MetadataPastPaper.sampleTitle

First, calculate the water potential of the plant cell using the formula: \(\psi_{cell} = \psi_s + \psi_p\). This gives \(\psi_{cell} = -800\text{ kPa} + 200\text{ kPa} = -600\text{ kPa}\). The water potential of the surrounding solution is \(-500\text{ kPa}\). Since \(-500\text{ kPa}\) is higher (less negative) than \(-600\text{ kPa}\), water moves down a water potential gradient from the solution into the cell by osmosis.

Award 1 mark for the correct option (B).
- Reject other options because they either suggest water moves out of the cell (A) or state incorrect reasoning regarding water potential gradients (C, D).

Collagen is a fibrous protein composed of three polypeptide chains wrapped around one another to form a tight triple helix (tropocollagen), providing high tensile strength. Haemoglobin is a globular protein containing four polypeptide chains (two alpha and two beta chains), each associated with a heme prosthetic group containing an iron ion, allowing it to bind reversibly to oxygen.

Award 1 mark for identifying the correct structural differences between collagen and haemoglobin (B).

A competitive inhibitor binds reversibly to the active site of the enzyme, competing directly with the substrate. At very high substrate concentrations, the substrate molecules completely outcompete the inhibitor molecules, allowing the reaction to reach the same maximum rate (\(V_{max}\) is unchanged). However, because more substrate is required to achieve half of the maximum velocity, the Michaelis-Menten constant (\(K_m\)) is increased.

Award 1 mark for identifying that competitive inhibitors leave \(V_{max}\) unchanged but increase the \(K_m\) value (B).

Cyclic photophosphorylation involves only photosystem I (PSI). Electrons from PSI are excited and return to PSI via electron carriers, generating ATP by chemiosmosis, but no reduced NADP or oxygen is produced. Non-cyclic photophosphorylation involves both PSI and PSII. Photolysis of water occurs at PSII, releasing oxygen and protons. The protons and electrons are used to reduce NADP to form reduced NADP, and ATP is also synthesised.

Award 1 mark for the option showing cyclic photophosphorylation yields ATP only and non-cyclic yields ATP, reduced NADP, and oxygen (C).

Proton pumps (H+-ATPase) in the cell surface membrane of companion cells actively pump protons (\(H^+\)) out of the cytoplasm into the cell wall. This creates a higher concentration of protons outside the cell, establishing an electrochemical gradient. Protons then diffuse back into the companion cells down this gradient through co-transporter proteins, bringing sucrose molecules with them against their concentration gradient.

Award 1 mark for identifying active transport of protons via proton pumps / ATP-hydrolase as the primary gradient-establishing step (B).

Sodium ions are actively pumped out of the epithelial cell across the basolateral membrane into the tissue fluid and blood, maintaining a low concentration of sodium ions inside the cell. This concentration gradient drives the co-transport (secondary active transport) of sodium ions and glucose from the lumen of the tubule across the apical membrane into the cell. Glucose then leaves the cell down its concentration gradient into the tissue fluid by facilitated diffusion.

Award 1 mark for option A, which correctly identifies co-transport across the apical membrane (lumen into cell) and facilitated diffusion/active transport across the basolateral membrane (cell into tissue fluid/blood).

When lactose is present, it is converted into allolactose, which acts as an inducer by binding to the repressor protein. This binding alters the tertiary structure of the repressor, preventing it from binding to the operator. As a result, the operator is clear, allowing RNA polymerase to bind to the promoter and transcribe the structural genes (lacZ, lacY, and lacA).

Award 1 mark for the option showing lactose binds to the repressor, changing its shape so it detaches from the operator, allowing transcription of structural genes (B).

Cholera is caused by the bacterium Vibrio cholerae and is primarily transmitted via contaminated drinking water and food (water-borne). Malaria is caused by Plasmodium, which is a protoctist (not a virus). Tuberculosis is caused by Mycobacterium tuberculosis (a bacterium) and is transmitted via airborne droplets. HIV/AIDS is caused by the Human Immunodeficiency Virus and is transmitted by sexual contact or infected blood, not through contaminated water.

Award 1 mark for the correct pairing of cholera with its bacterial pathogen and water-borne transmission (A).

Amylose is an unbranched polymer of \alpha-glucose monomers linked by \alpha-1,4-glycosidic bonds, which coils into a helical shape. Cellulose is a polymer of \beta-glucose monomers linked by \beta-1,4-glycosidic bonds, forming straight, unbranched chains. Many hydrogen bonds form between neighboring cellulose chains to provide high tensile strength.

A is correct because it correctly describes the glycosidic bonds and structural features of both amylose and cellulose. B, C, and D contain incorrect statements regarding glycosidic bond types or structural configurations.

During non-cyclic photophosphorylation, light is absorbed by photosystem II (PSII), exciting electrons which are passed along an electron transport chain to photosystem I (PSI). This flow of electrons generates a proton gradient that drives ATP synthesis. The electrons lost from PSI are excited by light and are used to reduce NADP\(^{+}\). Electrons lost from PSII are replaced by those from the photolysis of water.

B is correct. A is incorrect because photolysis of water provides electrons to PSII, not PSI. C describes cyclic photophosphorylation, not non-cyclic. D is incorrect because reduction of NADP\(^{+}\) occurs after PSI, not directly at PSII.

The movement of the air bubble in a potometer measures water uptake, which is closely related to the rate of transpiration. Transpiration is highest when stomata are open (stimulated by high light intensity), the rate of evaporation is high (stimulated by high temperature), the water potential gradient is steep (enhanced by low relative humidity), and the water vapor boundary layer is removed (enhanced by moving air).

C is correct as all four conditions (high light, high temperature, low humidity, moving air) maximize the rate of transpiration and therefore water uptake. Options A, B, and D contain one or more conditions that would decrease the rate of transpiration.

The ascending limb of the loop of Henle actively pumps sodium and chloride ions into the tissue fluid of the medulla. Because the ascending limb is impermeable to water, this creates a high concentration of solutes in the medullary tissue fluid (low water potential) without water following it. This hypertonic environment is essential for the osmotic reabsorption of water from the collecting duct down a water potential gradient, regulated by ADH.

B is correct. A is incorrect because the active transport occurs out of the ascending limb, not directly affecting the descending limb's internal filtrate directly via active transport. C is incorrect because solute export decreases (lowers) water potential. D is incorrect because water is never actively pumped.

During metaphase of mitosis, there are 8 chromosomes lined up at the equator, each consisting of 2 sister chromatids, giving 16 chromatids. During anaphase, sister chromatids separate and are pulled to opposite poles. Once they separate, they are defined as individual chromosomes. Therefore, there are now 16 chromosomes in the cell, and since the sister chromatids have separated, they are no longer referred to as chromatids (0 chromatids).

A is correct. In metaphase, chromosomes are replicated (2 chromatids each), so 8 chromosomes have 16 chromatids. In anaphase, the centromere splits and chromatids separate to become 16 single-chromatid chromosomes, leaving 0 chromatids. Options B, C, and D incorrect due to miscalculating chromosome or chromatid counts at these stages.

In a dihybrid cross of two double heterozygotes (\(RrTt \times RrTt\)) with unlinked genes, the expected phenotypic ratio in the offspring is 9 red-tall : 3 red-dwarf : 3 white-tall : 1 white-dwarf. Therefore, the probability of obtaining red-flowered and dwarf offspring (\(R\)_\(tt\)) is \(\frac{3}{16}\).

B is correct. \(\frac{9}{16}\) (A) is the probability of the double dominant phenotype (red, tall). \(\frac{1}{16}\) (C) is the probability of the double recessive phenotype (white, dwarf). \(\frac{1}{4}\) (D) is incorrect.

When lactose is present, some of it is converted to allolactose inside the cell. Allolactose acts as an inducer by binding to the lac repressor protein. This causes a conformational change in the repressor so that it dissociates from, or can no longer bind to, the operator. This allows RNA polymerase to bind to the promoter and proceed with the transcription of the structural genes.

B is correct. A is incorrect because lactose (or allolactose) does not bind to the promoter. C is incorrect because when the repressor binds to the operator, transcription is blocked, not allowed. D is incorrect because lactose does not bind to RNA polymerase.

A competitive inhibitor competes with the substrate for the active site of the enzyme. This effectively reduces the affinity of the enzyme for its substrate, thereby increasing the Michaelis-Menten constant (\(K_m\)), as a higher substrate concentration is needed to reach half the maximum velocity. Since the inhibitor can be fully outcompeted at very high substrate concentrations, the maximum rate of reaction (\(V_{max}\)) remains the same.

A is correct. Competitive inhibitors increase \(K_m\) but do not affect \(V_{max}\). B describes a non-competitive inhibitor's effect on \(V_{max}\) but incorrectly states \(K_m\) decreases. C is typical of non-competitive inhibition where \(K_m\) is unchanged and \(V_{max}\) is reduced.

Statement 1 is correct because a triglyceride consists of one glycerol molecule ester-bonded to three fatty acid residues, while a phospholipid contains two fatty acid residues and a phosphate-derived group. Statement 2 is correct because triglycerides are entirely non-polar (hydrophobic), whereas phospholipids are amphipathic (possessing a hydrophilic phosphate head and hydrophobic fatty acid tails). Statement 3 is incorrect because both triglycerides and phospholipids can contain any combination of saturated and unsaturated fatty acids depending on the source organism.

Award 1 mark for selecting option A, which correctly identifies statements 1 and 2 as correct and statement 3 as incorrect.

In non-cyclic photophosphorylation, photolysis of water provides electrons to Photosystem II (PSII). These electrons pass down an electron transport chain to Photosystem I (PSI), and are finally accepted by NADP+. As electrons flow down the transport chain, energy is used to pump protons from the stroma into the thylakoid lumen, resulting in the highest concentration of protons within the thylakoid lumen.

Award 1 mark for identifying the correct electron flow (PSII to PSI) and the correct site of proton accumulation (thylakoid lumen).

Active loading involves companion cells using ATP-powered proton pumps to actively transport protons (H+ ions) out of the companion cell cytoplasm into the cell wall space. This creates a high proton concentration in the cell wall. Protons then diffuse back into the companion cells down their concentration gradient via co-transporter proteins, which simultaneously transport sucrose into the companion cells against its concentration gradient.

