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In a population in Hardy-Weinberg equilibrium, the frequencies of alleles and genotypes are represented by the equation:

\(p^2 + 2pq + q^2 = 1\) and \(p + q = 1\)

Let \(q\) be the frequency of the allele \(C^W\). The frequency of the homozygous genotype \(C^W C^W\) (white flowers) is represented by \(q^2\).

\(q^2 = \frac{45}{500} = 0.09\)

Therefore, the frequency of allele \(C^W\) is:

\(q = \sqrt{0.09} = 0.3\)

Since \(p + q = 1\), the frequency of allele \(C^R\) is:

\(p = 1 - 0.3 = 0.7\)

The expected frequency of heterozygous pink-flowered plants (\(C^R C^W\)) is represented by \(2pq\):

\(2pq = 2 \times 0.7 \times 0.3 = 0.42\)

Award 1 mark for selecting the correct heterozygous frequency (0.42).
- Correct frequency of the homozygous recessive genotype: 45 / 500 = 0.09
- Correct frequency of the q allele: 0.30
- Correct frequency of the p allele: 0.70
- Correct frequency of the 2pq heterozygote: 2 * 0.7 * 0.3 = 0.42

A competitive inhibitor binds reversibly to the active site of an enzyme, competing directly with the substrate. Because the inhibitor competes with the substrate, a higher substrate concentration is required to reach half the maximum velocity of the reaction, which means the Michaelis-Menten constant (\(K_m\)) increases. However, if the substrate concentration is increased sufficiently, it can completely overcome the effect of the inhibitor, allowing the reaction to reach its original maximum velocity (\(V_{max}\)). Therefore, \(V_{max}\) remains unchanged.

Award 1 mark for the correct identification that Km increases and Vmax remains unchanged.

During active loading of sucrose, the proton pump (\(H^+\)-ATPase) in the cell surface membrane of the companion cell uses energy from the hydrolysis of ATP to actively pump hydrogen ions (protons) out of the companion cell's cytoplasm into the surrounding cell wall (apoplast). This creates a high electrochemical gradient of protons outside the cell. Protons then diffuse back into the companion cell down this gradient through a co-transporter protein, carrying sucrose with them against its concentration gradient.

Award 1 mark for identifying the correct direct role of the proton pump as actively pumping hydrogen ions out of the companion cell to create an electrochemical gradient.

- S: No change in cell volume indicates that the external solution is isotonic to the initial cell state, meaning its water potential is equal to the initial cell water potential (\(-600\text{ kPa}\)).
- P: Incipient plasmolysis occurs when turgor pressure is zero, so the water potential of the cell equals its solute potential (\(\psi_s\)). Since the cell was initially turgid at \(-600\text{ kPa}\) with a positive pressure potential (\(\psi_p\)), its solute potential must be more negative than \(-600\text{ kPa}\) (e.g., \(-800\text{ kPa}\)). At equilibrium at incipient plasmolysis, the external solution must match this solute potential, so P is \(-800\text{ kPa}\).
- Q: Fully turgid cells have absorbed water, meaning the external solution must have a higher (less negative) water potential than the cell (e.g., \(-200\text{ kPa}\)).
- R: Highly plasmolysed cells have lost a significant amount of water, meaning the external solution must have a very low (more negative) water potential (e.g., \(-1200\text{ kPa}\)).

Award 1 mark for the correct combination of water potentials corresponding to the observed physiological states of the cells.

Consuming a large volume of water increases blood water potential. This is detected by osmoreceptors in the hypothalamus, which signal the posterior pituitary gland to decrease the release of antidiuretic hormone (ADH). A lower concentration of ADH leads to fewer aquaporins being inserted into the luminal membranes of the collecting duct cells. This decreases the permeability of the collecting ducts to water, meaning less water is reabsorbed by osmosis. Consequently, a large volume of dilute urine is produced, restoring blood water potential to normal.

Award 1 mark for the correct row showing decreased permeability, decreased number of aquaporins, and production of a large volume of dilute urine.

Myelin acts as an electrical insulator, preventing ion movements across the membrane where it is present. Consequently, depolarisation can only occur at the nodes of Ranvier, where myelin is absent and where sodium voltage-gated channels are highly concentrated. This causes local currents to jump from one node of Ranvier to the next (saltatory conduction), significantly speeding up the transmission of the action potential along the axon.

Award 1 mark for identifying that depolarisation occurs only at the nodes of Ranvier because sodium voltage-gated channels are concentrated there.

Increased carbon dioxide concentration causes blood pH to drop (due to carbonic acid formation). This decrease in pH reduces the affinity of haemoglobin for oxygen, which shifts the oxygen-haemoglobin dissociation curve to the right (the Bohr effect). A lower affinity at any given partial pressure of oxygen means that haemoglobin releases (unloads) oxygen more readily to the actively respiring tissues that need it.

Award 1 mark for selecting the option stating that the curve shifts to the right, decreasing affinity, and making it easier to unload oxygen.

One molecule of glucose yields two molecules of pyruvate during glycolysis, which are converted into two molecules of acetyl-CoA in the link reaction. Therefore, the Krebs cycle turns twice per molecule of glucose.

Each single turn of the Krebs cycle produces:
- 1 ATP (via substrate-level phosphorylation)
- 3 reduced NAD
- 1 reduced FAD

Multiplying these yields by two for one molecule of glucose gives a net yield of:
- 2 ATP
- 6 reduced NAD
- 2 reduced FAD

Award 1 mark for the correct combination of products per glucose molecule in the Krebs cycle (2 ATP, 6 reduced NAD, and 2 reduced FAD).

Active transport requires ATP produced by aerobic respiration, so its rate drops to near zero in anaerobic conditions (Substance X). Simple diffusion occurs down a concentration gradient and does not involve membrane proteins, so its rate increases linearly with external concentration (Substance Y). Facilitated diffusion is passive (unaffected by lack of oxygen) but relies on transport proteins, which can become saturated at high external concentrations, leading to a plateau (Substance Z).

1 mark for the correct option B.

Inhibitor P increases the Km value but does not alter the Vmax, which is characteristic of a competitive inhibitor that competes with the substrate for binding at the active site. Inhibitor Q decreases the Vmax value but does not alter the Km, which is characteristic of a non-competitive inhibitor that binds to an allosteric site, decreasing the catalytic activity of the enzyme without affecting substrate binding.

1 mark for the correct option C.

Protons (H+) are actively pumped out of the companion cell cytoplasm into the cell wall apoplast using ATP. This creates an electrochemical gradient. Protons diffuse down this gradient back into the companion cell through co-transporter proteins, bringing sucrose with them against its concentration gradient. Once inside the companion cell, sucrose diffuses down its concentration gradient into the sieve tube element via plasmodesmata (the symplast pathway).

1 mark for the correct option D.

Disruptive selection occurs when environmental conditions favour individuals at both phenotypic extremes over those with intermediate phenotypes. This leads to a bimodal distribution and increases the overall phenotypic variance within the population.

1 mark for the correct option B.

During glycolysis, 4 ATP are produced and 2 are used, resulting in a net yield of 2 ATP by substrate-level phosphorylation. The link reaction produces no ATP. The Krebs cycle produces 1 ATP (via GTP intermediate) per turn, which equals 2 ATP per glucose molecule. Thus, the total net yield of ATP from substrate-level phosphorylation is 2 + 0 + 2 = 4 molecules.

1 mark for the correct option C.

During repolarisation, the membrane potential must return to its resting negative state. This is achieved by closing (and inactivating) the voltage-gated sodium channels to stop the influx of Na+ ions, and opening the voltage-gated potassium channels to allow K+ ions to diffuse out of the axon down their electrochemical gradient.

1 mark for the correct option A.

Under Hardy-Weinberg equilibrium, the frequency of homozygous recessive genotype (tt) is q^2 = 0.09. Therefore, q = 0.3. Since p + q = 1, the frequency of the dominant allele (T) is p = 1 - 0.3 = 0.7. The frequency of the heterozygous genotype (Tt) is 2pq = 2 * 0.7 * 0.3 = 0.42.

1 mark for the correct option D.

ADH is a peptide hormone that cannot pass through the phospholipid bilayer. It binds to specific receptors on the basolateral membrane of collecting duct cells. This activates a G-protein, which in turn activates adenylyl cyclase to catalyze the formation of cAMP (second messenger). Rising cAMP levels trigger a signaling cascade that causes vesicles containing aquaporin water channels to undergo exocytosis and fuse with the luminal (apical) membrane, increasing its permeability to water.

1 mark for the correct option B.

