire length of the axon membrane rather than just at the nodes of Ranvier.
2. This eliminates saltatory conduction, which dramatically slows down transmission.
3. Furthermore, without the myelinated barrier, there is an increased leakage of ions (specifically sodium ions, \(\text{Na}^+\)) out across the axon membrane.
4. Consequently, local currents decrease in amplitude and may fail to depolarize the adjacent region of the membrane to the threshold potential, completely blocking the generation of a new action potential.

(c)
1. When the action potential depolarizes the presynaptic membrane, voltage-gated calcium ion channels open, and \(\text{Ca}^{2+}\) ions diffuse down their electrochemical gradient into the presynaptic knob.
2. The rise in intracellular \(\text{Ca}^{2+}\) concentration triggers synaptic vesicles containing acetylcholine (ACh) to move toward and fuse with the presynaptic membrane, releasing the neurotransmitter into the synaptic cleft by exocytosis.

**Part (a)** [Max 4 marks]
1. Myelin acts as an electrical insulator / composed of Schwann cells; [1]
2. Prevents the movement of ions / depolarization through the myelinated parts of the axon membrane; [1]
3. Voltage-gated sodium/potassium channels are localized only at the nodes of Ranvier / gaps in myelin; [1]
4. Depolarization / action potentials can only occur at the nodes of Ranvier; [1]
5. Local circuits are elongated / action potential "jumps" from node to node (saltatory conduction); [1]

**Part (b)** [Max 4 marks]
1. Loss of myelin means saltatory conduction is lost / conduction becomes continuous along the axon membrane; [1]
2. Continuous conduction is much slower than saltatory conduction; [1]
3. Loss of insulation allows sodium ions (\(\text{Na}^+\)) to leak across the axon membrane; [1]
4. Local currents become weaker / dissipate before reaching the next segment; [1]
5. Depolarization fails to reach the threshold potential, preventing the propagation of further action potentials / causing a nerve block; [1]

**Part (c)** [Max 2 marks]
1. (Depolarization causes) voltage-gated calcium channels to open and calcium ions (\(\text{Ca}^{2+}\)) diffuse into the presynaptic cytoplasm / knob; [1]
2. Calcium ions cause synaptic vesicles (containing acetylcholine) to move towards and fuse with the presynaptic membrane; [1]
3. Acetylcholine is released into the synaptic cleft by exocytosis; [1]

(a)
1. NAD and FAD act as hydrogen (and electron) carriers.
2. They are reduced (to NADH and \(\text{FADH}_2\)) when they accept protons and electrons during glycolysis, the link reaction, and the Krebs cycle.
3. They transfer these electrons and protons to the electron transport chain (ETC) located on the inner mitochondrial membrane, driving the chemiosmotic synthesis of ATP.

(b) (i)
1. DNP allows protons to diffuse freely through the lipid bilayer of the inner mitochondrial membrane back into the matrix.
2. This dissipates/reduces the concentration gradient of protons (the pH and electrical gradient) between the intermembrane space and the matrix.
3. As a result, the proton-motive force is diminished, meaning fewer protons flow down their electrochemical gradient through the channels of ATP synthase, reducing ATP phosphorylation.

(b) (ii)
1. Because the proton gradient is constantly being run down by DNP, the electron transport chain (ETC) continues to run at an accelerated rate to try to pump protons back out and restore the gradient.
2. Since oxygen is the terminal electron acceptor, it receives electrons from the end of the ETC and combines with protons to form water; therefore, faster ETC turn-over consumes more oxygen.

(c)
1. It is folded into cristae to provide a large surface area for containing many electron transport chain carriers and ATP synthase complexes.
2. It is impermeable to protons, preventing unregulated passage of \(\text{H}^+\) ions and thus maintaining the established proton gradient.

**Part (a)** [Max 3 marks]
1. Coenzymes act as hydrogen carriers / electron carriers; [1]
2. Reduced (to NADH and \(\text{FADH}_2\)) by dehydrogenation during glycolysis / link reaction / Krebs cycle; [1]
3. Transfer electrons and protons to the electron transport chain (ETC) on the inner mitochondrial membrane; [1]
4. Regenerated / oxidized to allow glycolysis / link reaction / Krebs cycle to continue; [1]

**Part (b)(i)** [Max 3 marks]
1. Protons (\(\text{H}^+\)) bypass ATP synthase / flow through the membrane lipid bilayer directly into the matrix; [1]
2. Dissipates / reduces the proton / pH / electrochemical gradient between the intermembrane space and matrix; [1]
3. Reduces the proton-motive force; [1]
4. Less flow of protons through ATP synthase means less ADP is phosphorylated to ATP; [1]