Award 1 mark for recognizing that proton pumps actively move protons out of the companion cell to create the electrochemical gradient necessary for sucrose co-transport.

Statement 1 is correct because myelin acts as an insulator, restricting depolarization to the nodes of Ranvier. Statement 2 is correct because the action potential jumps from node to node (saltatory conduction), meaning that local circuits are established over a much longer distance than the short continuous circuits of unmyelinated axons. Statement 3 is incorrect because myelin does not affect the physical or chemical properties of the cytoplasm, so cytoplasmic resistance is unaffected.

Award 1 mark for option A. Statements 1 and 2 are correct; Statement 3 is incorrect.

In metaphase, there are 8 chromosomes. Each chromosome consists of two sister chromatids joined at the centromere, giving a total of 16 chromatids. During anaphase, the centromeres split and the sister chromatids separate to become individual chromosomes. Therefore, there are now 16 chromosomes in the cell during anaphase.

Award 1 mark for identifying Row A as correct, showing 8 chromosomes and 16 chromatids at metaphase, and 16 chromosomes at anaphase.

When the water potential of the blood decreases, the hypothalamus triggers the release of ADH from the posterior pituitary. ADH binds to specific cell surface receptors on the collecting duct cells (since peptide hormones cannot cross the phospholipid bilayer). This activates G-proteins, stimulating the production of cyclic AMP (cAMP) as a second messenger. cAMP activates a cascade of enzymes, causing vesicles containing aquaporins to move to and fuse with the luminal (apical) membrane, increasing water reabsorption.

Award 1 mark for identifying the correct sequential pathway: cell surface receptor binding, G-protein activation, cAMP synthesis, vesicle fusion, and aquaporin insertion.

A competitive inhibitor has a complementary shape to the active site and competes directly with the substrate. Because it can be outcompeted at very high substrate concentrations, the maximum velocity (\(V_{\max}\)) is unchanged. However, because the inhibitor reduces the frequency of substrate binding at lower concentrations, the apparent affinity decreases, resulting in a higher Michaelis constant (\(K_{\text{m}}\)).

Award 1 mark for selecting option A, linking competitive inhibition to active site binding when \(V_{\max}\) is unchanged and \(K_{\text{m}}\) is increased.

One molecule of glucose produces two molecules of pyruvate via glycolysis. During the link reaction, 2 pyruvates are decarboxylated and oxidized to yield 2 \(CO_2\), 2 reduced NAD, and 2 acetyl-CoA. These 2 acetyl-CoA molecules enter the Krebs cycle, which yields 4 \(CO_2\), 6 reduced NAD, 2 reduced FAD, and 2 ATP. Combining these, the link reaction and Krebs cycle produce a total of 6 \(CO_2\), 8 reduced NAD, 2 reduced FAD, and 2 ATP.

Award 1 mark for the correct combined tally of products per glucose molecule (6 \(CO_2\), 8 reduced NAD, 2 reduced FAD, and 2 ATP).

Amylose is an unbranched polymer of \(\alpha\)-glucose containing only \(\alpha\)-1,4-glycosidic bonds, which coils into a helical shape. Amylopectin is a branched polymer containing both \(\alpha\)-1,4 and \(\alpha\)-1,6-glycosidic bonds. Glycogen has a structure very similar to amylopectin but features much more frequent branching due to a higher proportion of \(\alpha\)-1,6-glycosidic bonds.

Award 1 mark for correctly identifying the glycosidic bond types and branching patterns of amylose, amylopectin, and glycogen.

In non-cyclic photophosphorylation, photolysis of water provides electrons to Photosystem II (PSII). These electrons flow through an electron transport chain to Photosystem I (PSI). Energy released during this transfer is used to actively pump protons from the stroma into the thylakoid lumen, establishing a proton gradient.

Award 1 mark for identifying the correct pathway of electron flow and the direction of proton accumulation during non-cyclic photophosphorylation.

The apoplast pathway involves movement through the cell walls and intercellular spaces. At the endodermis, the waterproof Casparian strip (made of suberin) blocks this pathway. This forces water to cross the selectively permeable plasma membrane of the endodermal cells to enter the cytoplasm (symplast pathway), allowing the plant to regulate which ions enter the xylem.

Award 1 mark for identifying that the Casparian strip blocks the apoplast pathway at the endodermis, forcing water into the symplast pathway.

Upon arrival of an action potential, depolarization of the presynaptic membrane causes voltage-gated calcium channels to open. Calcium ions diffuse down their electrochemical gradient into the synaptic knob. This influx causes synaptic vesicles containing neurotransmitter to move to and fuse with the pre-synaptic membrane, releasing the neurotransmitter into the synaptic cleft by exocytosis.

Award 1 mark for the correct chronological sequence of events involved in synaptic transmission, specifying calcium influx and presynaptic vesicle fusion.

During anaphase, the centromeres divide and the sister chromatids are pulled apart toward opposite poles of the cell by the spindle fibers. Once separated, each chromatid is considered an individual chromosome.

Award 1 mark for identifying the mitotic stage as anaphase and correctly describing the separation of sister chromatids into individual chromosomes.

A low blood water potential causes water to leave osmoreceptor cells in the hypothalamus by osmosis, causing them to shrink. This triggers the posterior pituitary gland to secrete more antidiuretic hormone (ADH) into the blood. ADH binds to receptors on the collecting duct cells, stimulating vesicles carrying aquaporins to fuse with the luminal (apical) membranes. This increases the permeability of the collecting duct to water, allowing more water to be reabsorbed back into the blood.

Award 1 mark for the complete, correct homeostatic response mechanism involving osmoreceptors shrinking, ADH release from the posterior pituitary, and aquaporin insertion on luminal membranes.

A test cross involves crossing an individual of unknown or heterozygous genotype with a homozygous recessive individual (\(aabb\)). Since genes A and B assort independently, the heterozygous parent (\(AaBb\)) produces four types of gametes in equal proportions: \(AB\), \(Ab\), \(aB\), and \(ab\). The homozygous recessive parent only produces \(ab\) gametes. Combining these gametes results in four offspring genotypes (\(AaBb\), \(Aabb\), \(aaBb\), \(aabb\)) in equal proportions, giving a 1:1:1:1 phenotypic ratio.

Award 1 mark for calculating the correct phenotypic ratio of 1:1:1:1 for a dihybrid test cross with independent assortment.

When lactose is present, some of it is converted to allolactose, which acts as an inducer. Allolactose binds to the repressor protein, altering its tertiary structure. This conformational change prevents the repressor from binding to the operator region. Consequently, RNA polymerase can bind to the promoter and transcribe the structural genes (\(lacZ\), \(lacY\), and \(lacA\)).

Award 1 mark for identifying the correct interaction of allolactose with the repressor protein and its effect on transcription.

The apoplast pathway involves the movement of water and solutes through the non-living parts of the plant, mainly the cell walls and intercellular spaces. However, when the water reaches the endodermis, the waterproof Casparian strip (composed of suberin) blocks the apoplast pathway. Therefore, water and solutes must cross the selectively permeable cell surface membrane of the endodermal cells to enter the symplast pathway, allowing the plant to regulate which substances enter the xylem.

Award 1 mark for selecting option A. Option B is incorrect because plasmodesmata are part of the symplast pathway. Option C is incorrect because vacuoles are part of the vacuolar pathway. Option D is incorrect because the Casparian strip is a hydrophobic barrier that prevents free apoplastic movement into the vascular bundle.

When the water potential of the blood decreases, water leaves the osmoreceptors in the hypothalamus by osmosis, causing them to shrink. This triggers the posterior pituitary gland to release more antidiuretic hormone (ADH) into the blood. ADH binds to receptors on the collecting duct cells, initiating a cascade that causes vesicles containing aquaporins to fuse with the luminal cell surface membrane. This increases the water permeability of the collecting duct, allowing more water to be reabsorbed into the blood.

Award 1 mark for selecting option B. Osmoreceptors shrink when water potential decreases (reject swell). ADH is released from the posterior pituitary (reject anterior). More ADH leads to increased aquaporins in the luminal membrane (reject removal of aquaporins).

In cyclic photophosphorylation, excited electrons from photosystem I (PSI) are passed to an electron acceptor and then recycled back to the electron transport chain, which pumps protons to generate ATP via chemiosmosis, eventually returning the electrons to PSI. No reduced NADP or oxygen is produced. In non-cyclic photophosphorylation, electrons from PSI are used to reduce NADP, and electrons are not returned to PSI (they are replaced by electrons from PSII).

Award 1 mark for selecting option C. Excitation of PSI and ATP synthesis occur in both cyclic and non-cyclic photophosphorylation. Photolysis only occurs in non-cyclic. Thus, only C is unique to cyclic.

The parental genotype is \(C^R C^W Bb\). Crossing the alleles for flower color: \(C^R C^W \times C^R C^W \rightarrow 1/4\) Red (\(C^R C^R\)), \(1/2\) Pink (\(C^R C^W\)), \(1/4\) White (\(C^W C^W\)). Crossing the alleles for leaf shape: \(Bb \times Bb \rightarrow 3/4\) Broad (\(B-\)), \(1/4\) Narrow (\(bb\)). Using the product rule for independent assortment: Pink, broad: \(1/2 \times 3/4 = 3/8 = 6/16\); Red, broad: \(1/4 \times 3/4 = 3/16\); White, broad: \(1/4 \times 3/4 = 3/16\); Pink, narrow: \(1/2 \times 1/4 = 1/8 = 2/16\); Red, narrow: \(1/4 \times 1/4 = 1/16\); White, narrow: \(1/4 \times 1/4 = 1/16\). This gives the ratio: 6 : 3 : 3 : 2 : 1 : 1.

Award 1 mark for selecting option B. Standard dihybrid crosses (9:3:3:1) assume complete dominance for both genes. Because flower color is codominant / incompletely dominant, the heterozygote has a distinct phenotype, splitting the 12 (9+3) broad and 4 (3+1) narrow categories into distinct ratios.