One mole of sucrose is hydrolysed to produce one mole of glucose and one mole of fructose. During glycolysis, each mole of hexose produces a net yield of 2 moles of ATP and 2 moles of pyruvate. Therefore, two moles of hexose yield a total of 4 moles of ATP in both yeast and muscle cells. Under anaerobic conditions, yeast converts the 4 moles of pyruvate into 4 moles of ethanol and 4 moles of carbon dioxide, whereas mammalian muscle converts the 4 moles of pyruvate into 4 moles of lactate.

Award 1 mark for the correct option (A).

The increase in carbon dioxide concentration decreases the affinity of haemoglobin for oxygen (the Bohr effect). This shifts the oxygen-haemoglobin dissociation curve to the right. At a given tissue \( \text{pO}_2 \) of \( 4.0\text{ kPa} \), the saturation drops from \( 60\% \) to \( 35\% \). The difference (\( 60\% - 35\% = 25\% \)) represents additional oxygen that is unloaded/released to help meet the demands of actively respiring tissues.

Award 1 mark for the correct option (A).

A non-competitive inhibitor binds to an allosteric site on the enzyme. This reduces the number of fully functional enzyme molecules available, thereby decreasing the maximum rate of reaction (\( V_{\max} \)). However, because it does not compete with the substrate for the active site, the affinity of the remaining active enzymes for the substrate remains unchanged. Therefore, the Michaelis-Menten constant (\( K_{\text{m}} \)) remains constant.

Award 1 mark for the correct option (B).

Active transport of hydrogen ions (protons) out of the companion cells and into the cell wall space relies on ATP-driven proton pumps. Inhibiting ATP synthesis stops these pumps, resulting in a decreased concentration of hydrogen ions in the cell wall space (less acidic/higher pH). Without the proton concentration gradient, the sucrose-proton cotransporter cannot function effectively, leading to reduced sucrose accumulation in both the companion cells and the adjacent sieve tube elements.

Award 1 mark for the correct option (A).

At lower temperatures, phospholipid membranes pack together more tightly, making them less fluid (more rigid). To counteract this, cells increase the proportion of unsaturated fatty acid tails, which have double bonds ('kinks') that prevent close packing of lipids. At the same time, increasing cholesterol content helps disrupt the close packing of phospholipids at low temperatures, acting as a fluidity buffer.

Award 1 mark for the correct option (C).

Action potential conduction in myelinated axons is faster due to saltatory conduction (1), where depolarization leaps between nodes of Ranvier. This is possible because the myelin sheath acts as an electrical insulator preventing ion loss (2), and voltage-gated channels are concentrated specifically at these nodes (4). The internal axoplasmic resistance (3) is determined by axon diameter, which is specified as being the same for both axons, so this is not a contributing factor.

Award 1 mark for the correct option (B).

Mammal X is adapted to a desert environment where water conservation is critical, as shown by its extremely long loop of Henle, thick renal medulla, and highly concentrated urine (\( 9000 \text{ mOsm dm}^{-3} \)). The long loops of Henle function as countercurrent multipliers, actively transporting more sodium and chloride ions into the medulla. This builds up a steeper osmotic gradient in the medulla tissue, driving greater water reabsorption from the collecting ducts by osmosis.

Award 1 mark for the correct option (A).

Under Hardy-Weinberg equilibrium, the frequency of homozygous recessive individuals (\( ff \)) is given by \( q^2 \). Given \( q^2 = 0.16 \), we find the frequency of the recessive allele \( q = \sqrt{0.16} = 0.4 \). Since \( p + q = 1 \), the frequency of the dominant allele \( p = 1 - 0.4 = 0.6 \). The frequency of heterozygous individuals (\( Ff \)) is represented by \( 2pq \). Calculating this gives \( 2 \times 0.6 \times 0.4 = 0.48 \).

Award 1 mark for the correct option (C).

An increase in \(K_m\) indicates that a higher substrate concentration is needed to reach half of \(V_{max}\), meaning the enzyme's affinity for its substrate has decreased. Since \(V_{max}\) is unchanged, inhibitor X is a competitive inhibitor. When \(V_{max}\) decreases but \(K_m\) remains unchanged, the affinity for the substrate is unaffected, and inhibitor Y is non-competitive. Therefore, option A is correct.

A: 1 mark for identifying inhibitor X as competitive with decreased affinity, and inhibitor Y as non-competitive with no change in affinity.

Water potential (\(\psi\)) is calculated using the formula \(\psi = \psi_s + \psi_p\). For Cell P: \(\psi = -800 + 300 = -500\text{ kPa}\). For Cell Q: \(\psi = -900 + 200 = -700\text{ kPa}\). For Cell R: \(\psi = -600 + 200 = -400\text{ kPa}\). Water moves from a region of higher (less negative) water potential to a region of lower (more negative) water potential. Thus, water moves from R (\(-400\text{ kPa}\)) to P (\(-500\text{ kPa}\)), from P to Q (\(-700\text{ kPa}\)), and from R to Q. This matches option A.

A: 1 mark for correctly calculating the water potentials of cells P, Q, and R, and identifying the direction of net water movement from high to low water potential.

The 'fluid' part of the fluid mosaic model refers to the ability of phospholipid molecules to move laterally within their own monolayer, and the lateral movement of some proteins. The 'mosaic' part refers to the scattered, asymmetrical pattern of different protein molecules. Flip-flop movement is rare (B is incorrect). Cholesterol increases fluidity or maintains it at low temperatures by preventing close packing (C is incorrect). Peripheral proteins do not span the entire membrane (D is incorrect).

A: 1 mark for identifying the correct description of lateral movement of phospholipids and scattered distribution of proteins in the fluid mosaic model.

During vigorous exercise, high CO2 and lower pH cause a shift of the oxygen dissociation curve to the right (the Bohr effect), facilitating oxygen unloading to active tissues (1 is shifted to the right). Fetal haemoglobin has a higher affinity for oxygen than adult haemoglobin to allow oxygen transfer from maternal blood, shifting its curve to the left (2 is shifted to the left). Therefore, option A is correct.

A: 1 mark for correctly identifying that exercise shifts the curve to the right (Bohr effect) and fetal haemoglobin is shifted to the left.

One molecule of glucose produces two molecules of pyruvate via glycolysis. In the link reaction, 2 pyruvates produce 2 carbon dioxide and 2 reduced NAD molecules. In the Krebs cycle, the 2 acetyl CoA molecules produce 4 carbon dioxide, 6 reduced NAD, and 2 reduced FAD molecules. Combining these, the total is 6 carbon dioxide, 8 reduced NAD, and 2 reduced FAD molecules. Option A is correct.

A: 1 mark for calculating the combined products of the link reaction and the Krebs cycle per glucose molecule.

The frequency of the homozygous recessive genotype (rr) is \(q^2 = 80 / 500 = 0.16\). Therefore, \(q = \sqrt{0.16} = 0.4\). Since \(p + q = 1\), the frequency of the dominant allele R is \(p = 1 - 0.4 = 0.60\). The frequency of heterozygotes is \(2pq = 2 \times 0.60 \times 0.40 = 0.48\), which is 48%. Option A is correct.

A: 1 mark for using Hardy-Weinberg equations to calculate the dominant allele frequency as 0.60 and heterozygous percentage as 48%.

The sequence of synaptic transmission is: arrival of the action potential causes depolarization of the presynaptic membrane, opening voltage-gated calcium ion channels (4). Calcium ions diffuse in (1), stimulating synaptic vesicles to fuse with the presynaptic membrane and release neurotransmitter (5). Acetylcholine diffuses across the cleft and binds to ligand-gated sodium channels on the postsynaptic membrane (2), opening them so sodium ions diffuse in (3) to depolarize the membrane. Thus, the order is 4 to 1 to 5 to 2 to 3. Option A is correct.

A: 1 mark for the correct sequence of events during synaptic transmission.

Normally, placing a plant in the dark stops the light-dependent stage, preventing the production of ATP and reduced NADP. This stops the conversion of GP to TP and the regeneration of RuBP, causing GP to accumulate and RuBP to decrease. However, if ATP and reduced NADP are continuously supplied artificially alongside carbon dioxide, the reactions of the Calvin cycle can proceed fully. Therefore, both RuBP and GP will continue to be converted and regenerated, meaning their concentrations will initially remain stable. Option A is correct.

A: 1 mark for recognizing that supplying the products of the light-dependent stage (ATP and reduced NADP) allows the Calvin cycle to continue, keeping RuBP and GP levels stable.

A non-competitive inhibitor decreases the maximum rate of reaction (\(V_{\text{max}}\)) because it binds to an allosteric site (a site other than the active site) and alters the active site's shape, preventing some enzyme molecules from working regardless of substrate concentration. The remaining uninhibited enzyme molecules function normally, which is why the Michaelis-Menten constant (\(K_{\text{m}}\)), representing substrate affinity, remains unchanged.

1 mark for identifying the correct description of a non-competitive inhibitor which reduces \(V_{\text{max}}\) without altering \(K_{\text{m}}\).

