2025 Cambridge IAL Biology (9700) 模擬試題連答案詳解

Water moves down a water potential gradient from a region of higher water potential (less negative) to a region of lower water potential (more negative). Therefore, the correct sequence must show increasingly negative water potential values: Soil (-0.1 MPa) -> root xylem (-0.5 MPa) -> leaf xylem (-0.8 MPa) -> leaf air spaces (-100 MPa).

A is correct because it is the only option showing a continuous decrease in water potential from the soil to the atmosphere.

DNA polymerase requires a primer to initiate synthesis because it can only add nucleotides to an existing 3'-OH group. Options A, C, and D are incorrect: Taq polymerase still requires primers; 95 degrees Celsius breaks hydrogen bonds (denaturation), not phosphodiester bonds; and 55 degrees Celsius is the annealing temperature for primers, not for DNA polymerase binding.

B is the correct option because primers provide the necessary free 3'-OH group for DNA extension.

During depolarisation, voltage-gated sodium channels open, and sodium ions diffuse down their electrochemical gradient into the axon. During repolarisation, voltage-gated sodium channels close and voltage-gated potassium channels open, allowing potassium ions to diffuse out of the axon.

B is the only correct description of channel states and ion movements during depolarisation and repolarisation.

Rubisco is found in chloroplasts (fraction W). Aerobic respiration using oxygen and pyruvate occurs in mitochondria (fraction X). Hydrolytic enzymes at acidic pH are found in lysosomes (fraction Y). Ribosomes (fraction Z) are non-membrane bound and made of RNA and protein.

A is correct because all organelle identifications match their biochemical or physical profiles.

During non-cyclic photophosphorylation, NADP+ is reduced to reduced NADP by NADP reductase on the stromal side of the thylakoid membrane. Option A is incorrect because photolysis electrons go to PSII. Option B is incorrect because proton pumping uses energy from the electron transport chain, not directly photolysis. Option D is incorrect because protons diffuse from the thylakoid lumen into the stroma to make ATP.

C is correct as it accurately details the location of NADP reduction.

Amylopectin is branched with alpha-1,4 and alpha-1,6 glycosidic bonds. Glycogen is also branched and contains both alpha-1,4 and alpha-1,6 glycosidic bonds. Cellulose consists of unbranched chains of beta-glucose held by hydrogen bonds.

C is correct because it accurately defines all three structural properties of the carbohydrates.

The cross is CRCW Ll x CRCW Ll. The flower color cross (CRCW x CRCW) yields 1/4 red, 1/2 pink, 1/4 white. The leaf shape cross (Ll x Ll) yields 3/4 broad, 1/4 narrow. Multiplying these gives: Red, broad = 1/4 * 3/4 = 3/16; Red, narrow = 1/4 * 1/4 = 1/16; Pink, broad = 2/4 * 3/4 = 6/16; Pink, narrow = 2/4 * 1/4 = 2/16; White, broad = 1/4 * 3/4 = 3/16; White, narrow = 1/4 * 1/4 = 1/16. This matches the ratio 3:1:6:2:3:1.

B is correct because independent assortment of these independent genes yields a phenotypic ratio of 3:1:6:2:3:1.

In double-stranded DNA, cytosine equals guanine, so Guanine is 22%. Together they make up 44% of the bases. The remaining 56% must be shared equally between adenine and thymine, which means Adenine is 28% and Thymine is 28%.

A is correct because it aligns with Chargaff's base-pairing rules (%C = %G and %A = %T) totaling 100%.

Path 1 describes the apoplast pathway, where water travels through the non-living parts of the plant, including cell walls and intercellular spaces. Path 2 describes the symplast pathway, where water moves through the cytoplasm via plasmodesmata. Path 3 describes the vacuolar pathway, where water crosses plasma membranes and tonoplasts into vacuoles. Path 4 describes the Casparian strip, which is a band of waterproof suberin in the endodermis. Therefore, option A is correct.

Award 1 mark for identifying the correct combination of root water pathways and the Casparian strip (Option A).

Myelin acts as an electrical insulator because it is made of lipid-rich membranes of Schwann cells. It prevents the movement of ions across the axon membrane. Depolarisation can only occur at the nodes of Ranvier, where there is no myelin and where voltage-gated sodium channels are concentrated. This causes the action potential to jump from node to node (saltatory conduction), which is much faster than continuous conduction. Therefore, option B is correct.

Award 1 mark for identifying the role of myelin sheaths as electrical insulators in saltatory conduction (Option B).

In PCR: 1. Heating to 93-95 degrees C breaks the hydrogen bonds (not covalent bonds) between complementary base pairs to separate the DNA strands. 2. Cooling to around 55 degrees C (annealing) allows short DNA primers to bind to their complementary sequences on the single-stranded DNA templates. 3. Heating to 72 degrees C (extension) allows Taq polymerase to synthesise the complementary strands. Therefore, option B is correct.

Award 1 mark for identifying the correct temperature and purpose for the primer annealing step in PCR (Option B).

Parent 1 genotype is CRCW bb (pink, narrow). Gametes produced: CR b (0.5) and CW b (0.5). Parent 2 genotype is CRCW Bb (pink, broad). Gametes produced: CR B (0.25), CR b (0.25), CW B (0.25), and CW b (0.25). Crossing these gametes: 1/8 CRCR Bb (red broad), 1/8 CRCR bb (red narrow), 2/8 CRCW Bb (pink broad), 2/8 CRCW bb (pink narrow), 1/8 CWCW Bb (white broad), 1/8 CWCW bb (white narrow). This gives a ratio of 1 red broad : 1 red narrow : 2 pink broad : 2 pink narrow : 1 white broad : 1 white narrow. Therefore, option B is correct.

Award 1 mark for calculating the correct phenotypic ratio of the offspring from the dihybrid cross with codominance (Option B).

Plant mesophyll cells are eukaryotic cells and contain membrane-bound organelles. They have mitochondria and chloroplasts, both of which contain circular DNA and 70S ribosomes as part of their endosymbiotic origin. Additionally, plant mesophyll cells are enclosed by a cellulose cell wall. Therefore, all three features are present (+, +, +), and option A is correct.

Award 1 mark for correctly identifying that a plant mesophyll cell contains circular DNA, 70S ribosomes, and a cellulose cell wall (Option A).

The total cross-sectional area of the capillaries is much greater than that of the arteries or veins due to the vast network of millions of capillaries. Velocity of blood flow is inversely proportional to the total cross-sectional area. Therefore, the blood flow slows down significantly in the capillaries, which is highly advantageous as it allows sufficient time for the exchange of substances between blood and tissues. Option B is correct.

Award 1 mark for identifying the correct explanation for the low blood flow velocity in capillaries (Option B).

A high Simpson's Index of Diversity (D closer to 1.0) indicates a high level of biodiversity, meaning the habitat is stable, species richness and evenness are high, and the ecosystem is characterized by complex food webs with many ecological niches. Field X (D = 0.85) is highly diverse and therefore likely to have complex food webs. Field Y (D = 0.21) has low diversity, meaning it is dominated by one or a few species and is more vulnerable to abiotic changes. Option D is correct.

Award 1 mark for interpreting the Simpson's Index of Diversity correctly (Option D).

Amylose is a linear polymer of alpha-glucose joined by 1,4-glycosidic bonds, which coils into a helical shape. Amylopectin is a branched polymer of alpha-glucose joined by 1,4- and 1,6-glycosidic bonds. Cellulose is a linear polymer of beta-glucose joined by 1,4-glycosidic bonds, where alternate monomers are rotated 180 degrees, forming straight, unbranched chains. Therefore, option A is correct.

Award 1 mark for correctly identifying the structural features of amylose, amylopectin, and cellulose (Option A).

Active loading of sucrose into the phloem companion cells depends on the active transport of protons (\(\text{H}^+\)) out of the companion cells into the apoplast by a proton-pumping ATPase. This pump creates a proton gradient (making the apoplast acidic, i.e., lower pH) and hyperpolarises the membrane. If this ATPase is inhibited: (1) protons are no longer pumped out, so the apoplast pH increases (becomes less acidic); (2) the proton gradient is lost, stopping sucrose-proton co-transport, which decreases sucrose concentration in the companion cell; (3) the membrane potential depolarises (becomes less negative) because positive charges are no longer being actively expelled.