When an action potential reaches the presynaptic knob, voltage-gated calcium channels open, and calcium ions (\(\text{Ca}^{2+}\)) diffuse into the presynaptic neurone. This influx causes synaptic vesicles containing acetylcholine to move towards and fuse with the presynaptic membrane, releasing the neurotransmitter by exocytosis. Acetylcholine diffuses across the synaptic cleft and binds to ligand-gated receptor proteins on the postsynaptic membrane, causing sodium channels to open. Sodium ions (\(\text{Na}^+\)) diffuse into the postsynaptic neurone, causing depolarisation.

Award 1 mark for selecting option A. Candidates must know the specific locations and directions of movement for both ions. \(\text{Ca}^{2+}\) enters presynaptically, while \(\text{Na}^+\) enters postsynaptically.

The breakdown of the nuclear envelope occurs at the end of prophase (prometaphase). If the nuclear envelope cannot disassemble, the chromosomes remain trapped inside it and cannot align along the equator of the spindle (metaphase). Thus, the cell remains arrested in prophase. At this stage, the chromosomes have already condensed, and each consists of two sister chromatids held together by a centromere.

Award 1 mark for selecting option A. Cell arrest occurs at prophase because spindle fibers cannot interact with the centromeres without nuclear envelope breakdown. Chromosomes at this stage are fully condensed into sister chromatid pairs.

In collagen, every third amino acid is glycine, which is the smallest amino acid (having only a hydrogen atom as its R-group). This allows three polypeptide chains to pack tightly together to form a triple helix. Hemoglobin has no such regular repeating primary sequence. Collagen has a unique helical structure (not a standard alpha-helix) and is insoluble in water. Hemoglobin is a globular, soluble protein with alpha-helices and consists of four polypeptide chains (two alpha-globin, two beta-globin), which are not all identical.

Award 1 mark for selecting option A. Reject B because hemoglobin contains alpha-helices, not beta-sheets, and collagen has a unique tight triple-helix secondary/tertiary structure. Reject C because collagen has 3 chains and hemoglobin has 4 chains (which are not all identical). Reject D because collagen is insoluble and hemoglobin is soluble.

The operator is the binding site for the repressor protein. In a normal cell, when lactose is absent, the active repressor binds to the operator, blocking RNA polymerase from transcribing the structural genes. When lactose is present, allolactose binds to the repressor, changing its shape so it cannot bind to the operator, allowing transcription. If a mutation prevents the repressor from binding to the operator, there is nothing to block RNA polymerase, so transcription occurs continuously (constitutively) both in the presence and absence of lactose.

Award 1 mark for selecting option C. A mutation in the operator that prevents repressor binding eliminates the negative control mechanism, resulting in continuous (constitutive) expression of the operon structural genes.

(a) \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\)

\(\text{Image size} = 45\text{ mm} = 45\,000\ \mu\text{m}\)

\(\text{Magnification} = \frac{45\,000}{6.0} = \times 7500\)

(b) The resolution of a microscope is limited by the wavelength of the radiation used. Visible light has a relatively long wavelength (\(400\text{ to }700\text{ nm}\)), which limits the resolution of a light microscope to about \(200\text{ nm}\). An electron beam has a much shorter wavelength than light, allowing a much higher resolution (up to \(0.5\text{ nm}\)). Ribosomes (about \(20\text{ to }30\text{ nm}\)) and centrioles are much smaller than the limit of resolution of a light microscope, so they can only be resolved using an electron microscope.

(c) Comparative structural differences include:
- Nucleus / Membrane-bound organelles: Prokaryotic cells have no nucleus or membrane-bound organelles (such as mitochondria), whereas eukaryotic cells have a distinct nucleus and membrane-bound organelles.
- Ribosomes: Prokaryotic cells contain smaller 70S ribosomes, whereas eukaryotic cells contain larger 80S ribosomes in their cytoplasm.
- DNA: Prokaryotic DNA is circular, naked, and lies free in the cytoplasm (nucleoid), whereas eukaryotic DNA is linear and associated with histone proteins inside a nucleus.
- Cell Wall: Prokaryotic cells have a cell wall made of peptidoglycan (murein), whereas eukaryotic animal cells lack a cell wall entirely.

(a) [2 marks]
- 1 mark for correct conversion of units (\(45\text{ mm} = 45\,000\ \mu\text{m}\)) or correct rearrangement of the formula.
- 1 mark for correct calculation of magnification: \(\times 7500\) (accept \(7500\)).

(b) [3 marks]
- 1 mark for stating that resolution is limited by the wavelength of the radiation used.
- 1 mark for stating that light has a much longer wavelength than an electron beam (or that an electron beam has a much shorter wavelength).
- 1 mark for stating that the size of ribosomes/centrioles is below the resolution limit of a light microscope (approx. \(200\text{ nm}\)) but within that of an electron microscope.

(c) [5 marks]
- Max 5 marks for comparative statements (1 mark per point):
- Nucleus: Prokaryotic cell has no nucleus/nuclear membrane vs Eukaryotic cell has a nucleus.
- Organelles: Prokaryotes lack membrane-bound organelles vs Eukaryotes have membrane-bound organelles.
- Ribosomes: Prokaryotes have 70S ribosomes vs Eukaryotes have 80S ribosomes (in cytoplasm).
- DNA structure: Prokaryotic DNA is circular vs Eukaryotic DNA is linear.
- Association with proteins: Prokaryotic DNA is naked vs Eukaryotic DNA is associated with histones.
- Cell wall: Prokaryotes have peptidoglycan/murein cell wall vs Eukaryotic animal cells have no cell wall.

(a) Primary structure: The specific sequence of amino acids in a polypeptide chain, held together by covalent peptide bonds.
Secondary structure: The folding or coiling of the polypeptide chain into an \(\alpha\)-helix or \(\beta\)-pleated sheet, stabilized by hydrogen bonds between the \(\text{N-H}\) group of one amino acid and the \(\text{C=O}\) group of another.
Tertiary structure: The further folding of the polypeptide chain into a complex three-dimensional shape, stabilized by interactions between the R-groups, including hydrogen bonds, ionic bonds, disulfide bridges, and hydrophobic interactions.

(b) Differences between collagen and haemoglobin:
- Collagen consists of three polypeptide chains wound around each other to form a tight, triple helix, whereas haemoglobin consists of four polypeptide chains (two \(\alpha\) and two \(\beta\) subunits) arranged in a quaternary structure.
- Collagen forms long, thin, fibrous molecules, whereas haemoglobin is folded into a spherical, compact, globular shape.
- Collagen contains a highly repetitive amino acid sequence with glycine at every third position, whereas haemoglobin has a complex, non-repetitive sequence.
- Collagen has no prosthetic groups, whereas each of the four polypeptide chains in haemoglobin is associated with a haem prosthetic group containing an iron ion (\(\text{Fe}^{2+}\)).
- Collagen is insoluble in water, whereas haemoglobin is soluble in water because its hydrophilic R-groups face outwards while hydrophobic R-groups are tucked inside.

(c) The test used to confirm the presence of protein is the Biuret test. A positive result is indicated by a colour change from blue to purple (or lilac/violet).

(a) [4 marks]
- 1 mark for defining primary structure as sequence of amino acids linked by peptide bonds.
- 1 mark for secondary structure as \(\alpha\)-helix or \(\beta\)-pleated sheet, held by hydrogen bonds between NH and CO groups of the polypeptide backbone.
- 1 mark for tertiary structure as further folding into a 3D shape.
- 1 mark for identifying R-group interactions (hydrogen, ionic, disulfide, hydrophobic) stabilizing tertiary structure.

(b) [4 marks]
- Max 4 marks for comparative statements (1 mark each):
- Shape: collagen is fibrous/elongated vs haemoglobin is globular/spherical.
- Quaternary structure: collagen is a triple helix vs haemoglobin has 4 subunits (2 \(\alpha\), 2 \(\beta\)).
- Solubility: collagen is insoluble vs haemoglobin is soluble.
- Prosthetic groups: collagen lacks prosthetic groups vs haemoglobin contains 4 haem groups (containing \(\text{Fe}^{2+}\)).
- Amino acid repetition: collagen has glycine at every 3rd position vs haemoglobin does not have a simple repeating sequence.

(c) [2 marks]
- 1 mark for naming the Biuret test / Biuret reagent.
- 1 mark for describing the color change from blue to purple / lilac / violet (reject pink / red).

(a) Competitive inhibitors have a molecular shape similar to the substrate and bind directly to the active site, blocking the substrate. Non-competitive inhibitors bind to an allosteric site (a site away from the active site), which changes the shape of the active site so the substrate can no longer fit.
With competitive inhibitors, increasing the substrate concentration increases the chance of a substrate molecule binding rather than an inhibitor. Therefore, at high substrate concentrations, the maximum rate of reaction (\(V_{\max}\)) can still be reached. With non-competitive inhibitors, increasing substrate concentration has no effect on the degree of inhibition, and \(V_{\max}\) is significantly reduced.

(b) Lock-and-key hypothesis states that the active site is fully complementary in shape to the substrate before binding occurs, like a key fitting into a lock. In contrast, the induced-fit hypothesis states that the active site is flexible and undergoes a conformational change (moulds around the substrate) when the substrate binds, putting strain on substrate bonds and lowering activation energy.

(c) The Michaelis-Menten constant (\(K_m\)) is the substrate concentration at which the rate of an enzyme-catalysed reaction is half of the maximum velocity (\(\frac{1}{2} V_{\max}\)). A high \(K_m\) value indicates a low affinity of the enzyme for its substrate, as a high concentration of substrate is required to half-saturate the enzyme.

(a) [5 marks]
- 1 mark for stating competitive inhibitor binds to the active site (due to similar shape to substrate).
- 1 mark for stating non-competitive inhibitor binds to an allosteric site, altering active site shape.
- 1 mark for explaining that increasing substrate concentration overcomes competitive inhibition / \(V_{\max}\) is eventually reached.
- 1 mark for explaining that increasing substrate concentration does not overcome non-competitive inhibition / \(V_{\max}\) is reduced.
- 1 mark for identifying that competitive inhibition is reversible by adding substrate, whereas non-competitive inhibition is not.

