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At very high substrate concentrations, substrate molecules outcompete the competitive inhibitor for the enzyme's active sites. This allows the rate of reaction to reach the same maximum velocity (\(V_{\max}\)) as the reaction without inhibitor, although a higher substrate concentration is required to achieve this.

Award 1 mark for the correct option (B) which states that the reaction reaches the same maximum rate because substrate outcompetes the inhibitor.

At high temperatures (above \(45\text{ }^\circ\text{C}\)), the increased kinetic energy causes the atoms within the enzyme to vibrate more violently. This breaks weak bonds, such as hydrogen bonds and ionic bonds, that hold the tertiary structure of the protein in place. Consequently, the active site changes shape (denatures) and the substrate can no longer fit.

Award 1 mark for the correct option (C) which identifies the disruption of hydrogen and ionic bonds in the tertiary structure of the enzyme leading to denaturation.

A competitive inhibitor binds to the active site and can be outcompeted by high substrate concentrations, so \(V_{\max}\) remains unchanged but a higher substrate concentration is needed to reach half \(V_{\max}\), increasing \(K_m\). A non-competitive inhibitor binds to an allosteric site, decreasing the number of functional enzymes and lowering \(V_{\max}\), but the affinity of the remaining active enzymes for the substrate remains unchanged, so \(K_m\) is unchanged.

Award 1 mark for the correct option (A) which matches the impact of both competitive and non-competitive inhibitors on both \(V_{\max}\) and \(K_m\).

The apoplast pathway is the movement of water through the cell walls and intercellular spaces. It relies on the cohesive forces between water molecules to pull water along. The Casparian strip blocks the apoplast pathway (not the symplast pathway), forcing water into the symplast pathway.

Award 1 mark for the correct option (C) which describes the apoplast pathway's reliance on cohesive forces within the cell walls.

Active loading of sucrose is achieved by companion cells using energy. Proton pumps (3) actively transport hydrogen ions out of the companion cell, creating a gradient. Mitochondria (1) produce the ATP required for this active transport. Sieve tube elements, not companion cells, lack nuclei and ribosomes (4).

Award 1 mark for the correct option (B) which correctly identifies features 1 and 3 as supporting active sucrose loading.

Still, humid air (1) creates the lowest rate of transpiration because the concentration gradient of water vapor is very low. Still, dry air at \(15\text{ }^\circ\text{C}\) (3) has a higher rate than (1). Increasing the temperature of still, dry air to \(25\text{ }^\circ\text{C}\) (4) increases transpiration because the water molecules have more kinetic energy. Moving, dry air (2) has the highest rate of transpiration because wind removes the boundary layer of saturated air, maintaining a steep water potential gradient.

Award 1 mark for the correct option (A) which correctly sequences the transpiration rates based on environmental factors.

In double-stranded DNA, the percentage of thymine (T) equals adenine (A), so A = \(32\%\). Together, A + T = \(64\%\). The remaining bases, cytosine (C) and guanine (G), make up \(100\% - 64\% = 36\%\). Since C = G, the percentage of guanine is \(36\% / 2 = 18\%\).

Award 1 mark for the correct option (A) which correctly calculates the percentage of guanine based on complementary base pairing.

DNA polymerase catalyses the formation of phosphodiester bonds between the sugar and phosphate groups of adjacent nucleotides on the new strand. DNA ligase joins the Okazaki fragments on the lagging strand together by forming phosphodiester bonds.

Award 1 mark for the correct option (A) which correctly states the role of both DNA polymerase and DNA ligase during replication.

A competitive inhibitor has a similar structure to the substrate and binds reversibly to the active site. Because its binding is reversible, its effect can be overcome by increasing the substrate concentration. As a result, the maximum rate of reaction (\(V_{\text{max}}\)) remains unchanged, but a higher substrate concentration is required to reach half of \(V_{\text{max}}\), meaning the Michaelis-Menten constant (\(K_{\text{m}}\)) increases.

1 mark for identifying that competitive inhibitors increase \(K_{\text{m}}\), leave \(V_{\text{max}}\) unchanged, and bind reversibly to the active site.

In temperature investigations, it is vital that both the enzyme and substrate solutions are at the exact target temperature when the reaction starts (time zero). Mixing solutions at different temperatures would mean the reaction begins at an intermediate, changing temperature, leading to invalid rate measurements.

1 mark for explaining that pre-incubation ensures both reactants are at the target temperature prior to starting the reaction, controlling temperature.

High temperatures provide excessive thermal kinetic energy, causing atoms in the enzyme to vibrate violently. This breaks the relatively weak hydrogen bonds, ionic bonds, and hydrophobic interactions that stabilize the specific tertiary structure of the enzyme. As a result, the enzyme denatures and the shape of the active site is altered, meaning it is no longer complementary to the substrate.

1 mark for identifying that high temperatures break tertiary-stabilizing bonds, altering the active site structure.

The apoplast pathway involves the movement of water and solutes through the non-living cell walls and intercellular spaces of the plant. This pathway is completely free and unhindered until water reaches the endodermis, where the suberin-containing Casparian strip blocks the cell walls, forcing water to pass through the cell surface membrane into the symplast pathway.

1 mark for identifying the apoplast pathway as movement through cell walls and intercellular spaces, blocked at the endodermis.

Active loading of sucrose begins with the active transport of protons (\(\text{H}^+\)) out of the companion cell into the cell wall using energy from ATP. This creates an electrochemical proton gradient. Protons then diffuse back into the companion cell down their concentration gradient through a co-transporter protein, which co-transports sucrose into the cell against its concentration gradient.

1 mark for describing active proton pumping out of the cell followed by co-transport of protons and sucrose back into the cell.