ATP hydrolysis is directly used by proton pumps (\(\text{H}^+\)-ATPase) to pump protons out of the companion cell against their concentration gradient (Step 1). The return of protons down their concentration gradient (Step 2) coupled with sucrose entry (Step 3) is a form of secondary active transport (co-transport) which does not directly hydrolyse ATP. Sucrose movement into the sieve tube element through plasmodesmata (Step 4) occurs by passive diffusion. Thus, only Step 1 directly requires ATP hydrolysis.

1 mark for correctly identifying that only the active pumping of hydrogen ions out of the companion cell directly uses ATP.

At high temperatures, cholesterol decreases membrane fluidity by restricting the lateral movement of phospholipid molecules and their fatty acid chains, preventing the membrane from becoming too fluid or disintegrating. At low temperatures, it prevents the close packing of fatty acid chains, keeping the membrane fluid.

1 mark for identifying that cholesterol decreases fluidity at high temperatures by restricting the movement of phospholipid fatty acid chains.

Ultrafiltration is driven by net filtration pressure, which is glomerular hydrostatic pressure minus the opposing forces (glomerular osmotic pressure and Bowman's capsule hydrostatic pressure). An increase in the hydrostatic pressure of the fluid inside the Bowman's capsule directly opposes the movement of filtrate out of the glomerulus, reducing the rate of ultrafiltration.

1 mark for selecting the correct physiological change that opposes and reduces net filtration pressure.

The sequence of events is as follows: arrival of action potential opens voltage-gated calcium channels (5); calcium ions diffuse into the presynaptic neurone (1); calcium entry triggers synaptic vesicles to fuse with the presynaptic membrane (4), releasing acetylcholine; acetylcholine binds to postsynaptic receptors (2); ligand-gated sodium channels open and sodium ions enter the postsynaptic neurone (3). This gives the sequence \(5 \rightarrow 1 \rightarrow 4 \rightarrow 2 \rightarrow 3\).

1 mark for identifying the correct sequential order of synaptic transmission events.

In respiring tissues, carbon dioxide diffuses into the red blood cells, where carbonic anhydrase converts it to carbonic acid. Carbonic acid dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)). Hydrogencarbonate ions diffuse out of the cell into the plasma down their concentration gradient, and chloride ions (\(\text{Cl}^-\)) diffuse in to maintain electrical neutrality (the chloride shift).

1 mark for identifying that carbonic acid dissociates into hydrogen and hydrogencarbonate ions, with hydrogencarbonate ions exiting in exchange for chloride ions.

In the mitochondrial matrix, the link reaction converts 2 pyruvates to 2 acetyl-CoA, producing 2 NADH. The Krebs cycle then processes 2 acetyl-CoA molecules, producing 6 NADH and 2 FADH2. Therefore, the total produced in the matrix is 8 NADH (2 from link reaction + 6 from Krebs cycle) and 2 FADH2 (from Krebs cycle). The 2 NADH from glycolysis are produced in the cytoplasm, not the mitochondrial matrix.

1 mark for identifying the correct numbers of NADH and FADH2 produced specifically in the mitochondrial matrix during aerobic respiration of one glucose molecule.

Let \(q^2\) be the frequency of homozygous recessive white-flowered plants (rr). Thus, \(q^2 = 0.16\), which gives \(q = 0.4\). Since \(p + q = 1\), \(p = 1 - 0.4 = 0.6\). The frequency of heterozygous individuals (Rr) is given by \(2pq = 2 \times 0.6 \times 0.4 = 0.48\), which is 48%.

1 mark for calculating the correct percentage of heterozygous individuals using the Hardy-Weinberg equations.

**(a)**
- Magnification is defined as the number of times larger an image is compared to the actual size of the specimen.
- Resolution is the ability of a microscope to distinguish between two points that are close together, determining the level of detail visible.

**(b)**
- Formula: \(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}}\)
- Convert measured length from cm to \(\mu\text{m}\):
\(4.5\text{ cm} = 45\text{ mm} = 45\,000\ \mu\text{m}\)
- Calculation:
\(\text{Actual size} = \frac{45\,000\ \mu\text{m}}{15\,000} = 3\ \mu\text{m}\)

**(c)**
- **(i)** Nucleolus: Synthesises ribosomal RNA (rRNA) and assembles ribosome subunits.
- **(ii)** Rough endoplasmic reticulum: Studded with ribosomes; site of protein synthesis and transport of proteins within the cell.
- **(iii)** Smooth endoplasmic reticulum: Lacks ribosomes; synthesises and metabolises lipids, steroids, and phospholipids.

**(d)**
- Prokaryotic cell walls are composed of peptidoglycan (murein).
- Plant cell walls are composed of cellulose (containing \(\beta\)-glucose microfibrils).

**(a) [Max 2 marks]**
- 1 mark: Magnification is the ratio of image size to actual size / number of times larger.
- 1 mark: Resolution is the ability to distinguish between two adjacent points / detail shown.

**(b) [Max 3 marks]**
- 1 mark: Correct conversion of \(4.5\text{ cm}\) to \(45\,000\ \mu\text{m}\) (or \(45\text{ mm}\)).
- 1 mark: Correct calculation formula/working: \(45\,000 / 15\,000\).
- 1 mark: Correct final answer of \(3\ \mu\text{m}\) (units required for mark, accept 3 without units if the unit is pre-printed, but accept any correct path to 3).

**(c) [Max 3 marks]**
- **(i)** 1 mark: Synthesises ribosomal RNA / rRNA / ribosomal subunits.
- **(ii)** 1 mark: Protein synthesis / translation / protein folding / transport of proteins.
- **(iii)** 1 mark: Lipid / cholesterol / steroid synthesis / transport of lipids.

**(d) [Max 2 marks]**
- 1 mark: Prokaryotic cell wall is made of peptidoglycan / murein.
- 1 mark: Plant cell wall is made of cellulose.

**(a)**
- Primary structure is the specific linear sequence of amino acids in a polypeptide chain, linked by covalent peptide bonds.
- Tertiary structure is the overall three-dimensional folding of a single polypeptide chain, held together by interactions (hydrogen, ionic, disulfide, and hydrophobic interactions) between R-groups.

**(b)**
- Shape: Haemoglobin is **globular**; Collagen is **fibrous**.
- Solubility: Haemoglobin is **soluble**; Collagen is **insoluble**.
- Number of chains: Haemoglobin has **4** (two alpha, two beta); Collagen has **3** (forming a triple helix).
- Prosthetic group: Haemoglobin **yes** (haem group containing iron); Collagen **no**.

**(c)**
- Three polypeptide chains wind around each other to form a tight, stable triple helix.
- Individual collagen molecules lie parallel to one another and form covalent cross-links between lysine residues on adjacent molecules.
- These molecules are staggered along their length, forming fibrils which aggregate into fibres, preventing any weak points and resisting pulling forces.

**(d)**
- The reagent is Biuret reagent.
- A positive result is indicated by a colour change from blue to purple (or lilac/violet).

**(a) [Max 2 marks]**
- 1 mark: Primary structure is the sequence/order of amino acids in a polypeptide chain.
- 1 mark: Tertiary structure is the 3D shape held by interactions between R-groups (accept examples of bonds: disulfide, ionic, hydrogen, hydrophobic interactions).

**(b) [Max 4 marks - 1 mark per correct row]**
- Row 1: Globular AND Fibrous.
- Row 2: Soluble AND Insoluble.
- Row 3: 4 (or two \(\alpha\) and two \(\beta\)) AND 3.
- Row 4: Yes (or haem / contains \(\text{Fe}^{2+}\)) AND No.

**(c) [Max 3 marks]**
- 1 mark: Three polypeptide chains wind together to form a triple helix (held by hydrogen bonds).
- 1 mark: Covalent cross-links form between adjacent collagen molecules/helices.
- 1 mark: Molecules are staggered / form fibrils and fibres to distribute load and prevent weak points.

**(d) [Max 1 mark]**
- 1 mark: Biuret (test/reagent) AND turns (from blue to) purple / violet / lilac (both reagent and positive colour change required).

**(a)**
- Malonate has a molecular shape similar/complementary to the active site of succinate dehydrogenase (similar to succinate).
- It competes with succinate for binding to the enzyme's active site.
- It binds temporarily to the active site, blocking succinate from entering, which reduces the rate of enzyme-substrate complex (ESC) formation.

**(b)**
- **(i)** In the presence of a competitive inhibitor, increasing succinate concentration increases the rate of reaction. This is because the substrate molecules outcompete the inhibitor for the active sites. At very high succinate concentrations, the maximum rate of reaction (\(V_{max}\)) can still be achieved.
- **(ii)** In the presence of a non-competitive inhibitor, increasing succinate concentration has little or no effect on restoring the rate. This is because the inhibitor binds to an allosteric site, altering the shape of the active site so that succinate can no longer bind. The maximum rate of reaction (\(V_{max}\)) is reduced and cannot be reached.