(b) [3 marks]
- 1 mark for describing lock-and-key: rigid active site, fully complementary to substrate before binding.
- 1 mark for describing induced-fit: active site is flexible and changes conformation upon substrate binding.
- 1 mark for explaining that the conformational change in induced-fit puts strain on chemical bonds in the substrate, lowering activation energy.

(c) [2 marks]
- 1 mark for definition: substrate concentration at half the maximum rate (\(\frac{1}{2} V_{\max}\)).
- 1 mark for stating that high \(K_m\) means low affinity (requires more substrate to reach \(\frac{1}{2} V_{\max}\)).

(a)(i) Cholesterol regulates membrane fluidity. At high temperatures, it stabilizes the membrane and prevents it from becoming too fluid; at low temperatures, it prevents phospholipids from packing too closely, avoiding rigidity. It also reduces the permeability of the membrane to small, water-soluble molecules.
(a)(ii) Glycoproteins act as receptor molecules for hormones and neurotransmitters, allowing cell signalling. They act as antigens for cell recognition by the immune system, and they help cells adhere to one another to form tissues.

(b) Active transport requires metabolic energy in the form of ATP, whereas facilitated diffusion is a passive process that relies solely on the kinetic energy of particles. Active transport moves substances against their concentration gradient (from a region of low concentration to high concentration), whereas facilitated diffusion moves substances down their concentration gradient. Furthermore, active transport utilizes carrier proteins (pumps), while facilitated diffusion utilizes both channel proteins and carrier proteins.

(c) When a plant cell is placed in a concentrated sucrose solution, water moves out of the cell down a water potential gradient (from high to low water potential) by osmosis. The protoplast shrinks and pulls away from the cell wall, resulting in plasmolysis (the cell becomes plasmolysed/flaccid).
When an animal cell is placed in pure water, water enters the cell down a water potential gradient by osmosis. Since animal cells lack a rigid cell wall, the cell swells and eventually bursts (lysis/haemolysis).

(a)(i) [2 marks]
- 1 mark for regulating membrane fluidity (prevents it from becoming too fluid at high temp or too rigid at low temp).
- 1 mark for mechanical stability / reducing permeability of polar/hydrophilic molecules.

(a)(ii) [2 marks]
- 1 mark for cell-to-cell signaling / receptor for hormones/transmitters.
- 1 mark for cell-to-cell adhesion OR cell recognition / acting as antigens.

(b) [3 marks]
- Max 3 marks for comparative points:
- Energy: active transport requires ATP, facilitated diffusion does not / is passive.
- Gradient: active transport is against concentration gradient, facilitated diffusion is down concentration gradient.
- Proteins: active transport requires carrier proteins only, facilitated diffusion can use channel or carrier proteins.

(c) [3 marks]
- 1 mark for plant cell: water leaves by osmosis, protoplast shrinks / plasmolysis occurs (cell becomes plasmolysed).
- 1 mark for animal cell: water enters by osmosis, cell swells and bursts / lyses.
- 1 mark for explanation: animal cell has no cell wall to prevent bursting, whereas plant cell has a rigid cell wall that exerts pressure potential (preventing bursting but allowing plasmolysis when water leaves).

(a) Xylem vessel adaptations:
- Hollow tubes: Cells are dead and lack cytoplasm, nuclei, and end walls. This allows an uninterrupted, continuous column of water to flow with minimal resistance.
- Lignification: The cell walls are thickened with lignin, which provides high tensile strength, preventing the vessels from collapsing inward under the tension created by transpiration. Lignin also provides structural support to the whole plant.
- Pits: Non-lignified areas of the wall (pits) allow the lateral movement of water into adjacent vessels or surrounding tissues.

(b) Water enters root hair cells from the soil via osmosis down a water potential gradient. Once inside, water crosses the root cortex to the endodermis using two pathways:
- The apoplast pathway: water travels through the non-living cell walls and intercellular spaces.
- The symplast pathway: water travels through the living cytoplasm and plasmodesmata.
At the endodermis, the apoplast pathway is blocked by the Casparian strip (a waterproof band of suberin). Water is forced to enter the symplast pathway by crossing the cell surface membranes of endodermal cells. Active transport of ions into the xylem lowering its water potential then drives the movement of water into the xylem vessels via osmosis.

(c) Features of xerophytic leaves:
- Sunken stomata (stomata in pits): Traps a layer of moist air outside the stomata, reducing the water potential gradient between the inside of the leaf and the outside air, thus slowing down transpiration.
- Rolled leaves: Creates a microenvironment inside the roll that traps water vapour, keeping the humidity high around the stomata and reducing the water potential gradient.
- Thick, waxy cuticle: Creates an impermeable barrier on the upper epidermis, increasing the diffusion distance for water vapor and preventing cuticular transpiration.

(a) [4 marks]
- Max 4 marks for adaptations paired with functions:
- No cytoplasm / no organelles / dead cells: forms a hollow tube to allow continuous water column / low resistance.
- No end walls: forms continuous tubes for bulk flow of water.
- Lignified walls: prevents collapse under tension / negative pressure.
- Lignin: provides mechanical support to the plant.
- Pits: allows lateral movement of water to bypass blockages / supply surrounding cells.

(b) [4 marks]
- 1 mark for entry into root hair cells by osmosis down water potential gradient.
- 1 mark for defining apoplast pathway (through cell walls) and symplast pathway (through cytoplasm/plasmodesmata).
- 1 mark for stating that the Casparian strip (suberin) blocks the apoplast pathway at the endodermis.
- 1 mark for stating that water is forced into the symplast pathway / crosses selectively permeable membrane of endodermal cells into the xylem.

(c) [2 marks]
- Max 2 marks (1 mark for each adaptation paired with correct explanation):
- Sunken stomata / stomata in pits: traps moist air, reducing water potential gradient.
- Rolled leaves: traps moist air, reducing water potential gradient.
- Thick waxy cuticle: increases diffusion distance / waterproof barrier, reducing transpiration.
- Hairs (trichomes): traps moist air, reducing water potential gradient.
- Reduced leaf surface area (spines): reduces surface area available for evaporation.

(a) Carbon dioxide (\(CO_2\)) is transported in three main ways:
- Dissolved directly in blood plasma (about 5%).
- Bound to amine groups of haemoglobin to form carbaminohemoglobin (about 10%).
- As hydrogencarbonate ions (\(HCO_3^-\)) in the plasma (about 85%).
Inside red blood cells, \(CO_2\) reacts with water to form carbonic acid (\(H_2CO_3\)), a reaction catalysed by the enzyme carbonic anhydrase. Carbonic acid then dissociates into hydrogen ions (\(H^+\)) and hydrogencarbonate ions (\(HCO_3^-\)). The \(HCO_3^-\)

(a) [5 marks]
- 1 mark for stating that carbon dioxide is transported as hydrogencarbonate ions dissolved in plasma, dissolved carbon dioxide, and carbaminohemoglobin.
- 1 mark for writing the reaction: \(CO_2 + H_2O \rightarrow H_2CO_3\) (carbonic acid).
- 1 mark for identifying that carbonic anhydrase is the enzyme inside red blood cells that catalyses this reaction.
- 1 mark for stating that carbonic acid dissociates into \(H^+\) and \(HCO_3^-\).
- 1 mark for mentioning the chloride shift (chloride ions enter red blood cells as hydrogencarbonate ions leave).

(b) [3 marks]
- 1 mark for stating that high \(pCO_2\) / high \(H^+\) concentration leads to hydrogen ions binding to haemoglobin (forming haemoglobinic acid).
- 1 mark for explaining that this causes a conformational change in haemoglobin, lowering its affinity for oxygen.
- 1 mark for stating that the oxygen-haemoglobin dissociation curve shifts to the right, facilitating oxygen unloading/release at respiring tissues.

(c) [2 marks]
- 1 mark for describing correct distribution: cartilage is in trachea/bronchi but absent in bronchioles; smooth muscle is in trachea/bronchi/bronchioles.
- 1 mark for stating correct functions: cartilage prevents collapse of airways / keeps them open; smooth muscle contracts/relaxes to adjust airway diameter / control airflow.

(a) (i)
- 0.8 mol dm\(^{-3}\): 16 cm\(^3\) S, 4 cm\(^3\) W
- 0.6 mol dm\(^{-3}\): 12 cm\(^3\) S, 8 cm\(^3\) W
- 0.4 mol dm\(^{-3}\): 8 cm\(^3\) S, 12 cm\(^3\) W
- 0.2 mol dm\(^{-3}\): 4 cm\(^3\) S, 16 cm\(^3\) W

(ii) Results table constructed with appropriate column headings (including units):
- Concentration of sucrose / mol dm\(^{-3}\)
- Time taken, \(t\) / s
- Rate of reaction (\(1/t\)) / s\(^{-1}\)
Raw data should show that at higher sucrose concentrations, respiration is faster, resulting in smaller times (e.g., 1.0 mol dm\(^{-3}\) rises fastest) and higher rate values (calculated to 3 d.p.).

(b) (i) Graph features:
- x-axis: Copper sulfate concentration / mmol dm\(^{-3}\), y-axis: Time taken for starch to be completely hydrolysed / s.
- Linear scales covering at least 50% of the grid.
- Points plotted accurately and joined with straight lines or a smooth curve.

(ii) Description: As copper sulfate concentration increases, the time taken for starch hydrolysis increases, meaning the rate of amylase activity decreases.
Explanation: Copper ions act as inhibitors. They bind to the amylase enzyme (at the active site or an allosteric site), altering its tertiary structure and the conformation of the active site. This prevents starch substrates from binding, resulting in fewer enzyme-substrate complexes.

(c) (i) Key errors:
1. Non-uniform size or volume of yeast-alginate beads, causing differences in yeast cell count and surface area.
2. Inability to maintain constant temperature during the trials.

(ii) Modifications:
1. Use a syringe pump or calibrated dropper to prepare beads of uniform size, or screen them using a caliper.
2. Place the tubes in a thermostatically controlled water bath to keep temperature constant.