1 mark for the correct answer A.

The resting membrane potential is determined primarily by the diffusion of \(\text{K}^+\) out of the cell down its concentration gradient through open leak channels. Increasing extracellular \(\text{K}^+\) reduces this concentration gradient, so less \(\text{K}^+\) leaves the cell, making the resting membrane potential less negative (depolarised). Since the resting potential is now closer to the threshold potential, a smaller depolarising stimulus is required to trigger an action potential, making it easier to reach the threshold.

1 mark for the correct answer A.

Standard E. coli DNA polymerase is not thermostable. During the first denaturation step of PCR (typically \(94\text{--}95\ ^\circ\text{C}\)), the enzyme is permanently denatured and inactivated. Therefore, no DNA synthesis can occur in any of the subsequent cycles, regardless of the primers present.

1 mark for the correct answer C.

In a dihybrid cross of heterozygous parents (AaBb x AaBb), the expected phenotypic ratio of offspring genotypes is 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb. Genotypes with A_B_ (9/16) produce the pink intermediate and convert it to purple pigment (phenotype: purple). Genotypes with A_bb (3/16) produce the pink intermediate but cannot convert it to purple (phenotype: pink). Genotypes with aaB_ (3/16) cannot produce the intermediate, so no substrate is available for enzyme B (phenotype: white). Genotypes with aabb (1/16) cannot produce the intermediate or enzyme B (phenotype: white). Thus, the combined phenotypic ratio is 9 purple : 3 pink : 4 white.

1 mark for the correct answer A.

Chloroplasts and mitochondria are both semi-autonomous eukaryotic organelles surrounded by a double membrane (envelope) and contain their own circular DNA and 70S ribosomes. The nucleus is surrounded by a double membrane (nuclear envelope) but does not contain 70S ribosomes. The rough endoplasmic reticulum is bounded by a single membrane and has 80S ribosomes attached to its outer surface.

1 mark for the correct answer A.

At this moment: (1) The pressure in the left ventricle (10 kPa) is greater than the pressure in the left atrium (1.2 kPa), which causes the left atrioventricular valve to shut to prevent backflow. (2) The pressure in the aorta (11 kPa) is greater than the pressure in the left ventricle (10 kPa), which keeps the aortic semilunar valve closed. Therefore, both valves are closed (this corresponds to the isovolumetric contraction phase of ventricular systole).

1 mark for the correct answer A.

First, calculate the total number of individuals, N = 10 + 6 + 4 = 20. Next, calculate (n/N)^2 for each species: For Species P: (10/20)^2 = 0.25. For Species Q: (6/20)^2 = 0.09. For Species R: (4/20)^2 = 0.04. Sum of these values: \sum (n/N)^2 = 0.25 + 0.09 + 0.04 = 0.38. Finally, calculate Simpson's Index of Diversity: D = 1 - 0.38 = 0.62.

1 mark for the correct answer C.

Disulfide bonds are covalent bonds formed between the sulfur atoms of cysteine R-groups. These bonds help stabilize the tertiary structure (3D conformation of a single polypeptide) and quaternary structure (arrangement of multiple polypeptides). Primary structure is held together by peptide bonds, and secondary structure is stabilized by hydrogen bonds between peptide groups of the backbone; neither of these is stabilized by disulfide bonds. Therefore, only tertiary and quaternary structures can be disrupted by an agent that specifically breaks disulfide bonds.

1 mark for the correct answer A.

During phloem loading, proton pumps actively transport hydrogen ions (\(\text{H}^+\)) out of the companion cell cytoplasm into the cell wall space, utilizing ATP. This establishes a high electrochemical gradient of protons outside the cell. Protons then diffuse back into the companion cell down their gradient through a co-transporter protein (symport). This movement is coupled to the transport of sucrose molecules into the companion cell against their concentration gradient. Therefore, \(\text{H}^+\right. is pumped out of the companion cell by active transport, and both \)\text{H}^+\right. and sucrose enter the companion cell together via the co-transporter.

Award 1 mark for the correct option (A).
- Reject other options because they do not correctly state the direction of active transport or co-transport of protons and sucrose.

During the repolarization phase of an action potential:
- Voltage-gated sodium channels close (or are inactivated) to stop the influx of sodium ions (Statement 1 is correct).
- Voltage-gated potassium channels open (Statement 2 is correct).
- Potassium ions (\(\text{K}^+\)) diffuse out of the axon (from high concentration inside to low concentration outside) down their electrochemical gradient, not into the axon (Statement 3 is incorrect).
- The efflux of positive potassium ions causes the inside of the axon to become more negative, returning it toward the resting potential (Statement 4 is correct).
Thus, statements 1, 2 and 4 are correct.

Award 1 mark for the correct option (B).
- Reject other combinations because Statement 3 is incorrect (potassium diffuses out of, not into, the axon).

At 95 °C, double-stranded DNA denatures as hydrogen bonds between complementary base pairs are broken by thermal energy. At 55 °C, primers anneal to the single-stranded DNA template by forming hydrogen bonds with complementary bases. At 72 °C, Taq DNA polymerase synthesizes new complementary strands by catalyzing the formation of phosphodiester bonds between adjacent nucleotides. Therefore, Row A is correct.

Award 1 mark for the correct option (A).
- Reject other options because they misidentify the types of bonds being broken/formed at each respective stage.

The heterozygous plant has alleles in coupling (cis linkage): \(PL\) on one chromosome and \(pl\) on the other. The test-cross partner is homozygous recessive (\(ppll\)) and only produces gametes with \(pl\). The non-recombinant (parental) offspring phenotypes are Purple/Long (412) and Red/Round (388). The recombinant offspring phenotypes (due to crossing over in the heterozygous parent) are Purple/Round (104) and Red/Long (96).
Total offspring = \(412 + 388 + 104 + 96 = 1000\).
Number of recombinants = \(104 + 96 = 200\).
Recombination frequency = \(\frac{200}{1000} \times 100\% = 20.0\%\).

Award 1 mark for the correct option (B).
- Reject A (10.0%), C (40.0%), and D (80.0%) as they result from incorrect division or summing of parental instead of recombinant classes.

Lysosomal enzymes (which are proteins) are synthesized on ribosomes bound to the Rough Endoplasmic Reticulum (RER). The polypeptide enters the RER lumen, where it undergoes initial folding. It is then transported via transport vesicles to the Golgi body, where it is modified (such as by phosphorylation of mannose residues) and sorted. Finally, the sorted enzyme is packaged into lysosomal vesicles (or vesicles that fuse with endosomes) to become a functional lysosome. Free ribosomes (option B) do not target proteins into the RER. Secretion to the cell surface membrane (option C) is for secretory proteins, not lysosomal enzymes. The Smooth Endoplasmic Reticulum (option D) is not involved in protein translation.

Award 1 mark for the correct option (A).
- Reject B because translation of lysosomal proteins does not complete on free ribosomes.
- Reject C because lysosomal enzymes are not secreted via the cell surface membrane.
- Reject D because the SER is not involved in translation.

In actively respiring tissues, carbon dioxide (\(\text{CO}_2\)) levels are high. \(\text{CO}_2\) diffuses into red blood cells, where carbonic anhydrase catalyzes its reaction with water to form carbonic acid (\(\text{H}_2\text{CO}_3\)). Carbonic acid dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)). The hydrogencarbonate ions diffuse out of the cell down their concentration gradient into the blood plasma. To maintain electrical neutrality, chloride ions (\(\text{Cl}^-\)) diffuse into the red blood cell from the plasma (the chloride shift). Thus, Row B is correct.

Award 1 mark for the correct option (B).
- Reject A because hydrogencarbonate leaves the cell and chloride enters, not vice-versa.
- Reject C and D because they state incorrect directions of ion diffusion and incorrect carbonic anhydrase activities.

To synthesize one molecule of glucose, the Calvin cycle must fix 6 molecules of \(\text{CO}_2\). The fixation of 6 \(\text{CO}_2\) molecules yields 12 molecules of glycerate 3-phosphate (GP). The reduction of 12 GP to 12 triose phosphate (TP) requires 12 ATP and 12 reduced NADP. Out of these 12 TP, 2 are used for the synthesis of one glucose molecule, leaving 10 TP to regenerate 6 molecules of RuBP, which requires an additional 6 ATP. Thus, the total energy requirements are: \(12 + 6 = 18\) ATP, and 12 reduced NADP. This matches Row C.