Cutting the stem under water and assembling the potometer under water are critical steps. They prevent air from entering the cut xylem vessels (which would break the continuous water column and cause cavitation) and prevent accidental air bubbles from forming in the capillary tube, ensuring that the movement of the air bubble accurately reflects water uptake.

1 mark for identifying that cutting the shoot and assembling the apparatus under water is essential to prevent air bubbles from entering xylem vessels.

1. Since the DNA is double-stranded and has 1200 nucleotides, there are 600 base pairs. 2. If there are 360 adenine (A) nucleotides, there must also be 360 thymine (T) nucleotides due to complementary base pairing. This makes 360 A-T base pairs. Each A-T pair has 2 hydrogen bonds: \(360 \times 2 = 720\) hydrogen bonds. 3. The remaining nucleotides are guanine (G) and cytosine (C): \(1200 - (360 + 360) = 480\) nucleotides, which corresponds to 240 G-C base pairs. Each G-C pair has 3 hydrogen bonds: \(240 \times 3 = 720\) hydrogen bonds. 4. Total hydrogen bonds = \(720 + 720 = 1440\).

1 mark for correctly calculating the number of A-T and G-C base pairs and determining the total hydrogen bonds as 1440.

DNA contains deoxyribose, thymine (and cytosine, adenine, guanine), and contains hydrogen bonds. tRNA contains ribose, uracil (and cytosine, adenine, guanine), and despite being single-stranded, it folds into a 3D shape held together by complementary hydrogen bonds between base pairs.

1 mark for identifying the correct pentose sugars, pyrimidine bases, and that both DNA and tRNA contain hydrogen bonds.

A competitive inhibitor binds to the active site of the enzyme, directly competing with the substrate. At very high substrate concentrations, the substrate molecules outcompete the inhibitor, meaning the maximum rate of reaction (\(V_{max}\)) can still be reached. However, a higher concentration of substrate is required to achieve half of this maximum rate, which increases the value of the Michaelis-Menten constant (\(K_m\)). Non-competitive inhibitors bind to an allosteric site, altering the shape of the active site, which decreases the \(V_{max}\) regardless of substrate concentration.

1 mark for identifying that the inhibitor is competitive and binds to the active site (Option A).
- Reject B: Competitive inhibitors do not bind to allosteric sites.
- Reject C and D: Non-competitive inhibitors lower the \(V_{max}\).

At high temperatures, the increased kinetic energy causes the atoms within the enzyme to vibrate more violently. This breaks the weaker hydrogen and ionic bonds stabilizing the tertiary structure. As the tertiary structure is disrupted, the shape of the active site changes, meaning the substrate can no longer fit (denaturation). Peptide bonds are covalent and are not broken by high physiological temperatures; they require strong chemical hydrolysis or peptidase enzymes to break.

1 mark for identifying the correct explanation of denaturation due to the breaking of hydrogen and ionic bonds (Option B).
- Reject A: Kinetic energy increases, not decreases, at higher temperatures.
- Reject C: Peptide bonds are not hydrolyzed by high temperatures.
- Reject D: Hydrophobic interactions and other structural features are disrupted rather than strengthened to cause a locked inactive state.

Active transport (using ATP) is used to pump protons (\(\text{H}^+\)) out of the companion cell cytoplasm into the cell wall space (apoplast), creating a proton gradient. Protons then diffuse back into the companion cell down their concentration gradient through a co-transporter protein, carrying sucrose along with them (symport). The high concentration of sucrose in the companion cell then allows sucrose to diffuse into the sieve tube element through the plasmodesmata.

1 mark for the correct pathway sequence (Option A).
- Reject B: Protons are actively pumped out of, not into, the companion cell.
- Reject C: Sucrose is not directly actively pumped into the cell wall.
- Reject D: Protons are pumped out of the companion cells, not the sieve tube elements.

The apoplast pathway involves water moving through the non-living parts of the plant, primarily the cellulose cell walls and intercellular spaces. The symplast pathway involves water moving through the living cytoplasm, connected between cells by plasmodesmata. At the endodermis, the waterproof suberin band called the Casparian strip blocks the apoplast pathway, forcing all water to cross the selectively permeable cell membrane into the symplast pathway.

1 mark for the correct table row (Option A).
- Reject B: Transposes the descriptions of apoplast and symplast pathways.
- Reject C: The symplast is not restricted to vacuoles, and the Casparian strip does not block the symplast.
- Reject D: The apoplast pathway does not go through vacuoles.

Initially (generation 0), both strands of all DNA molecules contain \({}^{15}\text{N}\).
- After the first division (generation 1), each of the 2 DNA molecules has one old \({}^{15}\text{N}\) strand and one new \({}^{14}\text{N}\) strand (100% hybrid/intermediate).
- After the second division (generation 2), there are 4 DNA molecules: 2 are hybrid (\({}^{15}\text{N}/{}^{14}\text{N}\)) and 2 are light (\({}^{14}\text{N}/{}^{14}\text{N}\)) (50% hybrid).
- After the third division (generation 3), there are 8 DNA molecules: 2 are hybrid (\({}^{15}\text{N}/{}^{14}\text{N}\)) and 6 are light (\({}^{14}\text{N}/{}^{14}\text{N}\)).

The percentage of hybrid DNA molecules is \(\frac{2}{8} \times 100\% = 25\%\).

1 mark for calculating the correct percentage (Option B).
- Reject A: \(0\%\) would be after many more generations or if replication were conservative.
- Reject C: \(50\%\) is the percentage of hybrid molecules after two divisions.
- Reject D: \(75\%\) is the percentage of light DNA molecules (\({}^{14}\text{N}/{}^{14}\text{N}\)), not hybrid.