Maximum 20 marks total:

(a)(i) [3 marks]
- Correct volume of S for all concentrations (16, 12, 8, 4 cm\(^3\)) [1]
- Correct volume of W for all concentrations (4, 8, 12, 16 cm\(^3\)) [1]
- Table headers clear with appropriate units (e.g., Volume of S / cm\(^3\)) [1]

(a)(ii) [6 marks]
- Organized table with appropriate column headings and units [1]
- Raw times recorded consistently in whole seconds [1]
- Rates calculated correctly as \(1/t\) to 3 decimal places [1]
- All 5 concentrations tested and recorded [1]
- Correct trend: higher sucrose concentration gives shorter times / higher rates [1]
- At least one repeat trial shown or standard layout showing replicate clarity [1]

(b)(i) [5 marks]
- x-axis: Copper sulfate concentration / mmol dm\(^{-3}\) AND y-axis: Time taken / s [1]
- Scale: Linear and occupies at least half of the grid in both directions [1]
- Plotting: All 6 points plotted accurately to within half a small square [1]
- Line: Points joined from point-to-point with straight ruled lines or a smooth curve of best fit, no extrapolation [1]
- Quality: Thin, sharp lines, with no double lines or excessively thick points [1]

(b)(ii) [2 marks]
- Correct description: increasing inhibitor concentration increases time taken / decreases rate of reaction [1]
- Correct explanation: inhibitor binds to enzyme (amylase) causing a change in active site shape, reducing enzyme-substrate complexes [1]

(c)(i) [2 marks]
- Identify variation in bead size / mass / cell number [1]
- Identify lack of temperature control [1]

(c)(ii) [2 marks]
- Suggest using a syringe/calibrator to make uniform beads [1]
- Suggest using a thermostatically controlled water bath to maintain constant temperature [1]

(a) (i) Low-power plan diagram of the stem sector:
- Drawn with clean, continuous pencil lines; no individual cells visible.
- Represents the arrangement of tissues: epidermis outer layer, cortex, vascular bundles in a ring, and central pith.
- One vascular bundle is shown completely, with adjacent ones partially shown.
- Labels correctly indicate xylem (on the inner aspect of the bundle) and phloem (on the outer aspect).

(ii) Working:
- 4 divisions of the stage micrometer = \(4 \times 0.1\text{ mm} = 0.4\text{ mm}\)
- \(0.4\text{ mm} = 400\ \mu\text{m}\)
- 40 eyepiece divisions = \(400\ \mu\text{m}\)
- 1 eyepiece division = \(400 / 40 = 10\ \mu\text{m}\)

(iii) Working:
- Diameter = \(12\text{ eyepiece units} \times 10\ \mu\text{m / unit} = 120\ \mu\text{m}\)

(b) (i) Drawing of three adjacent cortical parenchyma cells:
- Exactly three cells are drawn, representing their relative positions.
- Cell walls are drawn as double lines to represent thickness.
- Cells show realistic, non-spherical shapes (polygonal or rounded) with intercellular spaces clearly depicted.
- No shading or stippling is used.

(ii) Comparison Table:
| Feature | Stem (Figure 2.1) | Root (Figure 2.2) |
| :--- | :--- | :--- |
| Distribution of vascular tissue | Separated into distinct vascular bundles arranged in a peripheral ring. | Grouped into a single, central vascular cylinder. |
| Arrangement of xylem | Positioned on the inner side of each vascular bundle. | Located in the center, forming a star/cross shape. |
| Arrangement of phloem | Positioned on the outer side of each vascular bundle. | Located in groups between the arms of the star-shaped xylem. |
| Presence of central pith | Pith is present in the center. | No pith is present (the center is occupied by xylem). |
| Endodermis | No distinct endodermis layer surrounding bundles. | A distinct, continuous endodermis ring surrounds the central vascular bundle. |

Maximum 20 marks total:

(a)(i) [6 marks]
- Size: Large diagram occupying at least half of the space provided [1]
- Quality: Clean, continuous lines, no shading, no sketching, no individual cells drawn [1]
- Layout: Shows correct sector of the stem with one complete vascular bundle and parts of adjacent bundles [1]
- Proportions: Cortex width, vascular bundle size, and pith ratio are realistic [1]
- Structures: Shows separate regions for epidermis, cortex, xylem, and phloem [1]
- Labeling: Correctly labels 'xylem' (pointing to the inner region of the bundle) and 'phloem' (pointing to the outer region of the bundle) [1]

(a)(ii) [3 marks]
- Convert stage micrometer divisions to mm/\(\mu\text{m}\): \(4 \times 0.1 = 0.4\text{ mm}\) or \(400\ \mu\text{m}\) [1]
- Shows division of length by 40 eyepiece units: \(400 / 40\) [1]
- Correct final answer: \(10\ \mu\text{m}\) [1]

(a)(iii) [2 marks]
- Shows multiplication of eyepiece units by the calibration factor: \(12 \times 10\) [1]
- Correct final answer: \(120\ \mu\text{m}\) [1]

(b)(i) [4 marks]
- Count: Exactly three adjacent cells drawn [1]
- Detail: Double lines representing cell walls drawn consistently for all three cells [1]
- Shape: Realistic, non-circular polygonal parenchyma shapes with correct cell junctions/intercellular spaces [1]
- Quality: Thin, sharp lines, no shading/stippling, and occupying a reasonable size [1]

(b)(ii) [5 marks]
- Format: Table constructed with headings (Feature, Stem/Fig 2.1, Root/Fig 2.2) [1]
- Comparison 1: Vascular tissue arrangement (bundles in ring in stem vs central star/core in root) [1]
- Comparison 2: Location of xylem (inner side of separate bundles in stem vs center of root/star shape) [1]
- Comparison 3: Location of phloem (outer side of separate bundles in stem vs between xylem arms in root) [1]
- Comparison 4: Pith (present in stem vs absent/xylem in center of root) OR Endodermis (absent/unclear in stem vs present as a distinct ring in root) [1]

(a) Accessory pigments (such as chlorophyll b and carotenoids) absorb light wavelengths that chlorophyll a cannot absorb well, broadening the action spectrum of photosynthesis. They are organized into light-harvesting systems (antenna complexes) in the thylakoid membrane. When these pigments absorb photons of light, they become excited and pass this excitation energy from one pigment molecule to another via resonance energy transfer. Eventually, this energy reaches the reaction centre, exciting a pair of specialised chlorophyll a molecules. (b)(i) Blocking electron flow from PSII to plastoquinone stops the electron transport chain (ETC). This prevents the active pumping of protons (H+ ions) across the thylakoid membrane into the thylakoid lumen, meaning no proton gradient is established. Without the proton motive force, ATP synthase cannot synthesise ATP via chemiosmosis. Additionally, because electrons cannot reach photosystem I (PSI), NADP reductase cannot reduce NADP+ to reduced NADP. (ii) In the Calvin cycle, GP requires both ATP and reduced NADP from the light-dependent reaction to be converted into triose phosphate (TP). In the presence of herbicide X, ATP and reduced NADP levels drop. Consequently, the conversion of GP to TP stops, causing GP to accumulate. Since TP is needed to regenerate RuBP (which also requires ATP), RuBP cannot be regenerated, leading to a rapid decrease in RuBP levels.

(a) [Max 3 marks] 1. Accessory pigments (e.g. chlorophyll b/carotenoids) absorb light of different wavelengths (not absorbed by chlorophyll a); 2. Broadens the absorption spectrum / increases light-absorption efficiency; 3. Energy is passed from pigment to pigment via resonance transfer / energy transfer; 4. Energy is funnelled towards the primary pigment/reaction centre/chlorophyll a. (b)(i) [Max 3 marks] 1. No electron transport along the electron transport chain (ETC); 2. No/reduced pumping of protons (H+) into the thylakoid lumen; 3. No/reduced proton gradient / proton motive force; 4. No phosphorylation of ADP to ATP / no ATP synthesis (via ATP synthase); 5. No electrons reach PSI to reduce NADP+ to reduced NADP. (b)(ii) [Max 4 marks] 1. GP requires ATP and reduced NADP to be reduced/converted to triose phosphate (TP); 2. (With herbicide X) GP is not converted to TP, so GP levels rise/accumulate; 3. TP is required for the regeneration of RuBP; 4. Regeneration of RuBP also requires ATP (which is unavailable); 5. RuBP levels decrease as it continues to combine with CO2 but is not regenerated.

(a) Both pathways begin with glycolysis, producing pyruvate and a net of 2 ATP molecules, and both regenerate NAD+ in the cytoplasm. However, in yeast, anaerobic respiration (ethanol fermentation) is a two-step process where pyruvate is first decarboxylated by pyruvate decarboxylase to form ethanal, releasing CO2, and then ethanal is reduced to ethanol by alcohol dehydrogenase. In mammalian muscle cells (lactate fermentation), it is a one-step process where pyruvate is directly reduced to lactate by lactate dehydrogenase, and no CO2 is produced. Ethanol fermentation in yeast is irreversible, whereas lactate fermentation is reversible. (b) Pyruvate is converted to ethanal in a decarboxylation reaction, releasing carbon dioxide (CO2). This reaction is catalysed by the enzyme pyruvate decarboxylase. Ethanal is then reduced to ethanol by the enzyme alcohol dehydrogenase. In this step, reduced NAD (NADH) acts as the reducing agent, donating hydrogen atoms (electrons and protons) to ethanal, becoming oxidized to NAD+. This pathway is essential because glycolysis is the only source of ATP under anaerobic conditions. Glycolysis requires a continuous supply of oxidized NAD (NAD+) to accept hydrogen atoms during the triose phosphate to pyruvate conversion. Without oxygen, the link reaction, Krebs cycle, and oxidative phosphorylation cannot occur to regenerate NAD+. The reduction of ethanal to ethanol regenerates NAD+, allowing glycolysis to continue and the cell to survive by producing 2 ATP per glucose molecule.