Award 1 mark for the correct option (C).
- Reject A (this is the requirement for 1 molecule of triose phosphate, not glucose).
- Reject B (incorrect ATP count).
- Reject D (incorrect reduced NADP count).

1. A DNA fragment of 120 base pairs has a total of \(120 \times 2 = 240\) bases.
2. 35% of the total bases are adenine (A), which equals \(240 \times 0.35 = 84\) adenine bases.
3. By complementary base pairing (Chargaff's rules), thymine (T) also comprises 35% of the bases, meaning there are 84 thymines. Thus, there are 84 A-T base pairs in total.
4. Since each A-T base pair is held together by 2 hydrogen bonds, the A-T pairs contribute \(84 \times 2 = 168\) hydrogen bonds.
5. The remaining base pairs are guanine-cytosine (G-C) pairs. Number of G-C base pairs = \(120 - 84 = 36\) pairs.
6. Each G-C base pair is held together by 3 hydrogen bonds, so they contribute \(36 \times 3 = 108\) hydrogen bonds.
7. Total hydrogen bonds = \(168 + 108 = 276\).

Award 1 mark for the correct option (B).
- Reject A (assumes 2 hydrogen bonds per pair for all 120 pairs).
- Reject D (swaps the bond counts, calculating 3 bonds for A-T and 2 for G-C).
- Reject C (incorrect calculation).

Active transport of hydrogen ions (\(H^+\)) out of companion cells into the cell wall apoplast creates a proton gradient. This gradient drives the co-transport of sucrose into the companion cells against its concentration gradient. At the sink, sucrose is unloaded, which increases the water potential inside the sieve tube element, causing water to leave by osmosis and lowering the hydrostatic pressure.

Correct option is A (1 mark).
- Option A is correct because proton pumps actively transport \(H^+\) out of companion cells into the apoplast.
- Option B is incorrect because sucrose is transported against (not down) its concentration gradient.
- Option C is incorrect because unloading sucrose increases (not lowers) the water potential of the sieve tube, causing water to leave (not enter).
- Option D is incorrect because water leaves (not enters) the sieve tube at the sink, maintaining the pressure gradient.

During the repolarization phase of an action potential, the voltage-gated sodium channels close (or become inactivated), stopping the influx of \(Na^+\) ions. Simultaneously, the voltage-gated potassium channels open, and potassium ions (\(K^+\)) diffuse out of the axon down their electrochemical gradient to restore the negative resting potential.

Correct option is D (1 mark).
- Sodium channels must be closed to stop depolarization.
- Potassium channels are open, allowing potassium ions to diffuse out of the axon (repolarization).

The correct sequence of temperatures and events in a PCR cycle is:
1. Heating to approximately 94 °C to denature the double-stranded DNA by breaking the hydrogen bonds between complementary bases.
2. Cooling to around 55 °C to allow primers to anneal (form hydrogen bonds) to their complementary sequences on the single strands.
3. Heating to about 72 °C, which is the optimum temperature for the heat-stable Taq polymerase to synthesize the complementary strands.

Correct option is A (1 mark).
- Denaturation occurs at ~94 °C by breaking hydrogen bonds (not covalent bonds).
- Primer annealing occurs at ~55 °C.
- Extension by Taq polymerase occurs at ~72 °C.

This is an example of duplicate recessive epistasis. The self-pollination of a dihybrid plant (\(AaBb \times AaBb\)) produces a standard 9:3:3:1 phenotypic ratio of genotypes:
- \(9/16\) will have at least one dominant allele of both genes (\(A\_B\_\)), which results in purple flowers.
- \(3/16\) will have genotype \(A\_bb\) (white flowers).
- \(3/16\) will have genotype \(aaB\_\) (white flowers).
- \(1/16\) will have genotype \(aabb\) (white flowers).
Summing the white phenotypes gives \(3 + 3 + 1 = 7\). Therefore, the expected phenotypic ratio is 9 purple : 7 white.

Correct option is B (1 mark).
- Standard dihybrid cross ratio is 9:3:3:1.
- Only the 9/16 with genotype \(A\_B\_\) are purple.
- All other genotypes (7/16) are white, giving a 9:7 ratio.

A typical bacterial cell is prokaryotic and contains circular DNA (plasmid and chromosomal) and 70S ribosomes. A eukaryotic plant cell contains 80S ribosomes in its cytoplasm and has a cellulose cell wall, but it also contains mitochondria and chloroplasts, both of which contain circular DNA and 70S ribosomes. Therefore, circular DNA (1) and 70S ribosomes (2) are found in both types of cells. 80S ribosomes and cellulose cell walls are not found in bacteria.

Correct option is A (1 mark).
- Statement 1 (circular DNA) is found in bacteria and plant organelles (mitochondria, chloroplasts).
- Statement 2 (70S ribosomes) is found in bacteria and plant organelles.
- Statement 3 (80S ribosomes) is not found in bacteria.
- Statement 4 (cellulose cell wall) is not found in bacteria (which have peptidoglycan).

During ventricular systole, the pressure in the left ventricle increases. When the pressure in the left ventricle exceeds the pressure in the left atrium, the left atrioventricular (bicuspid) valve closes. As contraction continues, the ventricular pressure continues to rise until it exceeds the pressure in the aorta, which forces the aortic semi-lunar valve open, allowing blood to be pumped into the systemic circulation.

Correct option is C (1 mark).
- Atrioventricular valve closes earlier in systole when ventricular pressure exceeds atrial pressure.
- Semi-lunar valve opens when ventricular pressure exceeds aortic pressure.

Simpson's Index of Diversity (\(D\)) ranges from 0 (no diversity) to 1 (infinite diversity). A high value close to 1 indicates high biodiversity, which is characterized by high species richness (number of different species) and high species evenness (similar relative abundance of each species). High biodiversity leads to complex ecological networks with alternative food sources, making the community more stable and resilient to environmental changes (such as climate change or disease).

Correct option is B (1 mark).
- High value of \(D\) means high species richness and evenness.
- Higher diversity is associated with greater community stability.

Collagen is a fibrous protein. A single collagen molecule is made of three polypeptide chains wound together into a tight triple helix. Due to its hydrophobic R-groups on the outside and fibrous structure, it is insoluble in water. Haemoglobin is a globular protein consisting of four polypeptide chains (two alpha and two beta chains), each associated with an iron-containing haem prosthetic group. Hydrophilic R-groups on its outer surface make it soluble in water.

Correct option is A (1 mark).
- Collagen: 3 chains, triple helix, fibrous, insoluble.
- Haemoglobin: 4 chains, globular, soluble.

(a) Hydrogen ions (protons) are actively pumped out of the companion cell cytoplasm into the cell wall (apoplast) using energy from ATP hydrolysis, via proton pumps. This establishes a high hydrogen ion concentration gradient (or electrochemical gradient) outside the cell. Hydrogen ions then diffuse back down their concentration gradient into the companion cell through co-transporter proteins. Sucrose is co-transported along with the hydrogen ions against its own concentration gradient.

(b) The accumulation of sucrose inside the companion cell lowers its water potential. Water moves from the xylem or surrounding tissues into the companion cell/sieve tube element by osmosis, down a water potential gradient. This entry of water increases the hydrostatic pressure within the sieve tube element at the source. At the sink, sucrose is unloaded, raising the water potential, so water leaves, lowering the hydrostatic pressure. This difference in hydrostatic pressure between the source and the sink creates a pressure gradient that drives the mass flow of phloem sap.

(c) Sieve tube elements have minimal cytoplasm and organelles to reduce resistance to the flow of phloem sap, allowing unobstructed and more rapid mass transport of solutes.