Total number of bases in double-stranded DNA = \(120 \times 2 = 240\) bases.
- Adenine (A) = \(35\%\) of \(240 = 84\) bases.
- Due to complementary base pairing, Thymine (T) must also equal \(84\) bases.
- This accounts for \(84\) A-T base pairs. Each A-T pair contains 2 hydrogen bonds: \(84 \times 2 = 168\) hydrogen bonds.
- The remaining bases are Cytosine (C) and Guanine (G): \(240 - (84 + 84) = 72\) bases.
- This accounts for \(72 / 2 = 36\) G-C base pairs. Each G-C pair contains 3 hydrogen bonds: \(36 \times 3 = 108\) hydrogen bonds.
- Total hydrogen bonds = \(168 + 108 = 276\).

1 mark for the correct calculation of hydrogen bonds (Option B).
- Reject A: Incorrect calculation assuming only 2 hydrogen bonds for all pairs (\(120 \times 2 = 240\)).
- Reject C and D: Incorrect base pairing calculations.

Changes in pH change the concentration of hydrogen ions (\(\text{H}^+\)) in the solution. These extra protons or lack thereof alter the charge on the R-groups (side chains) of the amino acids making up the enzyme. This disruption breaks the ionic and hydrogen bonds holding the tertiary structure together, changing the shape of the active site so that the substrate can no longer bind.

1 mark for identifying the correct structural disruption caused by pH changes (Option B).
- Reject A: Disulfide bonds are covalent and are typically broken by reducing agents, not mild-to-moderate pH changes.
- Reject C: Peptide bonds of the primary structure require extreme conditions or enzyme hydrolysis to break.
- Reject D: Trans/cis configurations of peptide bonds are not directly altered by pH changes in a way that causes denaturation.

A continuous column of water is pulled up the xylem vessels under tension due to the transpiration pull. This mechanism relies on: (1) cohesion, where water molecules stick to each other via hydrogen bonds, and (2) adhesion, where water molecules adhere to the hydrophilic components of the xylem walls, preventing the column from breaking or slipping down under gravity. Statement 3 describes root pressure, which contributes to water movement in some plants but is not the mechanism that maintains the high tension-driven continuous columns over long distances in tall trees; that is driven by transpiration pull.

1 mark for selecting statements 1 and 2 only (Option B).
- Reject A, C, and D: Statement 3 is incorrect as root pressure is not sufficient nor the primary mechanism responsible for maintaining and lifting a continuous water column to the top of tall trees.

A non-competitive inhibitor binds to an allosteric site (a site other than the active site), changing the tertiary structure of the enzyme and reducing its catalytic activity. Because it does not compete with the substrate for the active site, the affinity of the enzyme for the substrate is unaffected, so the \(K_m\) remains unchanged. However, because the inhibitor reduces the rate of product formation, the maximum velocity (\(V_{max}\)) is decreased.

Award 1 mark for selecting the option that correctly identifies non-competitive inhibition and describes how it affects the enzyme structure and kinetic parameters (reduced \(V_{max}\) and unchanged \(K_m\)).

The Casparian strip is made of suberin, a waxy, waterproof material found in the cell walls of the endodermis. It blocks the apoplast pathway (movement through cell walls and intercellular spaces). This forces water and dissolved mineral ions to pass through the selectively permeable cell surface membrane of endodermal cells into the symplast pathway, which allows the plant to control and select which mineral ions enter the vascular tissue (xylem).

Award 1 mark for identifying the apoplast as the blocked pathway and explaining that the consequence is the regulation of mineral ion entry across the selectively permeable cell surface membrane of endodermal cells.

If 30% of the bases are thymine (T), then by base-pairing rules, 30% must be adenine (A). This means A-T pairs account for 60% of the base pairs. Since the segment has 140 base pairs, the number of A-T base pairs is \(140 \times 0.60 = 84\). The remaining 40% of the base pairs must be G-C pairs, which is \(140 \times 0.40 = 56\) base pairs. Since A-T base pairs are held together by 2 hydrogen bonds and G-C base pairs are held together by 3 hydrogen bonds, the total number of hydrogen bonds is: \((84 \times 2) + (56 \times 3) = 168 + 168 = 336\) hydrogen bonds.

Award 1 mark for the correct calculation: 84 A-T pairs (2 bonds each) and 56 G-C pairs (3 bonds each) resulting in 336 hydrogen bonds.

Replacing a charged amino acid (carboxyl R-group) with a non-polar hydrophobic amino acid (hydrocarbon R-group) disrupts an ionic bond that normally stabilizes the enzyme's tertiary structure near the active site. Because the stability and flexibility of the tertiary structure are impaired, the active site is less stable and may not hold its precise shape as effectively. This results in decreased catalytic activity across all pH values and could shift or broaden the optimum pH curve.

Award 1 mark for selecting the correct explanation: disrupting the stabilizing ionic bond reduces the stability of the active site/tertiary structure, lowering activity and potentially altering the optimum pH.

Xylem vessel elements are dead cells aligned end-to-end with no cross-walls (or end walls) and have no cytoplasm, which provides a continuous, hollow tube offering extremely low resistance to the upward mass flow of water and mineral ions. Phloem sieve tube elements are living cells (allowing them to maintain their cell surface membranes to contain the pressurized sap) but they lack a nucleus, ribosomes, and have very little cytoplasm, which together with sieve plates minimizes resistance to the mass flow of organic solutes.

Award 1 mark for identifying the correct structural modifications for both xylem vessel elements (dead, no end walls, low resistance) and phloem sieve tube elements (living, sieve plates, minimal cytoplasm) to support their respective transport functions.