(a) [Max 4 marks] 1. Both pathways start with glycolysis / both produce pyruvate / both occur in cytoplasm; 2. Yeast fermentation produces ethanol AND CO2, whereas mammalian fermentation produces lactate only / no CO2 in mammal; 3. Yeast pathway involves decarboxylation (pyruvate to ethanal), mammalian does not; 4. Yeast pathway involves two enzymes (pyruvate decarboxylase and alcohol dehydrogenase) whereas mammalian involves one (lactate dehydrogenase); 5. Yeast pathway is irreversible whereas mammalian pathway is reversible. (b) [Max 6 marks] [Process of ethanol production (max 3)]: 1. Pyruvate is decarboxylated / carbon dioxide (CO2) is released; 2. Catalysed by pyruvate decarboxylase to form ethanal; 3. Ethanal is reduced to ethanol; 4. Catalysed by alcohol dehydrogenase; 5. Reduced NAD (NADH) is oxidised to NAD+ (by donating hydrogen/electrons to ethanal). [Significance (max 3)]: 6. Regenerates oxidized NAD / NAD+; 7. Allows glycolysis to continue / prevents glycolysis from stopping; 8. Glycolysis is the only source of ATP under anaerobic conditions (2 ATP net per glucose); 9. Prevents cell death by maintaining essential cellular processes requiring ATP.

(a) In the absence of lactose, the regulatory gene (lacI) is transcribed and translated to produce an active repressor protein. This repressor protein binds to the operator region of the operon, physically blocking RNA polymerase from binding to the promoter and transcribing the structural genes (lacZ, lacY, and lacA). When lactose is present, some of it is converted to allolactose, which acts as an inducer. Allolactose binds to the allosteric site of the repressor protein. This binding causes a conformational change in the tertiary structure of the repressor, making it inactive so it can no longer bind to the operator. As a result, the operator is free, and RNA polymerase can bind to the promoter and transcribe the structural genes. (b) The lac repressor system in prokaryotes acts primarily as a negative control mechanism where the regulatory protein (repressor) physically blocks transcription unless inhibited by an inducer (lactose/allolactose). In contrast, eukaryotic transcription factors are highly complex and can act as both activators (promoting transcription) and repressors. Eukaryotic transcription factors bind to specific promoter regions, enhancers, or silencers, often recruiting other co-activators or RNA polymerase II, or modifying chromatin structure (such as histone acetylation or DNA methylation) to make the DNA accessible. Furthermore, eukaryotic gene expression is not organized into operons; genes are transcribed individually rather than as a single polycistronic mRNA, requiring coordinated transcription factors across different chromosomes.

(a) [Max 5 marks] 1. Regulatory gene (lacI) produces active repressor protein; 2. Lactose / allolactose acts as an inducer; 3. Inducer binds to the repressor protein at its allosteric site; 4. Causes a conformational change in the repressor shape; 5. Repressor can no longer bind to the operator; 6. RNA polymerase binds to the promoter; 7. Transcription of structural genes (lacZ, lacY, lacA) occurs, producing polycistronic mRNA. (b) [Max 5 marks] 1. Prokaryotes (lac operon) use a single repressor blocking a single operator, while eukaryotes use multiple transcription factors; 2. Eukaryotes use activators (which increase transcription) and repressors, whereas the lac system is primarily negative control/repressive; 3. Transcription factors in eukaryotes can bind to enhancers or silencers far from the promoter, whereas the lac repressor binds adjacent to the promoter (at the operator); 4. Eukaryotic transcription factors often alter chromatin structure (histone modification/acetylation), whereas prokaryotes lack histones/nucleosomes; 5. Prokaryotes transcribe genes together as an operon (polycistronic), while eukaryotes transcribe genes individually (monocistronic) requiring coordinated transcription factors.

(a) Primers are short, single-stranded DNA sequences (usually 18–25 nucleotides long) that are complementary to the regions flanking the target DNA sequence to be amplified. They provide a free 3'-hydroxyl group, which is required for DNA polymerase to begin synthesizing the new strand, and they determine the specificity of the amplification. Taq polymerase is a thermostable DNA polymerase isolated from the bacterium Thermus aquaticus. It synthesises the new DNA strand by adding complementary nucleotides to the 3' end of the primer. Taq polymerase is used because it does not denature at the high temperatures (95 degrees C) used to separate the DNA strands during the denaturation step of PCR. This eliminates the need to add new enzyme after each cycle, allowing the process to be automated in a thermocycler. (b) DNA has a negative charge due to the phosphate groups in its sugar-phosphate backbone. When placed in an agarose or polyacrylamide gel matrix and subjected to an electrical field, the DNA molecules migrate towards the positive electrode (anode). The gel matrix acts as a molecular sieve. Smaller DNA fragments move more easily through the pores of the gel, so they migrate faster and further than larger, heavier fragments. This separates the DNA fragments based on their size (length in base pairs). To visualise the separated fragments, a fluorescent dye (such as ethidium bromide or GelRed) that intercalates between DNA base pairs is added to the gel. When exposed to ultraviolet (UV) light, the dye fluoresces, revealing distinct bands corresponding to the separated DNA fragments.

(a) [Max 5 marks] 1. Primers are short, single-stranded DNA sequences complementary to the start/end of the target DNA; 2. Primers bind/anneal to target DNA and provide a starting point / free 3'-OH group for DNA polymerase; 3. Taq polymerase synthesises new DNA strands by adding complementary nucleotides; 4. Taq polymerase is thermostable / does not denature at high temperatures (e.g. 95 degrees C); 5. This means the enzyme does not need to be replaced after every heating cycle / allows automation of PCR. (b) [Max 5 marks] 1. DNA is negatively charged due to its phosphate groups; 2. DNA fragments migrate towards the positive electrode / anode (in an electric field); 3. Agarose gel acts as a molecular sieve/mesh; 4. Smaller fragments move faster / migrate further than larger fragments (separation by size/molecular mass); 5. Visualised using a fluorescent dye (e.g. ethidium bromide / GelRed) that binds/intercalates to DNA; 6. Bands are revealed under ultraviolet (UV) light.

(a) When blood glucose concentration rises above the set point, this change is detected by the beta-cells in the islets of Langerhans in the pancreas. The beta-cells act as receptors and coordinators; glucose enters the beta-cells via GLUT2 transporters, is metabolised to produce ATP, which closes ATP-sensitive K+ channels, leading to depolarization of the membrane. This opens voltage-gated Ca2+ channels, causing an influx of calcium ions, which triggers the exocytosis of insulin stored in vesicles into the blood. (b) Insulin travels in the blood and binds to specific tyrosine kinase receptors on the cell surface membranes of target cells (such as muscle and liver cells). This binding activates the receptor, initiating an intracellular signalling cascade via secondary messengers. This cascade triggers vesicles containing glucose transporter proteins (specifically GLUT4) to move to and fuse with the cell surface membrane. This increases the permeability of the cell membrane to glucose, allowing more glucose to enter the cell by facilitated diffusion. Once inside the cell, insulin activates key enzymes: Glucokinase phosphorylates glucose to glucose-6-phosphate, maintaining a steep concentration gradient for glucose entry. Glycogen synthase is activated to convert glucose to glycogen for storage (glycogenesis). Enzymes involved in glycolysis are stimulated to increase glucose breakdown. Enzymes for gluconeogenesis are inhibited to prevent the synthesis of new glucose.

(a) [Max 3 marks] 1. beta-cells act as receptors/detectors for high blood glucose; 2. Glucose enters beta-cells (via GLUT2) and is respired to produce ATP; 3. ATP closes K+ channels, causing depolarization and opening of Ca2+ channels / entry of Ca2+; 4. Stimulates exocytosis of vesicles containing insulin into the bloodstream. (b) [Max 7 marks] 1. Insulin binds to specific (tyrosine kinase) receptors on the cell surface membrane of target cells; 2. Triggers an intracellular signalling cascade / second messenger activation; 3. Vesicles containing GLUT4 glucose transporter proteins move towards the cell surface membrane; 4. Vesicles fuse with the cell surface membrane, increasing the number of GLUT4 transporters; 5. Increases rate of facilitated diffusion of glucose into the cell; 6. Insulin activates glucokinase/hexokinase to phosphorylate glucose (trapping it in the cell/maintaining concentration gradient); 7. Insulin activates glycogen synthase to convert glucose to glycogen (glycogenesis); 8. Accept reference to inhibition of enzymes for glycogenolysis or gluconeogenesis.

(a) The resting potential (around -70 mV) is established and maintained by the active transport of ions and the selective permeability of the neurone membrane. The sodium-potassium pump actively pumps three sodium ions (Na+) out of the axon for every two potassium ions (K+) pumped in, consuming ATP. This creates electrochemical gradients for both ions. The membrane is much more permeable to potassium ions than to sodium ions because there are many open 'leak' potassium channels and very few open sodium channels. Consequently, potassium ions diffuse out of the axon down their concentration gradient much faster than sodium ions can diffuse in. This net loss of positive charges from inside the axon results in a negative potential difference across the membrane, polarising it. (b) When an action potential arrives at the presynaptic membrane, it causes depolarization, which opens voltage-gated calcium channels. Calcium ions (Ca2+) diffuse rapidly into the presynaptic knob down their electrochemical gradient. The influx of Ca2+ causes synaptic vesicles containing the neurotransmitter acetylcholine (ACh) to move towards and fuse with the presynaptic membrane, releasing ACh into the synaptic cleft by exocytosis. ACh diffuses across the synaptic cleft (a short distance of about 20 nm) and binds to specific ligand-gated receptor proteins on the postsynaptic membrane. This binding causes a conformational change in the receptor proteins, opening sodium ion channels. Sodium ions (Na+) diffuse into the postsynaptic neurone, depolarising the membrane (producing an excitatory postsynaptic potential, EPSP). If this depolarization exceeds the threshold potential, voltage-gated sodium channels open, generating a new action potential in the postsynaptic neurone.

(a) [Max 4 marks] 1. Sodium-potassium pump actively pumps 3 Na+ out and 2 K+ into the axon using ATP; 2. Creates concentration gradients for both ions (Na+ high outside, K+ high inside); 3. Membrane is more permeable to K+ than to Na+ / there are more open K+ leak channels; 4. K+ diffuses out of the axon down its concentration gradient; 5. Inside becomes negatively charged relative to the outside / potential difference of -70 mV established. (b) [Max 6 marks] 1. Action potential depolarises presynaptic membrane, opening voltage-gated calcium channels; 2. Calcium ions (Ca2+) diffuse into the presynaptic knob; 3. Calcium ions cause synaptic vesicles to fuse with the presynaptic membrane; 4. Acetylcholine (ACh) is released into the synaptic cleft by exocytosis; 5. ACh diffuses across the synaptic cleft and binds to ligand-gated receptors on the postsynaptic membrane; 6. Sodium ion channels open, and Na+ diffuses into the postsynaptic neurone; 7. Depolarisation / excitatory postsynaptic potential (EPSP) is generated; 8. If threshold is reached, an action potential is initiated in the postsynaptic neurone.