(a) Max 4 marks:
1. Active transport / pumping of hydrogen ions / protons (H+) out of companion cell [1]
2. via proton pump / using ATP [1]
3. creating a proton / H+ concentration gradient (higher outside) [1]
4. H+ diffuse back into companion cell through co-transporter protein [1]
5. sucrose enters companion cell with H+ / against its concentration gradient [1]

(b) Max 4 marks:
1. high sucrose concentration lowers water potential of companion cell / sieve tube [1]
2. water moves in by osmosis from xylem / surrounding tissue [1]
3. down a water potential gradient [1]
4. increases hydrostatic pressure inside sieve tube (at source) [1]
5. difference in hydrostatic pressure between source and sink causes mass flow [1]

(c) Max 2 marks:
1. reduces resistance / friction to flow [1]
2. allows unobstructed / rapid movement of sap / ease of transport [1]

(a) The alginate matrix traps/anchors the enzyme molecules, restricting their movement. This structural support makes the tertiary structure of the enzyme more rigid, meaning that hydrogen bonds and other non-covalent interactions are less likely to break at higher temperatures. Consequently, the active site is protected from denaturation and retains its complementary shape to the substrate at higher temperatures.

(b) Variables to control: 1. Concentration of lactose (substrate). 2. Volume of lactose solution. 3. Enzyme concentration (or mass of alginate beads containing the enzyme). 4. pH of the substrate and buffer solutions.

(c) Competitive inhibitors have a similar shape to the substrate and bind to the active site, preventing substrate binding; their inhibitory effect can be overcome by increasing the substrate concentration. Non-competitive inhibitors have a different shape to the substrate, bind to an allosteric site (a site other than the active site), causing a change in the shape of the active site so the substrate can no longer bind; their effect cannot be overcome by increasing the substrate concentration.

(a) Max 3 marks:
1. Alginate matrix holds/anchors enzyme in place / restricts movement [1]
2. Prevents/reduces vibrational damage to hydrogen bonds / ionic bonds [1]
3. Prevents/reduces denaturation / maintains tertiary structure [1]
4. Maintains shape of active site at higher temperatures [1]

(b) Max 3 marks (any three of):
1. Concentration of substrate / lactose [1]
2. Concentration of enzyme / mass of alginate beads [1]
3. pH of the reaction mixture [1]
4. Volume of substrate solution / reaction volume [1]

(c) Max 4 marks:
1. Competitive inhibitor has a similar structure/shape to substrate; non-competitive does not / binds to allosteric site [1]
2. Competitive binds to active site; non-competitive binds to allosteric site / site other than active site [1]
3. Competitive does not alter the shape of active site; non-competitive alters active site shape [1]
4. Competitive inhibition can be overcome by increasing substrate concentration; non-competitive cannot be overcome [1]

(a)(i) Cholesterol regulates membrane fluidity (prevents the membrane from becoming too fluid at high temperatures and prevents crystallisation/freezing at low temperatures) and increases membrane mechanical stability, preventing the leakage of polar molecules.
(a)(ii) Glycoproteins act as receptors for cell-signalling molecules (e.g., hormones), act as antigens/cell-recognition markers, and help to stabilise the membrane structure by forming hydrogen bonds with the surrounding water molecules.

(b) High temperatures increase the kinetic energy of phospholipids, causing them to move faster and further apart, which increases the fluidity and creates larger gaps in the bilayer. At 60 degrees C, transport proteins in the membrane denature, further disrupting the membrane structure and creating large pathways, leading to a significant increase in membrane permeability.

(c) Facilitated diffusion is a passive process that does not require ATP, whereas active transport is an active process that requires ATP. Facilitated diffusion moves substances down a concentration gradient, whereas active transport moves substances against a concentration gradient. Facilitated diffusion uses both channel and carrier proteins, whereas active transport uses only carrier proteins/pumps.

(a)(i) Max 2 marks:
1. Regulates/controls membrane fluidity (makes membrane less fluid at high temps / prevents freezing at low temps) [1]
2. Stabilises membrane structure / decreases permeability to polar/charged molecules [1]

(a)(ii) Max 2 marks:
1. Cell signalling / act as receptors for hormones or neurotransmitters [1]
2. Cell-to-cell recognition / act as antigens [1]
3. Help stabilise membrane structure by forming hydrogen bonds with water [1]

(b) Max 3 marks:
1. Phospholipids gain kinetic energy and move more rapidly [1]
2. Phospholipid bilayer becomes more fluid / gaps form between phospholipids [1]
3. Membrane proteins (transport proteins) denature [1]
4. Permeability of the membrane increases [1]

(c) Max 3 marks:
1. Facilitated diffusion is passive/no ATP; active transport requires energy/ATP [1]
2. Facilitated diffusion is down a concentration gradient; active transport is against a concentration gradient [1]
3. Facilitated diffusion uses channel or carrier proteins; active transport uses carrier proteins only [1]

(a) Transcription begins when the DNA double helix is unwound and unzipped, breaking the hydrogen bonds between complementary bases to expose the template strand. Free RNA nucleotides align by complementary base pairing opposite the exposed bases on the template strand (Adenine pairs with Uracil, and Cytosine pairs with Guanine). RNA polymerase then joins these RNA nucleotides together by catalyzing the formation of phosphodiester bonds to form the sugar-phosphate backbone, producing the mRNA molecule.

(b) Each tRNA molecule has a specific amino acid attached to its amino acid binding site. It also carries a specific triplet of bases called an anticodon. During translation, the tRNA anticodon binds to a complementary codon on the mRNA molecule inside the ribosome through temporary hydrogen bonds. This brings the specific amino acids into the correct sequence, allowing peptide bonds to form between adjacent amino acids to synthesise the polypeptide chain.

(c) A gene mutation is a change in the sequence of nucleotides or bases in a DNA molecule/gene.

(a) Max 4 marks:
1. DNA double helix unwinds/unzips / hydrogen bonds break [1]
2. One DNA strand acts as a template [1]
3. Free RNA nucleotides align by complementary base pairing (A-U, T-A, C-G) [1]
4. RNA polymerase joins RNA nucleotides together [1]
5. Formation of phosphodiester bonds [1]

(b) Max 4 marks:
1. tRNA carries a specific amino acid [1]
2. tRNA has a specific anticodon [1]
3. anticodon binds to complementary codon on mRNA [1]
4. via hydrogen bonding [1]
5. aligns amino acids in correct sequence for peptide bond formation [1]

(c) Max 2 marks:
1. Change in the nucleotide/base sequence [1]
2. in a gene / molecule of DNA [1]

(a) The tunica media of an artery contains a thick layer of elastic fibres (elastin), which allows the artery wall to stretch when blood enters under high pressure during ventricular systole, and then recoil during ventricular diastole to maintain a high blood pressure. It also contains smooth muscle, which can contract (vasoconstriction) or relax (vasodilation) to control blood flow and pressure. The overall thickness of the wall provides strength to withstand high pressures without bursting.

(b) At the arteriole end of the capillary bed, the hydrostatic pressure of the blood is higher than the hydrostatic pressure of the tissue fluid outside, and this outwards force exceeds the inwards osmotic force (water potential gradient). This net outward pressure forces water and small dissolved solutes (like glucose, oxygen, and amino acids) out through the small gaps in the capillary endothelium. Large proteins and blood cells remain behind because they are too large to pass.

(c) Some fluid remains because not all of it is reabsorbed at the venule end of the capillary, as the osmotic pressure gradient is not large enough to overcome the remaining hydrostatic pressure completely. This excess tissue fluid drains into the lymphatic capillaries/vessels, where it is called lymph. The lymphatic vessels eventually transport this lymph back to the subclavian veins near the heart, returning it to the blood circulatory system.

(a) Max 4 marks:
1. Tunica media contains elastic fibres / elastin [1]
2. Elastic fibres stretch to accommodate high pressure / surge of blood [1]
3. Elastic fibres recoil to maintain pressure (during diastole) [1]
4. Smooth muscle contracts / relaxes to control lumen size / blood flow [1]
5. Thick wall / collagen in tunica externa prevents bursting [1]

(b) Max 3 marks:
1. Hydrostatic pressure of blood at arteriole end is high [1]
2. Hydrostatic pressure is greater than osmotic pressure / water potential gradient [1]
3. Net outward force forces water and small solutes (glucose, ions) out [1]
4. Through gaps/pores in the capillary wall / endothelium [1]
5. Large proteins / blood cells cannot pass through / remain in blood [1]

(c) Max 3 marks:
1. Hydrostatic pressure still opposes reabsorption at venule end / osmotic pull not sufficient [1]
2. Excess tissue fluid drains into lymphatic capillaries / lymph vessels [1]
3. Fluid is returned to the blood system via veins / subclavian veins near the heart [1]

(a) Active immunity involves the production of antibodies by the individual's own plasma cells, whereas passive immunity involves receiving antibodies produced by another organism. Active immunity results in the production of memory cells and provides long-term immunity, whereas passive immunity does not produce memory cells and is only short-term. Active immunity takes time to develop, whereas passive immunity provides immediate protection.