**Part (b)(ii)** [Max 2 marks]
1. Electron transport chain runs faster / uncoupled from ATP synthesis to try to re-establish the gradient; [1]
2. Oxygen acts as the final / terminal electron acceptor; [1]
3. Oxygen combines with electrons and protons to form water / \(\text{O}_2 + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O}\) (leading to higher oxygen consumption); [1]

**Part (c)** [Max 2 marks]
1. Folded into cristae to increase the surface area; [1]
2. Contains / holds many electron transport chain carriers / proteins / cytochrome complexes; [1]
3. Contains ATP synthase complexes; [1]
4. Impermeable to protons (\(\text{H}^+\)) to maintain the proton gradient; [1]

**(a) (i)**
* Independent variable: Temperature (of the reaction mixture).
* Dependent variable: Rate of cellobiose hydrolysis / concentration of glucose produced per unit time (or absorbance at 10 minutes as a proxy).

**(a) (ii)**
* Mix equal volumes (e.g., \(5.0\text{ cm}^3\)) of the \(\beta\)-glucosidase enzyme solution and the \(2.0\%\) sodium alginate solution thoroughly in a beaker.
* Draw the mixture into a syringe and express it dropwise from a constant height into a beaker of \(0.1\text{ mol dm}^{-3}\) calcium chloride solution.
* Leave the beads in the calcium chloride solution for a fixed time (e.g., 10 to 15 minutes) to fully harden, then strain and rinse them thoroughly with distilled water to remove any unbound enzyme.

**(a) (iii)**
* **Independent Variable:** Maintain at least 5 different temperatures (e.g., \(20, 30, 40, 50, 60^{\circ}\text{C}\)) using thermostatically controlled water baths.
* **Control of pH:** Add a fixed volume of pH 6.0 buffer to the cellobiose substrate solution.
* **Pre-incubation:** Keep the substrate solution (with buffer) and the enzyme beads in separate test tubes in the respective water baths for 5 minutes before mixing, allowing them to reach the target temperature.
* **Standardisation:** Use a constant volume of cellobiose solution (e.g., \(10\text{ cm}^3\) of \(10.0\text{ g dm}^{-3}\)) and a constant number of beads (e.g., 20 beads) for each reaction tube.
* **Measurement:** After exactly 10 minutes of incubation, withdraw a fixed-volume sample of the supernatant, mix with a fixed volume of the GOD-POD colorimetric reagent, and measure the absorbance at \(505\text{ nm}\) using a colorimeter.
* **Replication & Safety:** Repeat the entire procedure 3 times at each temperature to calculate a mean. Wear safety goggles and gloves when handling chemicals.

**(b) (i)**
* Anomalous result: \(0.22\) at \(35^{\circ}\text{C}\) (Trial 3).
* Handling: The student should exclude this value from the calculation of the mean.
* Corrected mean: \(\frac{0.38 + 0.41}{2} = 0.395\) (rounds to \(0.40\) to match the decimal precision of other entries).

**(b) (ii)**
* **Variable 1:** Variation in the volume/size of the individual calcium alginate beads. *Control:* Use a syringe pump or an automated dropper with a standard needle size to ensure uniform droplet volume.
* **Variable 2:** Temperature fluctuation while taking samples and transferring to the colorimeter. *Control:* Perform the reactions and colorimetric readings in a room with stable ambient temperature, or use a heated cuvette holder in the colorimeter.

**(a) (i) [2 marks]**
* 1 mark: Identifies temperature as the independent variable.
* 1 mark: Identifies rate of hydrolysis / rate of glucose production / absorbance at 10 minutes as the dependent variable.

**(a) (ii) [3 marks]**
* 1 mark: Mixes sodium alginate and enzyme solutions thoroughly.
* 1 mark: Drops mixture from a constant height / uniform speed into calcium chloride solution.
* 1 mark: Leaves beads to harden for a set period of time AND rinses with distilled water.

**(a) (iii) [6 marks]**
* 1 mark: Uses at least 5 different temperatures (specified with values and units).
* 1 mark: Describes pre-incubation of enzyme beads and substrate solutions separately before mixing.
* 1 mark: Uses a buffer solution to keep pH constant.
* 1 mark: Keeps concentration and volume of substrate constant AND number/mass of beads constant.
* 1 mark: Describes how the colorimeter is calibrated and used (using GOD-POD reagent and reading at \(505\text{ nm}\)).
* 1 mark: Describes replicates (at least 3 at each temperature to calculate a mean) OR describes a relevant safety measure (e.g., eye protection / gloves).

**(b) (i) [2 marks]**
* 1 mark: Correctly identifies \(0.22\) at \(35^{\circ}\text{C}\) as the anomaly and states it must be excluded from calculations.
* 1 mark: Correctly calculates corrected mean as \(0.395\) or \(0.40\).