By analyzing the base compositions:
- Sample 1: Contains Thymine (no Uracil), indicating DNA. Because %A = %T (28%) and %G = %C (22%), it is double-stranded DNA.
- Sample 2: Contains Uracil (no Thymine), indicating RNA. Because %A = %U (21%) and %G = %C (29%), it is double-stranded RNA.
- Sample 3: Contains Uracil (no Thymine), indicating RNA. Because %A (35%) is not equal to %U (30%), and %G (20%) is not equal to %C (15%), it is single-stranded RNA.
- Sample 4: Contains Thymine (no Uracil), indicating DNA. Because %A (18%) is not equal to %T (32%), and %G (18%) is not equal to %C (32%), it is single-stranded DNA.

Award 1 mark for analyzing base composition and matching: Double-stranded DNA (Sample 1), Single-stranded RNA (Sample 3), Double-stranded RNA (Sample 2), and Single-stranded DNA (Sample 4).

By definition, the Michaelis-Menten constant (\(K_m\)) is the substrate concentration at which the rate of reaction is exactly half of the maximum velocity (\(1/2\,V_{max}\)). Since the \(K_m\) is given as \(0.05\text{ mmol dm}^{-3}\), the substrate concentration at which the rate is half of \(V_{max}\) is indeed \(0.05\text{ mmol dm}^{-3}\).

Award 1 mark for identifying that \(K_m\) is the substrate concentration that corresponds to half of the maximum velocity.

Active loading begins when hydrogen ions/protons (\(\text{H}^+\)) are actively pumped from the cytoplasm of companion cells into the cell wall using ATP. This creates a high proton concentration in the cell wall (apoplast). Protons then diffuse back down their electrochemical gradient into the companion cells through specialized co-transporter proteins. This flow of protons provides the energy needed to bring sucrose molecules with them against the sucrose concentration gradient into the companion cells.

Award 1 mark for selecting the correct explanation of active sucrose loading, detailing active proton pumping out of the cell followed by co-transport of protons and sucrose back into the companion cells.

At pH 4.5 (which is below the optimum pH of 6.8), the high concentration of hydrogen ions (\(H^+\)) affects the ionic and hydrogen bonds holding the tertiary structure of the enzyme. Specifically, it alters the charge of the R-groups of the amino acids within the active site. This prevents the substrate (starch) from binding properly to the active site, decreasing the rate of reaction. This effect is often reversible and does not immediately destroy the secondary structure.

1 mark for identifying that excess hydrogen ions alter active site R-group charges, preventing substrate binding.

An unchanged \(V_{max}\) combined with an increased \(K_m\) is the characteristic signature of a competitive inhibitor. Competitive inhibitors have a similar shape to the substrate and bind reversibly to the active site of the enzyme. At high substrate concentrations, the substrate outcompetes the inhibitor, allowing the reaction to reach the original \(V_{max}\).

1 mark for identifying that competitive inhibitors bind reversibly to the active site, resulting in unchanged \(V_{max}\) and increased \(K_m\).

The initial rate of reaction is calculated during the first time interval where the rate is at its highest (from 0 to 30 seconds). The change in volume of oxygen is \(12.0\ cm^3 - 0.0\ cm^3 = 12.0\ cm^3\). The time interval is 30 seconds. Therefore, the rate is \(12.0\ cm^3 / 30\ s = 0.40\ cm^3\ s^{-1}\).

1 mark for dividing the change in product volume in the first 30 seconds by the time interval to get \(0.40\ cm^3\ s^{-1}\).

Water potential (\(\psi\)) is calculated as \(\psi = \times_s + \times_p\). For Cell 1: \(-800 + 300 = -500\ kPa\). For Cell 2: \(-900 + 500 = -400\ kPa\). For Cell 3: \(-700 + 100 = -600\ kPa\). For Cell 4: \(-1000 + 400 = -600\ kPa\). Water moves net down a water potential gradient (from higher water potential, i.e., less negative, to lower water potential, i.e., more negative). Since Cell 2 has a water potential of \(-400\ kPa\) and Cell 1 has a water potential of \(-500\ kPa\), water will move from Cell 2 to Cell 1.

1 mark for calculating correct water potentials (\(-500\), \(-400\), \(-600\), \(-600\ kPa\)) and identifying net flow from Cell 2 to Cell 1.

Sieve tube elements are adapted for transport by losing many of their organelles (including the nucleus, ribosomes, and vacuole) to reduce resistance to the mass flow of sap (1 is correct). They also have sieve plates with pores to allow sap to flow easily between elements (3 is correct). However, sieve tube elements do not have highly folded cell surface membranes (this is a feature of some companion/transfer cells) (2 is incorrect), nor do they have plentiful mitochondria; they rely on companion cells for ATP (4 is incorrect).

1 mark for identifying statement 1 and 3 as correct adaptations of sieve tube elements, while excluding companion cell features (2 and 4).

An increase in wind speed removes the boundary layer of water vapor around the leaf, increasing the water potential gradient (1 is correct). An increase in temperature increases the kinetic energy of water molecules, increasing evaporation (2 is correct). An increase in relative humidity reduces the water potential gradient, which decreases transpiration (3 is incorrect). An increase in light intensity stimulates stomatal opening, increasing transpiration (4 is correct).

1 mark for correctly identifying that wind speed, temperature, and light intensity increase transpiration, while humidity decreases it.