(b) A vaccine contains dead, weakened, or attenuated pathogens, or isolated surface antigens, which do not cause disease. These antigens are engulfed by macrophages, which present the antigens on their cell surface membrane (becoming antigen-presenting cells, or APCs). Helper T-lymphocytes with complementary receptors bind to these APCs and release cytokines. These cytokines stimulate specific B-lymphocytes (with complementary membrane-bound antibodies) to divide by mitosis (clonal expansion) and differentiate into plasma cells, which secrete large amounts of specific antibodies, and memory B-cells, which persist in the body to provide long-term protection.

(c) Reasons why vaccination programmes fail include: 1. The pathogen may undergo antigenic mutation/antigenic shift, changing its surface antigens so that existing memory cells no longer recognise them. 2. Some individuals may have a poor immune response to the vaccine or not develop immunity. 3. Vaccine hesitancy/low uptake of the vaccine, preventing herd immunity. 4. Pathogens may hide inside host cells (like tuberculosis) or have complex life cycles (like malaria), making them difficult for antibodies to target.

(a) Max 3 marks:
1. Active immunity involves antibody production by host cells; passive involves receiving pre-made antibodies [1]
2. Active produces memory cells; passive does not [1]
3. Active is long-term; passive is short-term / temporary [1]
4. Active has a lag phase / takes time to develop; passive is immediate [1]

(b) Max 5 marks:
1. Vaccine contains harmless/dead/attenuated pathogen or antigens [1]
2. Antigens recognized as foreign / non-self [1]
3. Antigen-presenting cells (APCs) / macrophages present antigen [1]
4. T-helper cells bind and release cytokines [1]
5. Cytokines stimulate B-cells to divide by mitosis / undergo clonal expansion [1]
6. B-cells differentiate into plasma cells (which secrete antibodies) and memory cells (which persist) [1]

(c) Max 2 marks (any two of):
1. Antigenic mutation / antigenic shift / antigenic drift [1]
2. Low vaccine uptake / lack of herd immunity [1]
3. Pathogen lives inside host cells / has a complex life cycle [1]
4. Poor immune response in some individuals [1]

(a)
1. Extract mRNA from both normal lung cells and cancerous lung cells.
2. Synthesize complementary DNA (cDNA) from the mRNA using reverse transcriptase.
3. Label the cDNA samples with different fluorescent dyes (e.g., green for normal cells, red for cancerous cells).
4. Apply both labelled cDNA samples to the microarray chip, where they hybridize with complementary single-stranded DNA probes fixed to the wells.
5. Wash the microarray to remove any unhybridized cDNA.
6. Scan the microarray with a laser scanner to detect fluorescence.

(b)
1. The intensity and color of the fluorescence at each spot on the microarray are analyzed (red = gene overexpressed in cancer, green = gene overexpressed in normal cells, yellow = equal expression in both, dark/black = gene not expressed in either).
2. Clinicians can use these profiles to classify the subtype of cancer, predict tumor behavior, and choose targeted drug therapies tailored to the patient's specific gene expression profile (personalized medicine).

(c)
1. Microarrays are highly specialized for detecting the expression of a vast, pre-defined set of genes simultaneously at a lower cost per sample compared to whole-transcriptome RNA-sequencing (RNA-Seq).
2. They produce smaller, more manageable datasets that require simpler bioinformatic pipelines for rapid clinical decision-making.

Part (a) [Max 5 marks]:
- Extract mRNA from both cell types (1)
- Use reverse transcriptase to synthesize cDNA (1)
- Label cDNA samples with distinct fluorescent dyes / tags (e.g., green/red) (1)
- Hybridize cDNA to the single-stranded DNA probes on the microarray (1)
- Wash to remove unbound cDNA (1)
- Detect fluorescence using a laser scanner / microarray reader (1)

Part (b) [Max 3 marks]:
- Describe color analysis: red = cancer-specific expression, green = normal-specific expression, yellow = equal expression (1)
- Identify/detect which genes are mutated/upregulated/downregulated (1)
- Use for personalized medicine / targeted therapy / prognosis based on molecular profile (1)

Part (c) [Max 2 marks]:
- Lower cost per sample for screening known gene targets (1)
- Faster turnaround time / less complex bioinformatics/data analysis required (1)

(a)
1. Osmoreceptors are located in the hypothalamus of the brain.
2. When blood water potential decreases, water moves out of the osmoreceptor cells by osmosis down a water potential gradient.
3. This causes the osmoreceptor cells to shrink / lose turgidity, which triggers the generation of action potentials (nerve impulses) that travel to the posterior pituitary gland (stimulating ADH release).

(b)
1. ADH binds to complementary receptors on the cell surface membrane of the collecting duct cells (on the basal / blood-facing side).
2. This binding activates a G-protein, which in turn activates the enzyme adenylyl cyclase.
3. Adenylyl cyclase catalyzes the conversion of ATP to cyclic AMP (cAMP), which acts as a second messenger.
4. cAMP activates a kinase enzyme (protein kinase A), which initiates a phosphorylation cascade.
5. This cascade causes vesicles containing aquaporin water-channel proteins to move towards and fuse with the luminal (apical / urine-facing) membrane via exocytosis.

(c)
1. Condition: Nephrogenic diabetes insipidus.
2. Explanation: Without ADH binding, the second-messenger signaling cascade is not initiated, meaning aquaporins remain sequestered in intracellular vesicles and are not inserted into the luminal membrane. Consequently, the collecting duct remains impermeable to water, resulting in a failure to reabsorb water and the excretion of a high volume of dilute urine.

Part (a) [Max 3 marks]:
- Location: Hypothalamus (1)
- Water moves out of osmoreceptors by osmosis down a water potential gradient (1)
- Osmoreceptors shrink / dehydrate, initiating action potentials (to the posterior pituitary) (1)

Part (b) [Max 5 marks]:
- ADH binds to specific/complementary receptors on the basal/basolateral membrane (1)
- Activates G-protein (1)
- Activates adenylyl cyclase, converting ATP to cAMP (1)
- cAMP acts as a second messenger, activating protein kinase A (1)
- Triggers a signaling/phosphorylation cascade (1)
- Vesicles containing aquaporins move to and fuse with the luminal/apical membrane by exocytosis (1)

Part (c) [Max 2 marks]:
- Name: Nephrogenic diabetes insipidus (1)
- Mechanism: Aquaporins are not inserted into the luminal membrane, preventing water reabsorption from the collecting duct (1)

(a)
1. Photosystems: Cyclic photophosphorylation involves only Photosystem I (PSI / P700), whereas non-cyclic photophosphorylation involves both Photosystem II (PSII / P680) and Photosystem I (PSI / P700).
2. Electron donor/loop: In cyclic, electrons from PSI return to the electron transport chain (PSI is the donor and acceptor); in non-cyclic, electrons from photolysis of water replenish PSII, and electrons from PSII pass to PSI and ultimately to \(NADP^+\).
3. End-products: Cyclic photophosphorylation produces ATP only. Non-cyclic photophosphorylation produces ATP, reduced NADP (NADPH), and oxygen.

(b)(i)
1. Oxygen production stops because the photolysis of water at the oxygen-evolving complex of PSII is inhibited (electrons cannot be transferred from water if PSII cannot release its excited electrons to the blocked plastoquinone pool).
2. Reduced NADP production ceases because the linear flow of electrons along the electron transport chain is blocked, meaning there are no electrons reaching the stroma-facing ferredoxin-\(NADP^+\) reductase to reduce \(NADP^+\) to reduced NADP.

(b)(ii)
1. The Calvin cycle stops because it depends on the products of the light-dependent stage (ATP and reduced NADP).
2. The conversion of glycerate 3-phosphate (GP) to triose phosphate (TP) is blocked because this reaction requires both ATP (for phosphorylation) and reduced NADP (as a reducing agent).
3. As a result, GP accumulates, and the regeneration of ribulose bisphosphate (RuBP) from TP ceases, stopping the fixation of carbon dioxide.