**(b) (ii) [2 marks]**
* 1 mark: Identifies variation in bead size/mass and suggests using a syringe pump or standard needle to control it.
* 1 mark: Identifies temperature drop during assay and suggests using a heated cuvette holder / working quickly.

**(a) (i)**
* Cut the stem of the leafy shoot under water to prevent air from entering the xylem vessels and breaking the continuous water column.
* Cut the stem at an angle to increase the surface area available for water uptake.
* Assemble the potometer completely under water (and insert the stem into the rubber tubing underwater) to ensure that no air bubbles are trapped inside the capillary tube.
* Apply grease/petroleum jelly (e.g., Vaseline) to all joints to ensure they are airtight.

**(a) (ii)**
* **Varying and measuring wind speed:** Position a hair dryer at a fixed distance from the shoot and adjust its speed settings. Measure the actual wind speed near the leaves using a digital anemometer to ensure accuracy.
* **Control of temperature:** The hair dryer may emit heat. Place a clear glass sheet (heat shield) between the hair dryer and the shoot to prevent heat from affecting transpiration, or use the 'cool' setting on the hair dryer.
* **Control of light intensity:** Perform the experiment in a room with a lamp set at a fixed distance from the leafy shoot to maintain constant light intensity.

**(a) (iii)**
* Formula: \(V = \pi r^2 d\) (or \(V = \pi \left(\frac{D}{2}\right)^2 d\))
* Where \(r\) is the radius of the capillary tube (or \(D\) is the internal diameter) and \(d\) is the distance moved by the bubble in one minute.

**(b) (i)**
* Mean distance moved in 5 minutes at \(3.0\text{ m s}^{-1}\) is \(46.0\text{ mm}\).
* Mean distance moved per minute: \(d = \frac{46.0}{5} = 9.2\text{ mm min}^{-1}\).
* Internal diameter = \(1.0\text{ mm}\), so radius \(r = 0.5\text{ mm}\).
* Volume of water uptake per minute: \(V = \pi \times (0.5)^2 \times 9.2\)
* \(V = 3.14159 \times 0.25 \times 9.2 = 7.2257\text{ mm}^3\text{ min}^{-1}\).
* Rounded to two significant figures: \(7.2\text{ mm}^3\text{ min}^{-1}\).

**(b) (ii)**
* Statistical test: Spearman’s rank correlation coefficient (or Pearson’s linear correlation coefficient).
* Find the critical value in tables for \(n = 5\) pairs of data at the \(p = 0.05\) (or \(5\%\)) significance level.
* If the calculated correlation value is greater than the critical value, reject the null hypothesis and conclude there is a significant correlation between wind speed and transpiration rate.

**(b) (iii)**
* The potometer measures water uptake, but not all water taken up is transpired.
* Some water is used in photosynthesis to produce organic molecules.
* Some water is used to maintain cell turgidity (turgor pressure) in the plant cells.

**(a) (i) [3 marks]**
* 1 mark: Cuts the stem underwater.
* 1 mark: Cuts stem at an angle.
* 1 mark: Assembles apparatus underwater / seals joints with grease to make it airtight.

**(a) (ii) [4 marks]**
* 1 mark: Varies wind speed by changing hairdryer settings or distance, and measures it using an anemometer.
* 1 mark: Controls temperature using a glass shield / cold-air setting.
* 1 mark: Controls light intensity using a lamp at a fixed distance.
* 1 mark: States that humidity or carbon dioxide concentration should be kept constant.

**(a) (iii) [1 mark]**
* 1 mark: Correct formula: \(V = \pi r^2 d\) (or \(V = \pi \left(\frac{D}{2}\right)^2 d\)) where symbols are defined (\(r\) = capillary radius, \(d\) = distance per minute).

**(b) (i) [2 marks]**
* 1 mark: Calculates distance per minute correctly (\(9.2\text{ mm min}^{-1}\)) OR shows the formula containing \(\pi \times 0.5^2 \times 46.0\) divided by 5.
* 1 mark: Correct final calculation of \(7.2\text{ mm}^3\text{ min}^{-1}\) (must be to 2 significant figures).

**(b) (ii) [3 marks]**
* 1 mark: Identifies Spearman’s rank correlation (or Pearson's correlation).
* 1 mark: States the need to compare calculated value to critical table values at \(p = 0.05\) (or \(5\%\) significance level).
* 1 mark: Explains that if the calculated value exceeds the critical value, the correlation is statistically significant (null hypothesis is rejected).

**(b) (iii) [2 marks]**
* 1 mark: Explains that some water is used in photosynthesis.
* 1 mark: Explains that some water is stored/used to keep cells turgid (turgor pressure).