If 30% of the 1200 nucleotides are adenine, then there are 360 adenine nucleotides. In double-stranded DNA, the percentage of thymine is also 30% (360 nucleotides), making 360 A-T base pairs. The remaining 40% of nucleotides are guanine and cytosine (480 nucleotides total, or 240 G-C base pairs). Each A-T pair has 2 hydrogen bonds: \(360 \times 2 = 720\) bonds. Each G-C pair has 3 hydrogen bonds: \(240 \times 3 = 720\) bonds. Total hydrogen bonds = \(720 + 720 = 1440\).

1 mark for calculating 360 A-T pairs and 240 G-C pairs, multiplying by their respective hydrogen bond numbers, and summing to 1440.

Initially, all DNA is heavy (\(^{15}N^{15}N\)). After one round of replication in light (\(^{14}N\)) medium, all DNA molecules are hybrid (\(^{14}N^{15}N\)) due to semi-conservative replication. After the second round of replication in light medium, the two hybrid molecules produce two hybrid molecules (\(^{14}N^{15}N\)) and two light molecules (\(^{14}N^{14}N\)). This gives a ratio of 2 hybrid : 2 light, which simplifies to 1 hybrid : 1 light.

1 mark for tracing semi-conservative replication over two cycles to determine a 1:1 ratio of hybrid to light DNA.

(a) Activation energy is the minimum energy required to start a chemical reaction / break existing bonds and form transition states. Enzymes increase the rate of chemical reactions by lowering this activation energy barrier. They do this by providing an alternative pathway, holding substrates close together in the active site in the correct orientation, or putting strain on bonds within the substrate molecule.

(b) (i) Type of inhibition: Non-competitive inhibition. Reason: The \(V_{\text{max}}\) is reduced (from 45 to 18 \(\mu\text{mol dm}^{-3}\text{ s}^{-1}\)) because the inhibitor decreases the number of functional active sites, but the affinity of the enzyme for its substrate remains unchanged, which is why the \(K_m\) remains unchanged at \(2.5\text{ mmol dm}^{-3}\).

(ii) A non-competitive inhibitor binds to an allosteric site (a site other than the active site) of the enzyme. This binding alters the tertiary structure of the enzyme, causing a conformational change in the shape of the active site. Consequently, the substrate can either no longer bind, or the catalytic reaction cannot occur even if the substrate binds.

(c) The addition of a chelating agent would increase the rate of reaction (restoring it toward the uninhibited rate). This is because the chelating agent binds to and removes the free lead ions (\(\text{Pb}^{2+}\)) from the solution. This shifts the equilibrium, causing lead ions to dissociate from the enzyme's allosteric sites, allowing the enzymes to return to their active conformations.

**Part (a): [3 marks max]**
- Definition: Minimum energy required to start a chemical reaction / reach the transition state [1]
- Role: Lowering activation energy [1]
- Mechanism: Substrates held close together / in correct orientation / bonds strained [1]

**Part (b)(i): [3 marks max]**
- Identification: Non-competitive inhibition [1]
- Explanation of Vmax: \(V_{\text{max}}\) is reduced because the maximum catalytic rate is decreased regardless of substrate concentration [1]
- Explanation of Km: \(K_m\) remains unchanged, indicating substrate affinity is unaffected / inhibitor does not compete with substrate for active site [1]

**Part (b)(ii): [2 marks max]**
- Inhibitor binds to allosteric site / site other than active site [1]
- Causes change in tertiary structure / conformation of active site, preventing catalytic action [1]

**Part (c): [2 marks max]**
- Rate of reaction increases / is restored [1]
- Chelating agent binds to lead ions, removing them from the system, causing them to dissociate from the enzyme's allosteric sites / active site shape is restored [1]

(a) Lactase enzyme is mixed with a solution of sodium alginate. This mixture is then added dropwise, using a syringe or pipette, into a solution of calcium chloride. The sodium ions are replaced by calcium ions, forming insoluble calcium alginate beads that trap the lactase enzymes within their matrix. The beads are then rinsed with distilled water to remove any free lactase from the surface.

(b) (i) At temperatures above the optimum (\(40^\circ\text{C}\)), the molecules have high kinetic energy, causing intense vibrations. This breaks the weak hydrogen bonds, ionic bonds, and hydrophobic interactions holding the enzyme's tertiary structure in place. This results in denaturation, changing the shape of the active site, meaning it is no longer complementary to the substrate (lactose), and enzyme-substrate complexes cannot form.

(ii) Immobilisation restricts the movement of the enzyme molecules because they are held or trapped within the alginate matrix. This physical support stabilizes the tertiary structure and prevents the enzyme from unfolding / denaturing as easily, requiring more thermal energy to disrupt the bonds holding the active site shape.

(c) 1. The enzyme can be easily recovered and reused, reducing production costs.
2. The final product (lactose-free milk) is not contaminated with the enzyme, eliminating the need for downstream purification processes.

**Part (a): [3 marks max]**
- Mix enzyme with sodium alginate solution [1]
- Add dropwise into calcium chloride solution (to form beads) [1]
- Filter and wash/rinse beads with distilled water [1]

**Part (b)(i): [3 marks max]**
- High temperature increases kinetic energy, causing molecules to vibrate excessively [1]
- Breaks hydrogen bonds / ionic bonds / hydrophobic interactions (retaining tertiary structure) [1]
- Leads to denaturation / change in active site shape so it is no longer complementary / no enzyme-substrate complexes can form [1]

**Part (b)(ii): [2 marks max]**
- Matrix/support restricts movement of enzyme molecules [1]
- Stabilises the tertiary structure / prevents unfolding / makes active site more rigid and resistant to thermal disruption [1]

**Part (c): [2 marks max]**
- Any two from:
- Easy recovery / separation of enzyme from product [1]
- Any enzyme can be reused multiple times / cost-effective [1]
- Continuous flow process is possible [1]
- No contamination of product / higher purity [1]

(a) Water is absorbed by osmosis into root hair cells. It moves across the cortex via the symplast pathway (through cytoplasm and plasmodesmata) or apoplast pathway (through cell walls). When it reaches the endodermis, the Casparian strip (made of suberin) blocks the apoplast pathway, forcing all water to cross the cell surface membrane and enter the symplast pathway before entering the xylem.