Part (a) [Max 4 marks]:
- Cyclic photophosphorylation uses Photosystem I / PSI only AND non-cyclic uses both PSI and PSII (1)
- Cyclic does not involve photolysis of water AND non-cyclic involves photolysis of water (1)
- End-products of cyclic: ATP only (1)
- End-products of non-cyclic: ATP, reduced NADP, and oxygen (1)

Part (b)(i) [Max 3 marks]:
- Oxygen production ceases because photolysis of water is blocked/inhibited at PSII (1)
- No electron flow from PSII to PSI/ETC (1)
- Reduced NADP production stops because no electrons are available to reduce \(NADP^+\) (1)

Part (b)(ii) [Max 3 marks]:
- The Calvin cycle / light-independent stage stops due to lack of ATP and reduced NADP (1)
- Glycerate 3-phosphate (GP) cannot be reduced to triose phosphate (TP) (1)
- Amount of GP increases and TP/RuBP decreases (1)

(a)
1. Protons (\(H^+\) ions) are pumped from the mitochondrial matrix across the inner mitochondrial membrane into the intermembrane space.
2. This active transport is powered by energy released from electrons passing down the electron transport chain.
3. This creates an electrochemical/proton concentration gradient across the inner mitochondrial membrane (higher concentration in the intermembrane space than the matrix).
4. Protons diffuse down this gradient back into the matrix by facilitated diffusion through ATP synthase.
5. The passage of protons through ATP synthase provides the proton motive force that drives the rotation of the enzyme, catalyzing the phosphorylation of ADP + inorganic phosphate (\(P_i\)) to form ATP.

(b)(i)
1. UCP1 provides an alternative pathway for protons to leak back into the matrix, bypassing the ATP synthase channel.
2. Because the protons do not flow through ATP synthase, ADP is not phosphorylated to ATP.
3. The potential energy of the electrochemical proton gradient is instead released directly as heat energy (thermogenesis).

(b)(ii)
1. Oxygen consumption rate increases.
2. Because the proton gradient is constantly dissipated by UCP1, the electron transport chain (ETC) operates at its maximum rate to continuously pump protons out of the matrix.
3. To feed the ETC, the oxidation of reduced coenzymes (NADH and \(FADH_2\)) must occur at an accelerated rate.
4. Since oxygen is the terminal electron acceptor of the ETC (combining with electrons and protons to form water), a highly active ETC directly increases the rate of oxygen consumption.

Part (a) [Max 4 marks]:
- Protons pumped from matrix to intermembrane space (1)
- Powered by energy from electron transport chain (1)
- Establishes a proton concentration / electrochemical gradient (1)
- Protons diffuse down gradient back into matrix through ATP synthase (1)
- Proton flow / proton motive force drives phosphorylation of ADP and Pi to ATP (1)

Part (b)(i) [Max 3 marks]:
- Protons diffuse through UCP1 bypassing ATP synthase (1)
- Energy stored in the proton gradient is released/dissipated as heat (rather than chemical energy) (1)
- ATP synthesis decreases / no ATP is produced via chemiosmosis (1)

Part (b)(ii) [Max 3 marks]:
- Oxygen consumption increases (1)
- Uncoupling causes the electron transport chain to work faster/continuously to pump protons (1)
- More oxygen is needed as the final electron/proton acceptor to form water (1)

(a)
1. Unidirectional light shining on a shoot tip is detected by photoreceptors (phototropins).
2. Auxin (indol-3-acetic acid / IAA) is synthesized in the shoot tip and transported downwards.
3. The light stimulus causes auxin to be transported laterally from the illuminated side to the shaded side of the shoot.
4. There is a higher concentration of auxin on the shaded side of the shoot.
5. This high concentration of auxin stimulates cell elongation on the shaded side, while the lower concentration on the illuminated side results in less elongation.
6. This unequal growth rate causes the shoot to bend towards the light source (positive phototropism).

(b)
1. Auxin binds to an auxin-binding protein (ABP1) on the cell surface membrane of target cells.
2. This stimulates proton pumps (\(H^+\)-ATPases) in the cell surface membrane to actively transport protons (\(H^+\) ions) from the cytoplasm into the cell wall space.
3. The accumulation of protons lowers the pH of the cell wall (making it more acidic, pH approx 4.5).
4. The low pH activates specific wall-loosening enzymes/proteins called expansins.
5. Expansins break the hydrogen bonds holding cellulose microfibrils to hemicelluloses in the cell wall matrix.
6. This loosens the cell wall, making it flexible and plastic.
7. At the same time, the cell takes up ions (such as potassium ions, \(K^+\)), which lowers the solute potential and thus the water potential inside the cell.
8. Water enters the cell by osmosis down a water potential gradient, increasing turgor pressure.
9. The high turgor pressure pushes against the loosened cell wall, causing the cell to expand/elongate.

Part (a) [Max 4 marks]:
- Auxin (IAA) is made in the shoot tip and moves downwards (1)
- Unidirectional light causes auxin to move / migrate to the shaded side of the shoot (1)
- Higher concentration of auxin on the shaded side than the illuminated side (1)
- Auxin stimulates cell elongation on the shaded side, causing the shoot to bend towards the light (1)

Part (b) [Max 6 marks]:
- Auxin binds to receptors and stimulates proton / \(H^+\) pumps (1)
- Protons are actively pumped from cytoplasm into the cell wall (1)
- Acidification / lowering of cell wall pH occurs (1)
- Low pH activates expansins (1)
- Expansins break hydrogen bonds between cellulose microfibrils and other cell wall components (1)
- Cell wall becomes loose / plastic (1)
- Solutes / potassium ions enter the cell, lowering water potential (1)
- Water enters by osmosis, and high turgor pressure causes the loosened cell wall to stretch (1)

(a)
1. An operon is a cluster of genes that are transcribed together under the control of a single promoter.
2. It includes structural genes and regulatory sections of DNA (promoter and operator) that coordinate gene expression.

(b)(i)
1. The regulator gene (*lacI*) is transcribed and translated to produce an active repressor protein.
2. It is constitutively expressed, meaning it is continually active regardless of the presence of lactose.

(b)(ii)
1. The operator (*lacO*) is the site on the DNA where the active repressor protein binds.
2. Binding of the repressor to the operator physically blocks RNA polymerase from binding to the promoter and transcribing the structural genes.

(c)
1. Lactose enters the cell and some is converted into allolactose, which serves as an inducer.
2. Allolactose binds to the allosteric site of the repressor protein.
3. This binding alters the three-dimensional (tertiary) structure of the repressor protein.
4. The conformational change prevents the repressor from binding to the operator site.
5. RNA polymerase can now freely bind to the promoter region and transcribe the structural genes *lacZ* (for \(\beta\)-galactosidase) and *lacY* (for lactose permease).

Part (a) [Max 2 marks]:
- Cluster of structural genes (1)
- Transcribed together / under control of a single promoter and regulatory elements (1)

Part (b)(i) [Max 2 marks]:
- Transcribed/translated to produce active repressor protein (1)
- Constitutively expressed / always 'on' (1)

Part (b)(ii) [Max 2 marks]:
- Site where active repressor protein binds (1)
- Prevents RNA polymerase from binding to promoter / transcribing structural genes (1)

Part (c) [Max 4 marks]:
- Lactose/allolactose acts as an inducer (1)
- Binds to repressor protein (at allosteric site) (1)
- Alters the tertiary structure of the repressor protein (1)
- Repressor can no longer bind to operator (1)
- RNA polymerase can bind to promoter and transcribe structural genes / lacZ and lacY (1)

(a)
1. Allopatric speciation requires geographic isolation of populations by a physical barrier (e.g., mountains, rivers, lava flows).
2. Sympatric speciation occurs in the same geographic location without physical separation, often due to ecological, behavioral, or polyploidic isolation.

(b)
1. Geographical isolation: The lava flow prevents individuals from crossing, preventing gene flow between the two populations.
2. Mutations: Random, independent mutations occur in the DNA of lizards in both populations, generating new alleles.
3. Selection pressures: The environments on either side of the lava flow may differ (e.g., vegetation, temperature, predators), presenting different selection pressures.
4. Natural selection: Lizards with advantageous alleles are more likely to survive, reproduce, and pass on these alleles to their offspring.
5. Changes in allele frequencies: Over time, the allele frequencies of the two gene pools diverge significantly.
6. Genetic drift: In small isolated populations, random changes in allele frequencies can occur independently.
7. Reproductive isolation: Accumulation of genetic differences eventually prevents successful interbreeding (due to changes in courtship behavior, mating seasons, or gamete incompatibility) even if they meet again.