**(c)**
- The Michaelis-Menten constant (\(K_m\)) is defined as the substrate concentration at which the reaction rate is half of the maximum velocity (\(\frac{1}{2} V_{max}\)).
- It is an inverse measure of enzyme-substrate affinity.
- A low \(K_m\) indicates high affinity of the enzyme for its substrate (meaning the enzyme reaches maximum catalytic efficiency at low substrate concentrations), while a high \(K_m\) indicates low affinity.

**(a) [Max 3 marks]**
- 1 mark: Inhibitor has a similar/complementary shape to the substrate / fits into the active site.
- 1 mark: Binds to the active site (blocking it).
- 1 mark: Prevents substrate from binding / fewer successful collisions / fewer enzyme-substrate complexes (ESCs) formed per unit time.

**(b) [Max 4 marks]**
- **(i)** [2 marks]
- 1 mark: Rate of reaction increases (as substrate concentration increases).
- 1 mark: At high substrate concentrations, \(V_{max}\) is achieved / inhibition is overcome because substrate outcompetes the inhibitor.
- **(ii)** [2 marks]
- 1 mark: Rate of reaction does not reach \(V_{max}\) / rate remains low.
- 1 mark: Non-competitive inhibitor binds to an allosteric site / active site remains permanently altered, so substrate cannot compete.

**(c) [Max 3 marks]**
- 1 mark: \(K_m\) is the substrate concentration at half of \(V_{max}\) (or \(\frac{1}{2} V_{max}\)).
- 1 mark: Indicates the affinity of the enzyme for the substrate.
- 1 mark: Low \(K_m\) indicates high affinity AND high \(K_m\) indicates low affinity.

**(a)**
- A phospholipid molecule consists of a polar, hydrophilic phosphate head and two non-polar, hydrophobic fatty acid tails linked by a glycerol backbone.
- In an aqueous environment, phospholipids spontaneously align into a double layer.
- The hydrophilic heads face outwards towards the aqueous cytoplasm and external fluid, while the hydrophobic tails face inwards, away from water. This bilayer is held together and stabilised by hydrophobic interactions between the fatty acid tails.

**(b)**
- **(i)** Carrier proteins have a specific binding site for a particular polar molecule or ion. Upon binding of the molecule, the carrier protein undergoes a conformational (shape) change, which moves the molecule across the membrane and releases it on the other side down its concentration gradient.
- **(ii)** Active transport moves substances against (up) their concentration gradient (from a low concentration to a high concentration). This is thermodynamically unfavourable and requires energy from ATP hydrolysis to alter the shape of the protein pump to transport the solute.

**(c)**
- Increasing temperature increases the kinetic energy of the phospholipids and proteins in the cell membrane.
- The phospholipids vibrate and move more rapidly, increasing membrane fluidity and creating gaps in the bilayer.
- At high temperatures (typically above \(40^\circ\text{C}\)), the proteins (including those in the tonoplast and plasma membrane) denature as hydrogen and ionic bonds break. This destroys the selective permeability of the membranes, allowing betalain pigment to leak out.

**(a) [Max 3 marks]**
- 1 mark: Phospholipid has hydrophilic phosphate head AND hydrophobic fatty acid tails.
- 1 mark: Hydrophilic heads face towards the watery/aqueous environments (inside and outside the cell) AND hydrophobic tails face inwards/away from water.
- 1 mark: Stabilised by hydrophobic interactions between fatty acid tails.

**(b) [Max 4 marks]**
- **(i)** [2 marks]
- 1 mark: Carrier protein has a specific binding site complementary to the polar molecule.
- 1 mark: Binding of molecule triggers a conformational / shape change in the protein to release it on the other side.
- **(ii)** [2 marks]
- 1 mark: Transport is against / up a concentration gradient.
- 1 mark: Hydrolysis of ATP provides the energy required to change the shape of the pump / move the molecule.

**(c) [Max 3 marks]**
- 1 mark: High temperature increases kinetic energy of phospholipids, causing them to move more / increasing membrane fluidity.
- 1 mark: Membrane proteins (carrier/channel/structural) denature due to loss of hydrogen/ionic bonds.
- 1 mark: Loss of membrane integrity / selective permeability allows pigment to leak out by diffusion.

**(a)**
- Hydrogen ions (protons, \(\text{H}^+\)) are actively pumped out of the cytoplasm of the companion cell across its plasma membrane into the cell wall space.
- This active transport is mediated by proton pumps and requires energy from the hydrolysis of ATP.
- This establishes a high concentration gradient of \(\text{H}^+\)-ions in the cell wall compared to the inside of the companion cell.
- \(\text{H}^+\)-ions diffuse back into the companion cell down their electrochemical gradient through co-transporter proteins.
- Sucrose is co-transported (carried simultaneously) with the \(\text{H}^+\)-ions against its concentration gradient into the companion cell.
- Sucrose then diffuses from the companion cell into the sieve tube element through interconnecting plasmodesmata.

**(b)**
- The accumulation of sucrose inside the sieve tube element lowers its solute potential (and thus its water potential, \(\psi_w\)).
- Water enters the sieve tube element from the adjacent xylem vessels or surrounding tissues by osmosis, down a water potential gradient.
- The entry of water increases the volume of liquid inside the rigid sieve tube, generating a high hydrostatic pressure at the source. This drives mass flow towards the sink.

**(c)**
- **Xylem vessel elements:** have lignified cell walls; have no cell contents/cytoplasm (are dead cells); have no end walls (form continuous tubes); have pits.
- **Phloem sieve tube elements:** have cellulose cell walls (non-lignified); have a thin peripheral layer of cytoplasm with no nucleus/ribosomes (are living cells); have end walls modified into sieve plates with sieve pores.

**(a) [Max 5 marks]**
- 1 mark: Proton pumps actively transport \(\text{H}^+\)/protons out of the companion cell into the cell wall.
- 1 mark: This process requires ATP.
- 1 mark: A proton/\(\text{H}^+\) concentration gradient is created (high in wall, low in cell).
- 1 mark: protons/\(\text{H}^+\) diffuse back into the companion cell through a co-transporter protein.
- 1 mark: Sucrose is transported into the companion cell against its concentration gradient along with the protons.
- 1 mark: Sucrose diffuses into the sieve tube element via plasmodesmata.

**(b) [Max 3 marks]**
- 1 mark: Loading of sucrose decreases the water potential (\(\psi_w\)) in the sieve tube element.
- 1 mark: Water moves into the sieve tube element by osmosis (from xylem/surrounding cells).
- 1 mark: This entry of water increases the volume and generates high hydrostatic pressure.

**(c) [Max 2 marks - any two differences]**
- Xylem wall is lignified / phloem wall is cellulose / non-lignified.
- Xylem has no end walls (continuous tube) / phloem has sieve plates.
- Xylem elements are dead / have no cytoplasm / phloem elements are living / have peripheral cytoplasm.
- Xylem has pits / phloem has plasmodesmata.

**(a)**
- An antigen is injected into a mouse (or other mammal) to stimulate an immune response.
- Plasma cells (B-lymphocytes) that produce the specific complementary antibody are harvested from the mouse's spleen.
- These plasma cells are fused with cancerous tumor cells (myeloma cells) to form hybridoma cells.
- Polyethylene glycol (PEG) or electrofusion is used to facilitate the fusion.
- The hybridoma cells are cultured and screened to identify those producing the desired antibody. The selected hybridomas are cloned to produce a large, genetically identical population that secretes the monoclonal antibodies in large quantities.

**(b)**
- During the primary immune response, memory B-lymphocytes and memory T-lymphocytes are produced and remain in circulation.
- Upon second exposure to the same pathogen, these memory cells immediately recognize the specific antigen.
- Memory B-cells rapidly divide by mitosis (clonal expansion) and differentiate into antibody-secreting plasma cells.
- This bypasses the time-consuming stages of antigen presentation and clonal selection, resulting in rapid antibody production.

**(c)**
- **Active immunity** is achieved when an individual's own immune system is stimulated to produce antibodies and memory cells in response to an antigen (either naturally from infection or artificially via vaccination).
- **Passive immunity** is achieved by transferring antibodies from an external source (e.g., across the placenta, via colostrum, or by injection). No memory cells are produced by the host.
- Passive immunity is temporary/short-term because the injected/transferred antibodies are foreign proteins that are eventually broken down and cleared from the blood, and the body cannot produce more because there are no memory cells.

**(a) [Max 4 marks]**
- 1 mark: Inject antigen into mouse to stimulate plasma cells / B-lymphocytes.
- 1 mark: Harvest B-lymphocytes / plasma cells from the spleen of the mouse.
- 1 mark: Fuse plasma cells with myeloma / tumor cells using PEG / electrofusion to form hybridomas.
- 1 mark: Screen hybridoma cells to find the ones producing the correct antibody, and clone them to produce monoclonal antibodies.