(b) (i) Still air: \(\frac{18\text{ mm}}{10\text{ min}} = 1.8\text{ mm min}^{-1}\).
Moving air: \(\frac{45\text{ mm}}{5\text{ min}} = 9.0\text{ mm min}^{-1}\).

(ii) Moving air sweeps away the water vapour that accumulates on the leaf surface (the boundary layer). This decreases the humidity of the air immediately outside the stomata, maintaining a steep water vapour potential gradient between the inside of the leaf and the external atmosphere, resulting in a faster rate of diffusion of water vapour out through the stomata.

(c) 1. Some water is used in photosynthesis to produce glucose.
2. Some water is used to maintain cell turgidity / storage within vacuole or cytoplasm.

**Part (a): [3 marks max]**
- Water moves via symplast (cytoplasm) or apoplast (cell walls) pathways [1]
- Casparian strip is made of waterproof suberin in endodermal cell walls [1]
- Blocks apoplast pathway, forcing water to pass through cell membranes / into symplast pathway [1]

**Part (b)(i): [2 marks max]**
- Still air: \(1.8\text{ mm min}^{-1}\) [1]
- Moving air: \(9.0\text{ mm min}^{-1}\) [1]
*(Deduct 1 mark if unit is missing or working is completely absent, but both answers are correct)*

**Part (b)(ii): [3 marks max]**
- Moving air removes the humid boundary layer of air on leaf surface [1]
- Decreases external water vapour potential / makes external air drier [1]
- Maintains/steepens the water vapour potential gradient between air spaces inside leaf and outside [1]
- Increases rate of diffusion of water vapour through stomata [1]

**Part (c): [2 marks max]**
- Any two from:
- Water is used in photosynthesis [1]
- Water is used to keep cells turgid / cell expansion [1]
- Water is stored in vacuoles/cells [1]

(a) Xylem vessel elements are dead, hollow tubes with no cytoplasm, organelles, or end walls, which reduces resistance to water flow and allows an uninterrupted column of water. Their walls are thickened with lignin, which provides high tensile strength to prevent the vessel from collapsing inward under the strong negative pressure (tension) of transpiration. Pits in the lignified walls allow lateral movement of water between adjacent vessels.

(b) Sieve tube elements have very little cytoplasm, no nucleus, no ribosomes, and no vacuole, leaving a large open space for flow, and have sieve plates with pores at their ends. In contrast, companion cells have a dense cytoplasm, a large nucleus, many mitochondria to generate ATP for active transport, and numerous ribosomes, performing metabolic functions for both cells. They are connected by numerous plasmodesmata.

(c) Hydrogen ions (\(\text{H}^+\)) are actively pumped out of the companion cell into the cell wall using ATP. This creates a proton gradient. Hydrogen ions then diffuse back down their electrochemical gradient into the companion cell through a co-transporter protein, bringing sucrose along with them against its concentration gradient (co-transport). Sucrose then diffuses from the companion cell into the sieve tube element through plasmodesmata.

**Part (a): [4 marks max]**
- Dead / hollow / no cytoplasm / no organelles [1]
- No end walls / continuous tube to allow uninterrupted water column [1]
- Walls reinforced with lignin to prevent collapse under tension/negative pressure [1]
- Pits allow lateral water movement [1]

**Part (b): [3 marks max]**
- Sieve tube element has little/peripheral cytoplasm, no nucleus/tonoplast, whereas companion cell has dense cytoplasm, nucleus, and organelles [1]
- Sieve tube element has sieve plates with pores, whereas companion cell does not [1]
- Companion cell has many mitochondria (for ATP production) and high metabolic activity compared to sieve tube element [1]
- Presence of plasmodesmata linking them [1]

**Part (c): [3 marks max]**
- Active pumping of protons / \(\text{H}^+\) ions out of companion cells into the cell wall using ATP [1]
- Protons diffuse back into companion cell through co-transporter proteins, carrying sucrose with them [1]
- Sucrose diffuses into the sieve tube element via plasmodesmata [1]

(a) Nucleotides are linked together by condensation reactions that form covalent phosphodiester bonds. The phosphate group attached to the 5' carbon of the pentose sugar of one nucleotide binds to the 3' hydroxyl group of the pentose sugar of the adjacent nucleotide. This forms an alternating sugar-phosphate backbone.

(b) (i) Generation 0: One heavy band at the bottom of the tube.
(ii) Generation 1: One intermediate band (hybrid DNA containing one \(^{15}\text{N}\) strand and one \(^{14}\text{N}\) strand) in the middle of the tube.
(iii) Generation 2: Two bands - one intermediate band and one light band (containing only \(^{14}\text{N}\)) near the top of the tube.

(c) The conservative model predicts that after one division, there would be two separate bands: one completely heavy band (\(^{15}\text{N}\)) and one completely light band (\(^{14}\text{N}\)). The presence of a single intermediate band in Generation 1 proved that each new DNA molecule consists of one old (heavy) strand and one new (light) strand, ruling out conservative replication.

(d) (i) DNA helicase: Unwinds the double helix and breaks the hydrogen bonds between complementary base pairs to separate the two strands.
(ii) DNA polymerase: Catalyses the formation of phosphodiester bonds to link free activated nucleotides together in a 5' to 3' direction, complementary to the template strand.