(c)
1. Attempt to interbreed individuals from both populations under controlled conditions.
2. If they cannot mate, or if mating does not produce viable and fertile offspring, they are classified as separate species (according to the biological species concept).

Part (a) [Max 2 marks]:
- Allopatric involves geographical isolation / physical barrier (1)
- Sympatric occurs in the same geographic area / without physical barrier (1)

Part (b) [Max 6 marks]:
- Geographical barrier / lava flow prevents gene flow between populations (1)
- Different environmental conditions / selection pressures in each habitat (1)
- Random mutations occur independently in both populations (1)
- Natural selection favors different alleles in each population (1)
- Change in allele frequencies occurs over generations (1)
- Genetic drift may play a role (especially if populations are small) (1)
- Development of reproductive isolation (e.g., structural, behavioral, physiological changes) (1)

Part (c) [Max 2 marks]:
- Attempt to interbreed individuals from both populations (1)
- If they cannot produce fertile offspring, they are separate species (1)

(a)
1. Small populations have a small gene pool and low genetic diversity.
2. This leads to inbreeding depression, which increases the likelihood of offspring inheriting two copies of harmful recessive alleles, decreasing fitness/survival rates.
3. The population has lower evolutionary plasticity / adaptive capacity, making it vulnerable to environmental changes (e.g., climate change, habitat loss) or disease outbreaks.
4. Random events (genetic drift) can cause the loss of beneficial alleles, leading to a further decline in population size (extinction vortex).

(b)(i)
1. Oocytes (egg cells) are retrieved from females, sometimes utilizing hormone therapy to trigger superovulation.
2. Sperm is collected from males, which can be fresh or cryopreserved.
3. The eggs and sperm are mixed in a controlled laboratory environment (in vitro) to allow fertilization to take place.
4. The fertilized eggs are cultured until they form early-stage embryos.

(b)(ii)
1. The developed embryos are transferred/implanted into the prepared uterus of a surrogate mother.
2. The surrogate is typically a female of a closely-related, more common/non-endangered species, which carries the embryo to full term and gives birth, increasing the reproductive rate of the endangered species without placing the endangered females at risk.

(c)
1. Concept: A facility where genetic materials (such as semen, oocytes, embryos, and somatic/stem cells) from endangered animals are stored at extremely low temperatures (cryopreservation, using liquid nitrogen at \(-196^\circ C\)).
2. Importance: It acts as a genetic repository, preserving the genetic diversity of threatened species long-term. This material can be used in the future via ART to reintroduce genetic diversity into inbred populations, preventing extinction.

Part (a) [Max 3 marks]:
- Small gene pool / low genetic diversity (1)
- Inbreeding depression / increased risk of inheriting harmful recessive alleles (1)
- Lack of adaptation to changing environments / high risk from a single disease outbreak (1)
- Impact of genetic drift / extinction vortex (1)

Part (b)(i) [Max 3 marks]:
- Oocytes / eggs harvested from females (1)
- Sperm collected from males / thawed from frozen storage (1)
- Fertilization occurs in laboratory glassware / in vitro (1)
- Embryos are cultured to early stages of development (1)

Part (b)(ii) [Max 2 marks]:
- Embryo implanted into uterus of surrogate mother (1)
- Surrogate is often a closely-related, non-endangered species (1)
- Reduces risk to endangered breeding females / increases reproductive output (1)

Part (c) [Max 2 marks]:
- Cryopreservation / freezing (in liquid nitrogen) of gametes, embryos, or tissue samples (1)
- Preserves genetic diversity long-term for future reintroduction / breeding programs (1)

(a) Genomic DNA is identical across all physiological states and cell types, meaning it does not reflect dynamic changes in gene activity. Extracting mRNA allows researchers to isolate only the transcripts being actively synthesized in response to the drought stress stimulus. (b) The extracted mRNA must first be reverse-transcribed into complementary DNA (cDNA) using the enzyme reverse transcriptase and fluorescently tagged nucleotides. For example, control cDNA can be tagged with a green fluorophore and drought-stressed cDNA with a red fluorophore. The samples are mixed and applied to the microarray chip, where they hybridize with complementary single-stranded DNA probes. Unbound cDNA is washed away, and a laser scanner is used to measure the intensity of the green and red fluorescence at each spot on the microarray. (c) Highly upregulated genes can be cloned and introduced into crop genomes via genetic modification under the control of stress-inducible or constitutive promoters. Alternatively, genome-editing tools like CRISPR-Cas9 can be used to target and enhance the expression of these endogenous drought-response genes.

(a) [Max 2 marks] 1. mRNA shows which genes are actively being transcribed/expressed under specific environmental conditions; 2. Genomic DNA is identical in all cells and does not change in response to environmental stimuli; 3. Only genes relevant to drought response will produce mRNA. (b) [Max 5 marks] 1. Reverse transcribe mRNA from both samples to produce cDNA; 2. Use reverse transcriptase; 3. Label cDNA from control and drought-stressed plants with different/distinct fluorescent dyes; 4. Mix the cDNA samples and hybridize them to the single-stranded DNA probes on the microarray; 5. Wash the microarray to remove any unhybridized/unbound cDNA; 6. Scan the microarray with lasers/detectors to measure fluorescence intensity at each spot. (c) [Max 3 marks] 1. Identify specific drought-responsive genes (such as transcription factors, osmoprotectants, or dehydrins); 2. Clone/insert these genes into target crop plants using recombinant DNA technology / genetic engineering; 3. Use an appropriate constitutive or stress-inducible promoter; 4. Select plants showing high expression or test for improved performance under drought conditions; 5. Use gene-editing (CRISPR) to activate/upregulate native drought-tolerance genes.

(a) When blood water potential decreases, water leaves osmoreceptor cells in the hypothalamus by osmosis down a water potential gradient. This causes the osmoreceptors to shrink, depolarizing their membranes and generating action potentials. These action potentials travel down the axons of neurosecretory cells into the posterior pituitary gland, triggering calcium influx and the exocytosis of ADH into nearby capillaries. (b) ADH acts as a first messenger, binding to specific G-protein-coupled receptors on the basolateral membrane of collecting duct epithelial cells. This activates adenylyl cyclase, which converts ATP to cyclic AMP (cAMP). cAMP acts as a second messenger, activating protein kinase A (PKA). This phosphorylation cascade triggers the translocation of intracellular vesicles containing aquaporin-2 water channels to the apical (luminal) membrane, where they fuse. Water can then freely flow down its osmotic gradient from the collecting duct lumen into the hypertonic medulla. (c)(i) Cranial diabetes insipidus involves a failure of the hypothalamus to synthesize ADH or the posterior pituitary to secrete it, often due to brain injury, tumors, or genetic defects. Nephrogenic diabetes insipidus involves normal ADH secretion, but the kidneys cannot respond to it, usually due to mutations in the ADH receptor gene or the aquaporin gene. (c)(ii) Because the kidney receptors or signaling pathways are non-functional, providing exogenous ADH cannot initiate the signal transduction pathway required to insert aquaporins into the apical membrane.

(a) [Max 3 marks] 1. Osmoreceptors in the hypothalamus detect a decrease in blood water potential / increase in blood osmolarity; 2. Water moves out of osmoreceptor cells by osmosis, causing them to shrink; 3. This triggers action potentials/nerve impulses along neurosecretory cells; 4. ADH is released from the posterior pituitary gland by exocytosis into the blood. (b) [Max 4 marks] 1. ADH binds to specific receptors on the cell surface membrane of collecting duct cells; 2. Activates G-protein, which activates adenylyl cyclase; 3. Adenylyl cyclase converts ATP to cAMP (second messenger); 4. cAMP triggers a kinase cascade causing vesicles containing aquaporins to move to the apical/luminal membrane; 5. Vesicles fuse with the apical/luminal membrane, inserting aquaporins; 6. Increases permeability of membrane to water, allowing water to move out by osmosis down a water potential gradient. (c)(i) [Max 2 marks] 1. Cranial: pituitary gland cannot produce/secrete ADH; 2. Nephrogenic: kidneys/collecting ducts do not respond to ADH / have faulty receptors or aquaporins. (c)(ii) [Max 1 mark] 1. The defect is in the target kidney cells/receptors, so they cannot respond to any concentration of ADH.