**(b) [Max 3 marks]**
- 1 mark: Primary response leaves memory (B and T) cells in the body.
- 1 mark: Memory cells recognise the antigen instantly upon reinfection.
- 1 mark: Memory cells undergo rapid clonal expansion / mitotic division to form plasma cells (bypassing the slow clonal selection phase), producing high concentrations of antibodies quickly.

**(c) [Max 3 marks]**
- 1 mark: Active immunity involves the host's own immune system producing antibodies and memory cells.
- 1 mark: Passive immunity is the acquisition of pre-made antibodies from an external source (no host antibody production or memory cells).
- 1 mark: Passive immunity is short-lived because antibodies are proteins that are eventually broken down/metabolized by the host.

(a)(i) Dilution table calculations are based on \(C_1 V_1 = C_2 V_2\). For a total volume of 10 cm\(^3\) using a stock of 1.0%: 0.8% requires (0.8/1.0)*10 = 8.0 cm\(^3\) stock and 2.0 cm\(^3\) water. (ii) Method: Filter paper discs are cut to equal size, soaked in yeast suspension, and drained uniformly. They are dropped into a boiling tube containing the hydrogen peroxide and specific concentration of inhibitor. The oxygen produced by catalase gets trapped in the disc, causing it to float. The time taken from hitting the bottom to reaching the surface is measured using a stopwatch. Results are recorded in a table with headings: Copper sulfate concentration (%) and Time taken for disc to rise / s. (b) Graph plotting: x-axis = Inhibitor concentration / mol dm\(^{-3}\), y-axis = Rate of oxygen production / cm\(^3\) min\(^{-1}\). Scale should be linear and occupy at least half the grid. Points are plotted accurately and joined with a ruled straight line or smooth curve. (c) Errors: Variations in yeast saturation; manual stopwatch delay. Improvements: Standardize soaking time and blotting; electronic timing or multiple replicates.

(a)(i) [3 marks] 1 mark for correct concentrations listed; 1 mark for correct volumes of S and W for all concentrations; 1 mark for appropriate table headers with units. (a)(ii) [6 marks] 1 mark for detailing soaking and standardized draining of discs; 1 mark for specifying a constant volume of hydrogen peroxide; 1 mark for measuring time using a stopwatch; 1 mark for repeating each concentration at least 3 times; 1 mark for presenting results in a table with correct headings and units; 1 mark for showing how rate (1/t) is calculated. (a)(iii) [2 marks] 1 mark for independent variable (concentration of copper sulfate); 1 mark for dependent variable (time taken for disc to rise / s). (b) [6 marks] 1 mark for axes labeled with units; 1 mark for correct linear scale using >50% of grid; 1 mark for accurate plotting of all points; 1 mark for drawing a smooth line of best fit or joining points with ruled straight lines; 2 marks for describing the relationship (rate decreases as inhibitor concentration increases with quantitative data from the graph). (c) [3 marks] Max 3 marks: 1 mark for identifying error 1 (e.g., disc size/yeast absorption variation) + 1 mark for corresponding improvement (e.g., use a single hole punch for identical discs, standardize blotting time); 1 mark for identifying error 2 (e.g., temperature changes during reaction) + 1 mark for corresponding improvement (e.g., use a thermostatically controlled water bath).

(a)(i) Cut a small square of red onion epidermis. Use forceps to peel a single cell layer. Place the peel on a clean microscope slide in a drop of the specific sucrose solution. Lower a coverslip at an angle using a mounted needle to prevent air bubbles. Leave for 5-10 minutes to allow equilibration before viewing. (ii) Drawings must show clear continuous outlines with no shading. Fully turgid cell: cell wall, cell membrane pressed tightly against cell wall, large central vacuole filling the cytoplasm. Fully plasmolysed cell: cell wall, shrunken protoplast/plasma membrane pulled away from cell wall, clearly labeled space between cell wall and plasma membrane filled with sucrose solution. (iii) Incipient plasmolysis is the point where 50% of the cells are plasmolysed. It is found by locating 50% on the y-axis of the graph of percentage plasmolysis against sucrose concentration and reading the corresponding concentration on the x-axis. (b)(i) 18/50 * 100 = 36%; 38/50 * 100 = 76%. (b)(ii) Plot points: (0, 0), (0.2, 8), (0.4, 36), (0.6, 76), (0.8, 98). Drawing a smooth sigmoidal curve gives a value of approx 0.47-0.49 mol dm\(^{-3}\) at 50% plasmolysis. (b)(iii) At incipient plasmolysis, the turgor pressure (pressure potential, \(\Psi_p\)) is 0. Since \(\Psi = \Psi_s + \Psi_p\), the water potential (\(\Psi\)) equals the solute potential (\(\Psi_s\)) of the cell, which is in dynamic equilibrium with the external solution's osmotic potential (-1300 kPa).

(a)(i) [4 marks] 1 mark for peeling a single layer of onion epidermis; 1 mark for placing it in a droplet of sucrose solution on a slide; 1 mark for lowering the coverslip gently with a needle/forceps to avoid bubbles; 1 mark for allowing a set time (e.g., 5-10 minutes) for plasmolysis to occur before observation. (a)(ii) [4 marks] 1 mark for clear, thin, continuous lines and no shading; 1 mark for turgid cell showing protoplast completely occupying the cell volume; 1 mark for plasmolysed cell showing shrunken protoplast/vacuole pulled away from corners of cell wall; 1 mark for correct labels of cell wall, plasma membrane, vacuole, and the space containing sucrose solution. (a)(iii) [2 marks] 1 mark for defining incipient plasmolysis as the state where 50% of cells are plasmolysed / turgor pressure is zero; 1 mark for explaining that it is found by reading the x-axis (sucrose concentration) where the y-axis (percentage plasmolysis) is at 50%. (b)(i) [2 marks] 1 mark for 36% (0.4 mol dm\(^{-3}\)); 1 mark for 76% (0.6 mol dm\(^{-3}\)) showing basic working. (b)(ii) [5 marks] 1 mark for x-axis labeled 'Sucrose concentration / mol dm\(^{-3}\)' and y-axis labeled 'Percentage plasmolysis / %'; 1 mark for linear scale using more than half of the grid space; 1 mark for accurate plotting of all 5 coordinates; 1 mark for drawing a smooth sigmoidal curve; 1 mark for correct estimation of concentration at 50% plasmolysis (range 0.47 to 0.49 mol dm\(^{-3}\)). (b)(iii) [3 marks] 1 mark for correct value (-1300 kPa); 1 mark for stating that turgor pressure is zero at incipient plasmolysis; 1 mark for stating that cell water potential is equal to the external solute/osmotic potential at equilibrium.

(a) Geographical isolation prevents gene flow because populations are physically separated. Over time, different selection pressures in their respective environments favor different alleles, and new mutations accumulate. This changes allele frequencies, eventually leading to reproductive isolation where they can no longer interbreed to produce fertile offspring. (b) Small populations have a very small gene pool. The random loss or fixation of alleles by chance has a much greater proportional impact on their allele frequencies than in large populations, where random changes are diluted. (c) mtDNA is inherited solely from the maternal parent and does not undergo recombination. Its mutation rate is relatively high and steady, serving as a molecular clock. By comparing mtDNA sequences, populations with fewer sequence differences are shown to have a more recent common ancestor.

(a) Max 4 marks: 1. physical/geographical barrier prevents gene flow; 2. different environmental selection pressures; 3. selection of different advantageous alleles/mutations; 4. change in allele frequencies; 5. development of reproductive isolation. (b) Max 3 marks: 1. small gene pool; 2. random/chance survival has a larger proportional effect; 3. rapid loss of alleles or fixation of alleles; 4. large populations buffer random fluctuations. (c) Max 3 marks: 1. maternal inheritance/no crossing over; 2. mutation rate acts as a molecular clock; 3. fewer differences mean closer phylogenetic relationship / more recent common ancestor.

(a) C4 plants show Kranz anatomy. The vascular bundle is surrounded by a ring of large bundle sheath cells, which are themselves surrounded by a ring of mesophyll cells. In contrast, C3 plants have distinct palisade and spongy mesophyll layers and lack highly developed bundle sheath cells containing large chloroplasts. (b) PEP carboxylase catalyzes the reaction between CO2 and phosphoenolpyruvate (PEP) to form oxaloacetate. Unlike Rubisco, PEP carboxylase has no affinity for oxygen. In hot, dry environments, plants partially close their stomata to save water, causing internal CO2 concentrations to drop. PEP carboxylase can still efficiently capture CO2, allowing photosynthesis to continue and avoiding wasteful photorespiration. (c) The Calvin cycle occurs in the stroma of the chloroplasts within the bundle sheath cells. High rates of Rubisco activity are maintained because malate is transported from mesophyll cells into bundle sheath cells and decarboxylated, releasing high concentrations of CO2 directly around Rubisco. This keeps the CO2:O2 ratio high, favoring carboxylation over oxygenation.