**Part (a): [3 marks max]**
- Condensation reaction [1]
- Formation of phosphodiester bonds [1]
- Bond is between phosphate group on 5' carbon and hydroxyl group on 3' carbon of adjacent deoxyribose [1]

**Part (b): [3 marks max]**
- (i) Generation 0: Heavy band only [1]
- (ii) Generation 1: Intermediate / hybrid band only [1]
- (iii) Generation 2: Light and intermediate bands [1]

**Part (c): [2 marks max]**
- Conservative model would yield two bands: one heavy and one light [1]
- The single intermediate band indicates hybrid DNA / one strand \(^{14}\text{N}\) and one strand \(^{15}\text{N}\), which is impossible in conservative replication [1]

**Part (d): [2 marks max]**
- (i) DNA helicase: Unzips/unwinds DNA by breaking hydrogen bonds [1]
- (ii) DNA polymerase: Synthesises new strand / forms phosphodiester bonds / joins nucleotides [1]

(a) Hydrogen bonds form between polar R-groups (or between the oxygen of carbonyl groups and hydrogen of amine groups). Ionic bonds form between positively and negatively charged R-groups. Hydrophobic interactions occur between non-polar R-groups, which cluster together in the interior of the protein to avoid contact with water. These interactions fold the polypeptide chain into its final precise, compact 3D shape.

(b) 1. Secondary structure: Hemoglobin polypeptides contain regions of \(\alpha\)-helices, whereas collagen consists of three polypeptide chains, each in a loose, left-handed helix (collagen helix) that do not form standard \(\alpha\)-helices.
2. Quaternary structure: Hemoglobin is a tetramer consisting of four polypeptide chains (two \(\alpha\) and two \(\beta\) chains), each associated with a prosthetic haem group. Collagen is a triple helix (tropocollagen) made of three polypeptide chains wound tightly together, which then assemble into fibrils with covalent cross-links between molecules to form fibers.

(c) The substitution of glutamic acid (hydrophilic) with valine (hydrophobic) on the surface of the \(\beta\)-globin chain creates a hydrophobic patch. At low oxygen concentrations, this hydrophobic patch interacts with a complementary hydrophobic site on another hemoglobin molecule. This causes hemoglobin molecules to polymerize into long, insoluble fibers, which distorts the shape of the red blood cells into a sickle shape and reduces oxygen solubility/carrying capacity.

**Part (a): [3 marks max]**
- Hydrogen bonds form between polar R-groups / -NH and -CO groups [1]
- Ionic bonds form between oppositely charged / acidic and basic R-groups [1]
- Hydrophobic interactions occur between non-polar R-groups which cluster in the center of the protein [1]

**Part (b): [4 marks max]**
- Secondary structure comparison:
- Hemoglobin has \(\alpha\)-helix regions [1]
- Collagen has a unique, tighter collagen helix / left-handed helix [1]
- Quaternary structure comparison:
- Hemoglobin has 4 polypeptide chains and 4 haem groups [1]
- Collagen has 3 polypeptide chains wound into a triple helix (tropocollagen) [1]

**Part (c): [3 marks max]**
- Polar glutamic acid replaced by non-polar valine on the exterior of the protein [1]
- Creates a hydrophobic area / patch on the protein surface [1]
- At low oxygen concentration, hemoglobin molecules stick together / polymerize / form insoluble fibers [1]

Part (a): Use the simple dilution formula \(C_1 V_1 = C_2 V_2\) where \(C_1 = 0.10\text{ mol dm}^{-3}\) and \(V_2 = 10.0\text{ cm}^3\).
- For \(0.08\text{ mol dm}^{-3}\): \(V_1 = (0.08 \times 10.0) / 0.10 = 8.0\text{ cm}^3\). Volume of water \(= 10.0 - 8.0 = 2.0\text{ cm}^3\).
- For \(0.06\text{ mol dm}^{-3}\): \(V_1 = 6.0\text{ cm}^3\). Volume of water \(= 4.0\text{ cm}^3\).
- For \(0.04\text{ mol dm}^{-3}\): \(V_1 = 4.0\text{ cm}^3\). Volume of water \(= 6.0\text{ cm}^3\).
- For \(0.02\text{ mol dm}^{-3}\): \(V_1 = 2.0\text{ cm}^3\). Volume of water \(= 8.0\text{ cm}^3\).
- For \(0.00\text{ mol dm}^{-3}\): \(V_1 = 0.0\text{ cm}^3\). Volume of water \(= 10.0\text{ cm}^3\).

Part (b): A detailed practical description ensuring fair testing and temperature control.

Part (c): Calculations of rates (\(1/t\)):
- \(0.00\text{ mol dm}^{-3}\): \(1 / 80 = 0.0125\text{ s}^{-1}\)
- \(0.02\text{ mol dm}^{-3}\): \(1 / 110 = 0.0091\text{ s}^{-1}\)
- \(0.04\text{ mol dm}^{-3}\): \(1 / 170 = 0.0059\text{ s}^{-1}\)
- \(0.06\text{ mol dm}^{-3}\): \(1 / 290 = 0.0034\text{ s}^{-1}\)
- \(0.08\text{ mol dm}^{-3}\): \(0.0000\text{ s}^{-1}\) (no reaction)

Part (d): Identifies key experimental limitations (subjective end-point, time intervals, temperature) and solutions.

Part (e): Explanation based on varying substrate (starch) concentration while keeping inhibitor concentration constant.

Part (a) [3 marks total]:
- 1 mark: All volumes of inhibitor **I** calculated correctly (8.0, 6.0, 4.0, 2.0, 0.0 cm3).
- 1 mark: All volumes of water **W** calculated correctly (2.0, 4.0, 6.0, 8.0, 10.0 cm3).
- 1 mark: All volumes recorded to 1 decimal place (e.g., 8.0 instead of 8) to show appropriate precision.