(a) According to the biological species concept, a species is a group of organisms that can interbreed with one another to produce fertile offspring, and are reproductively isolated from other such groups. This definition is difficult to apply to organisms that reproduce exclusively non-sexually (asexually), such as bacteria. It is also impossible to apply directly to extinct organisms known only from fossils. Additionally, many plants (and some animals) can form viable, fertile hybrids with other distinct species (hybridisation), blurring species boundaries. (b) Geographic isolation prevents gene flow between the coastal and inland populations of wild radish. The two environments have different selection pressures; the coastal environment has high salinity, wind exposure, and different soil nutrients, while the inland environment has low salinity and different competitors. In each population, random mutations occur, creating new alleles. In the coastal environment, alleles that confer salt tolerance (e.g., efficient ion pumps in roots) provide a selective advantage. Plants with these alleles are more likely to survive, reproduce, and pass these advantageous alleles to their offspring. Over generations, the frequency of these salt-tolerant alleles increases in the coastal population. Conversely, in the inland population, different alleles are selected for. Over time, the genetic composition of the two populations diverges significantly. This genetic divergence leads to reproductive isolating mechanisms, which prevent the two populations from interbreeding even if they meet again. These could include: Ecological/habitat isolation (flowering at different times or in different soil conditions), Temporal isolation (coastal plants might adapt to flower earlier or later to avoid dry seasons), Mechanical isolation (changes in flower structure/shape that adapt them to different pollinators), Post-zygotic isolation (genetic incompatibility meaning hybrid seeds are non-viable or sterile). Eventually, the populations become separate species (allopatric speciation).

(a) [Max 3 marks] 1. Species: A group of organisms that can interbreed to produce fertile offspring; 2. Difficulty 1: Cannot be applied to organisms that reproduce asexually; 3. Difficulty 2: Cannot be applied to fossils/extinct organisms; 4. Difficulty 3: Some distinct species can hybridise and produce fertile offspring (especially in plants). (b) [Max 7 marks] 1. Geographical isolation prevents gene flow / genetic exchange between the two populations; 2. Different environments exert different selection pressures (high salinity vs low salinity); 3. Mutation occurs in both populations, introducing new alleles; 4. Natural selection occurs: plants with alleles conferring survival advantage (e.g. salt tolerance in coast) are more likely to survive and reproduce; 5. These advantageous alleles are passed to the next generation, changing allele frequencies over time; 6. Genetic divergence occurs between the two populations; 7. Development of pre-zygotic isolating mechanisms (e.g. temporal isolation/flowering at different times, mechanical/pollinator isolation); 8. Development of post-zygotic isolating mechanisms (e.g. hybrid inviability / embryo cannot develop); 9. Leads to allopatric speciation (populations can no longer interbreed to produce fertile offspring).

(a) Species richness is the total number of different species present in a particular community or habitat. Species evenness is a measure of the relative abundance of the individuals of each species present in the habitat. Simpson’s Index of Diversity (D) is calculated using the formula D = 1 - sum(n/N)^2, where n is the number of individuals of a particular species and N is the total number of all individuals of all species. It is more useful than species richness alone because it takes into account both richness and evenness. A habitat with high species richness but dominated by just one species (low evenness) is highly vulnerable to environmental changes. Simpson’s Index correctly reflects this lower ecological stability, whereas species richness would falsely suggest high biodiversity. (b) Zoos and captive breeding programmes serve as vital ex situ conservation tools. Their primary roles include: Breeding endangered species in a controlled environment to increase population size; Rescuing species threatened by habitat loss, poaching, or disease in the wild; Conducting scientific research into species biology, veterinary medicine, and reproductive technologies (e.g., artificial insemination, IVF); Educating the public to raise awareness and funds for in-situ conservation. However, they face major challenges: Genetic drift and inbreeding depression due to small founder populations. To combat this, studbooks are used to coordinate breeding and maintain genetic diversity. Behavioral issues: animals raised in captivity may not learn crucial survival skills, such as hunting, avoiding predators, or social interactions. Disease transmission in confined zoo environments. Reintroduction difficulties: successfully releasing captive-bred animals into the wild requires suitable, protected habitats, which may no longer exist, and animals may struggle to adapt to wild diets and pathogens.

(a) [Max 4 marks] 1. Species richness is the number of different species in an area; 2. Species evenness is the relative abundance / proportion of individuals of each species; 3. Simpson’s Index of Diversity (D) takes both species richness and evenness into account; 4. A high value of D indicates a more stable/diverse community (less dominated by a single species); 5. Richness alone does not reveal if a community is dominated by one/few species (which makes it vulnerable). (b) [Max 6 marks] [Roles (max 3)]: 1. Captive breeding to increase population numbers of endangered species; 2. Research into reproductive biology / husbandry / veterinary care; 3. Education of public / raising funds for in-situ projects; 4. Protection of species from immediate threats in the wild (poaching/disease/habitat loss). [Challenges (max 3)]: 5. Small gene pool leading to inbreeding depression / accumulation of harmful homozygous recessive alleles; 6. Loss of genetic diversity / genetic drift (managed by studbooks/transferring animals); 7. Loss of natural behaviours (e.g. hunting, predator avoidance, socialization); 8. Difficulty in successful reintroduction (e.g. lack of suitable/safe habitat, susceptibility to wild pathogens).

\( \textbf{Part (a)} \)
The cytochrome \(bc_1\) complex (Complex III) is located in the inner mitochondrial membrane (or cristae).

\( \textbf{Part (b)} \)
Stigmatellin prevents the transfer of electrons from ubiquinone to Complex III and subsequently to cytochrome c and Complex IV. As a result, carriers upstream of the block remain fully reduced and cannot be reoxidised, stopping the entry of electrons from reduced NAD and reduced FAD. Because oxygen is the terminal electron acceptor of the electron transport chain, a complete block in electron flow means oxygen cannot accept electrons and protons to form water (\(\text{O}_2 + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O}\)). Therefore, oxygen uptake drops to near zero.

\( \textbf{Part (c)} \)
ATP synthesis ceases because active proton pumping across the inner membrane by the electron transport chain complexes ceases when electron flow stops. This destroys the proton gradient (or electrochemical gradient / proton motive force) between the intermembrane space and the matrix. Consequently, there is no flow of protons down their concentration gradient through ATP synthase, meaning ADP cannot be phosphorylated to ATP.

Adding an uncoupling agent permeabilises the membrane to protons, which normally allows electron transport to run at maximum speed without being limited by a proton gradient. However, because the electron transport chain itself is physically blocked by stigmatellin at Complex III, electrons still cannot flow through the chain to oxygen. Therefore, adding an uncoupler will have no effect on oxygen uptake; it will remain extremely low.

\( \textbf{Part (a)} \)
1. Inner mitochondrial membrane / cristae ; [1]

\( \textbf{Part (b)} \)
1. Stigmatellin blocks electron transfer at Complex III / prevents transfer from ubiquinone to cytochrome c ;
2. Electron carriers upstream of Complex III remain reduced / cannot be reoxidised ;
3. Reduced NAD / reduced FAD cannot donate electrons / cannot be reoxidised ;
4. Oxygen acts as the final/terminal electron acceptor ;
5. No electrons reach oxygen to react with protons to form water / \(\text{O}_2 + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O}\) ; [Max 4]

\( \textbf{Part (c)} \)
ATP synthesis:
1. No electron flow means active pumping of protons into the intermembrane space stops ;
2. Proton / electrochemical gradient / proton motive force is lost ;
3. No protons pass down the gradient through ATP synthase ;
4. No phosphorylation of ADP (to ATP) ;
Uncoupling agent:
5. Oxygen uptake remains low / does not increase ;
6. Because the block in the electron transport chain is still present / electrons still cannot flow to oxygen ; [Max 5]

\( \textbf{Part (a)} \)
In the absence of gibberellin (GA), DELLA proteins are stable and act as transcriptional repressors. They bind directly to transcription factors (such as Phytochrome Interacting Factors, or PIFs), preventing them from binding to the promoter regions of target genes. As a result, the transcription of genes required for cell elongation and growth is switched off.

\( \textbf{Part (b)} \)
When gibberellin is present, it binds to a soluble receptor protein (called GID1). This gibberellin-receptor complex then binds to the DELLA protein, forming a transient complex that is recognised by an enzyme complex (E3 ubiquitin ligase). This tags the DELLA protein with ubiquitin, targeting it for destruction/degradation by proteasomes. Once the DELLA protein is broken down, the transcription factors are released and become active. They bind to the promoter region of the \(\alpha\)-amylase gene, enabling RNA polymerase to bind and transcribe the gene.

\( \textbf{Part (c)} \)
\textbf{Prediction:} The mutant barley seeds will fail to germinate (or show dwarf/severely stunted growth) even when external gibberellin is added.
\textbf{Explanation:} Because the modified DELLA protein cannot bind to the GA-receptor complex, it cannot be targeted for ubiquitination or degraded by proteasomes. Thus, the DELLA proteins remain stable and continuously bound to the transcription factors, permanently inhibiting them. Consequently, the gene for \(\alpha\)-amylase is never transcribed, and no enzyme is synthesised. Without \(\alpha\)-amylase, starch in the endosperm cannot be hydrolysed to maltose or glucose, depriving the embryo of the respiratory substrates needed for energy and growth during germination.