(a) To vary the wind speed, the student can use an electric fan placed at different measured distances from the shoot (e.g., 10 cm, 20 cm, 30 cm, 40 cm, 50 cm), or use a fan with variable speed settings. An anemometer must be placed next to the shoot to measure and record the actual wind speed at each setting. To ensure reliability, the student should repeat the measurement at least three times at each wind speed and calculate a mean rate, identifying and omitting any anomalous results.

(b) To determine \( r \) (the internal radius): Use a light microscope fitted with a calibrated eyepiece graticule to measure the internal diameter of the capillary tube cross-section, then divide this value by 2. To determine \( d \) (the distance): Place a millimeter ruler directly adjacent to the capillary tube and record the starting and ending positions of the air bubble over the specified time interval, then calculate the difference.

(c) Remove all leaves from the shoot at the end of the experiment. Trace the outline of each leaf onto grid/graph paper of known square size (e.g., 1 mm by 1 mm). Count the total number of squares enclosed by each outline (counting squares that are more than half inside as full squares, and ignoring those that are less than half). Multiply the total number of squares by the area of one square to find the one-sided area, then multiply by 2 to obtain the total surface area (both upper and lower surfaces) of all leaves.

(d) The potometer measures water uptake rather than water lost directly through transpiration. Some of the absorbed water is retained by the plant cells to maintain turgidity, and some is used chemically during photosynthesis rather than being transpired into the atmosphere.

(a) Maximum 3 marks:
- 1 mark for varying wind speed using a fan at different distances / variable speed settings.
- 1 mark for measuring wind speed quantitatively using an anemometer.
- 1 mark for repeating the experiment at least 3 times at each wind speed to calculate a mean / identify anomalies.

(b) Maximum 2 marks:
- 1 mark for measuring \( r \) using an eyepiece graticule under a microscope to find the internal diameter and dividing by 2 (or using a manufacturer's specification sheet). Reject using a standard ruler to measure internal capillary diameter directly.
- 1 mark for measuring \( d \) using a millimeter ruler placed next to the tube to record initial and final positions.

(c) Maximum 3 marks:
- 1 mark for tracing leaf outlines onto graph/grid paper.
- 1 mark for counting the grid squares inside the outlines (and multiplying by the area of one square).
- 1 mark for multiplying by 2 to account for both surfaces of all leaves.

(d) Maximum 2 marks:
- 1 mark for stating that water is used to maintain cell turgidity / cell expansion.
- 1 mark for stating that water is consumed in metabolic reactions / photosynthesis.

(a) First, extract mRNA from both normal and cancerous prostate tissue samples. Use the enzyme reverse transcriptase along with free nucleotides to synthesize single-stranded complementary DNA (cDNA) from the mRNA templates. During this synthesis, incorporate fluorescently-labeled nucleotides: label the normal tissue cDNA with one color (e.g., green fluorescent dye) and the cancerous tissue cDNA with a different color (e.g., red fluorescent dye). Denature the double-stranded cDNA if necessary to ensure they are single-stranded and ready to hybridize with the microarray probes.

(b) Gene X is highly upregulated/overexpressed in cancerous prostate cells compared to normal cells. Gene Y is downregulated/underexpressed in cancerous prostate cells. Gene Z shows no change in expression (it is expressed equally in both normal and cancerous prostate cells).

(c) To perform a 10-fold serial dilution to obtain 100 cm^3 of each required concentration:
1. To prepare 2.0 mmol/dm^3: Pipette 10 cm^3 of the 20.0 mmol/dm^3 stock primer solution into a measuring cylinder or flask and dilute with 90 cm^3 of sterile distilled water. Mix thoroughly.
2. To prepare 0.2 mmol/dm^3: Pipette 10 cm^3 of the newly prepared 2.0 mmol/dm^3 solution into a second vessel and dilute with 90 cm^3 of sterile distilled water. Mix thoroughly.
3. To prepare 0.02 mmol/dm^3: Pipette 10 cm^3 of the newly prepared 0.2 mmol/dm^3 solution into a third vessel and dilute with 90 cm^3 of sterile distilled water. Mix thoroughly.

(a) Maximum 4 marks:
- 1 mark for extracting/isolating mRNA from both tissue types.
- 1 mark for using reverse transcriptase to synthesize cDNA from the mRNA template.
- 1 mark for using fluorescently-labeled nucleotides / tags.
- 1 mark for specifying distinct tags (e.g., green for normal, red for cancer) and mixing them / denaturing to single strands.

(b) Maximum 2 marks:
- 1 mark for stating Gene X is upregulated/overexpressed AND Gene Y is downregulated/underexpressed in cancerous cells.
- 1 mark for stating Gene Z shows identical/unchanged expression in both tissues.

(c) Maximum 4 marks:
- 1 mark for identifying that a 1:10 dilution factor (or 1 part solute to 9 parts solvent) is required at each step.
- 1 mark for the first dilution: 10 cm^3 of 20.0 mmol/dm^3 stock + 90 cm^3 water to make 2.0 mmol/dm^3.
- 1 mark for the second dilution: 10 cm^3 of 2.0 mmol/dm^3 solution + 90 cm^3 water to make 0.2 mmol/dm^3.
- 1 mark for the third dilution: 10 cm^3 of 0.2 mmol/dm^3 solution + 90 cm^3 water to make 0.02 mmol/dm^3 (and specifying thorough mixing at each stage).

(a) The control tube must contain: 1. An equal volume/mass of non-living material (e.g., boiled/dead peas or glass beads) to equal the volume of the germinating peas. 2. An identical volume and concentration of potassium hydroxide (KOH) or sodium hydroxide (NaOH) solution (or soda lime) as in the experimental tube. 3. The same total gas volume/air space as the experimental tube.

(b) To determine both gas volumes: Set up one respirometer with potassium hydroxide (KOH) solution in the tube to absorb all carbon dioxide produced. The movement of the manometer fluid measures oxygen uptake only. Set up a second, identical respirometer running concurrently but with water instead of KOH solution. The movement of the fluid in this second tube measures the net change in gas volume (oxygen uptake minus carbon dioxide release). Subtract the net volume change of the second tube from the oxygen uptake of the first tube to calculate the volume of carbon dioxide produced.

(c) \( \text{RQ} = \frac{\text{Volume of } CO_2 \text{ produced}}{\text{Volume of } O_2 \text{ absorbed}} = \frac{9.2}{12.6} = 0.73 \). An RQ value of approximately 0.7 to 0.73 indicates that lipids (or fatty acids) are the primary respiratory substrate being oxidized.

(d) Temperature affects the rate of enzyme activity in cellular respiration, so a fluctuating temperature would change the biological rate of respiration. Additionally, according to gas laws (Charles's Law), changes in temperature alter the volume and pressure of the gas within the sealed tubes, which would cause the manometer fluid to move independently of the gas exchange being measured, invalidating the results.

(a) Maximum 3 marks:
- 1 mark for dead peas / boiled peas / glass beads of the exact same volume.
- 1 mark for identical volume and concentration of alkali (KOH / NaOH / soda lime).
- 1 mark for identical air space / total volume inside the tube.

(b) Maximum 3 marks:
- 1 mark for stating that the first setup has alkali (KOH / NaOH) to absorb CO2 to measure O2 consumption only.
- 1 mark for stating that the second setup has water instead of alkali to measure net gas volume change (O2 consumed - CO2 produced).
- 1 mark for explaining that CO2 volume is calculated by finding the difference between the two measurements.

(c) Maximum 2 marks:
- 1 mark for the correct calculation: 9.2 / 12.6 = 0.73 (accept 0.7 or 0.73, reject 0.730 without rounding or incorrect values).
- 1 mark for identifying the substrate as lipids / fats / fatty acids.

(d) Maximum 2 marks:
- 1 mark for explaining that temperature affects the rate of respiration because respiratory enzymes are temperature-dependent.
- 1 mark for explaining that temperature changes alter gas volume/pressure (causing expansion/contraction of gas), which would displace the manometer fluid and cause systematic errors.