(a) Max 3 marks: 1. presence of Kranz anatomy; 2. bundle sheath cells form a ring around vascular bundles, surrounded by mesophyll; 3. bundle sheath cells have large/many chloroplasts; 4. absence of palisade/spongy layer distinction in C4. (b) Max 4 marks: 1. PEP carboxylase fixes CO2 to PEP to form oxaloacetate; 2. high affinity for CO2 / does not bind to O2; 3. works at very low CO2 concentrations; 4. prevents photorespiration under conditions of water stress / closed stomata. (c) Max 3 marks: 1. stroma of bundle sheath cell chloroplasts; 2. decarboxylation of malate/C4 acids releases CO2; 3. maintains high CO2 concentration around Rubisco to prevent oxygenase activity.

(a) Reduced NAD and reduced FAD release hydrogen atoms, which split into protons (H+) and electrons. Electrons are transferred along a series of electron carriers in the inner mitochondrial membrane. As electrons pass down energy levels, energy is released. This energy is used by the electron carriers to pump protons from the mitochondrial matrix into the intermembrane space, creating a high concentration of protons there relative to the matrix. (b)(i) Thermogenin provides an alternative pathway for protons to leak down their electrochemical gradient back into the matrix, bypassing ATP synthase. Consequently, the proton motive force is dissipated, and ATP synthesis decreases or stops. (b)(ii) The energy stored in the electrochemical proton gradient is released as heat energy rather than chemical energy. This heat is transferred to blood vessels in the brown adipose tissue to warm the body, preventing hypothermia in newborns. (c) Oxygen is the final electron acceptor. It accepts electrons from the final carrier in the ETC and combines with protons in the matrix to form water, preventing the ETC from backing up.

(a) Max 4 marks: 1. electrons from reduced NAD/FAD pass along carriers; 2. energy is released as electrons move down energy levels; 3. energy is used to actively pump protons; 4. from matrix into intermembrane space; 5. creates a high proton concentration/electrochemical gradient. (b)(i) Max 2 marks: 1. protons diffuse back without passing through ATP synthase; 2. less/no ATP is synthesized. (b)(ii) Max 2 marks: 1. energy of the gradient is dissipated as heat; 2. raises/maintains core body temperature in newborns. (c) Max 2 marks: 1. oxygen is the final electron acceptor; 2. combines with protons and electrons to form water / equation: \(O_2 + 4H^+ + 4e^- \rightarrow 2H_2O\).

(a) The arrival of an action potential depolarizes the presynaptic membrane, opening voltage-gated calcium channels. Calcium ions diffuse down their concentration gradient into the presynaptic neurone. This causes synaptic vesicles containing acetylcholine (ACh) to move to and fuse with the presynaptic membrane, releasing ACh into the synaptic cleft by exocytosis. ACh diffuses across the cleft and binds to specific receptors on the sarcolemma (postsynaptic membrane). This opens ligand-gated sodium channels, allowing sodium ions to enter and depolarize the sarcolemma. (b)(i) Blocking voltage-gated calcium channels prevents calcium ions from entering the presynaptic cytoplasm. Without calcium influx, acetylcholine-containing vesicles cannot fuse with the presynaptic membrane, so no ACh is released. Consequently, no ligand-gated sodium channels are opened on the sarcolemma, and no action potential can trigger muscle contraction. (b)(ii) It can be used as a muscle relaxant during surgical procedures to prevent reflex movements, or as a medication to alleviate severe muscle spasms, spasticity in conditions like cerebral palsy, or as an analgesic to block pain transmission.

(a) Max 5 marks: 1. depolarization of presynaptic membrane opens Ca2+ channels; 2. calcium ions enter presynaptic neurone; 3. vesicles fuse with presynaptic membrane; 4. ACh released by exocytosis and diffuses across cleft; 5. ACh binds to receptors on sarcolemma; 6. opening of ligand-gated Na+ channels and entry of Na+ causing depolarization. (b)(i) Max 3 marks: 1. no calcium entry; 2. no vesicle fusion / no ACh release; 3. no binding to sarcolemma receptors; 4. no depolarization / no action potential in muscle fiber. (b)(ii) Max 2 marks: 1. muscle relaxant / treatment for spasms; 2. management of chronic pain / anesthetic; 3. treatment for tetanus / overactive muscle conditions.

(a) The regulatory gene lacI is transcribed and translated continuously to produce the active lac repressor protein. In the absence of lactose, this repressor protein binds specifically to the operator region (lacO). This physical binding blocks RNA polymerase from binding to the promoter or moving along the DNA, preventing the transcription of the structural genes. (b) lacZ encodes beta-galactosidase, an enzyme that hydrolyzes lactose into glucose and galactose. lacY encodes lactose permease, a transport protein embedded in the cell membrane that facilitates the uptake of lactose into the bacterium. (c) If a mutation prevents the repressor from binding to the operator, the operator remains open. RNA polymerase can freely bind to the promoter and transcribe the operon. Thus, beta-galactosidase and permease will be synthesized constitutively (constantly) even when lactose is completely absent.

(a) Max 4 marks: 1. lacI is a regulatory gene that is constitutively expressed; 2. codes for repressor protein; 3. repressor protein binds to operator; 4. blocks RNA polymerase from binding to promoter / transcribing. (b) Max 3 marks: 1. lacZ codes for beta-galactosidase; 2. which hydrolyzes lactose to glucose and galactose; 3. lacY codes for lactose permease; 4. which increases cell permeability to / transports lactose. (c) Max 3 marks: 1. repressor cannot bind to operator; 2. RNA polymerase always transcribes the operon; 3. enzymes synthesized constitutively / in absence of lactose.

(a)(i) Heating to 95 degrees C denatures the double-stranded DNA by breaking the hydrogen bonds between complementary base pairs, separating it into two single strands. (a)(ii) Cooling to 55 degrees C allows the primers to bind/anneal to their complementary target sequences on the single-stranded template DNA via hydrogen bonding. (a)(iii) Heating to 72 degrees C provides the optimum temperature for the DNA polymerase (Taq polymerase) to extend the primers by adding free complementary nucleotides (dNTPs). (b) Taq polymerase is thermostable, meaning it can withstand the high temperature (95 degrees C) used in the denaturation step without denaturing. Human DNA polymerase would denature and become inactive, requiring fresh enzyme to be added at each cycle. (c) DNA fragments have a negative charge due to the phosphate groups in their sugar-phosphate backbones. When an electric current is applied, they migrate towards the positive electrode (anode). The agarose gel acts as a molecular sieve; smaller fragments move faster and further through the pores of the gel than larger fragments, separating them by size.

(a)(i) 1 mark: breaks hydrogen bonds / separates double-stranded DNA into single strands. (a)(ii) Max 2 marks: 1. allows primers to anneal; 2. by complementary base pairing / hydrogen bonding. (a)(iii) Max 2 marks: 1. optimum temperature for Taq polymerase; 2. to synthesize new strand / add complementary dNTPs. (b) Max 2 marks: 1. Taq polymerase is thermostable; 2. does not denature at high temperature / does not need replacement each cycle. (c) Max 3 marks: 1. DNA has negative charge (phosphate group) so moves to anode/positive electrode; 2. gel acts as a molecular sieve/mesh; 3. smaller fragments move faster/further through pores than larger fragments.

(a) When blood water potential decreases, water leaves the osmoreceptor cells in the hypothalamus by osmosis down a water potential gradient. This causes the osmoreceptor cells to shrink, which generates nerve impulses that are transmitted to the posterior pituitary gland to stimulate the release of ADH. (b) ADH binds to specific receptor proteins on the cell surface membranes of the collecting duct cells. This activates a G-protein, which stimulates adenylyl cyclase to convert ATP to cyclic AMP (second messenger). cAMP activates a kinase cascade, leading to the movement of vesicles containing aquaporins towards the luminal membrane. These vesicles fuse with the membrane, inserting the aquaporins. Water then moves out of the collecting duct lumen by osmosis into the hypertonic tissue fluid of the medulla, resulting in concentrated urine. (c)(i) The volume of urine will increase and its concentration will decrease (it becomes dilute) because aquaporins are not inserted, preventing water reabsorption. (c)(ii) Blood pressure will decrease because less water is reabsorbed into the blood, leading to a reduced blood volume.