Part (b) [6 marks total]:
- 1 mark: Mixes amylase and inhibitor before adding starch (to allow inhibitor binding).
- 1 mark: Uses a constant volume of amylase and starch for each trial.
- 1 mark: Control of temperature using a water bath / specifies temperature (e.g., 35-40 °C).
- 1 mark: Serial sampling at regular intervals (e.g., every 10 or 20 seconds).
- 1 mark: Identifies the end-point as the first sample that does not turn blue-black / remains yellow-brown.
- 1 mark: Replicates each concentration at least twice/three times and calculates mean.

Part (c) [5 marks total]:
- 1 mark: Table with fully enclosed borders and appropriate column headings with units: Inhibitor concentration / mol dm-3, Time / s, and Rate / s-1.
- 1 mark: Raw data correctly transcribed into the table.
- 1 mark: Rates calculated correctly (1/t):
- 0.00 mol dm-3 = 0.0125 (or 0.013)
- 0.02 mol dm-3 = 0.0091
- 0.04 mol dm-3 = 0.0059
- 0.06 mol dm-3 = 0.0034
- 1 mark: The rate for 0.08 mol dm-3 correctly designated as 0 or 'no reaction' / 'no hydrolysis' (since starch was not hydrolyzed after 600s).
- 1 mark: All rates calculated to a consistent number of significant figures (2 or 3 s.f.).

Part (d) [4 marks total]:
- 1 mark: Source of error: Subjective judgment of color change / end-point.
- 1 mark: Improvement: Use a colorimeter OR compare to a color standard chart.
- 1 mark: Source of error: Large time interval between testing (e.g., 20 s) leads to inaccuracy.
- 1 mark: Improvement: Test at smaller time intervals (e.g., every 5 s or 10 s).
- (Accept: Temperature fluctuations -> use a thermostatically controlled water bath instead of a beaker of water.)

Part (e) [2 marks total]:
- 1 mark: Measure the rate of reaction at increasing concentrations of starch (substrate) while keeping inhibitor concentration constant.
- 1 mark: If competitive, the maximum rate (Vmax) will be reached at high starch concentrations / if non-competitive, the maximum rate (Vmax) will remain lower.

Part (a): Plan diagrams must show the boundaries of tissue zones, not individual cells. The diagram must represent the wedge sector accurately, showing outer cuticle/epidermis, cortex, vascular bundles (containing xylem on the inside and phloem on the outside), and central pith.

Part (b):
- 1 stage micrometer division \(= 0.01\text{ mm} = 10\mu\text{m}\).
- 10 stage micrometer divisions \(= 10 \times 10\mu\text{m} = 100\mu\text{m}\).
- 40 eyepiece divisions \(= 100\mu\text{m}\).
- 1 eyepiece division \(= 100\mu\text{m} / 40 = 2.5\mu\text{m}\).

Part (c):
- Mean of measurements \(= (12 + 14 + 11 + 13 + 15) / 5 = 65 / 5 = 13\text{ divisions}\).
- Mean actual diameter \(= 13 \times 2.5\mu\text{m} = 32.5\mu\text{m}\).

Part (d): A table listing contrasting observable structures like wall thickness, lumen size, and associated cells (companion cells).

Part (e): Adaptations include lignified walls to prevent collapse under tension, empty lumen to minimize flow resistance, and perforation plates/open ends for continuous flow.

Part (a) [6 marks total]:
- 1 mark: Clean, continuous lines drawn with a sharp pencil (no shading, no feathering).
- 1 mark: Large drawing (takes up at least half of the available space).
- 1 mark: Correct plan diagram showing a wedge-shaped sector of the stem (epidermis, cortex, vascular bundles, pith) without drawing individual cells.
- 1 mark: Correct proportion of tissue layers (cortex width relative to vascular bundle size and pith).
- 1 mark: Xylem correctly labeled (inner region of the vascular bundle).
- 1 mark: Phloem correctly labeled (outer region of the vascular bundle).

Part (b) [3 marks total]:
- 1 mark: Correctly converts stage micrometer divisions to micrometers (10 divisions = 0.1 mm = 100 μm).
- 1 mark: Shows division of total stage micrometer length by the number of eyepiece divisions (100 / 40).
- 1 mark: Correct final answer with units: 2.5 μm.

Part (c) [3 marks total]:
- 1 mark: Calculates correct mean of eyepiece divisions (13 divisions).
- 1 mark: Multiplies mean eyepiece divisions by the calibration value (13 x 2.5).
- 1 mark: Correct final answer with units: 32.5 μm.

Part (d) [5 marks total]:
- 1 mark: Comparison presented as a clear table with matching/contrasting points on the same row.
- 1 mark: Wall thickness: Xylem has thick/lignified walls, Phloem sieve tubes have thin/cellulose walls.
- 1 mark: Lumen size: Xylem has large/wide lumen, Phloem sieve tubes have small/narrow lumen.
- 1 mark: Companion cells: Xylem is not associated with companion cells, Phloem sieve tubes are closely associated with companion cells.
- 1 mark: Sieve plates: Xylem has no sieve plates (open end walls), Phloem sieve tubes have sieve plates.

Part (e) [3 marks total]:
- 1 mark: Lignin in cell walls provides mechanical strength to prevent the vessel from collapsing under negative pressure/tension.
- 1 mark: Complete breakdown of end walls forms a continuous, uninterrupted column/tube for water transport.
- 1 mark: Loss of cytoplasm/organelles (empty lumen) minimizes resistance to water flow.