\( \textbf{Part (a)} \)
1. DELLA proteins act as repressors / inhibit transcription factors ;
2. DELLA binds to transcription factors / prevents them from binding to the promoter region ;
3. Gene transcription is switched off / blocked ;
4. Correct reference to a specific transcription factor (e.g. PIF) ; [Max 3]

\( \textbf{Part (b)} \)
1. Gibberellin binds to a soluble receptor (GID1) ;
2. Gibberellin-receptor complex binds to DELLA protein ;
3. DELLA protein is tagged with ubiquitin / targeted for degradation ;
4. DELLA is broken down / degraded by proteasomes ;
5. Transcription factors are released / free to bind to promoter ;
6. RNA polymerase binds to promoter and transcribes gene (for \(\alpha\)-amylase) ; [Max 4]

\( \textbf{Part (c)} \)
1. (Prediction) seeds fail to germinate / dwarf plants / no amylase produced ;
2. (Explanation) mutant DELLA cannot be degraded / remains stable (despite high GA) ;
3. transcription factors remain permanently inhibited / gene for \(\alpha\)-amylase is not expressed ;
4. starch is not hydrolysed to maltose/glucose / embryo has no respiratory substrate (for growth) ; [Max 3]

(a) Method:
1. Mix equal volumes (e.g., 10 cm\(^3\)) of Chlorella suspension and 3% sodium alginate solution in a small beaker.
2. Draw the mixture into a syringe and drop it slowly from a fixed height into a beaker of 2% calcium chloride solution to form spherical alginate beads.
3. Leave the beads in the calcium chloride solution for 15 minutes to harden, then strain them and rinse thoroughly with distilled water to remove residual calcium chloride.
4. Place 10 beads into a test tube containing a fixed volume (e.g., 20 cm\(^3\)) of 1% sodium hydrogencarbonate solution.
5. Wrap the test tube in one layer of colored cellophane (e.g., red) to act as a filter. Wrap other tubes with blue, green, and clear cellophane (to act as a control representing full white light).
6. Place the test tube at a fixed distance (e.g., 15 cm) from the LED bench lamp, using a ruler to measure the distance.
7. Switch on the lamp, start the stopwatch, and record the time taken (in seconds) for each bead to rise from the bottom of the tube to the surface of the solution.
8. Repeat the procedure at least 3 times for each color filter using fresh beads and fresh solution, and calculate the mean rate of photosynthesis (represented as 1 / mean time taken to rise).

(b) Variables:
(i) Independent variable: Wavelength/color of light (manipulated using different cellophane filters).
(ii) Dependent variable: Rate of photosynthesis (measured as the time taken for a set number of beads to rise to the surface, in seconds).
(iii) Controlled variables (choose any two):
- Light intensity: Controlled by keeping the distance of the lamp from the reaction vessel constant (e.g., at 15 cm) using a ruler.
- Carbon dioxide concentration: Controlled by using the same volume and concentration (1%) of sodium hydrogencarbonate solution for all trials.
- Temperature: Controlled by using a low-heat LED bulb and placing a clear water jacket/bath between the lamp and the tubes to absorb heat; monitor with a thermometer.
- Bead size: Controlled by using the same syringe size and dropping the mixture from the same height.

(c) Control experiment:
- Setup: Wrap a test tube containing Chlorella beads in 1% sodium hydrogencarbonate solution completely in aluminium foil (or place it in a dark cupboard) to exclude all light.
- Expected result: The beads will not rise to the surface, showing that the buoyancy change is due to photosynthesis (oxygen production) occurring in the light and not due to passive buoyancy or temperature.

(d) Statistical test:
- Test: Student's t-test (two-sample / unpaired t-test).
- Reason: It is used to compare the means of two independent, continuous, normally-distributed data sets (red light vs. blue light) with a large enough sample size (n = 15 per group).

Part (a) [Max 8 marks]:
1. Mixes Chlorella suspension and sodium alginate AND drops into calcium chloride solution.
2. Rinses/washes the beads with distilled/deionized water before use.
3. Uses a syringe/dropping pipette from a constant height to produce uniform bead sizes.
4. Specifies a fixed volume (e.g., 20 cm\(^3\)) and concentration (e.g., 1%) of sodium hydrogencarbonate solution.
5. Alters wavelength by wrapping test tubes in different colored cellophane sheets (red, blue, green).
6. Controls light source by placing a lamp at a fixed, measured distance (using a ruler).
7. Measures the time taken for a set number of beads to reach the surface using a stopwatch.
8. Replicates the experiment at least 3 times for each filter and calculates a mean.
9. Suggests a relevant safety precaution (e.g., keep liquids away from electrical lamp cords / wear safety goggles).

Part (b) [Max 3 marks]:
1. Correctly identifies the independent variable (wavelength/color of light) AND dependent variable (time taken for beads to float / rate of photosynthesis).
2. Identifies one key controlled variable and a valid method for controlling it (e.g., light intensity controlled by keeping lamp distance constant).
3. Identifies a second key controlled variable and its control method (e.g., carbon dioxide concentration controlled by using the same concentration of sodium hydrogencarbonate solution, OR temperature controlled by using a water screen/jacket).

Part (c) [Max 2 marks]:
1. Describes a suitable control: wrap the reaction container in aluminum foil / place in total darkness.
2. Explains the expected result: the beads do not rise because there is no light to drive photosynthesis/oxygen production.

Part (d) [Max 2 marks]:
1. Identifies the Student's t-test / unpaired t-test.
2. Explains that the test is used to compare the means of two continuous, normally-distributed data sets (red vs. blue light).

(a) Dilutions:
To prepare 20 cm\(^3\) of each solution, use the formula C\(_1\)V\(_1\) = C\(_2\)V\(_2\):
- 0.4 mol dm\(^{-3}\): 4.0 cm\(^3\) of stock glucose + 16.0 cm\(^3\) of distilled water.
- 0.8 mol dm\(^{-3}\): 8.0 cm\(^3\) of stock glucose + 12.0 cm\(^3\) of distilled water.
- 1.2 mol dm\(^{-3}\): 12.0 cm\(^3\) of stock glucose + 8.0 cm\(^3\) of distilled water.
- 1.6 mol dm\(^{-3}\): 16.0 cm\(^3\) of stock glucose + 4.0 cm\(^3\) of distilled water.
- 2.0 mol dm\(^{-3}\): 20.0 cm\(^3\) of stock glucose + 0.0 cm\(^3\) of distilled water.
Measure all volumes using a graduated syringe or a graduated pipette.

(b) Method:
1. Mix a constant volume (e.g., 10 cm\(^3\)) of 10% active yeast suspension with 10 cm\(^3\) of the prepared 0.4 mol dm\(^{-3}\) glucose solution in a test tube.
2. To ensure anaerobic conditions, carefully layer a small volume (e.g., 2 cm\(^3\)) of liquid paraffin or vegetable oil on top of the yeast-glucose mixture using a pipette to prevent oxygen from dissolving into the mixture.
3. Insert a rubber bung containing a delivery tube into the test tube. Ensure all joints are sealed airtight using petroleum jelly (Vaseline).
4. Connect the other end of the delivery tube to a 50 cm\(^3\) gas syringe.
5. Place the test tube in a thermostatically controlled water bath set at 35\(^\circ\)C. Allow 5 minutes for the mixture to equilibrate to the temperature.
6. Adjust the gas syringe plunger to zero. Start the stopwatch and record the volume of gas produced in the syringe at regular intervals (e.g., every 1 minute) for 10 minutes.
7. Repeat the process at least three times for each of the five glucose concentrations, using fresh yeast and glucose mixtures, to calculate a mean rate of CO\(_2\) production (cm\(^3\) min\(^{-1}\)).

(c) Variables:
- Temperature: Controlled by placing the reaction tube in a thermostatically controlled water bath (or a beaker of water manually adjusted with hot/cold water), and monitored using a thermometer.
- Yeast concentration/volume: Controlled by using the same stock 10% yeast suspension and measuring exactly 10 cm\(^3\) using a volumetric pipette.
- pH: Controlled by adding a set volume of buffer solution (e.g., pH 6.0) to the yeast-glucose mixture.

(d) Adaptations for Aerobic Respiration:
- Add a carbon dioxide absorbent (such as potassium hydroxide (KOH) solution, sodium hydroxide (NaOH), or soda lime) in a small container/vial suspended inside the reaction tube (physically isolated from the yeast suspension).
- Omit the liquid paraffin layer to allow the yeast cells access to oxygen.
- Since CO\(_2\) is absorbed, the consumption of oxygen by the yeast will decrease the gas volume inside the tube, pulling the syringe plunger inward. The rate of inward movement represents the rate of oxygen uptake.

Part (a) [Max 3 marks]:
1. Calculates correct volumes of stock glucose for all five concentrations (4, 8, 12, 16, and 20 cm\(^3\)).
2. Calculates correct volumes of distilled water for all five concentrations (16, 12, 8, 4, and 0 cm\(^3\)).
3. Mentions measuring these volumes using a graduated pipette or syringe to ensure accuracy.

Part (b) [Max 8 marks]:
1. Describes mixing fixed, equal volumes of yeast suspension and glucose solution.
2. Explains the method to maintain anaerobic conditions (adding a layer of liquid paraffin/oil, OR allowing yeast to respire to use up oxygen before sealing the syringe).
3. Mentions seal-checking/making the system airtight using petroleum jelly or grease on joints.
4. Uses a gas syringe to measure the volume of carbon dioxide gas produced.
5. Describes a temperature equilibration period (e.g., 5 minutes in a water bath before starting measurements).
6. Records the volume of gas at regular time intervals (e.g., every minute) for a specified duration (e.g., 5-10 minutes).
7. Replicates the experiment at least 3 times for each glucose concentration and calculates a mean.
8. Describes a safety precaution: wear safety goggles AND take care when inserting glass tubes into rubber bungs.

Part (c) [Max 2 marks]:
- Award 1 mark for each correct pair of variable + control & monitor method (max 2):
- Temperature: controlled using a water bath AND monitored with a thermometer.
- Yeast biomass/concentration: controlled by using the same stock 10% suspension and measuring with a volumetric pipette.
- pH: controlled by adding a buffer solution.

Part (d) [Max 2 marks]:
1. States that potassium hydroxide (KOH), sodium hydroxide (NaOH), or soda lime must be added to absorb carbon dioxide gas.
2. Explains that the liquid paraffin layer is omitted (for aerobic conditions) AND that the decrease in gas volume (movement of syringe plunger inward) measures oxygen uptake rate.