(a) Max 2 marks: 1. water moves out of osmoreceptors by osmosis; 2. cells shrink; 3. triggers impulses to posterior pituitary gland. (b) Max 5 marks: 1. ADH binds to receptors on collecting duct cells; 2. activates G-protein / adenylyl cyclase / cyclic AMP; 3. causes vesicles with aquaporins to move to / fuse with luminal membrane; 4. increases permeability of membrane to water; 5. water moves out of lumen by osmosis; 6. into hypertonic medulla tissue. (c)(i) Max 2 marks: 1. larger volume of urine; 2. dilute / lower concentration. (c)(ii) 1 mark: blood pressure decreases.

(a) In situ conservation involves protecting species within their natural habitats, such as establishing national parks or nature reserves. Ex situ conservation involves conserving species outside their natural habitats, such as in zoos, botanic gardens, or seed banks. (b) Oocytes (egg cells) can be harvested from the remaining northern white rhino females and fertilized in vitro using cryopreserved sperm collected previously from deceased males. The resulting embryos are cultured and then transferred into surrogate mothers of a closely related and more abundant subspecies, such as the southern white rhinoceros. This overcomes the lack of reproductive males and allows multiple offspring to be produced simultaneously. (c) Maintaining genetic diversity ensures a large pool of different alleles is present in the population. This variation allows the population to adapt to environmental changes, such as new pathogens or climate shifts, through natural selection. It also reduces the incidence of inbreeding depression, which can lead to the expression of harmful homozygous recessive genetic defects.

(a) Max 3 marks: 1. in situ is conservation in natural habitat; 2. ex situ is conservation outside natural habitat; 3. correct example for in situ (e.g. reserve) AND ex situ (e.g. zoo). (b) Max 4 marks: 1. harvesting oocytes from remaining females; 2. in vitro fertilization (IVF) using stored/frozen sperm; 3. embryo culture to suitable stage; 4. embryo transfer into surrogate mothers (e.g. southern white rhino); 5. allows reproduction when natural mating is impossible. (c) Max 3 marks: 1. provides a large pool of alleles; 2. allows adaptation to changing environments / diseases; 3. avoids inbreeding depression / homozygous recessive genetic disorders.

(a) When the water potential of the blood decreases below a set point, water leaves the osmoreceptors by osmosis down a water potential gradient. This loss of water causes the osmoreceptor cells to shrink. The shrinkage of these cells depolarizes their membrane, generating action potentials that are transmitted along axons through the pituitary stalk to the posterior pituitary gland. This triggers the release of ADH into the bloodstream via exocytosis.

(b) When ADH binds to its complementary receptor on the basolateral membrane of the collecting duct epithelial cells, it activates a G-protein. This active G-protein stimulates the enzyme adenylyl cyclase, which converts ATP to cyclic AMP (cAMP). cAMP acts as a second messenger, activating a cascade of protein kinases (specifically protein kinase A). These kinases trigger the movement of intracellular vesicles containing aquaporins towards the luminal (apical) membrane. The vesicles fuse with the luminal membrane, increasing the density of aquaporin channel proteins, which makes the membrane highly permeable to water.

(c) Tolvaptan acts as a competitive inhibitor of the ADH receptor, meaning ADH cannot bind. Consequently, there is no G-protein activation or cAMP production. Vesicles with aquaporins do not fuse with the luminal membrane, and water permeability remains very low. Thus, less water is reabsorbed by osmosis from the lumen of the collecting duct into the hypertonic medulla tissue fluid. This results in the production of a much larger volume of highly dilute (low concentration) urine.

Part (a) [Max 3]
1. Water potential of blood decreases, so water leaves osmoreceptors by osmosis down a water potential gradient.
2. Osmoreceptor cells shrink / decrease in volume.
3. Membrane depolarises / action potential is generated.
4. Action potentials travel along axons to posterior pituitary gland, triggering exocytosis / release of ADH.

Part (b) [Max 4]
1. ADH binds to specific receptor on basolateral membrane of collecting duct cell.
2. Activates G-protein, which activates adenylyl cyclase.
3. Adenylyl cyclase catalyses conversion of ATP to cyclic AMP (cAMP) / second messenger.
4. cAMP activates protein kinase (A) / intracellular signaling cascade.
5. Vesicles containing aquaporins / water channel proteins move towards the luminal / apical membrane.
6. Vesicles fuse with the luminal / apical membrane, increasing water permeability.

Part (c) [Max 3]
1. Urine volume increases AND concentration decreases / becomes more dilute.
2. Drug blocks / binds to ADH receptors, preventing ADH from binding / preventing cAMP activation.
3. Fewer aquaporins are inserted into the luminal membrane.
4. Less water is reabsorbed by osmosis down the water potential gradient.

(a) C4 leaves show Kranz anatomy, characterized by vascular bundles surrounded by a tight ring of large bundle sheath cells, which are themselves surrounded by an outer ring of mesophyll cells. This is unlike C3 plants, which have no specialized bundle sheath cells and instead have distinct layers of palisade and spongy mesophyll. Additionally, bundle sheath cells in C4 plants contain large chloroplasts with few or no grana, whereas mesophyll cells contain normal chloroplasts.

(b) In the mesophyll cells of C4 plants, carbon dioxide is first fixed by the enzyme PEP carboxylase (instead of Rubisco). PEP carboxylase has a very high affinity for carbon dioxide and does not bind oxygen, preventing competition. The carbon dioxide is combined with phosphoenolpyruvate (PEP) to form the 4-carbon compound oxaloacetate, which is then converted into malate. Malate is transported via plasmodesmata into the bundle sheath cells, where it is decarboxylated to release carbon dioxide and pyruvate. This mechanism continuously supplies a high concentration of carbon dioxide directly around Rubisco within the bundle sheath cells, keeping the carbon dioxide to oxygen ratio high, thus preventing photorespiration and maintaining high rates of carbon fixation.

(c) Magnesium is the central metal ion in the chemical structure of the chlorophyll molecule, which absorbs light energy during the light-dependent stage of photosynthesis. A deficiency in magnesium ions restricts chlorophyll synthesis (causing chlorosis), which significantly reduces light absorption. Consequently, less ATP and reduced NADP are synthesized, which limits the Calvin cycle, resulting in reduced production of organic nutrients (photoassimilates) and stunted plant growth.

Part (a) [Max 3]
1. Mention of Kranz anatomy.
2. Vascular bundles are surrounded by a ring of large bundle sheath cells.
3. Bundle sheath cells are surrounded by a ring of mesophyll cells (no clear distinction between palisade and spongy mesophyll).
4. Bundle sheath chloroplasts are large and contain few/no grana, while mesophyll chloroplasts are normal.

Part (b) [Max 5]
1. Carbon dioxide is first fixed in mesophyll cells by PEP carboxylase (instead of Rubisco).
2. PEP carboxylase has a very high affinity for carbon dioxide and does not bind oxygen.
3. Carbon dioxide reacts with phosphoenolpyruvate (PEP) to form oxaloacetate (a 4-carbon compound), which is converted to malate.
4. Malate is transported into bundle sheath cells (via plasmodesmata).
5. Malate is decarboxylated in bundle sheath cells to release carbon dioxide (and pyruvate).
6. This maintains a high concentration of carbon dioxide in the bundle sheath cells / high carbon dioxide to oxygen ratio.
7. Rubisco, located only in bundle sheath cells, fixes this carbon dioxide into GP, preventing photorespiration (binding with oxygen).

Part (c) [Max 2]
1. Magnesium is the central atom of the chlorophyll molecule, which absorbs light energy during the light-dependent stage.
2. A deficiency in magnesium causes chlorosis (yellowing of leaves), which reduces light absorption, leading to less production of ATP and reduced NADP, thereby reducing carbon fixation in the Calvin cycle and restricting plant growth due to lack of photoassimilates.

### Part (a) Method Design
To obtain reliable and valid data, a systematic procedure must be followed:
1. **Independent Variable (ONPG Concentration):** Use the simple dilution series prepared in (b) to obtain 5 distinct concentrations (2.0 to 16.0 mmol dm\(^{-3}\)).
2. **Dependent Variable (Rate of Reaction):** Mix a fixed volume of the substrate solution with buffer, inhibitor/water, and lactase. Immediately transfer to a cuvette and place in a colorimeter. Measure absorbance at 420 nm at 15-second intervals for 2 minutes. Plot absorbance against time and calculate the gradient of the initial linear portion of the graph to find the initial rate of reaction (absorbance units per second).
3. **Control Variables:**
- **Enzyme Concentration:** Use a constant volume (e.g., 1.0 cm\(^{3}\)) and concentration (e.g., 1.0%) of lactase for all trials.
- **pH:** Use a constant volume of buffer solution (e.g., pH 7.0) to maintain stable enzyme structure.
- **Temperature:** Place separate tubes of enzyme, substrate, and buffer into a thermostatically controlled water bath set at 37 °C. Allow them to equilibrate for 5 minutes before mixing to ensure temperature remains constant.
4. **Inhibitor Treatment:** Set up two parallel experiments. In the test set, add a constant volume (e.g., 1.0 cm\(^{3}\)) of galactose (inhibitor). In the control set, add 1.0 cm\(^{3}\) of distilled water instead of galactose to maintain the same total reaction volume.
5. **Replication:** Perform at least three replicates for each concentration with and without the inhibitor, and calculate the mean initial rate.
6. **Safety:** Lactase enzyme and ONPG powder/solutions can act as respiratory and skin irritants. Wear protective eyewear (goggles), a lab coat, and disposable gloves.

### Part (b) Dilution Calculations
Using the formula \(C_1 V_1 = C_2 V_2\), where \(C_1 = 20.0\text{ mmol dm}^{-3}\) and \(V_2 = 10.0\text{ cm}^3\):
- \(V_1 = \frac{C_2 \times 10.0}{20.0}\)
- Volume of distilled water = \(10.0 - V_1\)

| Desired Concentration / \(\text{mmol dm}^{-3}\) | Volume of 20.0 \(\text{mmol dm}^{-3}\) stock ONPG / \(\text{cm}^3\) | Volume of Distilled Water / \(\text{cm}^3\) |
|---|---|---|
| 2.0 | 1.0 | 9.0 |
| 4.0 | 2.0 | 8.0 |
| 8.0 | 4.0 | 6.0 |
| 12.0 | 6.0 | 4.0 |
| 16.0 | 8.0 | 2.0 |

### Part (c) Michaelis-Menten Interpretation
- **\(V_{max}\) Determination:** Locate the maximum plateau value on the y-axis of the rate-substrate concentration curve.
- **\(K_m\) Determination:** Calculate half of the \(V_{max}\) value (\(\frac{1}{2} V_{max}\)). Find this value on the y-axis, project horizontally to intersect the curve, and project down to the x-axis to read the substrate concentration. This is \(K_m\).
- **Effect of Competitive Inhibitor on \(V_{max}\):** \(V_{max}\) remains unchanged because high substrate concentrations can outcompete and overwhelm the competitive inhibitor.
- **Effect of Competitive Inhibitor on \(K_m\):** \(K_m\) increases because the inhibitor competes for active sites, reducing the enzyme's apparent affinity for the substrate, meaning a higher substrate concentration is required to reach half \(V_{max}\).

### Part (a) [Max 8 marks]
- **M1 (Independent):** Description of preparing and using at least 5 different concentrations of substrate (ONPG) [Accept: those from part b].
- **M2 (Inhibitor):** Clear statement of setting up two experimental lines: one with a constant concentration of galactose and one with distilled water (as control) keeping volumes equal.
- **M3 (Dependent - setup):** Mix enzyme and substrate and use a colorimeter with a blue filter / 420 nm.
- **M4 (Dependent - measurement):** Measure absorbance at regular specified intervals (e.g. every 15/30 seconds) and plot absorbance vs time to find the initial rate (gradient).
- **M5 (Control - Temperature):** Use a water bath set at a specific temperature (e.g. 37 °C) and allow reactants to equilibrate before mixing.
- **M6 (Control - pH):** Use a buffer solution of a specified pH (e.g., pH 7.0).
- **M7 (Control - Enzyme):** State that enzyme volume and concentration are kept constant across all trials.
- **M8 (Reliability):** State that each concentration should be repeated at least 3 times and a mean calculated.
- **M9 (Safety):** Identify enzyme/ONPG as irritants and suggest wearing eye protection/gloves.

### Part (b) [3 marks]
- **M10:** Correctly calculates volumes of stock for all concentrations (1.0, 2.0, 4.0, 6.0, 8.0 cm\(^3\)).
- **M11:** Correctly calculates volumes of distilled water for all concentrations (9.0, 8.0, 6.0, 4.0, 2.0 cm\(^3\)).
- **M12:** Table presented with clear, unambiguous headings and units for all columns.

### Part (c) [4 marks]
- **M13:** Explains \(V_{max}\) as the maximum plateau rate of the curve.
- **M14:** Explains \(K_m\) as the substrate concentration corresponding to \(\frac{1}{2} V_{max}\).
- **M15:** Correctly states that the competitive inhibitor does not change \(V_{max}\).
- **M16:** Correctly states that the competitive inhibitor increases \(K_m\) (decreases apparent affinity).

### Part (a) Variables
- **Independent Variable:** The independent variables are the strain of yeast used (Wild Type vs *mut-1* mutant) and the temperature (25 °C vs 40 °C).
- **Dependent Variable:** The dependent variable is the time taken for methylene blue decolorization, from which the rate of respiration (\(1000/t\)) is calculated.

### Part (b) Experimental Controls and Standardization
- **Temperature Control:** Use a thermostatically controlled water bath. Monitor the temperature using a thermometer to ensure it remains constant at 25 °C and 40 °C respectively throughout the incubation period.
- **End-point Standardization:** Visual determination of "colorless" is subjective. To reduce this error:
- Use a colorimeter to measure the transmittance of light through the tube until it reaches a stable plateaud maximum transmittance value.
- Or, set up a control tube (reference tube) containing only yeast and glucose without methylene blue to act as a visual standard of comparison for "completely decolorized".

### Part (c) Standard Error Calculation
- **Formula:** \(S_E = \frac{s}{\sqrt{n}}\)
- **Parameters:** For WT at 40 °C, standard deviation \(s = 1.58\). Number of replicates \(n = 10\).
- **Calculation:**
\[S_E = \frac{1.58}{\sqrt{10}} = \frac{1.58}{3.16227766} = 0.499642\dots\]
- **Rounded to 3 significant figures:** \(0.500\) (arbitrary units)

### Part (d) Statistical Analysis (t-test)
- **(i) Null Hypothesis:** There is no significant difference between the mean rate of respiration of Wild Type (WT) yeast and mutant (*mut-1*) yeast at 40 °C (any difference is due to chance alone).
- **(ii) Interpretation:**
- Since the calculated value of \(t\) (52.41) is significantly greater than the critical value of \(t\) (2.10),
- the probability (\(p\)) that the observed difference is due to chance is much less than 0.05 (\(p < 0.05\)).
- Therefore, the null hypothesis is rejected, and we can conclude that there is a highly statistically significant difference in respiration rates between the two yeast strains at 40 °C.

### Part (e) Limitations and Improvements
- **Limitations:**
1. Oxygen in the headspace of the test tube can re-oxidize the methylene blue back to its blue oxidized state, artificially prolonging the time recorded.
2. Methylene blue can be toxic to yeast at high concentrations, affecting their baseline cellular respiration rate.
- **Improvements:**
1. Add a thin layer of paraffin oil or mineral oil on top of the reaction mixture in the tube to exclude atmospheric oxygen.
2. Instead of a redox dye, use a gas-tight respirometer to measure carbon dioxide production directly, or use an oxygen probe/electrode connected to a data logger to record the rate of dissolved oxygen depletion directly and continuously.

### Part (a) [2 marks]
- **M1:** Identifies both independent variables: yeast strain AND temperature. [Reject: if only one is mentioned]
- **M2:** Identifies dependent variable as time taken to decolorize / rate of respiration.

### Part (b) [3 marks]
- **M3:** Suggests using a thermostatically controlled water bath to keep temperature constant.
- **M4:** Suggests standardizing end-point by using a colorimeter (to measure transmittance/absorbance).
- **M5:** Alternatively, suggests comparing with a reference standard tube containing only yeast and glucose (without dye) to determine the exact decolorization point.

### Part (c) [2 marks]
- **M6:** Correct substitution shown: \(1.58 / \sqrt{10}\) or \(1.58 / 3.16\).
- **M7:** Correct calculated value to 3 significant figures: **0.500** (Accept: 0.50).

### Part (d) [4 marks]
- **M8 (i):** States null hypothesis correctly (e.g., "no significant difference between the mean respiration rates of WT and mutant yeast at 40 °C").
- **M9 (ii):** Points out that the calculated \(t\)-value is greater than the critical value.
- **M10 (ii):** Connects this to probability: probability that the difference is due to chance is less than 0.05 / \(p < 0.05\).
- **M11 (ii):** Concludes that the null hypothesis is rejected and the difference is statistically significant.

### Part (e) [Max 4 marks]
- **M12 (Limitation):** Oxygen from air can re-oxidize methylene blue.
- **M13 (Limitation):** Methylene blue might be toxic to yeast / affect cell viability.
- **M14 (Improvement):** Add oil/paraffin layer to prevent entry of atmospheric oxygen.
- **M15 (Improvement):** Use a respirometer to measure CO2 produced / oxygen consumed directly.
- **M16 (Improvement):** Use an oxygen sensor / probe connected to a computer/data logger to measure the rate of oxygen decrease directly.